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# INTR MODRN PHYSCS I PHYS 321

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PHYSICS 321 CHAPTER 5 SCHRODINGER S THEORY OF QUANTUM MECHANICS 5 1 INTRODUCTION The foregoing experiments and models demonstrated waveparticle duality conclusively but they only dealt with speci c situations Schrodinger s equation o ers a general recipe to predict the behavior of the particles of any microscopic system For each system it specifies the equation that the wavefunction of the system must satisfy and the connection between the behavior of the wavefunction and the behavior of the particles Naturally it reduces to Newtonian physics in the macroscopic limit just as the theory of relativity reduces to Newtonian physics in the limit c gt 00 After introducing Schrodinger s equation we will identify some of its essential points and use them to treat several important situations such as transmission through and re ection from particle waves incident on various potential barriers and tunneling which is a purely quantum mechanical phenomenon Then we shall treat the structure and properties of atoms in detail to be followed by a relatively brief treatment of molecules Over time the student will find that she is developing a quantum mechanical intuition which will bear a striking resemblance to the intuition on EM phenomena for the obvious reason that in both cases we will are dealing with waves However the student will have to be patient as this exposition will unfold gradually We start from de Broglie s postulate which states that the motion of a microscopic particle is governed by the propagation of an associated wave but it does not provide the laws governing that propagation It does predict the wavelength it successfully but only when l is essentially constant Hence we first describe Schrodinger s equation 1925 which determines the behavior of any wave in terms of the wavefunction w if we provide the potential energy function for the particle associated with the wave and the initial and boundary conditions of that waveparticle Then we shall describe the relation between the wf and the associated particle developed by Max Born 1926 While particular solutions to Schrodinger s equation for some particular potential energy functions and boundary conditions are described in the next chapters some general features of the solutions are described in this chapter including quantization of energy and other properties To gain some clues as to the required nature of the equation we revisit the free particle In Chapter 3 we used a simple sinusoidal traveling wave Pxt sin27zx ti Vt 51 or a linear combination of such simple sinusoidal waves to describe the wf of a free particle But we arrived at this wf by guessing based on the constant linear momentum p which entailed a constant 1 But what do we do when a force is acting on a particle Generally p will change with position and time The problem is that l is not even welldefined if it changes very rapidly For example in Fig 51 the separation between adjacent maxima is not equal to that between adjacent minima The general solution to this problem is the Schrodinger equation which is a partial linear di erential equation whose solution is the wf I x t As a first step we practice our partial derivative skills Example 5 1 Evaluate the partial derivatives up to second order of the sinusoidal wf given in Eq 51 wrt x and t Solution We recall that a partial derivative wrt x is evaluated by treating all the other variables as fixed Hence if I x t sin2 72x t7 Vt sinkx 7 wt 52 then 6 1 xt6x kcoskx 7 wt 62 1 xt6x2 k2sinkx 7 mt 6 1 xt6t 7 acoskx 7 mt 621m t6t2 7 a2sinkx 7 mt 5 3 In solving Schrodinger s equation for different potential or boundary and initial conditions we shall often see that the solution is obtained by separation of variables If the solution can be written as the product of single variable functions then the partial differential equation separates into ordinary ie single variable equations 5 2 PLAUSIBILITY ARGUMENT LEADING TO SCHR DINGER S EQUATION We are now looking for the equation that is the quantum mechanical analog of Newton s equation of motion F 7 dpdt 7 dexd 5 4 or of Maxwell s equation 6Ex6x 6Ey6y 6Ez62 peo 55 which can also be written as VoE peo 56 and is equivalent to lEodA 7 gemgo 5 7 by Gauss s Theorem However while the wave equation for a stretched string can be derived from Newton s law and the EM wave equation can be derived from Maxwell s equations we cannot expect to derive the quantum mechanical wave equation from classical physics Yet we will get help from the de BroglieEinstein relations lhp and VEh 58 In other words the equation we seek must be consistent with these postulates at least for free particles It should be emphasized though that we are not deriving Schrodinger s equation only providing a plausibility argument for it that equation is a postulate L II 39 The desired equation must clearly satisfy the g or 1 1 i It must be consistent with the de BroglieEinstein postulates Eq 58 ii It must be consisten with the equation E 7 pzZm V 59 iii It must be linear in I x t so that if I lxt and I zxt are solutions of the equation then cl l l x t cz Pzx t is also a solution This linearity condition will ensure that we will be able to add wavefunctions to produce constructive and destructive interferences which are a basic characteristic of all waves and demonstrated by the various experiments such as the Davisson Germer experiment described above to be valid for particle waves as well iv The solution of the equation for a free particle where l hp and V Eh will be a linear combination of sinusoidal traveling wave functions We now consider the consequences of the foregoing assumptions From i and ii we get iii2mm Vx t 7 hv 5 10 or hzkzZm Vxt 7 ha 5 1 1 We now focus on the factors k2 and a In considering assumption iv we saw in Example 51 that taking the 2quotd derivative of the traveling sine wave wrt the spatial variable x results simply in the factor 7k2 and that the ISI derivative wrt time results in the factor 7amp2 Hence it appears that Schrodinger s equation must contain the 2quotd spatial and lSt temporal derivatives of I x t Since Vxt must also appear but every term in the equation must contain Pxt or its derivative the equation should have the form ott39iz l xt6x2 Vx t I x t B Pxt6t 512 with the constants a and to be determined While the form of Eq 512 appears reasonable there is a problem If we consider the constant potential Vxt V0 for which the particle is a free particle since F VV 0 and substitute the sinusoidal plane wave solution I x t sinkx 7 wt into 512 we get otsinkx 7 atk2 sinkx 7 mtV0 acoskx 7 mt 513 But this equation will be satis ed only for special values of x and t because a sine fuction is not proportional to a cosine function with the same phase for all phases Hence the problem obviously arises because differentiation changes sines to cosines and vice versa This suggests modifying the wave description of the free particle to be a combination of sine and cosine functions Pxt coskx 7 wt 7sinkx 7 wt 514 The partial derivatives of I x t are 62 1 xt6x2 k2coskx 7 wt k27sinkx 7 wt and 6 1 xt6t wsinkx 7 wt m7coskx 7 wt 515 Substituting these derivatives into 512 we get ozk2coskx 7 wt ak27sinkx 7 mt V0coskx 7 wt V07sinkx 7 wt wsinkx 7 wt mycoskx 7 wt or ak2 V0 m coskx 7 mt ak2y Voy msinkx 7 wt 7 0 516 But Eq 516 will be satisfied for all x and t only if the coefficients of the cos and sin are both zero ak2 V0 7 a 5 17 and ak2 V0 7 wy 518 Hence 39 wi WV Ie y lor y ii 519 Substituting Eq 519 into Eq 517 we get ozk2 V0 7 ii a Comparing with Eq 511 we get a 7 k22m 5 20 and 7 i144 5 21 Although there are two possible choices for the sign we will see that the choice is of no conse quence so by convention we choose the sign With that choice Schrodinger s Equation is W 62 6x2 Pxt Vx 01mm 7 ihg x t 53922 This equation satis es all of the four assumptions above on quantum mechanical wave motion Note the following 1 We were led to this equation by treating the special case of the free particle Hence we postulate it to be true for any Vx t Hence this is the quantummechanical wave equation whose solution I xt is the wavefunction associated with a particle of mass m subject to forces resulting from the potential energy Vx t The validity of the postulate is of course determined by the agreement between its predictions and observations 2 Schr dinger s equation is nonrelativistic In 1928 Dirac developed the relativistic quantum mechanics He utilized the same postulates as Schrodinger but replaced the classical expression for the energy with the relation E 02p2 mocz2 V As required in the lowvelocity limit the Dirac equation reduces to Schrodinger s equation The treatment of Dirac s equation is beyond the scope of this course but we will describe some interesting features of Dirac s theory qualitatively Indeed one feature namely pair production has already been described Example 5 2 Verify that Schrodinger s equation is linear in I xt ie that if I 1xt and I zxt are solutions of the equation so is any linear combination cl I 1xt cz I zxt where cl and c are any complex numbers Solution Straightforward In the sequel we shall at times pull solutions of the equation for a speci c Vxt out of the hat without showing how they were derived but once given it will be straightforward if tedious to verify that they do solve the equation for that potential The next example does just that for the harmonic potential Example 5 3 The wavefunction describing a particle of mass m acted on by a linear restoring force of force constant C in its lowest energy state as a SHO is LPOC t Ago2 cmzerb z xz where A is real Verify that this wavefunction is a solution to Schrodinger s equation for the harmonic potential Vx t Cx22 Solution Schrodinger s equation in this case is W 62 I xt 2m 6x2 The derivatives of the solution given above are sz 6 I xt z39h IJ xt 2 at Emmy apt and i Pxt ch2x P chx l so it 2 m 6x 2h h 62 Jcm 11L Jcm x lcm W lcm Cm a h h Pxt P x2 P h h if x2 Substuting these expressions into Schrodinger s equation we get 2 2 h cm 1 h cm xZ PxZ Pih P m 2mh 2mh2 0r 39 F P x 1 xz 1 ih i 9T 2 m 2 2 2 m which is obviously satis ed The general solution to the Schrodinger equation for the SH0 is given in the next chapter 5 3 BORN S INTERPRETATION OF WAVEFUNCTIONS If we now go back to Schrodinger s equation for a free particle we see that the wavefunction Pxt coskx 7 wt 7sinkx 7 wt coskx 7 wt isinkx 7 wt 523 In other words it is complex The reason is obvious the wave equation is a linear relation between the first time and second spatial derivatives of the wf This itself results from the linear relation between the energy and the square of the linear momentum The fact that the wf is complex implies that its physical interpretation is different from that of functions describing classical waves Hence we must explore the new physical meaning of this new beast Following Born s postulate 1926 the basic connection between Pxt and its associated particle is expressed in terms of the probability density Px t which is PW Pxt Pxt 524 where Pxt is the complex conjugate of I xt The postulate is the following If the location of the particle associated with I x t is measured at time t then the probability that it will be found in x x dx is Pxt l xtdx Note that Pxt l xt is real and nonnegative which is a necessary and almost but not quite sufficien condition for it to be a probability function The relation between the position of the particle and the wf is shown schematically in Fig 52 Example 5 4 Prove that 1 x t Px t Z 0 Proof Let I x t Rx t ix t 525a Then Ibk x t Rx t i1x t 525b and Px 01 t R2x t 12x t z 0 Note the analogy 7 and the distinction 7 between the probability density of the particle wf and the intensity of an EM wave which is proportional to E2x t where E is the amplitude of the electric eld Example 5 5 Evaluate the probability density of the ground state wf of the SH0 given above Solution Since TUCJ Aewz cmierb zr iyc2 we get POCJ 2 LP anLFOCJ Azeioz Cmte12 cmierWhx2 AzerUEhkz Note that this Pxt is independent of time even though I x t is a function of t We shall see that the wf is indeed time independent in any case in which the associated particle is in a single energy state i e its energy is well de ned i e its energy is a conserved quantity The probability density of the ground state of the SH0 Px given above is plotted vs x in the upper part of Fig 53 It is obviously a Gaussian which peaks at x 0 Hence quantum mechanics predicts that the particle will most likely be found in an element dx around the equilibrium point which is x 0 However there are no well de ned limits beyond which the probability is zero We shall now see that the probability function predicted by quantum mechanics is radically different from the classical prediction Example 5 6 Evaluate the classical probability density of the SH0 given above Solution Classically the particle has a definite momentum p and velocity v for any given displacement from equilibrium x The probability of finding it in element dx around x is proportional to the amount of time it spends in that element ie inversely proportional to its speed at x Hence P Bzv where B2 is a constant Hence to find Px we need to express v in terms of x This relation is contained in the expression for the energy E K V vaz szZ Rearranging we get 2 2 V ECx 31300 2 ll RC xi m 2 This Px is plotted vs x in the bottom part of Fig 53 It has a minimum at x 0 as expected since the speed is maximal at that point It also diverges at the turning points x im where the speed vanishes While the stark difference between the classical and quantum probabilities is obvious it is actually more subtle The quantum probability is that of the ground state of the SH0 and shows that while the particle will most likely be at the equilibrium point it has a nite probability of being elsewhere In the ground state of the classical SHO the particle has zero kinetic energy and zero potential energy and is at rest at the equilibrium point Hence the quantum mechanical result is an obvious manifestation of the uncertainty principle We shall see that the nonzero energy of the ground state of the quantum mechanical SHO aka its zero point motion is responsible for various phenomena such as the fact that at pressures less than 40 atms 4He does not freeze even at 0K However we will see that as the energy of the quantum mechanical SHO increases its probability density approaches that of the classical SHO as required by the correspondence principle The value of the constant B2 above is determined by requiring that the total probability that the particle be somewhere between the turning points is 1 Bl J2EC dx dex 1 w WWVE CxZ2 While this integral can be solved in closed form we shall not need it This determination of the constant B is called the normalization procedure and we will carry it out for the wavefunction of the ground state of the quantum mechanical SHO The foregoing example shows that quantum mechanics cannot predict that a particle in a given energy state will be in a precise location at a precise time but only the relative probabilities that it will be found in various locations at that time The obvious reason for this probabilistic interpretion is the uncertainty principle Example 5 7 Normalize the wavefunction of the ground state of the SH0 given in Example 53 Solution W W W dex IT delz yewMm foo Or A2 filmh dx 2A2 filmh dx 1 inc 0 From tables of integrals or Mathematica or Maple 90 1 8 fella393 2 alx 3 A so the normalized groundstate wavefunction of the 0 2 m 7239 18 s1mple harmon1c osc111ator 1s Twat Cm Wzhkzerczwmi h14 Again this normalization ensures that J39de J39 p gtxlt pdx 2 1 5 4 EXPECTATION VALUES We will now see that the wavefunction contains all of the information on the particle or system Consider a particle with wavefunction I x t The probability of nding it in the segment x x alx is by Bom s postulate Pxtdx 1 k x t I x tdx Note that this also means that if we have a very large number of identical noninteracting particles with that wavefunction the fraction that will most probably be in x x alx at time t is Pxtalx The average of observed values of a quantity e g x coordinate momentum energy etc is called the expectation value of that quantity In the case of the X coordinate it will be lt x gt JMOPxtxalx J39W Pquot x tx I x tdx Example 5 8 Find ltxgt of a particle in the lowest energy state of a simple harmonic oscillator Solution Figs 53 and 54 immediately show that ltxgt 0 Fig 55 shows the simple math proof Since 2 1 xt Px t elmhx is an even function of x 2 1 xtx Px t xe lmhlx is an odd function ofx so lt x gt J39pr gtxlt xatxpxtdx 0 This average value of x is called the expectation value of the position of the particle The expectation value of any physical quantity fx associated with the particle is lt fx gt 1 x t fx Px talx For example the expectation value of the potential energy of that SHO in the ground state is oo C we 2 4mm lt Vt gt 1 x tVx t Pxtdx 3 x e dx Even though it can readily be done we shall not evaluate this ltVtgt ltVgt at present since we wish to focus on other fundamental dynamical quantities namely the momentum p and energy E While the quantity oo lt p gt I xtp 1 xtdx is classically valid with px 2mE 7 Cx2212 it is ruled out as a quantum mechanical quantity due to the uncertainty principle So we turn to a new approach Consider the free particle with the wavefunction Pxt coskx 7 wt isinkx 7 wt eiaoquot 1 Then 6Tjk111392 11 j 2 p 1 i l 6x A h h In other words p Pxt ih6 Pxt6x Therefore there is an association between the linear momentum p anal the differential operator ihBBx We can derive a similar association for the energy 6 11 ia P i27239V P i lJ it h I e E Pxt iho Pxt6t Are these relations restricted to free particles NO Consider the general case where pZZm Vx t 7 E Replacing the dynamical quantities by the operators we get W 62 6x2 6 V xt ih at Z L ih3 Vxtih3 or 2m x it This is an operator equation so for any wavefunction I x t we have hz BZ P 6 1 Vxt Pih 39 39 quot39 39 2m 6x2 at which is Schrodinger s equation In summary pap mi E m 6x 0p 6t Returning to ltpgt we get lt p gt JW P xtp0p 1 xtdx 4hr x t63 1 xtdx is is x Similarly lt E gt 0 11 x tE0p Px tdx 139th x Og x tdx Indeed substituting Schrodinger s equation for ik PBt in the last expression we get hi 62 6x2 lt E gt 0 11 xt Vxt l x tdx Hence if fxpt is any dynamical quantity then its expectation value when the particle s wf is 1 t 39 m x m lt f gt j 1 mm x ih t l xtdx w x Example 5 9 Consider a ld free particle con ned to lxl lt a2 ie con ned to the region a2 a 2 We will see that the groundstate wf of the particle is moat ACOSeilEh for lxl lt 612 0 otherwise where A is a normalization constant and E is its energy Show that this Pxt satis es Schrodinger s equation and nd the lowest energy E Solution Since the particle is free within the box lxl lt a2 Vxt const which we can choose 0 Hence Schrodinger s equation is E a mag 2m 6x2 6t Substituting the given expression for Pxt we get zhz 2ma2 hi 6 P mil TET 2m 6x2 Z Pih i l 2ma h which is indeed satis ed ifE 7221922ma2 hzSmaz Example 5 10 Evaluate the expectation values a ltxgt b ltpgt c ltx2gt and d ltp2gt of the particleinabox described in the previous example when it s in its ground state Solution a As for the ground state of the SH0 we note that Pxt AcosmcaexpiEtJe is an even function of x and so is Pxt l xt Azcoszmca Therefore LI xtx l xt Azxcos2 Inca is an odd function of x so as expected aZ lt x gt A2 Ixcos2mcadx 0 02 a2 b lt p gt ih J 1 xt l xtdx Since the integrand is now proportional to x 702 cosmcasinmca it is also an odd function of x so as expected again ltpgt 0 m a2 a2 c lt x2 gt 1 x tx2 Pxtdx A2 1x2 cos2mcadx 2A2 1x2 cos2mcadx foo 02 0 2A2 3 2 cos2mcad 3lt x2 gt A2 as 1 1 a a 4 j n 0 As mentioned above A is the normalization constant which we now proceed to evaluate oo a2 7r2 1 pl xt Pxtdx A2 Icosz mean 2A2a7r Icosz mcadmca 702 0 7 21cos2t9 a 7 Ala 2A2a cos26dt92A2a Tdt92AZ 1O 2 2 0 0 3 Z 3 Z Z Z ltx2gtA2 a 1 2 a 1 quot 1 m0033a23xm ltx2gt018a 472392 6 0472392 6 272392 6 Returning to ltx2gt we get Hence xm is a measure of the uctuation of x about ltxgt 0 Generally ltx2gt ltxgt212 is a measure of the uctuation when ltxgt 0 Similarly for ltpgt ltp2gt and ltp2gt ltpgt212 d Now evaluate ltp2gt for the ld particle in the ld box a a 2 h 2 ltp2 gtO P zh2ampC de hzo l 6x2 dxh2o 1 1 dx7 j 3 lt p2 gt hZa which is precisely the de Broglie value ofp hl h2a So now we can see why ltx2gt12 Ax and ltpzgt12 Ap In addition AxAp ltx2gtmltpzgtm 018agtlt7rka 05714 2 142 5 5 THE TIME INDEPENDENT SCHRODINGER EQUATION We saw that Schrodinger s equation is h2 62 6x2 Px t Vx 01mm mgwx t As we mentioned in class one of most common techniques for solving partial differential equations is separation of variables searching for solutions which are products of functions each of which is a single variable function Let us now apply it to Schrodinger s equation We therefore express the wavefunction as a product of two functions one of x only the other of t only LI JCJ lx t We shall now see that this is the form of the solution to Schrodinger s equation if the potential energy function does not depend explicitly on t Indeed substituting this solution into the equation we get h2 6 2W I 6 0 2m 6x2 0tVxlxlt0t lhlIx at DiViding throughout by 1xpt we get Z Z 1 h6 liVxllx Gia yx 2m 6x2 gpt it where G is the called the separation constant First consider the equation for pt 6gp iG it h We have encountered this differential equation often enough and its solution is idsh W 6 This is an oscillatory function with V Gh But V Eh so G E Hence Schrodinger s equation for 1x becomes hz 6211 2m 6x2 wowx Ellx This is the time independent Schrodinger equation The functions Wx which satis this equation are eigenfunctions of the equation The full wavefunction can now be written as Tom wxequot Example 5 11 Provide a plausibility argument for the timeindependent equation Solution pz2m V E But phlkgtlt27rlkk so ksz2m V E Now consider 1x sin2mcl sinkx d2 1xdx2 k2 1x 2 2 2 2 2 SO h alfVxlIxh k wVl p V wEwx 2m 6x 2m 2m 5 6 REQUIRED PROPERTIES OF WAVEFUNCTIONS Both 1x and d 1 dx must obviously satisfy the following conditions i They must be nite ii They must be singlevalued iiiThey must be continuous In the following analyses we shall often see that these conditions lead us to the unique solution for 1x for given initial and or boundary conditions 5 7 ENERGY QUANTIZATION IN SCHRODINGER S EQUATION We shall now seek insight into solutions from plots of Vx etc First recall that divELquot dx 2 Vx Ewx The properties of 1x depend on those of Vx since classicallyF dVdx First consider the example shown in Fig 59 which is typical for an atom bound to a similar atom in a diatomic molecule e g Hz N2 02 etc At the minimum in Vx F 0 For closer x the atom will experience the hard core repulsion from the other atom For farther x the atom will experience attaction to the other atom When the molecule dissociates Vx V0 F 0 Now consider an atom with a total energy E as shown in Fig 510 Classically x and x are the turning points so x S x S x Recall that 61211de tells us how the slope of zjis changing Hence if 61211de gt 0 and zgt 0 the slope is increasing and zjis concave upwards and vice versa Now divide the x axis into three regions see Fig 513 In region I x lt x Vx 7E gt 0 so If zgt 0 61211de gt 0 and yis concave upwards if zlt 0 d2 11de lt 0 and yis concave downwards In both cases yis convex toward the x axis or concave away from the x axis In region II the behavior of yis opposite and it is concave toward the x axis In region III it s behavior is the same as in region I Fig 514 shows the consequences of these properties of yand its derivatives Assume that zgt 0 and dex lt 0 at some x lt x0 lt x Then dZLldx2 lt 0 and yis concave downwards If yis still gt 0 at x x it will start curving upwards at x gt x and grow without a bound diverging as x gt 00 This behavior is obviously unacceptable If ybecomes negative at some x gt x it will start curving downwards and diverge to w as x gt 00 which is equally unacceptable Hence only if yand d ydx have special values at x0 so that ycurves upwards at x gt x but at a slower and slower rate will z gt 0 as x gt co to yield an acceptable solution In general we will also have similar problems at x lt x Hence we will see that there are only special values of E for which 1 gt 0 as x gt ice 0 This results in energy quantization automatically so we find a set ofeigenfunctions 111 112 14 E2 E3 Fig 515 shows 111 112 and 14 for the lowest energies E1 E2 and E3 4 behaves as function with energy eigenvalues E1 3 of Fig 514 for large x and also gt 0 for small x zl2x0 14x0 but the magnitude of its second derivative ldzzjdle is larger so it crosses the x aXis at some point x lt x lt x0 Then the sign of d2 11de changes and it turns from concave downwards to concave upwards At x x it reverses again turning concave downwards and it approaches the x axis The basic observation that should be noted here is that I K W Similarly E3 gt E1 Note that E3 7E2 and E2 E1 are finite so the energy levels are discrete This is true as long as there are x and x such that Vx gt E for x lt x or x gt x see Fig 516 When the energy of the atom is E gt V the situation changes Classically the atom is unbound for all x gt x Vx E lt 0 so zjwill be concave towards the x aXis for all x In other words it will oscillate And since we can always cause z gt 0 as x gt 00 by an appropriate choice of an initial value of d de we will find an eigenfunction for any value of E Thus the set of allowed eigenvalues forms a continuum In summary when the relation between E and Vx is such that classically the particle is bound then Schrodinger s equation leads to quantized energies If classically the particle is not bound the set of eigenvalues forms a continuum Example 5 12 Use the foregoing arguments to describe the eigenfunctions of the SH0 Then compare the resulting Px with the classical SHO of the same E Solution Consider Figs 517 518 and 519 Vx is such that for any E the particle is classically bound between x lt 0 lt x with x x Since d 211 2m dx2 h Z Vx Ellx zjwill oscillate in the region where Vx E lt 0 However as we approach x and x in the bounded region Vx El decreases so oscillations weaken In other words the separation between the nodes of yincreases Hence the larger E is the more negative Vx E the more rapid are the oscillations the greater the number of nodes the shorter the e ective A the greater the momentum p W1 the greater the energy E pzZm V We have come full circle in understanding the relation between the nature of the wavefunction and the energy of the particle For the nth allowed energy Equot there are n 1 nodes As the SH0 particle approaches x or x the amplitude of the oscillations of yincreases since yitself must be larger when it bends over Since dz Mdle cx Vx 7EH 1 and Vx 7E is smaller My must be larger in order to bend over Hence Px zx 1x 1x is higher This situation resembles the behavior of the classical SHO Also note that as E increases the oscillations become more rapid so the decay of youtside the bounded region becomes more rapid and the overall behavior of Px approaches the classical limit 5 8 SUPERPOSITION OF WAVEFUNCTIONS For each allowed energy eigenvalue En we have an eigenfunction Woe and a wavefunction I nxt xexpiEnth n is called the quantum number If the system is described by I nxt it is said to be in state n Since the Schrodinger equation is linear I xt ch l nxt is also a solution of the equation In fact it is the general solution of the equation for a specific Vx and specific boundary conditions Example 5 13 Consider a particle in a state such that measurement of its total energy E can result in either of two values E1 or E2 e g an e39 making a transition from an excited state to the ground state Then I x t c1 Plx t cz l zx t c1111xexpiE1th c zIzxexpiEzth Show that Px oscillates with t and calculate the oscillation frequency Solution Px t 1 k x t I x t c1c1 111xz1x czczzzxzx czc1 zIzxL1xexpiEz 7E1th cc Since eXpi cost9 isint9 Px oscillates at V E 7E12 72h E2 7E1h Note that if an e39 in an atom is in an eigenstate Px does not oscillate which is a situation analogous to the classical stationary state Hence it should not radiate Chapter 7 Outline OneElectron Atoms Reduced Mass again The 3dimensional Schrodinger equation in spherical coordinates r 6 p separation of variables again quantum numbers n 1 ml degeneracy Eigenfunctions comparison of Bohr and Schrodinger treatments veri cation of typical eigenfunctions Probability Densities shells comparison with Bohr atom angular dependence Orbital Angular Momentum expectation values of z component geometrical description Eigenvalue Equations expectation value of a uctuating quantity Hamiltonian operator HW 7 Chapter 7 due 103106 questions 9 11 problems 3 7 11 14 17 20 21 22 23 Expansion of methods from previous chapter to three dimensions Focusing on the eigenfunctions we ll be able to investigate probability density functions 7 giving us a picture of atomic structure that does not violate the uncertainty principle as do the precise orbits of the Bohr model the orbital angular momenta 7 which were incorrectly predicted by the Bohr model the electron spin and other effects of relativity incorrectly predicted by the Bohr model the rates at which the atom makes transitions from its excited states to its ground state 7 measurable quantities not at all predictable by the Bohr model As we will see Schrodinger s theory of the one electron atom will form the foundation of our treatment of all multielectron atoms as well as that of molecules nuclei and bound states of elementary particles Although the single electron atom is the simplest bound system found in nature the fact that it consists of two particles and that it exists in 3 dimensions introduce some mathematical complications compared to the systems we ve looked at so far In Chap 4 we dealt With the two body prob1em by 7 quotr using the reduced mass j u M l l g V gquot M m l l mM v What s the physical picture for this model Midi What does the potential energy U depend on m1 391 C M What is requal to 7 1 1quot2 I 17 77172 R r How 1s the Lagrang1an de ned L T V r 1 1 2 1 2 2 m2 What1s1tequalt01n th1s case LEm1r1 Em2 lrzl U0 In general how do we find the center of mass 2 mi W If we now move the origin of our coordinate system to the CM what s the value of R I 0 gt a m Therefore Emir 711171 m31 3 0 Comb1n1ng th1s w1th r 1 1 1 2 1 I CM m m We nd r1 2 r 72 1 r Substltutlng these 1nto our 7111 m2 m1 m2 Lagranglan we nd R E 0 1392 l m2 l m2 L m1 22r 2 m2 12r2 U0 2 m1 m2 2 m1 m2 quot392 L 1 2 2 1 12 gt2 1 gt2 ampWl Ultrgt5Wm mmlrl Ultrgt3mr Ultrgt m1m Development of the Schrodinger Equation For an electron of reduced mass M and charge 6 moving under the in uence of the Coulomb potential Ze2 Vxy Z 4718Mle y2 22 The energy can be written as the sum of K V 1 2 2 2 h2 82 82 82 a VX Z E Vx 2 17 2 u px py p2 y 2y 8x2 ayz 822 y at h2 2 6 SO V II Vxyz1Plh 1P 2 8t 62 62 62 where V2 del squared aka the Laplacian operator is V2 E 6x2 ayz 622 Note the source of this term which is due to the operator identity V V 66xi aayj aazk 66xi aayj aazk Since Vxyz does not depend explicitly on r what technique can we use to solve Schrodinger s equation separation of variables POC y Z l 1 X y Degmm where 111 satis es the timeindependent Schrodinger equation hZ V2WxayzVxyazwxayaz EWxay Z Separation of the tim 39 equation However there s a problem with this 7 anyone see it yet Vxyz is not itself separable so we cannot write solutions of the form 1p1x1p2y1p3z Z How do we solve this In polar coordinates Vr L A 4na r 712 but what is V2 in and 7V2 V E 2 w rm 1 polar coordinates Z V2 Z a j71 a sinei 1 6 Z r7 z 2 r Br Br rsmS 66 66 rsmeaip and our solution takes the form 1pr 6 4p Rr 9ltlgt So the time independent Schrodinger equation becomes 2 2 1 7 iii rZaR qgtl 2 a 51516761 qgtL 2 l7 3 RezqgtlVrR lt1gtER lt1gt 2y r Br Br r 51116 06K 06 r sm 6 01 J h2 lt1gt a lam RltIgt a am R6 3 i 0i V R ltIgt ER ltIgt or Zyl r2 arkr 0r risine 06 ism 06 risinze 01 J r dividing by R ltIgth2n22u multiplying by risinze and rearranging we get 1azqgt mailedmm sin d 1m 2y22 lt1gt0q92 R drlr dr 9 dalmade hi 51 6E W0 Since the lefthand side is a function of p only and the righthandside is a function of r and 6 only we must have that 621 2 2 mylt1gt 809 1d2dR 1dd 2M2 m 6 E V and Rdrr dr sin6d6sm d6 h2r 0 sin26 ldzdR 2M2 mf 1ddo r r E Vr sma 111 or Rdr dr 122 O sin28 sin6d6 d6 l d d m2 gt sin6 391 lll sin8d9 d sin28 1 d 2arR 2y ll1 39 E V R R and r2 drr dr h2 0 r2 We will now see that we get acceptable solutions to the p equation only for certain values of ml to the 6 equation only for certain values of 1 l and with these values of ml and l 1 solutions to the r equation exist only for certain values of E hence the energy of the electron in that atom or ion is quantized Solutions of the equations 6 I The solution to the p equation is obv10us a 2 m2q3 CIgtp e ml p P What constraint do we have on d q0 13271 so we must have that eimlo eimla that is l cos2nml isin2nml So what can we say about ml lmll 0 l 2 3 This de nes the quantum number ml and we can de ne the functions d by this number Pm 90 em 9 The 9 equation is solved in Appendix N The solutions are acceptable nite only if I ml m1 l ml 2 ml 3 The solutions can be written in the form 9m 8 sin m l 8Flyml cos 8 where F m are polynomials in cos6 which depend on I and Im II The F m are usually referred to as the associated Legendre functions The r equation is also solved in Appendix N As predicted it is found that the bound state solutions are nite only if uZ2e4 136 EnW7 eV Wheren211l2l3and 0 1 47180712 72 GnZra0 a0 0529 R r e Zrnao n Zr 0 and GnlZra0 are polynomials in ZraOThese functions are closely related to the associated Laguerre polynomials E sze 136 T 39 quantum numbers and 39 quot 4nao2 h2 P n2 How does this expression for En compare to that derived from the Bohr model 7 Identical eV Note that the Coulomb potential also yields a minimum E E1 and zeropoint motion Comparing with the finite square well and the SH0 we see all have a zeropoint energy HUT gmmhvi mHt Energy 2V Cnulmnli well i The appearance of 3 quantum numbers n l and ml is a consequence of the 3dimensional nature of this Schrodinger equation We can rewrite the relations between these quantum numbers as follows The principal quantum number n l 2 3 The azimuthal quantum number I 0 l 2 n 7 l The magnetic quantum number ml l l l l 2 If 1 Eigenfunctions With the same n but different or ml are different ut their energy eigenvaluesEn are the same Different eigenfunctions With the same energy eigenvalue are said to be degenerate and the number of distinct eigenfunctions With the same energy is called the degree of degeneracy Degeneracy always results from a symmetry of the system the nature of the particles andor the potential energy function In the present case the simple Coulomb potential Vr is spherically symmetric in l 2 3 l I 0 l U ml 0 o t t Number of dcgcncl 2m ergcn lnclions for each Number of for each value of n there are n 7 1 values of l for each value of I there are 21 1 values of m Thus the level ofdegeneracy ofthe nLh energy level is 1 3 5 2n 71 n2l 2n 71 nZ However if eg an external magnetic eld H is also applied the spherical symmetry of the system is broken and the degeneracy is lifted L Rquot r z e mwa G Layan n TIME H n 1 71 2 n 3 Radial Functions lor the OneElectron Mom 8 7 27ml3 6 4 Z 70 awao 1 p 7 1 W as 2 1 7 9 26aJ v2 2 l 2 2 2 W3 7 p p f 3V3a 3 27 1 Eip3 rnz 91MB sm ml 6Fmlcose Spherical Hammics 0 m0 Ym 71 1 ml Yn m0 Yw mAl Yul 15 391 a 2 quot2 122 Esm c ml ll smacosgz39 7r 0 y l 5 3 19 1 m I 16quotcos 15 mv A1 Y2 8 sin49c0549tquot 77 7 15 sin gs 77 m We just found our Wavefunctions for Mr B W R 08 a I w 7 ltIgt 8 andrto be Quanmm Numbers n I m Eigcn lnclions Ic m 1 g quotV T 1 0 0 mg nmma m r w a f r 1 3 a Zam afmw mwm 2 0 0 W200 Z A E 4 quota u u 32 2 1 0 1pm 7 amulcos 39 4 27 u lt1 3quot22 7 2 1 ii WZALA if quotquot051n0yw 7 D U I z Wlt Zr 22 3 0 0 7 7 27718 2 W slaw quot0 a5 2 m r Zr V 3 0 7 7 7H lrJlln I In 81V to 6 an no L C05 Example 7 2Venfy that 1112 7 I Z 3 3 Zr Zr 7mg M andLhe associated eigenvalueE2 5 1 i1 v13 3w lt6 EEe 03m 0 y m satisfy Schr dinger s equation I u 3 2 2311 3 2 7 7 zrJnnj 2 0 W3 Slv C n Hi I cos 0 H M2 2 3 2 i1 W32t8quotlra7gt 2e z 3 quotsinaceswm w n I Z ml H r 3 2 i2 1132 if 7 Musinlnezw mm quota an Probability densities DE Who rem embers the Bohr radius To begin extracting info on our oneelectron r nza0 atom eigenfunctions we will study the form M Z of the probability den51ty functions ltrm1gt is clearly a re nement How is the probability density function Of le BOhI result defined 7 By Born s postulate P 1V 11quot For the oneelectron case we have 002 x10 PO 6 P Hal11 114mmquot 001 ml ml 91m 91m m1 m1 C First let s con51der the r dependence 0 DZ 7 u 5 that is the radial probability density Pr x10 defined such that Prdr is the probability of 0mquot finding the e at any location between V and dr 0 054 To integrate the Probability density 1 1P probabilityunit volume over the volume enclosed between the spheres of radii r and rdr start with Pnlrdr RmrRnlr x4mldr Most probable location of the 6 will be on shell of radius ltrnl gt m 2 lt rm gtfVPn1rdr n a 11E1Ll1 n 0 Z r W triangles ltrw gt Physically what are we seeing in these plots Shells Example 73 a Calculate r for which Pup is maximal for the ground state of the H atom Ignoring normalization we get P1 Or R1 OrR1 000x4712 exp 2raor2 How to find maximum To find its maximum find r for which the derivative is zero 61 r 10 e727anr2 2re727an 0 gt rmax a0 dr a0 b Now find ltrgt the expectation value of r 2 2 111lltl21gt 1lhw 26 39 Z 2 n l 2 l 2 c Why is ltr1 Ogt larger than rmax The reason is that PI Or is asymmetric about its maximum with more area under its curve in the r gt rmax region Example 74 Show that the size of the ground state H atom can be obtained from the uncertainty principle W The closer the e gets to the 17 the more negative V e24n39 0R becomes and the more the kinetic energy K increases Since Ap hR hZ 2 2M 2M 2MR2 2MR2 4713012 The atom will adjust its size to minimize its total E we therefore minimize E to find R ltrgt dE 2W 22 4mth 3 20 gt R 2 a0 dR ZMR 47180R Me Important points The structure of the radial probability density does depend on I For a given n the function has a single strong maximum when 1 takes its largest value additional weaker maxima develop when 1 takes smaller values Pm is significant near r 0 only for l 0 This latter point is due to the fact that for small r Ipnlr at r1 so Pnlr dill at r21 andfor r gt 0 r0gtgt r2gtgt r4gtgt so near r 0 1pquot 111 is relatively large only for l 0 and decreases very rapidly with increasing 1 We now turn to the angular dependence of 111 ml 1pm Note that thmqu g lml l glmw 1 so the probability density is independent of q Hence figures of Pnlr should be rotated about the Z axis Therefore Lml k Lml alone plays the role of a directionally dependent modulation factor The form of lmf 91ml is most conveniently represented in terms of polar diagrams 7 rotate around the Z axis to get full picture More realistically we have the polar l diagrams for z i z dependence on l l I mp 1 l 1 y r 3 Note that the regions ofhigh probability 1 7 l m 1 H density shi from the z axis to the xy plane as lmll increases jam 7 u and the polar diagrams for l 0 1 2 3 4 andml 2 lt H 7777 dependence on I 1 m 1 ll OLD 1 v ml I4n4 Artist s conception tm 0 fern 1 2 3 Ix2lly 110 n3l1 quot39l1 A 1 umquot dz n31139m o a quot312quot 2 r I x r V quot3yl2ynxl 41 A n3l239nxjo Since the potential is spherically symmetric how can the eigenfunctions have lower symmetries If the spherical symmetry of the potential is not broken as by eg a magnetic eld then the energies of the states with different values of Z and m are the same In addition the average or sum of these states is spherically symmetric Example 75 Show the spherical symmetry of the average of the states with E E2 1 Z 200 200 1P2L11J211 2101J210 110211110211 e ZVa j 2 sin2 9 1sin2 9 cos2 9 12871 10 10 2 2 10 It can be shown in general that the sum 21p l mfk Mm over the 11 subshell ie from ml l 0 ml 1 is spherically symmetric If we break the spherical symmetry of the potential by e g applying a uniform magnetic field the energy eigenvalues and eigenfunctions will change and they will no longer add up to a spherically symmetric sum By convention we de ne the preferred direction as the z axis We will now show that l and ml are related to the orbital angular momentum L and its Z component LZ through the relations L 144 L2 mlh Comparing the Bohr and Schrodinger models we note that in both cases the ground state corresponds to n l and they both give the same total energy But in the Bohr model the ground state orbital angular momentum is L nh h h M wereast ltls L lllh0 sincel0whennl vast experimental evidence shows zero angular momentum for the ground state Motion of the electron may be visualized as purely radial could achieve L 0 in Bohr model by letting the electron travel back and forth along a diameter through the nucleus Orbital angular momentum Want to show thatL 11 1 12 h L2 ml In doing so we shall encounter the angular momentum relations which are extremely important in atomic molecular nuclear and condensed matter physics The reason is that E and L are the only constants of atOms nuclei Recall that L rxp 16 L ypz Zpy Ly 219x x192 L2 xpy m This means that Lx op 4720662 zaay Ly op ihza6x xaaz LZ OP ihx66y yaax Or In polar coordinates see Appendix M 8 8 8 8 L 17 111 cotecos L 17 cos cot651n L ih xop p p yop p p j p 1 a a 1 82 i L2 L2 L2 L2 h2 1116 We also need Appendix M 0 W Mp up Sing 86 86 Sinz 6 W2 J What does this look like ie what is it proportional to r2hm1gular part of V2 0071271 First let s consider ltLZgtI lt L2 gtf11I LWIIIdr fff111 LWIIIr2 sin 6drd6d p f1p1mleopwnJmt d1 V 0 0 0 V where d1 is a volume element in 3d space Similarly lt L gtf1P LZPIPdI wamLipwmhmldr V V Example 76 Evaluate a ltLZgt and b lt L2gt a Since 10mm Rnzr zml 99quot 2 1pmml a aqua2 Lz0pwnm zh am Rnltrgt mllt9gt thRnltrgt mllt9gtEzhzmdgtmlon Lzopw4lml mlhwmlml Since Lop i Lipipm 111hzww m gt ltL gt111h2 y Since L1 is wellde ned LX and L are not de ned at all In other words 1p 1 ml is not an eigmfunction of LX 0i Ly As a matter offact ltLXgt ltLygt 0 The vector 11 1 lies anywhere on me cone at an angle ofcosquotmlll 1W withrespectto die 2 direction so ltLXgt ltLygt 0 Elm Oi quot11 die uncertainty in die direction ofL is cosquotmlll mm This uncmainty becomes vanishingly small in die classical limit where die quantum number is enormous Chapter 8 Outline Magnetic Dipole Moments Spin and Transition Rates Relation between magnetic dipole moment and angular momentum Orbital magnetic dipole moments Bohr magneton orbital g factor Larmor precession magnetic dipole moments in a uniform magnetic field effects of a nonuniform field The StemGerlach experiment and electron spin comparison with Schrodinger predictions PhippsTaylor experiment spin quantum numbers s and ms Zeeman effect Spin Orbit interaction Thomas precession Total Angular Momentum coupling between orbital and spin angular momenta conditions satisfied by the quantum numbers j and mj Spin Orbit interaction energy and the hydrogen energy levels Lamb shift hyperfine structure Hamiltonian operator Transition rates and selection rules electric dipole matrix element QED picture of stimulated and spontaneous emission symmetries HW 8 Chapter 8 due 119 questions 10 19 problems 2 3 5 6 10 16 17 18 We ll begin with a discussion of experiments which measure the orbital angular momentum L However we don t actually measure L what is the thing we actually measure The magnetic dipole moment u Shortly we ll develop a relation between L and M Our methods will use a combination of simple EampM pseudoclassical approaches such as the Bohr model and QM This procedure is justi ed in that the results agree with a full QM treatment which is beyond the scope of this course Orbital magnetic dipole moments Consider an electron of mass m and charge e moving with velocity v in a circular Bohr orbit of radius r E e 6V 2m T 4 2m From a magnetic point of view what does a current loop look like if you are a large distance away from it We have charge circulating in a loop what is the current i a dipole For a current i in a loop of area A what is the magnetic dipole moment anyone remember u 1A or in our case u 1A The quantity u speci es the strength of the magnetic dipole it equals the product of the poles strengths times their separation What is the angular momentum of our electron L mvr Is L parallel or antiparallel to W Since 9 is negative its magnetic dipole moment 11 is antiparallel to L Now putting our expression for i into our expression for Mt we nd ev 2 evr M1 11 if 7 2m 2 Combining this with our expression forL we see that sz L h which is usually written as M where M 1 0927 x10 23 amp mZ Bohr Magneton m and g 1 is called the orbital g factor later on we ll have cases where g a 1 guub Z h Writing our expression for W as a vector equation we have it By repeating this calculation for an elliptical orbit it can be shown that MlL is independent of the shape of the orbit Why Also note that MlL is independent of the details of the orbit 7 suggests that it might not even depend on the details of the mechanical theory used to evaluate it Indeed it can also be shown that this relation is quantum mechanically valid so Mlg175l1514 1hglyb lll and 12 L2 mlh 39gnubml h We now fall back to classical EampM and ask what happens to a dipole in a magnetic eld It experiences a torque How do we express this mathematically f l X B What does this do physically Tends to align the dipole with the eld What about potential energy for a given orientation between 11 and 3 what is the potential energy AE 1 E Example 81 Calculate the energy required to turn a moment M1 Mb from parallel to antiparallel to a eld B l T AE Eap EP ApB Mchosar M1Bcos0 ZMZB 2x0927x103923xl J 116x10394 eV This AE is very small AEk 134 K but it is the basis for electron paramagnetic resonance EPR also known as electron spin resonance ESR It is 2000 times larger than the corresponding AE s of nuclear magnetic moments which are the basis for nuclear magnetic resonance N MR Consider the case where there is no process by which a magnetic moment M1 in a magnetic eld B can dissipate its energy what do we expect will happen What happens to 6 6 cos391pBuB constant And what does it do u precesses about B since 1 uxB is perpendicular to L just like the precession of a spinning top which is due to the torque created by the gravitational force So what s the frequency of precession dL d5 d hitl Ti l wat 39l F ii n dt ana ogo e ons aw ma m dt dt hint 2 1 xB L xl 3 hint 3 From the gure what is dL equal to dL Lsin6udt him 4 stin6di1 g M LBsinG it h L armour Frequency gIMb w so h B or as a vector QM treatment gives same result but in terms of expectation values For the case where B is uniform is there a net translational force acting on u No the 939 simply executes a circular motion about an axis H B Suppose however that B is nonuniform converging is there now a net translational force Yes upward Clar39 I ation classical dipole in a eld I Consider a simple dipole in a eld Since AU in I B We have the lowest energy Whenm amp B line up And recalling that in geneml 17 U shouldn t there be a force to alignm amp B 7 A l a P A a A l 0 What s V in spherical coordinates 7 V rieii ff 0r r 06 rsln6 aq Hrnmm sure look like there s a force to alignm amp B However What do we expect if our magnet is spinning 7 Conservation of angular momentum won t let it simply line up with the eld Instead we get precession Looking at this another way let s consider the force acting on the poles of a magnetic dipole gtu In analogy with the behavior of an electric dipole in a converging Fx electric eld what can we say about the the force on the poles N 5 0E BB FN gt Fs 39 upward In generalcan show that F a a 2 4 R z z where z is in the direction of increasing eld strength Conclusion a magnetic dipole in a nonuniform magnetic eld experiences a torque causing precession and a force causing displacement in the direction of increasing B The StemGerlach ex eriment and electron s in 1922 Stem amp Gerlach sent a monoenergetic beam of Ag atoms through a nonuniform eld B Since Ag atoms are neutral the only force acting on them will be FZ dBdzmlz Classically what would we expect 7 Classically we expect that 2 can have any Value from H d to 141 and F2 accor ingly Quantummechanically what do we expect 7 Quantummechanically All g1pt m1 where m1 71 71 1 71 2 1 be 5 1711 Hence the positions of the Ag atoms on the screen wil discrete And what did the experiment show 7 Qualitatively the same for other 2 orientations and for ifferent atoms ie two or more discrete ands What s wrong with this picture 7 Observed classically predicted Possible Values ofptlz is 21 11e an odd number not 2 Hence the Schrodinger description must be wrong or incomplete 1927 Phipps amp Taylor using the StemGerlach technique observed two lines produced by a beam of H atoms where I 0 in the ground state simple single electron atom Who hasn t guessed the solution yet We assume that the 2 has an intrinsic spin magnetic moment 1 due to an intrinsic angular momen S 7 to rst order we can think of this as the electron producing the external magnetic eld ofa magnetic dipole due to the current loop associated with its spinning charge We further assume S rr 17z where g is called the sping factor S1 mjz Based on the Phipps amp Taylor experiment how many values can us assume 7 two And since the beam piit into r what can we say about these two values 0f Ms 7 They are equal in magnitude and opposite in sign Ifwe make the final assumption that the possible values of m should differ by one and range from 7x to x what can we conclude about two possible values of m 7 m 1 and we can also conclude that x 12 BB By analogy With our equation relating F2 and plzwe can write FZ a Z Mbgsm z And since BI and M are known this allows gxm to be experimentally determined Within experimental uncertainty it was found that gxm 1 Since we ve already concluded that m 1 2 we must have that gs 2 These values of and g have been confirmed many many times by the Zeeman e ct which is the splitting of spectral lines by a B field For example the Zeeman effect is demonstrated by H atoms in the ground state That energy is split into 2 levels symmetric about the value at B 0 E usB MZB gthmSB tgthBZ 2 i Measurements by Lamb showed that actually g 200232 This value ofg is accounted for by quantum electrodynamics developed by Feynman Schwinger Dyson and Tomonaga Example 82 A beam of H atoms is emitted from an oven at T 400 K and is sent through a Stem Gerlach magnet of lengthX l m The atoms experience a eld gradient of 6362 10 Tm Calculate the transverse de ection of a typical atom in each spin component of the beam due to the force exerted on its spin at the point where the beam leaves the magnet 1 What 221 relation Will you use to solve this 361th NOW what 7 TO nd a2 we need the force and to nd I we need vx we already have the magnet length 63 m aB 62 Au ng s 62 Mb What force do they experience Fz Now we need vx how do we nd it From the average kinetic energy which is The average kinetic energy of the atoms in the beam is 2kT rather than 3kT2 why va22 2kT 2 vx 4kT M 2 and how do we nd t tXvx XM4k 12 2 23 so Zlazt2l X2 idedzubX i10x093xl3 x1 alum 2 2M 4kT 8kT 8x138x1039 x400 The StemGerlach experiment was actually motivated by the predictions of Goudsmit amp Uhlenbeck 1925 who postulated the 6 spin of ms 12 amp gs 2 to explain the ne structure of alkali atom spectra The nature of the spin It is not due to any spinning motion of the e39 If that were the case the surface of the e39 would spin at speeds gtgt c And that s only one problem with that picture since it now appears that the e39 really is a pointlike particle with an upper bound of 103916 m on its size Finally Dirac showed that the existence of the e39 spin is postulated by the relativistic form of Schrodinger s equation k d39 pRL97 03030 2006 PHYSICAL REVIEW LETTERS New Measurement of the Electron Magnetic Moment Using a OneElectron Quantum Cyclotron B Odom D Hanneltc B D Ut so and G Gabl39itzlsct Department anltyxlrx Harrrrni University Cnmbridgn Mameinserts 02138 USA Received 17 May 2006 published 17 July 2006 new mensurement resolves cyclotron and spin levels for n singlereleclron quantum cyclotron to obtain nn electron magnetic moment given y g 2 1001 159552 13035 751mm ppl The n E q a a o E m s 2 r ncei39r s o tan in the post and g is shifted downward by 17 standard deviations The u g wrt n q tum electrotlynamics QED calculation detelnlines the line structure constant with n tomrrecoil determinations Remarkably this 100 inK measurement probes for intemal electron stnlcture at 130 GeV 2 I r E 2 E i S E in a E r a 3 DOI 10llOSPllysReVLett9703080I PACS numbers 062011 1220FV l340Em l460Cd PRLgmomgozlmom PHYSICAL REVIEW LETTERS week ending 21 JULY 2006 New Determination of the Fine Structure Constant front the Electron g Value and QED lt1 intuitle D Hunnote39 T ltinusliitu2 M NtnJ untl B tltlnni3939 Lrnrnn Lnlmmtm Hannnr Ulll39tleriin39 s Llllwmlur r lrnnnt tlrritlr PI rnnrll Ulllv IIIltit39tlA e11 lilt39k 4851 USA 3Tllgrlrm t39ml Pll Lnlmmmrv RIKE Wnka Sm39mmn Japan 35l0108 Recclvctl 17 May 1006 publisltutl 17 Jul 2006 Cilillhridgl rllasmt llllsen x 02 I353 U A 39 mum I t 39I of the election at and the titre structure constanl tort A qllanllltn cyclotron together willi A QED Cnlculnlinn involving 891 eighthorder Fcyllnmn tlingmlnsr tletenrrine trquot 10359997101901 070 ppb The uncertainties are 10 times smaller than those of inpal39imrw nl39 measured and culclllttlcd t1 ly and set a limit on internal electron structure III moment new utensurement of g using n otterelecu39on llcart sl rival llk llltX ltut include atom recoil measurements Cn test QED most stringent DOl lUllOJPllysR Lctt97HSUSlll PACS numbers 062011 ILECLFV 1340Em 46061 e77 S L E a trap cavity electron top endcap electrode quartz Spacer compensation electrode ring electrode x nickel rings 05 cm1 compensation 1 electrode bottom endcap field emission point GIGCtFOde Va 2 EBZwm ms3912 7 I 1 1 l E11 my gillms n gym IF 512511 E my N Va 2732f c vc f c362273lt2fcgt39 3139 20 uu uon39nu39o 39 39 39 ol39 39 Q 5 g 10 en E 0 gtN E 0 I I I Foo0005quot nu 00 Ion n0 0 20 4O 60 O 20 4O 60 time 5 time s C I l 9 g 020 c d 5 g 01 9 396 9 le E 3 005 C S 000 o o o U 1 0 1 2 0 1 frequency 7a in Hz frequency fC in kHz A new value for the electron magnetic moment g2 1001 159 652 180 85 76 076 ppt 9 40 s Platquot100 How well is on known a demon 9 Harvard 2006 H 99 i Rb 2008 elemron g uw 1937 39 En 55 90 95 100 105 1116 quot x10 v 39 b o 9mm 9 Hmvavd 2006 Rmuos m lt Mgmm H4 quamum Han 20011 neutron 1999 muonmmms 1999 gt an Josephsam em 0993 w eIecnon g UW11987 5 o 5 1o 15 20 25 4137 035 mam1039s FIG I color The leasl uncertain u determinations 356 In wilh older determinations l on 1 0 times larger scale b Measured g are convened lo I using current QED theory a 137035 99971090 33 066 ppb024 ppb 137035 999 710 96 070 ppb tat th th than ltbi irt H lan Vlt i VlVlcl lVldi FIG 3 Typical diagrams from each gauge invariant subgroup that contributes to the eighthorder electron nmgnetic moment Solid and wigeg curves represent the electron and photon respectively Solid horizontal lines represent the electron in an external magnetic eld 2 3 4 5 1 C2lt3gt C43gt C43 C8lt3gt 2 7T 77 W 7T am ahadronic weak The SpinOrbit Interaction u 7 L39 L 39 39 s ruin maonetic dipole moment and is orbital ma etic di ole moment 39 A 1 39 39 ie protons or iuiimi 39 39 39 neutrons is so strong it governs the periodic properties ofnuclei Recall the Bohr atom 7 in our usual picture the electron travels around nucleus This is certainly the view from the nucleus point ofview However in the reference frame ofthe e the nucleus moves about the 2 so the e is in effect inside a current loop which produces a magnetic field no L t JZev And what law from 222 gives us the B field from a current loop 7 21 0x7 77 3 gz iif 4 r 4 r BiotS avart law BiotSavart law 3 ampX7 zeMO x 411 r3 411 r3 I7 Ze 17 Recall class1cal EampM and ask what 3 1s E Field acting on the electron E 3 471780 r 4nso r s 5 E b c 1 g 1 E O 8 V X t so V X 3 0M0 u Vsoiuo c2 What do you expect the orientational potential energy of the equots spin magnetic dipole moment to be E M pLSB gSMbhSB However this is the energy in the equots reference ame it can be shown that in the nucleus s reference ame E M 12gSMbhS B see Appendix 0 Thomas precession It will be convenient to express this in terms of SL instead of SB to begin we ask what is the force on the electron due to the electric eld E of the nucleus F eE dV A dV And what is the relation between Force F and potential V F E E F 1 CW 1 1 1 1 dV u u So E d B 2vgtltE 2 e e dr r c ec r dr 1 1 dV NOWWhatlsLequalto Lrxprxmv mvxr so B emc r dr Note that the B eld experienced by the 6 due to its motiOn about the nucleus is 06 to L gsub 1dV E S o Putting this into our express10n for E M we have M Zemczh r dr gs 1 dV eh 1 E S L Recallin that 2 and e et E S 39L M Zemczhr dr g gs Mb 2m W g Note that this result is correct in the relativistic limit Example 83 Estimate E M for the e39 in the n 2 l 1 state of the H atom and compare it to the observed nestructure splitting 2 l dV 62 l Clearl we needVr V e y r 47180 r SO dr 47180 r2 Therefore E 52 35 Z Since the magnitude of S and L are each h M 4130x2mzc2 r 2 2 2 assume S L 3 h and for the n 2 l 1 state ltlr3gt z l3ao3 2 3 6 8 e 1 me 2 me M E w h 1023 J10394 6V 47180 x2mzc2 33 431906 544 804 2h4 so the splitting is 2gtlt10394 eV comparing this to the energy of the e in the n 2 l 1 state E2 34 eV we nd a nestructure splitting of 1 part in 104 same as discussed in chap 4 Example 84 Estimate the eld B created by the orbital magnetic moment acting on the spin magnetic dipole moment of the e39 in the n 2 l 1 state of the H atom This is big at the Since IEMI IMS B M33 and 115 Mb 103923 Amz B N IEMIMb N l JAm2 l T limit of an iron core electromagnet Total Angular Momentum If there were no spinorbit interaction L and S would be independent of each other But since there is a spinorbit interaction ie a strong internal magnetic field is acting on the spin of the e39 it exerm a torque on S This torque couples L and S causing them to precess about their sum with the orientation of each l dependent on the orientation ofthe other We find that they precess about their sum instead of lying on cones symmetric about the z axis Hence the values otLZ and S are not fixed If the atom is in free space subject to no net external torque the total angular momentum J L S has a constant magnitude and a constant JZ Similar toL and S the magnitude of is Note also that clearly J l 2 h andJZ mjh with m j jl j2 j lj JZ L SZ 2 In the absence of the spinorbit interaction we would have L2 quot117 and 32 msh We would still be able to de ne J as above and get m m m5 Then clearly mjmax l 2 This last result is actually valid even with spinorbit interaction because conservation of angular momentum prevents an interaction which is internal in the isolated atom from changing the vector J ie from changing JZ as well values of 1 am 11mquot W1 quot 1W 239 1 12 t 7 12732 7 52 but where does it terminate 7 quot 7 2 But since quot r L LS2HLSH wehave MAMM s LS s 1 s 1 mum L S L L L LS L L 7 Lb LS Must 1 AND22 1h7 ss1h 4L So satisfy this inequality 171 12 7 12 Notethatifl0s12 F amnlo X The possible values of J are 52 and 32 quot392 Q n N u 2 u u N v m N N u n N n N N 3 u For 52 m1 52 32 12 12 32 52 For 32 m1 32 A 12 32 12 12 32 SpinOrbit Interaction Energy and the Hydrogen Energy Levels We rst obtain an expression for the spinorbit interaction energy in terms of Vr l s and j Then we show how it predicts the detailed structure of the energy levels of the H atom 1 1 dV We saw that S L 2mzc2 r dr SinceJLS J JL LS SZS L so S LJ JL L SS2J2L2SZ2 or S L h2jj 1 1 1 Ss 12 ie r22 EM Wl lll Ssl In Example 83 we saw that for the n 2 l 1 state of the e39 in the H atom E M 10394 eV But that is also the order of the relativistic corrections to the energy levels The complete relativistic treatment yields 4 2 EFL 10 i 4222 152 n j12 4n 2 e l IS the ne structure constant 47180710 1 3 7 where O Splittings are exaggerated by a factor of 1372 188 x 104 Le as ifoc 1 corresponding mmerfel since those energies are degenerate if there is no external eld Energy EV Bum Summerield n3 quot2 ne2 Ram1 15 12Jo1 4 2 EM L 041 Schrodinger 39 animal 7 12 4n MZ 26AL1 ZZZ Li Sommerfeld 24nso2h2 n2 n he 4n Chap 4 1 quotquotquotquotquotquot quot What doyou notice about the case ofthe 181 x 10quot eV Hydrogen atom 7 eqns are quot51 j12i10 them e Since the Sommerfeld model is based on the Bohr model it is only a very rough approximation to rea i The Dirac theory on the other hand represents our most re ned view ofhoW the World Works How is it that these tWo models give the same results for the Hydrogen atom 7 Coincidence But this coincidence caused much con ision during the 1920s as the quantum theories Were developing What else do We notice about the Dirac energy level 7 For a givenj looks degenerate Wrtl Actually these levels are not degenerate either Hyperfine S gl Complete QED treatment predicts energy differences in the microwave radio region Lamb Shift was measured by Lamb in 1947 Can anyone guess Where this shi comes from 7 Where does the spinorbit coupling come from 7 Interaction of electron spin magnetic moment with internal magnetic eld of the atom due to motion of electron Are there any other sources ofmagnetic elds in an atom 7 Nuclear spin magnetic dipole moment toms kickedto aveguioe alternating E field h hv N 44 nev microwaves Some H metastable state n 21 0 w l Wit A 7 5 139 sensitive only to metastable atoms l2 n21o n1l0highlyinhibitedbyA1 selection rule and that all other states lie above except n 21 lj v2 which Dirac said was degenerate 4 Experiment rounalttn be he below39 lz1 12 10 X Grnul id stale Transition Rates and Selection Rules Recall Chap 4 excited atoms may decay to lower states by photon emission but not all conceivable transitions occur and those that do occur at widely different rates Photons are observed only with frequencies corresponding to transitions between energy levels whose quantum numbers satisfy the selection rules Ali1Aj0il If the wavefunction describing an electron in an atom is an eigenfunction of a single quantum state what can we say about the time dependence of the probability density Constant If however the wavefunction is a linear combination of eigenfunctions with different energies then what can we say about the time dependence P 1r will contain terms of the form expz39E2 E1t 2nh which oscillate with I at a equency a E2 E 1Zrth Note probability density of the e is also its charge distribution function Charge distribution of the e and consequently of the whole atom will oscillate between the states 2 and 1 But the atom is neutral so the simplest quantity related to its charge distribution that can oscillate with time is its electric dipole moment the product of the equots charge and the expectation value of its displacement from theessentiallyfixed nucleus We can actually use the classical formula for the rate of emission of energy by an oscillating electric dipole to obtain some factors for atomic transition rates Appendix B shows that a classical oscillating dipole radiates energy at an average rate R 4Jr3v4 2 3230c3 where p is the amplitude of the oscillating dipole moment and v is the frequency Since the energy is carried off by photons whose energies are of magnitude hv the rate of emission of photons is 43313 2 hv 39ch3 Note that this is the probability per second that a photon will be emitted and therefore the probability per second that the atom has undergone the transition Relative to the essentiallyfixed nucleus the electric dipole moment p of the l e39 atom is p er To obtain an expression for the oscillating electric dipole moment of the atom when it s in a state which is a mixture of two eigenstates we need to calculate the expectation value of p To that end we need to evaluate gtllt gtllt gtllt gtllt gtllt gtllt gtllt gtllt gtllt gtllt gtllt II l1 c1 lPl c2 l1 2 cllP1c21P2 c1 011111 lIllc2 021112 l112c2 011112 1Illcc where cc is the complex conjugate of the previous term Yet how do we determine c1 and c2 The answer is that we can t but since we ll see that the results are independent of c1 and c2 we set them to 1 c1 c2 1 Then l1 1I wf1llf 115 11 1p1pfexpiEl Eft 2nh cc 5 13 0lt fwjie 1pde fwfefwidr expEE Ef h11pel7wfdt cc If V is spherically symmetric 11f1Jf and 1111Jl will have even parity so wfer1Jf and 111er1pl will have odd parity so the rst two integrals will vanish It can be shown that even is V is not spherically symmetric those two integrals vanish Therefore the amplitude of the oscillating electric dipole moment is proportional to pf EUIP671id l7 This is called the matrix element of the electric dipole moment taken between the initial and nal states D0 Example 86 yourselves Chapter 6 Outline Solutions of the Timeindependent Schrodinger Equations Nonbinding and Binding potentials The Zero potential The Step potential energy lt step height energy gt step height The Barrier potential transmission coefficient electronatom scattering Ramsauer s effect size of resonances barrier potential Examples of barrier penetration 0c particlenucleus potential 0c emission GamowCondonGumey xdecay theory atomic clocks The Square well and infinite square well potentials The Simple Harmonic Oscillator potential zeropoint energy parity HW 6 Chapter 6 due 1021 questions 24 26 problems 11 12 20 21 24 25 30 31 32 Pxl 2 alum V 11 22 Zm 3x xJ xtl at I Recall the Schrodinger equation I As we look at different physical situations what is the only thing we expect to change 7 The potential Vxt We will start with the simplest potential that is no potential Vxt 0 and gradually add complexity What does it mean in terms of Vxt and E for a particle to be bound 7 The Zero Potential dV If Vxt 0 for our particle what force does it experience 7 F doc O x hz dzyx The timeindependent Schrodinger s equation for this case is 2 7 E 00 m x I 2 or with k we have Izzyx What are some solutions 7 x Most obvious 91x Acoskx Bsinkx we will pick WIOC elkquot coskx isinkx recall that the full solution is of the form 1Pxt lIxe E h So our full solution is lPx t 600quot What is this physically 7 traveling wave propagating in the x direction But a traveling wave propagating in the x direction is also a solution with energy E V2 x 6 k H gt lPOM e Since an ordinary 2nd order differential equation has two linearly independent solutions 1105 and 1205 are the two solutions so most general solution of the time independent equation is W A m x 31206 M Be lk For our traveling wave propagating in the x direction I xt should look like What will I xt I xt look like 7 Ix t Ixt e ZW mle m 1 We should now consider the physical interpretation of the solutions First let s evaluate ltpgt for the particlewave moving in the x direction what would you naively expect 7 p22mg p 2mg 3 Since pup iih i pap l 717137 Ae w quot may me Bx 3 ltpgt I P12mE de2mE Izmm Real padol i39lx r o Lm2Al x Alli 4 x 041 I lt p gt I I prde For a wave traveling in the x direction lt12 What about our particle s location in x clue 1 Our wavefunctions have a single value of k and therefore p since p h27tk clue 2 describe a beam of infinite length Px P P const The uncertainty principle is also satisfied here because as Ap gt 0 AX gt 00 Similarly for E and t Since there is no uncertainty in E we need an infinite At to measure the energy of the particle in the infinite beam Although these plane wave solutions are idealized wavefunctions many particle beams can be approximated as plane waves eg monoenergetic beams of 19 e n etc emerging from a source or accelerator However idealized plane waves are mathematically problematic 00 00 90 If we try to normalize the wavefunction we get I P I dx IA Adx A A Idx 1 One solution is the introduction of Dirae s delta metion 60c 60c 0 for all x i 0 and 5006116 1 This is however beyond the scope of this course quotm Thinking about the actual physical situation the solution is obvious theparti cle is clearly confined to some region 1 acceleratOr detethor a one dimensional box from L2 to quotL2 Bax normaliZatiOn which stipulates that 1 a plane wave for L2 S x S L2 and O elsewhere Although A is then dependent on L we will find that the result of calculating a measurable quantity will be independent of L As we discussed in Chap 3 we can obtain a more realistic wave which is really a wave packet or a bunch of wave packets a stream of overlapping y s by adding a large number of waves of the form r n ttnAr I i o h Ilxy 31bt70r each with a different value of k and a so their superposition will generate the group of waves or wave packer Chap 3 V fxt LNx I7 4 o Taking a simple example 1 packet moves in the x direction and spreads uncertainty principle 0 The location of the group at any time ris given by lt x gt I P x l dx d The constant velocity of the group is 3 VZEm m m Due to the large number of Fourier components the mathematical description of the wave packet is very complex Therefore wave packets are rarely used for modeling However we ll consider one very specific case where Vx 0 changes so slowly that it s almost constant over one 1 In that case it can be shown that d dltxgtltiiVxgt ltdvmdxgtltpmgt Schrodinger seqn m 3 dt d aNewton s Law in classical limit dtz m m New p s Law is a special case of schrodinger sequation l The Step potential energy lt step height m So what s a step potential 7 Wu 1 A real physical system with such a potential would be a charged particle moving along the axis of two cylindrical electrodes held at different potentials WA WA Consider a particle ofmass m and energy E lt VO moving in the x direction towards x O Classically what do we expect 7 The particle is free until it hits the wall where it rebounds Vrs and p gt 7p Mathematically we see that it has to bounce back into the amp x lt 0 region because ifit were to continue into the x gt O E 39 ld t Iago WE W011 g6 EVx ltVx gt p2 lt 0 m VC So how should we solve this quantum mechanically 7 vm Solve forx lt O solve for x gt 0 then join the two solutions atx O in such a way as to satisfy the requirements discussed 1n Chap 5 Both W and dex must be finite singlevalued Vm Va continuous 7 2 dill2 For xgt O 7 dxz V0yZxEyZx 7 7 We just solved this one who remembers the This can be rewritten as i 222 V0 7EyZx m general time independent solution 7 x 7 1 Wm Ae k Be C or with k2 2m0 E gt 2 kzwx Wham k1 What s the general solution 7 W200 Ce De39kzquot What can we say about C C 0 otherwise W Va co as x gt co And since Mx must be continuous atx 0 what additional constraint do we have 7 A B D d x z 4 x d x We also have that dex must be continuous atx O l Aek iBe 1 2 7k2De 1 Equating these derivatives atx O we get ik1A 78 ikzD or ikZk1D A 78 D ik D ik V 39 A 7 1 4 B i 17 J ButsrnceD AB 2 k1 2 k1 So only one of the four constants remains undetermined D W V and it is the normalization constant I D W w WXlzk2kle lizk2kle X50 vruo 7amp1 Derkzx x 2 O 39We ll quot 39 u btui39d p quot thr nWLhe 39 I l A m t u quotM P NR f fun 39 39 39 quantities the cancel out The complete wavefunction is Let s interpret this physically for x S 0 What does the 1st term describe 7 a particle moving in the x direction What does the 2 term describe 7 a particle moving in the 7x direction So we can interpret the 1st term as the wavefunction of the incidentparticle and 2 term as the wavefunction ofthe re ectedparticle The re ectian caefficient R is the probability that the incident particle will be re ected BB 1ik2k117ik2k171 in agreement with R AA liik2k1lik2k1 classical behavior D i 1 39k k W 14k k quotW x30 Now recall our time independent solution WW 2 2 lgte l 2 lgte substituting em cosk1xisin klx we find Dcosklx Dk2k1sink1x XS 7k e 2 W06 So Px t 11006 is not a traveling wave but is instead a standing wave since it has standing nodes at all points x which satisfy cosklx 7k2k1sink1x 0 So for x S 0 our solution is in complete agreement with the classical predictions However for x gt O we have a radical departure from classical physics W Dew mx u I l39 AH I Px r r DDe W Although a 39 quot 39 i 1 39 439 nonzero h parficle pehetrdt s the classicallyf0i idden re39giqn When we deposit additional atoms on a clean surface electrons will re ect from them and the wavefunctions will interfere with each other creating standing waves of electron densities Cu on Cu111 wwwspecsde This has been confirmed experimentally innumerable times over the past 70 years I 2 V E The penetration depth is defined to be Ax lk2 and 811106 k2 h Ax which decreases rapidly V 2quot100 E with increasing VO E In the classical limit 2mV0 E is so large relative to h that Ax is immeasurably small Example 61 Calculate the penetration depth for a small spherical dust particle of radius r 1 pm 10 6 m density p 104 kgm3 and speed 1 1 cms 10 2 ms impinging on a potential barrier V0 2E 2p22m h m 47zr3p3 4x1014 kg so E mv22 2x1014 J so Ax W M0 19m Consider an experiment designed to measure the coordinate x in x gt 0 Since Px is appreciable only in Ax above we must have Ap hAx 2mV0 E12 AE Ap22m V0 E So if a particle has penetrated the barrier the uncertainty in its energy is roughly equal to the additional energy required to pass over the barrier in the classical case However the phenomenon of tunneling which requires barrier penetration is very real and extremely important in understanding various systems It is also extremely important technologically Example 62 A conduction e in a block of Cu moves with energy E in a potential V z 0 inside the block which steps to V0 gt E outside The workfunction energy required to remove an electron from the block is VO 7 E 4 eV Estimate the penetration Ax of the 2 into the vacuum outside the metal 710 o V07E4eV6X10 19Jm91gtlt10 31kgso AX 2mltV0E10 m 1A Penetration of waves into forbidden regions e g EM waves into metals or into the forbidden region of the re ector in total internal re ection is well known so the QM phenomenon is just another manifestation of the wavelike properties of particles Before moving on let s see what the re ection from a step potential actually looks like for a group wavefunction describing a particle 5 m 1711 19A LllA1 Mfrl 12A Mm gum Obtaining such solutions involves considerable work on a computer 7 we were however able to learn much by exploring the behavior of simple wave functions Key pointszvparticle is re ected from the step with probability onebut there is somepenetration into the classically forbidden region The Step potential energy gt step height Vx First question what happens to a classical particle 7 E A particle moving in the x direction will simply slow W Vquot down at x 0 from v1 plm 2Em Z to V2 pZm 2E r V0m Z What happens to a wave 7 1 A wave traveling in the x direction and passing from the rst medium to the second medium will be partially re ected and partially transmit So the quantum mechanical interpretation for a particle would be 7 The particle has a certain probability T of being transmitted and a probability R 1 7 T ofbeing re ected The general solution here is of course the same as for the x lt 0 part of the E lt V0 case x AeW 32 x lt 0 h m 2mE Vo V Cgmzx D57u 2x xgt 0 W em k 7 and k2 77 Noting r the physical meaning of each term What can we say about D 7 7 that the particle is moving from x lt0 towards x gt 0 D 7 7 that there is no particle moving in x gt 0 towards x lt 0 And from the continuity of Mx at x 0 what constraint do we have 7 A B C What do we get fromthe continuity ofdede atx 0 7 ik1A iB ikZC A 7 B kZk1 C k C 1 C 2 Addingthesetwo equationsweget A1z2379k1 2 Ck1qsz k C kik C kik And subtracting Bliki33 B 1 2A 1 1 1 k1 k2 For S 0 what does the rst term represent 7 Ag W A k k2 W lt 0 incident particle Wm ki k2 I and the second term 7 k2 k e X2 0 re ected particle 2 For 2 0 what does the only term represent 7 w u n transmitted particle Choosing l 739 U 139 H N I k 2k So what does POM P Iquot look like 7 W I S 0 combination of a traveling plane wave oscillatory standing wave X2 0 traveling plane wave a Wq AlA x The re ection coef cient R is 41 C C 2 The transmission probability is T liR 2 3t 33 krkz 1wsz AA AA k k2 C C woops Why doesn t T equal 7 1sltlwhenkzltk1EgtVD AA The velocities are different in the two regions and the convention is that transmission and re ection coefficients are defined in terms of ratios of probability uxes A probability ux is the probability per second that a particle will befound crOssing some reference point traveling in a particular direction Since the probability per second that a particle will cross a given point is proportional to the velocity in addition to the intensity according to the strict definition v13 B B gtxlt B as before WA A AA 2 v CC v 2k p1 hkl P2 W T 2 2 1 and MA v1 k1 162 and since v1 m and v2 m m 2k 2 k 2k 2 4k k Tzv 2 1 2 2 1 221 R WealsoseethatRT1 V1 k1k2 k1 k1k2 k1k2 Note that R and T are unchanged if lt1 and k2 are reversed gt the same value of R and T will be obtained if the particle is incident from x gt 0 and travels in the x direction The wavefunction is partially transmitted and partially re ected due simply to the discontinuity in V and not because Vincreases This is just like waves on a string where re ection occurs if a wave hits a point where the density or tension on the string changes or like light passing from medium 1 to medium 2 whether to higher or lower n This is a property of all waves in optics it is sometimes called the reciprocity property To get a better feeling for this let s look at how R and T behave as functions of For E lt V0 R l T 0 as we saw in the previous section and as EV0 increases Tincreases and R decreases as we would expect Rewriting R in terms ofE and V17 we have 2 05 10 1 5 2 0 1214 w mi m 1 17V0 E V0 Question No real potential is a step function 7 when can a real potential Vx be approximated by a step 7 Answer If Vx changes Very gradually R will be negligible since change in A of particle is gradual 7 re ection arises from abrupt chan e in Speci cally ifthefractional change in Va is veiy small when x changes by one A gt R will be veiy small or particles in atomic or nuclear systems lcan be Very long relative to distance oVer which Vx changes 7 then the step approximation is Valid Example 63 A neutron n with E 5 MeV which is typical of a neutron generated by fission enters a nucleus where the bottom of the potential well I is at 750 MeV Estimate R lt 1 315121 ALA Take V0 50 MeV E 55 MeV 2 R 17 17V0E 14175055 203 1 14015 1t1750ss 39 The Step potential energy lt step height For xlt0 WK For xlt0 hi dzvy hi y 7 E 4 E m M WK W Vquot M ix V0 WAX 11105 AW Be W u vm u where 7 77quot From the continuity of Ml at 0 From continuity ofdyldx at 0 DAB iklAiB7kzD D x 1 x E1lk2k1elk 1 lk2k1e 6 M De k x 2 0 V416 The Step potential enegy gt step height Foxxlt0 V 1 W 2m 11 wa V OC Ae 7 3124 Wu 0 ZmE h Ftom the continuity of w at x 0 ABC Fox x lt 0 z E if d2 T yanmwx V00 w m y2xCeW De W x k2 7Vzmwo D 0 h Ftom the continuity of dwyiix at x 0 ik1ArBik2C gt ArBk2 c1C The Barrier potential vw What is a barrier potential 7 Classically what do we expect for particle traveling in the x direction towards x O 7 IfE lt VD particle bounces back ifE gt VD particle passes barrier slow down in O lt x lt a speed up x gt 1 Quantum mechanically we find tunneling through the barrier when E lt V0 with a probability which a increases as E VO increases and decreases as 1 increases b We will also find that inside the barrier Mx decays exponentially Solving Schrodinger s equation for Mx Looking rst at the regions x lt O andx gt a what are the general solutions 7 Ae k Bed 6 lt 0 VZmE x h k 7 V4 Cele Derzk xgt a W era 1 h And since there is no particle moving in x gt 1 towards x lt a what can we say about D 7 D 0 What info do we need before we can write down the general solution for the region 0 lt x lt a 7 Whether E lt VD or E gt VD We ve already examined both cases 7 let s start with E lt VD So what s the general solution in the region 0 lt x lt a with E lt V0 7 2mV0 7E W16 Film 68 where k2 h What are our continuity conditions 7 A B F G ik1A 7 B k27F G Ce k Feikz39z Ge w ilee k k2Gek2 Fe 2 As expected we have indeed 4 linear equations in the 5 unknowns A B C F and G To solve first express F and G in terms ofA and B then C in terms ofA and B then finally B in terms of A was t Wx Am What does WW look like 7 1 m 7m 2 2 ForT wefind TM1e e Hs nhil WA A 16EVo1EVo 4EVo175V0 E E and if kza gtgt 1 then T 1671117 Vje m o Tdecreases exponentially with increasing barrier width a The last two results provide the essence of barrier penetration or tunneling phenomena which are purely quantum mechanical 5112le El 2 2V E Note that since kza To 1V is extremely large for classical m a O and V0 the probability for tunneling of classical macroscopic particles is vanishingly small Now let s go back and look at the case where E gt V0 in the region 0 lt x lt a rzkzx lk2x k Vl2mE VO What s the general solution 7 M30 F e Ge where 2 h zkzai izkza Z 1 V Z 1 2 So TvlCCl1 6 e l11 sm kid I with kzu ZmQZVO 1 v1AA L 16EV0EVoilj 4EV0EV071 h V0 Example 64 An 2 is incident on a rectangular barrier V0 10 eV and thickness at 18gtlt10 10 m 18 A Evaluate T and R as functions of the energy E of the incident 2 zmazv0 2gtlt9gtlt103931gtlt10gtlt1i6gtlt103919 gtlt 1i8gtlt1039102 szmz9 Note that for E gt V0 T 1 whenever kza mri a result of destructive interference between re ections at x 0 and at x a This is reminiscent of the Ramsauer e ecz in which 2 of certain energies of a few eV pass through a nobel gas solid as if the solid wasn t there T 1 and of scattering of n with certain energies in the MeV range from nuclei 7 this is the size resanance 15 vi Finally we note that the time independent Schrodinger equation is of the same form as that governing classical wave motion For EM waves of frequency V passing through a medium with an index of refraction u 27W 2 WquotT j W 0 where llis either the electric field E or the magnetic field B Comparing to the time inde endent 2 2 Schrodinger equation V397TE Vltxgtw 0 we find 06 7mE Vx so the behavior of an optical system with ux should be identical to that of a mechanical system with Vx related to x One example is the coating of lenses to obtain very high light transmission Another example is in the imaginary ux encountered in total internal re ection which shows how we get frustrated tatal internal re ectian when the layer of air wedged between the two solids is very thin Detailed treatment shows that x along ABC is real on AB but imaginary on BC There are EM waves in BC but their amplitude decreases exponentially with distance into air They are also standing waves so their Poynting vector S 0 so the light is totally reflected But if we add another block of glass we find propagation of the wave through the 239 glass which is frustrated total internal reflection Frustrated total internal re ection was first observed by Newton ca 1700 step may And you can even see this with water waves Examples of Barrier penetration by paiticles e tunneling through a thin film of SiOZ A1203 or CuO2 eg from one wire to another Oxide layers are typically 20 7 30 A thick so without tunneling we would constantly face problems of connections becoming disconnected aemissiszmm a emitting nuclei Historically 06 s with E 88 MeV emitted from 212Po and scattered from 238U showed that Vr ZZe247wor for distances r 2 r 3gtlt103914 m where V0 88 MeV But 238U also emits Dis 42 MeV So classically aemission should not occur In 1928 the problem was solved by Gamow Condon and Gurney Recall that for E lt VO amp kza gtgt 1 we obtained U235 T 216EV017 E V0exp 72kg exp E2 2mV0 7 E7z2a 39 7 t t g VL K Liitircggeiw Gamow Condon and Gurney then approximated E 42 a i If v the coulomb potential as a sequence of adjacent g r m rectangular barriers of height Vr amp width Ar w t f z z I 7 In the limit Art gt 0 they found T n2 1 2mV0 7 Ew r Now Tis the probability that in one trial the awill penetrate the barrier If the Otis moving at speedv inside the nucleus the number of attempts per second it makes to penetrate the barrier is N vZr so the adecay rate should be H R exp721 2mV0 7Ehz r r t Using v as the speed of the Zafter it escapes and r 9 F 9gtlt10 15 m yields an agreement between the calculated and observed R However R varies tremendously among nuclei from 5gtlt10 18 sec 1 for 238U to 2gtlt106 sec 1 for 212Po This variation is due primarily to the E of the 06 lhsec ll The periodic inversion ofthe N atom in NH3 the oscillations of the N atom between the two minima occur at V 2quot 786gtlt1010 Hz when the molecule is in the ground state which is much lower than the vibrations without barrier penetration Due to its relatively low but fundamental value this vibration was used for atomic clocks Atomic Clockquot Aside on the Ammonia NHa atomic clock How do we measure time We count something typically something with a fixed period pendulum What do we typically call something with a fixed periodfrequency An oscillator So what are the main elements of any clock l Oscillator 2 Counting mechanism accumulates the number of cycles 3 Mechanism to display count The stability of the oscillator is of course critical 1600 s 1700 s 1920 s 1940 s Water Pendulum Springdriven Quartz Atomic gt gt gt gt clock clock watch clock clock 310 min 30 sday 10 sday 3 usyr noncloudy Let s look in some detail at the Ammonia clock we ll cover more modern atomic clocks when we discuss masers next semester Wt G What does the potential look like If we solve Schrodinger s Eqn what do we expect WI and V2 to look like ix 206 Assuming the molecule is in a superposition of its two lowest energy39 states what is the time dependent solution Px 1 il1xe iE1th W2xe iEzth J5 What does ltxgt look like oscillates m 0 i JED11008 What s the oscillation frequency 7 wont iiElth W2 xeiiEzth x E E iv v 2 1 ltxgt h So how to do this in practice 7 hint first appeared in the 1940 s During W Wll the evolution of radar required the development of very sophisticated microwave circuitry ie the ability to control and measure microwave with high precision Apply microwave beam near frequency v to Ammonia gas tune v What happens when v E2 7 E 1h 7 Apply Feedback circuit to v tuning to maintain resonance Ammonia molecules are used as an ultrasensitive regulator of the frequency of the microwave frequency transmitter What well known device is this the basis for 7 MASER 7 Microwave Amplification through Stimulated Emission of Radiation Atomic clocks are so accurate that minute adjustments must be made periodically to the length of the year to keep the calendar exactly synchronized with the Earth39s rotation which has a tendency to slow down There have been 17 adjustments made since 1972 adding a total of 20 seconds to the calendar In 1997 the northern hemisphere39s summer was longer than usual by one second An extra second was added to the world39s time at precisely 23 hours 59 minutes and 60 seconds on 30 June 1997 The adjustment was called for by the International Earth Rotation SerVice in Paris which monitors the difference between Earth time and atomic time W 139 The Square Well potential quotu V0 xlt a2 orxgtal2 V06 0 a2 lt x lt a2 What do we expect classically If E lt V0 then classically i the particle will be bound to the well if E 2 V0 particle is free Many systems can be approximated with a square well potential ie many closely space ions in a line 0Mquot The general solution of Schrodinger s equation inside the well is 100 2 Ad Be ik where k i39sz Thine m m m 1 h mm Alternatively the two independent solutions can be written as 00 B coskfx and 1100 2 A sinkfx so the general solution is N Manycwoseny spaced MX A sinklx B Cosklx AMMAMMAMmn Outside the well for E lt V0 we have 106 Cek Deik x lt a2 k 2mVo E n a 0 n2 106 Fek Ge k x gt W2 H 3 What can we immediately say about D and F 7 Both zero As usual the relations between A B C and G are determined by the continuity of lland d de at x iaZ So we now have 4 equations and only 4 unknowns 7 this is a problem Why 7 We need to leave one constant free for amplitude and so we can satisfy the normalization condition To solve this we treat the total energy as an additional constant that can be adjusted as necessary where we will find that E can only take on certain values The general solution to the squarewell potential is quite complex and involves a transcendental equation 7 see Appendix H three bound eigenvalues three bound eigenfunctions vquot J13 x 7 i 7 E3 0 9 42 X E2 0 iii E1 l O I a 2 0 a 2 a2 o a2 Note alternating parity r V0 becomes Very large Ime moderate height v i at ms Uzmifimx iniinite height u2 0 tz2 The Infinite Sguare Well potential m m w r xltra2 orxgta2 Vx 0 raZ ltxlta2 For 412 lt x lt aZ the general solution is still 702 0 u2 M15 Asinkx Bcoskx where What are the boundary conditions 7 M15 0 at x iaZ Asinrka2 BcosJaz2 0 Asinlaz2 Bcoska2 0 adding we nd ZBcosUazZ 0 subtracting we nd 2Asinka2 0 Since we require that A or B be different from zero we get two classes of solutions Class OneA 0 coska2 0 Class TWoB 0 sinlaz2 0 M15 Bcosk7c M15 Asink7c ka27V2370 257V2 lea2117 kn nila n odd kn nila n even 4 0 2 hzkz zzhznz 2m 2m Zma2 Equot n consider E1 This is the lowest energy 7 the particle cannot have zero energy 7 basically due to the uncertainty principle Equivalently there must be zeropoint energy because there must be zeropoint motion w i aZ 6 a2 Note alternating parity again pz 7 hik 7 zhznz 2m 2m Zma2 119339 J r w 77 a W 139 WHO l 7 if 7 1 ill xi t 77 iii a2 0 a2 compare classical dashed with QM A V0 The Simple Harmonic Oscillator potential This case is so important because the local motion about any stable equilibrium point x0 can be approximated by simple harmonic motion To show this we show that the potential energy can be approximated as that of a SHO dV 1 sz Vx Vc0lXE x7x03 dxz l xixo D Since the potential has a minimum at x0 the 1st derivative of Vx is zero at x0 and we can choose Vx0 0 so we get 1 d 2V 2 7 l 2 look Vx 162 x xo ECOHXO familiar Quantum mechanically Planck already predicted that W En MW 11 O l 2 3 m but the solution to Schrodinger s equation yields En n 12hV n 0 123 m where V Cm12271j is the classical frequency see Appendix 1 Therefore E0 12hv is the zeropoint energy Was Planck wrong 7 Yes although only the energy differences were important However in 1914 he published a speculation based on entropy U x considerations that the zero point energy should be hvZ HOW to draw energy levels r e m What should the wavefuntions look like 7 Cm14 7x it With it Quantum Number Eigcnfunctions 0 1 Anew 1 wlAI c39ulr 2 2 11 AZH Zuzlei39ml 3 I113 A3u luste ziz 4 114 2 A40 12141 t 414 43 5 195 15514 7 2013 4145197 Once again we see that the number of nodes 11 Note that Via contain two terms a polynomial responsible for the oscillations of V100 a term expu22 responsible for the decay of 150 in the classically forbidden region Also note that the parity of V100 switches from even for even 11 to odd for odd 1 Nunl of Physical Pulcnlilll lllll Probability Signilicunl Synlsm Example Tullll linurgl ca l Fculurc Zero Prolon III I lquot Results UMJJ pomminl beam l mm v for ullxer LVLlulrnn m quot yslems Slap Conduclinn l39m Penelmlinn polcnlial cleumn nullr ft l wmv orexclndcd cncrg surface nl 7 mglon below lopl melal O 0 slap Nculr n null pnnlal rc ccr potential lrylng to x m mm M lcnergy es pt pmem lal llhnve mp nucleus a quot lllswnlinuily nrrler 2 pnrmle Tunneling polenliul lrylng m energy escup below up 39ulnumh bllrricr arrier Elcclrnn my N0 re culion palcmizll lerin mm at ccrlnin cncrg ncgnlm y lquot energies ubme mpl ionled lnmn u inile Nelllrnn V Lnerg square bullnd in i b M quunlilalmn cll nuclcm V Y D I O n lnllnllc lolcculc Ix Approxlmnlinn square smcll 1 m inllc well cml ncnl llmll square 39 I pmanllul ln hm l v 0 O n n l lnl Simple Alum lvr Zeropalm lnnmnniu Vlbrllling oscillulol diatomic olcnllal molcculc Name of Physical Pulcnnul and Pmbuhllil Sigm cam Sym 1 Example Tutu Lner Dcmil Feature em on in E my pmcmiul bcum I39rmn I cyclotron Lquot SIL p vm Pcnelrauon pownlml 712 qnq ufexcludud cnm regxon below mp o 0 F 1quot l39 R rlial rc ccr pumnml rm 39 energy l uhnvc mm 0 4 quot il rricr Tunneling pmcmml o 1 cmrgy hcluw mm o n Name of prlem Burricr pmcnliul lcncrgw above mp7 well pmcnlial Simple Imrmomc 05 W l pommial Pmmbilily Signi cant Density Fcauurc Phwml Pnlcmiul and Emmplu ulul Energies Nn rcllccuun Elcclrnn scul ul ccruun lcnng rmm i5 negatively W39W energies mied ulom no 0 a Nculmn rm I gt hound m E my quunllmuon nuclcm 3 o a a n Moleculc quot1 Appnmmulion slricll a time mnfxncd E ql w square we m hm x o a o 1 WI Zcrnpninl Axum of cllcrg nmluculu Chapter 5 Outline Schrbdinger s Theory of Quantum Mechanics Need for a differential wave equation limitations of the de Broglie postulate Plausibility argument leading to Schrbdinger s equation Bom s Interpretation of Wave Functions complex character of wave functions probability density statistical predictions Expectation Values operators The TimeIndependent Schrbdinger Equation separation of variables Eigenfunctions Energy Quantization in the Schrbdinger Theory geometrical properties bound vs unbound HW 5 due 10907 questions 20 28 problems 3 4 9 11 12 13 15 27 29 Need for a differential wave equation We have discussed experiments which clearly show the existence of the waveparticle duality for microscopic particles 7 our classical equations Le Newton s laws are not sufficient Although we had specific procedures which applied to specific cases there was no general procedure Recall that the de Broglie postulate said that the motion of a microscopic particle is governed by a wave but it didn t tell us how the wave propagates It does predict the wavelength but only in cases where the wavelength is essentially constant As we discussed before waves are typically governed by a wave equation hm a V r Eijqzi fsjcmaai ggr 39 11 g Tc urw Schrodinger s theory goes on to specify the connection between the behavior of the wavefunction and the behavior of the particles recall Max Born And of course it reduces to Newtonian physics in the macroscopic limit just as the theory of relativity reduces to Newtonian physics in the limit c co Much of what we will do for the rest of the semester will involve the Schrodinger eqn In this chapter we will identify some of its essential points In chapter 6 we will use it to treat several important situations such as transmission and re ection of particle waves incident on various potential barriers and tunneling a purely quantum mechanical phenomenon In chapters 7 8 9 10 we will treat the structure and properties of atoms in detail Next semester molecules solids nuclei elementary particles Let s look at some of the problems with the application of the de Broglie postulate in hopes of getting some clues about what is needed to solve them attempt to motivate the Schrodinger eqn x Recall Chap 3 the wavefunction we guessed for a free particle was Px t 2 sm 27139 Vt or a linear combination of such we assumed constant linear momentum p which implied constant it w Fixed t But what do we do when a force is acting on a particle p will change with position and time A A A l x The problem is that it is not even welldefined if it changes very o I rapidly For example the separation between adjacent maxima I V V V is not even equal to that between adjacent minima The general solution to this problem is the Schrodinger equation which is a partial linear di erential equation whose solution is the wavefunction Px t Does everyone know what a partial di erential equation is 2 2 eqn containing terms of the form BTW D alpoc t a Tgc t a TC t 8x at 8x at Example 51 Evaluate the partial derivatives wrt x and t up to second order for our free particle wavefunction x Pxt 2 sm 27 Z Vt s1nkx 0t Depending on the potential or boundary and initial conditions solution may be obtained by separation of variables If the solution can be written as the product of single variable functions then the partial di erential equation separates into ordinary i e single variable equations Plausibility argument leading to Schrodinger s equation We are now looking for the equation that is the quantum dp m d 2x mechanical analog of Newton s equation of motion dz d aE 8E M or of Maxwell s equation x y Z g V 39E ax By dz 8 80 0 While the wave equation for a stretched string can be derived from Newton s law and the EM wave equation can be derived from Maxwell s equations we can t expect to derive the quantum mechanical wave equation from classical physics We will get some direction from the de BroglieEinstein relations xi hp and V Eh In other words the equation we seek must be consistent with these postulates at least for free particles Note we are not deriving Schrodinger s equation only providing a plausibility argument for it that equation itself is a postulate The text lists four assumptions which should be met by our equation 1 It must be consistent with the de BroglieEinstein postulates hp and V Eh 2 It must be consistent with the equation E p22m v 3 It must be linear in Pxt so that if P1x t and 1 2xt are solutions of the equation then cl l 1xt cz l 2x t is also a solution gt so we can add wavefunctions to produce constructive and destructive interferences basic characteristic of all waves and demonstrated by the various experiments ie Davisson Germer to be valid for particle waves as well 4 The solution of the equation for a free particle where xi hp and V Eh should be a linear combination of sinusoidal traveling wave functions Let s look at the consequences of these assumptions 2 2 puttinglinto 2 Ezp Vxt gt 2Vxthv 2m 2mt 2 2 introduce k 27V and w27w gt Vx t 2 M 2m Now recall Example 51 and assumption 4 note factors k2 and a 2nd derivative of the traveling sine wave wrt the spatial variable x factor of k2 lSt derivative wrt the time t factor of a Suspect that Schrodinger s equation must contain the 2nd spatial and 1st temporal derivatives of Pxt However Vxt must also appear and since every term in the equation must contain Pxt or its derivative to be linear the equation should have the form x t 05 8 2 V xt1rlxt 3 3399395 0 r X where the constants aand 6 are still to be determined Seems ok Let s test it with a constant potential Vxt 2 V0 free particle Putting in Pxt sinkx at we get asinkx 00k2 sinkx 00V 2 6anoskx at 0k ugh since sine is not proportional to cosine this only works for special values of x and t This suggests that we might try a combination of sine and cosine functions Pxt coskx at 7sinkx at The partial derivatives of Pxt are 82 11xtax2 k2coskx at k2 7sinkx at and Momat asinkx at wcoskx at Substituting these derivatives into our differential equation we get 0dlt2coskx at 0127Sinkx at V0coskx at V07sinkx at 6a1sinkx at 6wcoskx at or odc2 V0 Bw cosch at 0dc2y Voy 6wsinkx at 0 This only works if 0dlt2 V0 2 36ny and 0dlt2 V0 2 6w 7 So 6a76a 7 a 72 17 a 7ii Substituting 7 ii into 0dlt2 V 2 6w we get 0dlt2 V0 2 ii a 2 2 k V0 2 ha we conclude that a rr v and iii In Comparing with Although there are two possible choices for the sign we will see that the choice is of no consequence so by convention we choose the sign putting these values of oc and B into our differential equation we find 2 2 haP C 0Vx tPx t ih 3T0 Schrodinger s 2m ax at equation This equation satisfies all of the four assumptions above on quantum mechanical wave motion We were led to this equation by treating the special case of the free particle We postulate it to be true for any Vx t The validity of the postulate is of course determined by the agreement between its predictions and experiment SchrO dinger s equation is nonrelativistic In 1928 Dirac developed relativistic quantum mechanics He utilized the same postulates as Schrodinger but replaced the classical expression for the energy with the relation 2 E 02p2 mocz V As required in the lowvelocity limit the Dirac equation reduces to Schrodinger s equation The treatment of Dirac s equation is beyond our scope but we will describe some interesting features of Dirac s theory qualitatively One feature namely pair production has already been described Example 52 Verify that Schrodinger s equation is linear in lI x t ie that if P1x t and P2x t are solutions of the equation so is any linear combination cl l 1xt CZ I ZOC t where 61 and CZ are any complex numbers We will at times pull solutions of the equation for a specific Vx t out of the hat without showing how they were derived but once given it will be straightforward albeit tedious to verify that they do solve the equation for that potential Example 53 The wavefunction describing a particle of mass m acted on by a linear restoring force of force constant C in its lowest energy state as a SHO is Px 1 Ae uzwmte mnhyz where A is real Verify that this wavefunction is a solution to Schrodinger s equation for the harmonic potential Vx t Cx22 Born s Interpretation of Wave Functions If we now go back to Schrodinger s equation for a free particle we see that the wavefunction is Px t coskx wt 7sinkx wt 2 coskx wt isinkx wt It is complex The reason is obvious the wave equation is a linear relation between the first time and Second spatial derivatives of the wavefunction This itself results from the linear relation between the energy and the square of the linear momentum The fact that the wavefunction is complex implies that its physical interpretation is different from that of functions describing classical waves So what is the physical meaning of lPxt The postulate If the location of the particle associated with lI xt is measured at time t then the probability that it will be found in x x dx is lI gt xt l xtdx Note that lI gt Xt l xt is real and nonnegative which is a necessary and almost but not quite sufficient condition for it to be a probability function Example 54 Prove that lPxt 1 ct 2 0 Note the analogy and the distinction between the probability density of the particle wavefunction and the intensity of an EM wave which is proportional to 82xt where Sis the amplitude of the electric field Example 55 Evaluate the probability density of the ground state wavefunction of the simple harmonic oscillator given in Example 53 2 gloater dc nuth e l2Cmt Pay AZexC mthcz6i2Cmte 2hx2e i2Cmt Px Px 2 P 2 A26 JC mhx 39 E E 39 VZEC What s so special about this result P is independent of time even though Pxt is a function of t Later we ll see that Note also QM predicts that the particle Will most likely be found in an element dx around the equilibrium point x 0 However there are no welldefined limits beyond Which P 0 Example 56 Evaluate the classical probability density of our simple harmonic oscillator The probability of finding it in an element of the x axis is proportional to the amount of time it spends in that element and this is inversely proportional to its velocity when it passes through the element 82 V 82 some constant P P Px So we need to find v in terms of x how can we do that 2 2 EKVmv CX 2 2 C2 B2 v E x gt Px m 2 J3 ECx2 m 2 Note minimum at x O expected since the speed is maximal at that point It also diverges at the turning points x mZE C where V a O 12EC quot 12EC Consider the differences between the classical and quantum probabilities quantum probability is that of the ground state of the SHO particle is most likely be at the equilibrium point but has a finite probability of being elsewhere In the ground state of the classical SHO particle has zero kinetic energy and zero potential energy it s at rest at the equilibrium point quantum mechanical result is obviously a manifestation of the uncertainty principle 1 vs 1 1 given a wavefunction 1 Schrodinger equation determines exactly how it evolves for a given potential However initial form of 1 cannot be specified completely via measurements thus 1 1 gives only the relative probabilities This is because by initial measurement we learn only 1 1 but not 1 itself Born quote The motion of particles conforms tothe39law s of probability but the quotquot itself L L y in ac c ordancea with theglawsof causality Example 57 Normalize the wavefunction of the ground state of the simple harmonic oscillator given in Example 53 Tde Twde AZTe Ua de 1 an 1 1 is even 7 m 2 7 m 2 l x AZTe WW dx2A2Te WW dxl n 0 n 7 0 39 18 From tables of integrals fgi ahyzdx 2 mm gt A CH2M or Mathematica or Maple O 2 C 18 7 m 2 Modern formulation of quantum mechanics 1 Every well de ned observable in physics A has an associated operator A An experiment measuring this observable can yield only those values a that are the eigenvalues of the operator A A axa ax ax is eigenfunction of the operator A a is an eigenvalue of the operator A Examples of operators Position operator X x A 8 Momentum operator pX lh 2 Energy operator H h 1 Vx 2m 82x Xmmx m m 4 gm mm gmmvowm 2 A measurement of observable A that yields value 3 leaves the system in state aX Each immediate consecutive measurement of A will yield therefore value a 3 The state of the system at any moment is represented by wavefunction PXt This function and its first derivative are finite single valued and continuous This wavefunction containes amp physical information about the system If the system is in state PXt then an averaged value obtained in measurement of observable A is A w xtA Pxtdx ltAgt Is expectation value of observable A 4 Time evolution of the wavefunction associated with the system is described by equation d A zh I Oc t H lP06 t ampt Expanding the energy operator a if a2 1h Px t EE Pxt Vx Pxt What is this called Expectation Values We will now see that the wavefunction contains all of the information on the particle or system Consider a particle with wavefunction I xt The probability of finding it in the segment x x dx is by Born s postulate 7 Pxtdx I xt l xtdx Note that this also means that if we have a very large number of identical noninteracting particles with that wavefunction the fraction that will most probably be in x x dx at time t is Pxtdx The average of observed values of a quantity eg x coordinate momentum energy etc is called the expectation value of that quantity In the case of the x coordinate it will be lt x gt Pltxrx4x ijrxf39xz Pxtdx Example 58 Find ltxgt of a particle in the lowest energy state of our simple harmonic oscillator Pxt lxt elmW2 even 1quot xtx Pxz xei mmyz Odd SO ltXgt 0 What do we expect for lt7cgt in these cases 7 q l W i l a x z b l a x d symmetric centered at 0 lt7cgt0 symmetric centered at x1 gt 0 ltgtcgtcl asymmetric skewed slightly to the left centered at 0 ltgtcgt slightly to right of equilibrium as mmetric skewed more to the left centered at 0 ltgtcgt more to right of equilibrium More generally the expectation value of any physical quantity x associated with the particle is lt f 90gt xtx fPx tdx cgto For example the expectation value of the potential energy of our SHO in the ground state is lt Vt gt xtVxt Pxtdx g x2e hdx Other dynamical quantities of primary interest are the momentum p and energy E For p naively we would write lt p gt2 x tp Px tdx where p is some function of x ie px 2mE Cx2212 as we expect classically but such a function is not possible in QM why not Uncertainty principle So what to do to be able to evaluate ltpgt For a clue recall our free particle wavefunction lI xt coskx at isinkx at eia x at and note LT 39k P 212 i Z p P 2 1393 ax xi h h d This is typically written as p 1 x7t lha x Px7tl a gt there is an association between the linear momentum p and the differential operator lia x 3 E Similarly for the Energy we have 3 l 0 P 127IV P g f a Px7t which is typically written as E Pxt iha a gt there is an association between the Energy E and the differential operator 172 E We obtained these relations by considering the case of a free particle so we should ask whether they are restricted to the free particle case 7 2 Consider the general case 2P Vx t E and let s insert our operators for p and E m 1 a 2 a W a2 a ih V xt ih V xt ih 2m ax 8t 2m 8x2 8t This is an operator equation so for any wavefunction lI xt we have 2 2 h a l Vxt P ih 3 111 look familiar 7 2m ax at Now going back to our attempt to determine ltpgt which is what got us started on this we find lt 1 gt xtp0p Pxtdx ih xt Pxtdx and similarly for ltEgt lt E gt x tE0p Px tdx m x ogwx tdx It should be clear that if we substitute the other side of our Schrodinger equation into our expression for ltEgt we find 2 82 Vx t Px tdx 00 h lt E gt LIP xt W This is an example of the general property that if fx 9 t is any dynamical quantity then its expectation value is given by 00 a lt f gt2 Ixtf0p x zha t Pxtdx olt x We ve just seen that the wave function contains more info than just the probability density I P ie ltxgt ltVgt ltpgt ltEgt and any dynamical quantity ltfxptgt Key point The wavefunction contains all the information that the uncertainty principle will allow us to learn about about the associated particle Example 59 Consider a 1d free particle confined to le lt a2 ie confined to the region a2 a2 We will see later that the groundstate wf of the particle is 717C iEt h A C0 7 for le lt a2 I x t 0 otherwise where A is a normalization constant and E is its energy Show that this lI xt satisfies Schrodinger s equation and find the lowest energy E Since the panicle is free within the box M lt aZ Vxt const which we can choose 0 Hence Schrodinger s equation is W 3qu a W U Fm 3x2 1 E putting in our expression for Pxt we find 2 Z Z Z Z 2 h 3 f hzt11ih 45 LIJgt h2trEt11 2m Bx Zma h Zma 7rth which is indeed satisfied if E 2ma2 Example 510 Evaluate the expectation values a ltXgt b ltpgt c ltXzgt and d ltp2gt of the panicle inabox described in the previous example when it s in its ground state a As for the ground state of the SH0 we note that Pxt Acos7DaexpiEth is an even function of x and so is I xt 1 xt AZCOSZUZZXa Therefore I xtx 1 xt AZXCOSZUZZXa is an odd function of x so as expected M m lt x gt AZ Ixcosz7mladx 0 7112 b lt p gt2 ih 2P x 031 mix Since the integrand is propom39onal to cos7Dasin7Da it im Bx is also an odd function ofx so as expected again ltpgt 0 m a2 a2 c ltx2 gtjwxtx211xtdxA2 szcos2mcadx2A2 Ixzcos27Dcadx 7m raZ O 3a2 2 3 2 using 2A2 1 j 3 cos2 39xad E gt ltx2 gt Aza 2 1 tables 39 0 a a 47239 6 Before continuing we need to evaluate our normalization constant A we a2 7z2 1 I Pxt PxtdxA2 jeos2madx2A2a7z jeos2madma 7112 0 m mlcos 219 a 7 Aza 2 2A2a7139Jcos2 mm 2A2a7j d9 2A2 0 2 A F 0 0 2 7r 4 2 51 Therefore 3 2 3 2 2 2 ltx2gtA24 2 1 2 1 1 1z0033a2 2 xmzxltx2gt018a 7 6 E4rc2 6 2272 x is a measure of the uctuation of x about ltxgt 0 Generally ltx2gt ltxgt212 is a measure of the rms uctuation when ltxgt i 0 Similarly for ltpgt ltp2gt and ltp2gt ltpgt212 2 82111 8x2 am 2 8x2 2 00 2 1 lt172 gt ITih dx2 h2 ILP dx2h2 zjlptpdx h j m m a we a gt lt p2 gt h2a which is precisely the de Broglie value ofp hxl h2a If we define ltx2gt12 Ax and ltp2gt12 Ap we see that MA ltxzgt12ltp2gt12 018axn a 05725 2 252 Can we see the modulus of the wavefunction l xt l xt Yes well sort of example Scanning Tunneling Microscopy I u trajectory in ilruln lnquot httpWwwphysicsleidenuniVnl When we deposit additional atoms on a clean surface electrons will re ect from them and the wavefunctions will interfere with each other creating standing waves of electron densities Cu on Cu111 wwwspecsde 9K gt 12K wwwspecsde TimeIndependent Schrodinger Equation if a2 8 Recall Schrodinger s equation 2 Px t Vx t Px t 2 th Px t 2m ax at A common techniques for solving partial differential equations is separation of variables searching for solutions which are products of functions each of which is a single variable function Let s try this with Schrodinger s equation We express the wavefunction as a product of two functions one of x only the other of t only Pxt WOW We will find that this is the form of the solution to Schrodinger s equation if the potential energy function does not depend explicitly on t Substituting this solution into the equation we get hz 82 a VIM wowow zh woo 0 2m ax at Dividing throughout by Wxg0t we get 2 2 1 ha Vxyx G 13 yx 2m ax t at ago 10 where G is the called the separatton constant First consider the equation for 0t E 70 We ve seen this before and know its solution to be t Gh This is an oscillatory function with 1 Gh But 1 Elk so G E So Schrodinger s equation for 0 becomes hz a quot 75 wads1506 Ellvi Zm 3x This is the timeindependent Schrodinger equation The functions wx which satisfy this equation are eigenfunctions of the equation The full wavefunction can now be written as l x t Wxe iEht Required nronem es of F 3 inn Both 0 and d de must obviously satisfy the following conditions They must be finite They must be singlevalued They must be continuous for x I39J Nut smgle valued Not Contlhuous Nu nite 7 v l l 0 x 0 x 0 0 In the following analyses we shall often see that these conditions lead us to the unique solution for 5100 for given initial andor boundary conditions Energy 39 39 inthe quot quot quot Theory We shall now try to gain some intuitive understanding of the solutions of the Schrodinger equation by examining plots of Vx Px etc First recall that d2 2m 41 Vltxgt Ewgt The properties of 05 depend on those of Vx since classically F dVdxi This potential energy Vx is typical for an atom bound to a similar atom in a diatomic molecule erg H2 N2 02 etc At the minimum in Vx F 0 For closer x the atom will experience the hard core repulsion from the other atomi For farther x the atom will experience attraction to the other atomi am 7 7 When the molecule dissociates Vx 2 V0 F 0 i l f Ejng m Wx Energy t Now consider an atom with a total energy E Classically x and x are the tuniing points so x S x S x i Recall that d2 11de tells us how the m of 111139s changing Hence if allylair2 gt O and 4 O the slope is increasing and yis concave upwards and Vice versai dex is increasing becoming more positive therefore tillId gt 0 Now divide the x axis into three regions In region I x lt x V06 7 E gt 0 so If lgt 0 dezgt 0 and llis concave upwards if llt 0 Lil dez lt 0 and llis concave downwards In both cases llis convex toward the x axis In region II the behavior of llis opposite and it is concave toward the x axis In region III it s behavior is the same as in region I U dex is decreasing becoming more negative therefore tillId lt 0 WI Region I U RegIon 2 Region 3 x m I I HYU Egt o WU L lt0 I Hm El gt0 I l I l I I I I I I I l I I I I I I I I Let s see if we can put some of these together into a solution Assume that lgt 0 and dlldx lt 0 at some x ltx0 ltx Then LIZ dez lt 0 and llis concave downwards lf llis still gt 0 at x x it will start curving upwards at x gt x and grow without a bound diverging as x gt oo 1 unacceptable lf llbecomes negative at some x gt x it will start curving downwards and diverge to oo as x gt oo 2 also unacceptable if lland dex have special values at x0 so that llcurves upwards at x gt x but at a slower and slower rate so WA 0 asx gt 00 to yield an acceptable solution 3 In general we will also have similar problems at x lt x We will see that there are only special values ofE for which lgt 0 as x gt too 0 11 VIZ and 13 for the lowest energies E1 E2 and ET ul behaves as 3 for large x and also a 0 for small x uZx0 u1x0 but the magnitude of its second derivative Ia dezl is larger so it crosses the x axis at some point x lt x lt xoi Then the sign of till4d changes and it turns from concave downwards to concave upwards At x x it reverses again turning concave downwards and it approaches the x axis The basic observation that should be noted here is that Similarly E3 gt Eli Note that E3 7 E2 and E2 E1 are finite so the energy levels are discrete Will be true as long as there are x and x such that Vx gt Eforxltx orxgtx When the energy of the atom is E gt Vl the situation changes Classically the atom is unbound for all x gt x Vx E lt so uwill be concave towards the x axis for all x a it will oscillate And since we can always cause Vlgt 0 as x a 00 by an appropriate choice of an initial value of dex we will nd an eigenfunction for any value ofEi Thus the set of allowed eigenvalues forms a continuum bmm m mime W t im use 55ml stu that lm s 1 unearth equation energies o Recall the assumptions which led us to the Schrodinger s equation Important question Where did the energy quantization come from vm A ESE Chapter 10 Outline Multielectron Atoms Optical Excitations Analyzing a complicated system in a series of not too complicated steps Alkali atoms optically active electrons Hartree interpretation fine structure Atoms with several optically active electrons limitations of the Hartree aproximation residual interactions coulomb and spinorbit LS coupling geometrical representation spectroscopic notation experimental assignment of quantum numbers Energy levels of the carbon atom The Zeeman effect Tabulated properties less than half filled subshells more than half filled subshells HW 10 Chapter 10 Q 913 P 1 3 5 7 9 due 124 Recall the Hartree Philosophy First Assume each electron is in a spherically symmetric potential described by the average c0ulomb interactions with the nucleus and other electrons Then account for successively weaker interactions seen by the electron Typically l couplings between the orbital angular momenta of the electrons 2 couplings between the spin angular momenta of the electrons 3 couplings between the spin amp orbital angular momenta of the electrons weaker 4 interaction with external magnetic eld if present Zeeman effect Plan for this chapter Assume for the moment that only optically active e s can interact Atoms with one optically active e Atoms with several opticallv active es Can we have interaction 1 No can we haVe interaetien 1 7 Yes Can we have interaction 2 No can we haVe interaetien 2 7 Yes Can we have interaction 3 Yes can we haVe interaetien 3 7 Yes Can we have interaction 4 Yes can we haVe interaetien 4 7 Yes Which is largest It depends We ll conclude by putting it all together and look at how 14 complicates things Recall the Hartree Philosophy First Assume each electron is in a spherically symmetric potential described by the average coulomb interactions with the nucleus and 39other electrons Then account for succes siVely weaker interactions seen by the electron Typically l couplings between the orbital angular momenta of the electrons 2 couplings between the spin angular momenta of the electrons 3 couplings between the spin amp orbital angular momenta of the electrons weaker 4 interaction with external magnetic eld if present Zeeman effect What happens when an atom goes from its ground state to one of its lowenergy excited states An electron in one of the outer subshells is given a small amount of energy and jumps up to a higher state HOW might this happen 7 One atom collides with another coulomb eld of incident atom can act on our outer shell electron When the atom deexcites what do we eXpect A few lines Y s Of a few different energies In which part of the spectrum Optical What is different here from the X ray excitations we saw in the last chapter They involve inner subshells can almost think of it like bricks in the top and bottom of a building So why should we want to study the low energy states of atoms Excited states will give us a better understanding of the ground states Successive approximation procedure is used everywhere see Appendix J amp Phy480 Optical line spectra are excellent tools esp astronomy pattern of lines a composition of stars intensity of lines a temperature of stellar surfaces Doppler shift of lines a velocities of stars Zeeman effect a magnetic elds produced by stars We ll start the study of optical excitations with the simplest case Alkali atoms What is an Alkali atom Atoms which contain a completely filled set of shells highest being a p plus a single additional electron in the next s subshell What well studied atom should these be chemically similar to Hydrogen 1 Where are the Alkali s in the periodic table 7 I Z w mmm pI p2 p 4 ya 55 as D 2 54 55 5 s7 55 59 L II 4 Ce m Nd Pm 5m Eu ed M Dy HI I Tm vn Lu I m a I I II I II I I II I II I I I I I I I I I I I r I M I 4 so eI a2 93 94 as 96 7 923 99 100 IoI m2 m3 5 III Pa u ND Am Cm Bk u E Fm w Nu Lw I gt 1 mIIIIuIsI I I a I I II I I I I I a I I I I I I J g I g 1 I4 7 In 7 s a m In 11 H H I Recall from chapter 9 that the energy of an electron in a lledp shell is much more negative than the energy of the next s subshell so p electrons are not excited in the low energy processes which generate optical spectra Alkali s are basically an inert noble gas core plus a single electron in an external subshell gt the total energy of the core does not 1 change is constant and can be ignored I The total energy of the atom constant energy of our optically active electron Rede ne the zero of the total energy to be equal to that of the optically active electron Measured energy levels of H 3Li and 11Na Na ne structure is not shown So Where are the 2 Energy eV 3 As you might expect Hartree theory works very well in this case since Vr is spherically symmetric as is assumed for the rst step predicted energies agree with experiment The Alkali lines do show ne structure except for the l 0 states why 7 each line is really a double line With that hint and our earlier comments about additional interactions what is the likely source of this ne structure spinorbit interaction coupling between the magnetic dipole moment of the electron and the internal magnetic eld it feels because it moves through the electric eld of the atom Other relativistic effects are small since from the Bohr model Z 62 same as for 1H by contrast 0016 from Chap 4 spinorbit increases rapidly as we move up the periodic table V 4ns0nh So how do we calculate the size of the spinorbit splitting 1 1 dV where now Vr Back in chapter 8 we found Zmzcz o L obtainid from artreet eory Anyone remember how to calculate S 0 Z Ej11 111 ss1l gt 4 rdr As for the spinorbit coupling in the single electron case the relevant quantum numbers are n lj m1 and obey the same rules as beforej 1 12 mj j j l j 71j WhatAEdoweeXpectforl0 forl0jssoAE0 For I a 0 AE can have two different values depending on whetherj 1 12 orj If 12 So each energy level is split into two one slightly higher thanioriginal lower J This is equal to the amount of work to turn the electron magnetic dipole moment p gsubh s from one orientation to the other in the internal magnetic eld of the atom And we see that AE ltldV d gt which is equivalent to the strength of the magnetic eld 739 739 227 Argon Zre2 Recall from chapter 9 V 471501 So for larger Z dVdr is greater than for small Z So for larger Z we eXpect greater splitting And this is indeed seen moment Li M K Rb quotcs Subshcll 1 11 4p in on Sp quotquot39b l 04 X m 4 21x10 7 x 10 295 X loquot 187 x 1039 splitting in And the selection rules which we found for the single electron atoms are still good Does anyone remember what these are C7 A 1 Aj 0 1 Exarnnle 1071 The yellow light ofsodium vapor lamps frequently ernployeolin highway illumination is a spectral line arising from the Sp to 3s transitions in 11Na a Evaluate the waveleng 39 39 39 39 39 39 39 39 39 n yinorbit intera nn a Figure gives E3 7 Eh s 3 7 51 eV 21 eV A cV hcE 66x103934x3x10821x16gtlt10quot9 5900 A b Table gives dE 21x10393 eV Since A cV low advv7 mammy hchEZ s 57 A c The SF state of higher energy is that for the state with 1 12 1 12 32 the other state is that with 12 1 12 12 39ther case the 3slevelis1 0 Hence the transitions are N E 2 2 1 andAl 71 A 0 and both are allowed 4 may av I u 3 A V T 777777777777777 7 7 2n 31 Atoms with several opticallv active electrons We ll now look at the more typical case where we have a core of completely lled subshells plus several electrons in partially lled outer subshells Which electrons do we expect to be optically active all of those in the outer subshells Excited states are treated rst using Hartree to account for the stronger interactions and then by including the other interactions We will consider for the time being only cases where the outer subshell is less than half full if it is more than half full then we will talk about the behavior of holes ie X ray In the Hartree lSt approximation effective spherically symmetric Vr the energy of each 6 is determined only by n and 1 similar to that in the case of the single optically active 6 in the alkali atom with the same core I Hartree lSt approximation 9 22 l level degeneracy in each 111 I Much of this degeneracy is removed when weaker interactions are included The two primaries Residual Coulomb interaction due to the interaction with the other valence electrons which is not spherically symmetric Spinorbit interaction causes the spin of each 6 to interact with its own angular momentum in a way that their magnitude remains constant as they precess about J L S Other lesser corrections relativistic corrections corrections for interactions between spins of the optically active electrons So the big question will be 1 Do we consider the spinorbit Li Si coupling for each electron individually and then sum these OR 2 Do we sum the individual spins S ZSi and sum the individual orbital angular momenta L 2Li and consider the spinorbit coupling between S and L The answer will depend on which of the two interactions dominates More detail on the residual Coulomb interaction Source the charge distribution of the other optically active electrons is not spherically symmetric since the shell is only partly lled recall that full shells are spherically sym Hartree gives only the best spherically symmetric average A quantitative treatment requires calculating the expectation values of the residual Coulomb and spinorbit expectation values to the Hartree energies Don t worry we ll only look at this qualitatively What very important consequence of the Pauli exclusion principle must we consider Hint concerns the total wavefunction describing two electrons I total wavefunction must be antisymmetric under an exchange of labels I Th1s of course results 1n an exchange force Humps 1st rule recall S Energy electrons with symmetric spin wavefunctions triplet like to be far apart max lower asymmetric singlet close zero higher Coulomb repulsion is weaker in the triplet configuration so its energy is lower Optical electrons also exert torques on each other they don t change the magnitude of the L of each er but cause it to precess about the total L such that the magnitude of L is constant L The next question is Which of the possible values of L corresponds to the state of lowest energy This is best answered by remembering that the lowest energy of two e s in a Bohr atom is obtained when they are diametrically opposite each other But then their L s add ie they are parallel Lowest Energy gt hlghest L Hund s 2quot rule More detail on the spinorbit interaction Source spin of each 6 interacts with its own angular momentum in a way that their magnitude remains constant as they precess about J L S We also know that E is lowest for the state with the lowestJ see Sec 86 Residual Coulomb and spinorbit interactions are opposing interactions For small to medium Z residual Coulomb is stronger so it is treated first Ignoring the spinorbit coupling the individual spin angular momenta couple to form the total spin angular momenta S 2Si w ere s 515 1 Similarly the individual orbital angular momenta couple to form the total orbital angular momenta L 2Li where L lh We re now ready to consider the weaker spinorbit interaction how might we do this By considering the interaction between S and L J L S where J j j lh Hund s As we previously found for spinorbit coupling Lowest Energy 9 lowest J 3rd rule Although the name is rather misleading this procedure where Si couple first then Li couple and nally L and S couple is referred to as LS coupling also often called RussellSaunders coupling For largest Z the spinorbit interaction becomes larger than the residual Coulomb interaction Anyone want to guess the procedure for this case Si and Li of each valence 6 first couple to J i and finally the J i couple to the total J I This is referred to as JJ coupling JJ coupling completely dominates the spinorbit interaction of nucleons in the nuclei and quarks in Baryons and Leptons see Chaps 1518 LSCoup g Wequot r r t B CI quot L levelsofthe 39 Let s start by looking at the relevant vectors Where 1 1 and 12 2 and s 2 van one example ofmany Quantization off and J1 requiredrn absence ofextemal torques S ismaxs 1L ismaxl 3 m3 J 711 Fast precession around s and L slow precession around J So how many states with different quantum numbers 3quot l j can be formed by these two electrons where 11 1p and 2 2 d 7 How many Values of v combining 51 and 52 7 How many Values of l combining l1 and I2 7 2 V hi y 12139 1l2HHH 39 39 z I 11 u1 1 hlgt 39 quot1 3 1 How many Values off w combining v and l 7 113 L1 511 And each one contains H 2f1different 3 739 1 1M possiblevalues ofm l o 3 012392 30111 Or more algebraically and more easily s slis slisz lvlv2 515212 5 01 gt l 1llil2llilzl1ll12 11122 1 123 fs il Ms il Hlv l 9 j 01234 Example 102 Find the possible Values ofs l 3amp5 32 H and j for a con guration of three optically active electrons with 11 1 l2 2 and I3 4 Clearly s mm 12 s max 32 Hence 5 12 32 Similarly I 1 7 Hence mm I m 734 1 1 2 3 4 5 67 Hencej mm12j max172 quot 7 2 2 122 144 Now let s move on to the splitting I 1 ngle degenerate level oftwo Valence 2quot in the 3d 4p configuration 2 configuration as befor resi u spin Spemscopist coulomb orbit quotomon 27 Splitting ofasi am residual which splittings are due Lulomb to residual coulomb and Flu which are due to spin 7 314 p 012 Spectroscopist notation is in general given by ML J observations the 51 states have a lower energy than the s 0 states As expected 339 ilarly the energy ofa state decreases with increasing 1 To answer the question of Lowest Energy a highest 8 Lowest Energy a highest L vs J L s 7 Lowest Energy a lowest J Each ofHund rules refer to aspeci c angular momentum s L or Presentation of the explicit equations for the energies of all the levels is beyond our scope However we have already seen an equation which gives the j dependence of the spinorbit interaction energy ELS j 1 l l l s S l As expected ELS increases with Note that K is not simply proportional to ltlrdVdrgt because the potential is now more complicated However it is the same within the same multiplet ie for all of the states with the same s and l So the splitting between adjacent levels is given by AEJquot ELSU 1 ELSi Klti 1gtlti 2gtl ltl 1gt s lts 1gt1 0 1gtr1 1 mm 2K l Land interval rule same rule is used in molecular and nuclear physics Example 103 Consider the 3P0 3P1 and 3P2 states of the s l l l multiplet of the 3a 13a 1 con guration of 20Ca E3P0 lt E3P1 lt E3P2 and the measured splittings are E3P1 E3P0 167 meV a AEO E3P2 E3P1 333 meV a AEI Compare these A E s with the Lande interval rule AEI EKG40121 3 Excellent a reement L39ande39 rule can be used as a test AE0 2Kj391l0 1 g for the presence of LS coupling Lande rule tests for a triplet in one of the con gurations of ZEICa Rmm Contigumuun Luck Separation Levels Separation l 1 Thou EX103 mil Hair 107 y m 4w 13 1 333 x m 4w W 1 1 4 mp 15 1quot m m tr quot15315 1312 x In JCV 20 1 MI 1I 31 1w x In or JI ll 26 In W I59 3 3ll4p J1451 w x m c 3m 1 491 x 10 av Lin 3quot Tests on other con gurations in other atoms show LS coupling to be present in all atoms of small and intermediate Z indeed this is how to experimentally test which coupling dominates Example 104 Measurements on intermediate Z atoms yield which increases by a factor of 53 with j Use the Lande interval rule to determine s l and j 2 f Eaans 5 MN f2 1 135 H 15 1 77 jzi jzii alsilkaul7 I l 3 AEJ f1 2 2 2 2 2 2 jr i i In 12ands l 52thens 32andl l ii Ifs 7l l2ands l 52thenl 32 Energv Levels of the Carbon Atom What is the ground state ls22s22p2 How many optically active electrons are there 2 What are the quantum numbers for these two electrons M1 M2 2 I1 2 1 s1 32 12 What are the LS coupling quantum numbers What are the allowed values for s s O 1 2s lL What are the allowed values for l l O 1 2 Spectroscopic notation What are the allowed values for j s O l 0 j O 180 s 0l l j l 1P1 s 0l 2 j 2 1D2 s ll 0 j l 381 s ll l j 012 3P012 s 11 2 j 1 2 3 3D3 Now let s look at the energy levels Energy level diagram for 6C atom Energy EV 3 1P1 w I 1 a 3 Is the meaning clear 39 53 s n5 5 Q 55 I I 2p53 u 2p4d 2 4 i 294 p I 234 2p3p E O is defined such that the groundstate E tot is equal to the energy required to ionize the atom In this way the diagram may be compared directly with those of the H and alkali atoms Anything look suspicious 2 of our ground states are missing I What happened to them The Pauli Exclusion Principle Appendix P The Exclusion Principle in LS coupling is discussed brie y in Sec 105 and in detail in THE EXCLUSION PRINCIPLE IN LS COUPLING This is of course only an issue When our two or more electrons are in the same subshell have the same n an I Figuring out Which states are in violation is easy When We Work in the Hartree approximation Where the relevant quantum numbers are m and ms for each electron values of m and ms for each electron to determine the possible values m5 an m an from these We obtain the possible values ofl 571quot an m We use the possible of m so it s not always completely Table P1 Possmie Quantu rNumbers lor a7n npl Con gurahon 39 s 39o obvious which sta e v1 late Entry m m 1711 m m m m the Exclusion Principle 39 I 1 1 2 1 0 2 Is there an obvious failure f i 4 0 3 3 6 r u l inourCres w 4 7 7 7 7 7 An example relevant for the a 39t 2 2 I g 14 2 Si 3p con guration of 7 I W 7 I 0 0 Problem 109 nds 15 8 1 712 71 7i 7 different sew ofml and ms for 9 0 12 I 0 7i 1 the two electrons Which m 0 2 0 0 0 I n 7 I Q 7 1 7 1 0 satisfy the exclusion principle 12 0 I 2 7 1 0 7 I l3 7 l l 2 0 0 7 I A number ofother sets of 14 71 1 71 0 71 l5 7 I 71 J 7 i 7 7 course fai The hard part is now to identify the allowed and disallowed states in terms of l s j 39 Uquot 39 uf Eq 1014wi11 allow us to nd the allowed and disallowed sets of l s j Tahe P3 Numbers 1 3 Electrons m lhe Same Sunsneu T S 39 39 Fortunately the authors kindly prov1de us i 15 with this very helpful summary table nf 5 K I Which of our possible states p 1 1 1 3 3 3 HP39 2 sogt P1 DZ S1 P012 D113 rm 5 W quotI surv1ve if any 3 2P 1 3939 ls 1D 3P p 39 3 3 up4 s II p 0 2 012 lt P I 26 15 Which of these has the lowest energy think Hund ml 3 3P mt mil 5 11 139 miJ 4 ml 5 1 5 I I I 1115 7 r 5 11quot 5 39IL L mix 5 1 c Since 3D3 fails we can t have DZ or 3D1 either Note the following properties of the C atom In the ground state of the C atom the two 2 equots are in the I1 l 12 0 states since those give the highest s 12 12 1 so the ground state is 3P0 Note that in the 2172 configuration E3P0 lt E1D2 lt E1S0 so the s dependence of E is greater than the l dependence which is almost always the case In the 2p3s configuration E3P01112 lt E1P1 Deviations from the maXs Hund s 15 rule and from the maXl Hund s 2 rule are seen in highenergy configurations but in the C atom within a multiplet the state with the lowest E is the state withthe lowest j The absence of the 1P1 and 331 states in the 2172 configuration of the C atom is due to the exclusion principle The optical line spectrum of the C atom can be constructed from its energy level diagram by evaluating the splitting AE of photons emitted in all possible transitions consistent with the LS coupling selection rules 1 Transitions occur only between configurations which differ in the n and 1 numbers of a single e39 Hence all transitions are singlee39 transitions 2 Al 1 3 As 0Al 0 1Aj 0 1 butnotj 0 ej 0 9 a Use the periodic table of Figure 913 to determine the ground state con gurations for the atoms leg Al and 14Si h Then predict the LS coupling quantum numbers for the ground state of each atom Express your result in spectroscopic notation b So how do we determine the LS quantum numbers for 12Mg 1s22s22p63 s2 The two electrons close an s 10 subshell What are the values of1 and I2 112 0 And 1 l 0 What about s s 0 since HI H2 and 1 12 s 1 would Violate the exclusion principle 7 So what do we have for j j 0 Spectroscopic notation 150 What about the LS quantum numbers for 13A1 1s22s22p63s23p1 There is one electron in the 3p subshell What are the values ofs and l s s 12 l 1 1 So what do we have for j j 12 32 Which has lower energy j 12 Spectroscopic notation 2P1 2 9 a Use the periodic table of Figure 913 to determine the ground state con gurations for the atoms 12Mg Al and 14Si b Then predict the LS coupling quantum numbers for the ground state of each atom Express your result in spectroscopic notation What about the LS quantum numbers for 1 Si ls22s22p63 s23p2 There are two electrons in the 3p subshell What are the values of1 and 2 1112 1 And I 1 0 12 What about s s O 1 Which s has lower energy s 1 So in what follows we ll consider only s l Spectroscopic notation What are the allowed values for j s l l 0 j l 381 s 11 1 j 01 2 3PM2 s 11 2 j l 2 3 3D1923 Before we consider which state has the lowest energy what complication did almost everyone miss The Pauli Exclusion Principle I Appendix P The Exclusion Principle in LS coupling is IICIBJN discussed brie y in Sec 105 and in detail in LS COUPLING The Zeeman Effect In 1896 Pieter Zeeman noticed that When an atom is placed in a magnetic eld and then excited the spectml lines it emits in the deexcitation process are split into seveml components Transitions between any smglel Transitions Lienmen doublet states H atom with even numbev rst exalted slate and double oi gummy active Elections ground state m he sodium alum ls l ili No new ID V 7 Arrows show Weak new predictions of M M t SEEM Nuvma Anomalous Normal splitting could be explained in terms of classical theory Anomalous splitting could not be explained until QM and electron spin For elds lt 03 T splitting is proportional to the strength of the eld Zeeman splitting in such elds is smaller than the ne structure splitting Which is proportional to the more intense internal elds of the atom For this situation we have as usual AE wB So each energy level will be split into several discrete components corresponding to the different levels of EB associated with different orientations of u with respect to B To proceed we evaluate u by determining MI and us for each optically active 6 in terms of orbital and spin angular momenta and then sum over all electrons here we have set gugu gsugsu W Mzth Zthz 7 SI 7 S2 g11andgs2 11 12 2 1 2 If the atom obeys LS coupling then we can sum both the Li and Si to obtain L and S Z392 I We see that the total u is not antiparallel to the total angular momentum J L S However if S 0 then u is antiparallel to J this is the normal Zeeman effect Recall from Example 84 that the Let s now evaluate Zeeman splitting quantitatively internal magnetic eld was N1 T SO we want to evaluate E M39B HEB where MB is the component of u along B Since 1 precesses much faster about J than about B since the Larmor frequency is proportional to B we rst nd MI and then MB gjv Mb E 2 L 7 Hf JV I I39S v D 2 39 D 39 h J392 But 3L S 3J 2 7L 2 iS Z2 so Mb L 22S 23J 2 L 2 S 22J Z h J392 M 3J 2S 2 L 2 J h 21392 Mg B M 3J392S392 L392 h M2 SO E ME In the s l j m state S 2s s lh2L 2l l lhZJ 2 j j lh2 and j sm jsj j39j39l s39s39l Z39l39l 239j391 Land gfactor So EB 1me where g 1 Notethatggllwhens 0soj l and ggs2whenl 0soj s a g is a variable g factor that determines the ratio of M to j in states where j is due to both orbital and spin angular momenta Overall we see that an external B will split each level into 2 1 components the magnitude of the splitting will be different for different j Electron Spin Resonance ESR Zeeman splitting in Na m t ngZJ No external Weak external magquot em i e d magnetic eld What are the selection rules 7 Am 0 i1 Diagram doesn t show tmnsitions ofthe form AmJ i1 with Af 0 are they allowed 7 Yes Electricdipole transitions vs magneticdipole transitions magneticdipole tmnsition probability smaller by vcz spontaneous transition is iare 12 251 2 lt 12 Can we have this transition 7 Yes probability can be increased with increasedy density What s the energy ofthe photon 7 AEE MngAm J where g 71J39Q391s39s391rl39l391 39 What region ofthe frequency spectrum is our electron in 7 microwave put your sample in amw cavity in aB eld At resonance will be big drop in mw transmission Amongotherthingsallo JUI 39 39 39 X I I Forerunner ofNMIR laser maser Sumnth of Chapter 10 Importance in Determining nci gy Dominant interaction Must impoi lunl weaker inter action Slighin esk important Appmiabiy less important Least impoi ttiiii Nam Hai39ticc Rcddual Coulomb s in coupling Resitltitii Coulomb mbiltl coup ng SpinJrhil chmztn NALLII39C of Iiiieructmn Eicctm nvcrdgt potentittl Electric departures from ave 2 gs potential Magnetic internal eld Magnctic Dxicrntil licld Quunliim Ntlmhcn Dcici39mining uncrg a set of it I Encrg Lowest For Minimum n Minimum Maximum 339 Maximum 1 Minimum Most negatttc m39 Chapter 6 Outline Solutions of the Timeindependent Schrodinger Equations Nonbinding and Binding potentials The Zero potential The Step potential energy lt step height energy gt step height The Barrier potential transmission coef cient electronatom scattering Ramsauer s effect size of resonances barrier potential Examples of barrier penetration 0L particlenucleus potential 0L emission GamowCondonGumey ocdecay theory atomic clocks The Square well and in nite square well potentials The Simple Harmonic Oscillator potential zeropoint energy parity HW 6 Chapter 6 due 1017 questions 24 26 problems 11 12 20 21 24 25 30 31 7 12 6 21I x I 6x2 Recall the Schrodinger equation 2m Vxr1Pxr 272 I As we look at different physical situations what is the only thing we expect to change The potential Vxt We will start with the simplest potential that is no potential Vxt 0 and gradually add complexity What does it mean in terms of Vxt and E for a particle to be bound The Zero Potential dV If Vxt 0 for our particle what force does it experience F 7x 0 2 2 The timeindependent Schrodinger s equation for this case is E11 x 2m abc2 we have 2 or with k 2211 721 x What are some solutions Most obvious 1px Acoskx Bsinkx we will pick 1p1x em coskx isinkx recall that the full solution is of the form 1Px t 1p me Em So our full solution is 1I x I em quotU What is this physically traveling wave propagating in the x direction But a traveling wave propagating in the x direction is also a solution with energy E 1p2x 6quot 2 1I xt e quot quot Since an ordinary 2nd order differential equation has two linearly independent solutions 11500 and 1p2x are the two solutions so most general solution of the time independent equation is W05 1411506 Btp2x Ae m Be m Raal pmohwx n For our traveling wave propagating in the x direction lIJxt should look like What will lIJquotxt lIJxt look like lIJx I1Pxt equoth e h m l AM We should now consider the physical interpretation of the solutions 11 H W First let s evaluate ltpgt for the particlewave moving in the x direction lt p gt fly 1 de what would you naively expect 7 p2 2m E a p W W 6 Since p 4117 p 1y miAe lt gt may 42me up 6x op ax For a wave traveling in the x direction gt ltpgtjzlym dx42mE Pm What about our particle s location in x clue 1 Our wavefunctions have a single value of k and therefore 19 since 9 h2nk clue 2 describe a beam of in nite length PX l1 1I const The uncertainty principle is also satis ed here because as Ap a 0 AX a 00 Similarly for E and I Since there is no uncertainty in E we need an in nite At to measure the energy of the particle in the in nite beam Although these plane wave solutions are idealized wavefunctions many particle beams can be approximated as plane waves eg monoenergetic beams of 19 e n etc emerging from a source or accelerator However idealized plane waves are mathematically problematic If we try to normalize the wavefunction we get 1quot de fA Adx A Afdx 1 oops 00 00 One solution is the introduction of Dirac s delta function 6x Xx 0 for all x e 0 and f6 xdx 1 This is however beyond the scope of this course quotw Thinking about the actual physical situation the solution is obvious the particle is clearly con ned to some region ie accelerator detector a one dimensional box om L2 to L2 Box normalization which stipulates that 1P is a plane wave for L2 s x s L2 and 0 elsewhere Although A is then dependent on L we will nd that the result of calculating a measurable quantity will be independent of L As we discussed in Chap 3 we can obtain a more realistic wave which is really a wave packet or a bunch of wave packets a stream of overlapping y s m by adding a large number of waves of the form moat ezmmr each with a different value of k and a so their superposition will generate the group of waves L m Ar or wave packet Chap 3 1 won aw n o v P c Taking a simple example IJ W packet moves in r Li 2A the 1 direction and spreads uncertainty principle 0 7 The location ofthe group at any time t is given by lt x gt f lr39 X de The constant velocity of the group is 5 JZE m m m Due to the large number of Fourier components the mathematical description of the wave packet is very complex Therefore wave packets are rarely used for modeling However we ll consider one very speci c case where Vx at 0 changes so slowly that it s almost constant over one it In that case it can be shown that i dltxgt lt1 x gt ltdVX bcgtltpxgt Schrodinger seqn dz dz dx m 1 aNewton s Law in classical limit all m m Newton s Law is a special case ofSchrodinger s equation The Step potential energy lt step height So what s a step potential A real physical system with such a potential would be a charged particle moving along the axis of two cylindrical electrodes held at different potentials Consider a particle of mass m and energy E lt V0 moving in the x direction towards x 0 Classically what do we expect The particle is free until it hits the wall where it rebounds and p a p Mathematically we see that it has to bounce back into the x lt 0 region because if it were to continue into the x gt 0 re 39on we would et 2 g1 g EVxltVx p2lt0 m Wu WA Vt L V WU V0 VH So how should we solve this quantum mechanically vm Solve forx lt 0 solve for x gt 0 then join the two solutions at x 0 in such a way as to satisfy the requirements discussed 1 Chap 5 Both w and dipdx must be nite singlevalued continuous V0 7 W For xgt0 2 12 d2 T w2Vu 2xszx m dxz 2 12 121 For x lt 0 E dxl Ewlx Z Z We just solved this one who remembers the This can be rewritten as L d 112 VD EN X general time independent solution 2 dx 2 1p1x Ae k X Bei39k x or with k2 VzmV E d 111 h dx where k 739 2mE 1 h What s the general solution kZIPzOC 1p2x C2 De39k What can we say about C C 0 otherwise W Mpaooasxaoo And since 1px must be continuous at x 0 what additional constraint do we have A B D We also have that dip1x must be continuous at x 0 ddt ik 42 r 132quot I 017 JQDe W Equating these derivatives atx 0 we get ik1A iB kZD or ikZk1D A B D ik D ik 39 A714 371 72 ButsrnceD A B 2 kl 2 kl So only one of the four constants remains undetermined D and it is the normalization constant D x it x wx1ik2klek 1 1k2k1e 1 x50 W4 x 2 0 De 7k x Vm VIA Va We ll see that useful results can usually be obtained Without carrying through the normalization procedure that would specifyD for most interesting quantities the normalization cancels out The complete wavefunction is Let s interpret this physically forx s 0 What does the 1st term describe What does the 2 1 term describe a particle moving in the x direction a particle moving in the x direction So we can interpret the 1st term as the wavefunction of the ineidentpartiele and 2 1 term as the wavefunction of the reflectedpartiele The reflection eoe ieientR is the probability that the incident particle Will be re ected 33 1ik2 k11 ik2 k1 R AA 1 ik2k11ik2kl 1 in agreement with classical behavior Now recall our time independent solution 106 substituting e k x cosk1xisink1x we find Dcosklx Dk2k1sink1x x S 0 eikzx x 1M x 2 0 50 W06 I Ipx87 E h is not a traveling wave but is instead a standing wave since it has standing nodes at all points x Which satisfy cosklx 7k2k1sink1x 0 So for x s 0 our solution is in complete agreement with the classical predictions However forx gt 0 we have a radical departure from classical physics w x De k Px lr y DDe392kzx 1 112 ml 1 112 k12quotk De k wm wrm u AH Although Pxdecreases rapidly with increasing x it is name th epartiele pen etrates th e classically forbidden region This has been con rmed experimentally innumerable times over the past 70 years The penetration depth is de ned to be Ax 1162 and since k2 V2m0E h Ax which decreases rapidly V 2 quot V0 E with increasing V0 E In the classical limit 2mV0 E is so large relative to k that Ax is immeasurably small Example 61 Calculate the penetration depth for a small spherical dust particle of radius r l um 10 6 m density p 104 kgm3 and speed v l cms 10 2 ms impinging on a potential barrier V0 2E 2p22m h m 4W3p3 4x1014 kg so E mv22 2x1014 J so Ax 2x10 1 2mV0 E Consider an experiment designed to measure the coordinate x in x gt 0 Since Px is appreciable only in Ax above we must have Ap hAx 2mV0 E12 AE Ap22m V0 E So if a particle has penetrated the barrier the uncertainty in its energy is roughly equal to the additional energy required to pass over the barrier in the classical case However the phenomenon of tunneling which requires barrier penetration is very real and extremely important in understanding various systems It is also extremely important technologically Example 6 2 A conduction e in a block of Cu moves with energy E in a potential Va 0 inside the block which steps to V0 gt E outside The workfunction energy required to remove an electron from the block is V0 7 E 4 eV Estimate the penetration Ax of the 9 into the vacuum outside the metal 1 7 N 19 31 Ax z10391 m V0 E 4eV 6x10 Jm 91x10 kgso 2mltVOE 1A Penetration of waves into forbidden regions eg EM waves into metals or into the forbidden region of the re ector in total internal re ection is well known so the QM phenomenon is just another manifestation of the wavelike properties of particles Before moving on let s see what the re ection from a step potential actually looks like for a group wavefunction describing a particle 4k w i W 2mm 1 11m Obtaining such solutions involves considerable work on a computer 7 we were however able to learn much by exploring the behavior of simple wave functions Key points particle is re ected from the step with probability one but there is some penetration into the classically forbidden region The Step potential energy gt step height V1 First question What happens to a classical particle 7 g Apam39cle moving in the x direction Will simply slow W W down atx 0 from V1plm 2Em Z to v2 pzm 203 mm What happens to aWave 7 w x A wave traveling in the x direction and passing from the rst medium to the second medium Will be partially re ected and partially transmitted So the quantum mechanical interpretation for a particle Would be 7 The particle has a certain probability T of being transmitted and a probability R 1 r T of being re ected The general solution here is of course the same as for the x lt 0 part of the E lt VD case Who remembers 7 Ae k1x Be W x lt 0 J 2 E s V 1PX 1 m where klz and JQZM Ce 2 De 2 xgt 0 h h Noting the physical meaning of each term What can we say about D 7 tha the particle is moving from x lt0 towards x gt 0 that there is no particle moving in x gt 0 towards x lt 0 And from the continuity of 1px atx 0 What constraint do We have 7 A B C What do We get from the continuity of drpxdx atx 0 7 ik1A r B ikZC A r B kZkl C Adding Mfr k J39MC p l c LA kl 2 k1k2 k C k k C k k Andsubtracting B1kizilk 2 Bklk2A l l l 2 7 k Forx s 0 What does the rst term represent AZW A k 2 W x S 0 incident particle p x kl kz and the second term 7 LEW x z 0 k k2 For x z 0 What does the only term represent kw Vn39m r NH transmitted particle choosing Y k 2k2 So What does Pow 111W look like 7 r 169gt 4 A x s 0 combination of a traveling plane Wave oscillatory standin Wave x z 0 traveling plane Wave W 49 A m r The re ection coef cientR is Mk Cw C 39 39 iu uabilii i T17R7 2 RBBk k2 P 05 sz AM AA k k2 m C Woops Why doesn t Tequal 7 islt1Whenk2ltkEgtV0 AA The velocities are different in the two regions and the convention is that transmission and re ection coefficients are de ned in terms of ratios of probability uxes A probability ux is the probability per second that a particle will be found crossing some reference point traveling in a particular direction Since the probability per second that a particle will cross a given point is proportional to the velocity in addition to the intensity according to the strict definition V13 B B gtxlt B as before 11114 A A A 2 v C C 1 2k p1 illC1 p2 7162 T 2 2 1 and V A A V1 k1 k2 and s1nce V1 m m and v2 m m 2 2 2k k 2k 4k k TV 2 1 2 1 41 R WealsoseethatRTlduh v1 k1k2 k1 k1k2 k1k22 Note thatR and T are unchanged if k1 and k2 are reversed a the same value of R and T will be obtained if the particle is incident from x gt 0 and travels in the x direction The wavefunction is partially transmitted and partially re ected due simply to the discontinuity in V and not because Vincreases This is just like waves on a string where re ection occurs if a wave hits a point where the density or tension on the string changes or like light passing from medium 1 to medium 2 whether to higher or lower n This is a property of all waves in optics it is sometimes called the reciprocity property To get a better feeling for this let s look at how R and T behave as mctions of E V0 ForE lt V0 R 1 T 0 as we saw in the previous section and as EV0 increases Tincreases andR decreases as we would expect Rewriting R in terms ofE and V0 We have 2 l r l r V E R 17T z 7 1 J17 Vn E m No real potential is a step function When can a real potential Vx be approximated by a step Answer If Vx changes very gradually R will be negligible since change in A of particle is gradual re ection arises from abru t c an e in A Specifically iffhefracn39onal change in Vx is very smallwhenx changes by one A gt R will be very small Fo 39 39 39 clear systems A can be very long relative to distance over Which Vx changes then the step approximation is valid l0 EVu for gtl Vu Example 673 Aneutron n WithE 5 MeV Which is typical of a neutron generated by ssion enters a nucleus Where the bottom of the potential Well is at 750 MeV Estimate R Take V0 50 MeV E 55 MeV 2 14175055 Rlr lernE 11ernE 717 l1750SSJZ 0 3 The Barrier potential vm So what s a barrier potential Classically what do we expect for particle traveling in the x direction towards x 0 IfE lt V0 particle bounces back ifE gt V0 particle passes barrier slow down in 0 lt x lt a speed up x gt a Quantum mechanically we nd tunneling through the barrier whenE lt V0 with a probability which a increases as E V0 increases and decreases as a increases b We will also nd that inside the banier 1px decays exponentially In the next section we will consider several tunneling phenomena in detail in this section we solve Schrodinger s equation for 11 x Looking rst at the regions x lt 0 andx gt a what are the general solutions Ae k Be W x lt 0 JZmE woo Ceikix Derxk x gt a Where k1 7h And since there is no particle moving inx gt a towards x lt a what can we say aboutD D 0 What info do we need before we can write down the general solution for the region 0 lt x lt a Whether E lt Vg orE gt Vg We ve already examined both cases 7 let s start withE lt Vg So what s the general solution in the region 0 lt x lt a withE lt Vg x i x quot2 V E 00 Fe k2 Ge 9 where k2 What are our continuity conditions A B F G ik1A B k2F G Ce k F9719 Gelk ilee k k2 Ge 162 Fe kza As expected We have indeed 4 linear equations in the 5 unknowns A B C F and G To solve rst express F and G in terms ofA and B then Cin terrns ofA and B then nally B in terms ofA wax n ma t NIL What does WW look like 7 k 4 4 2a 2 2a 16EVnlrEVn sinhzkza 1 1 4EVnlrEVu ForT We nd T vAA E E 1 andifk2agtgt1then T5167 1 7 e a VB VB Tdecreases exponentially With increasing barrier Width a The last tWo results provide the essence of barrier penetration or funne ling phe name m1 which are purely quantum me chanical anmiV Ei 2 2V E Note that since kza To 17 is extremely large for classical m a 0 and V0 the probability for tunneling of classical macroscopic particles is vanishingly small Now let s go back and look at the case where E gt V0 in the region 0 lt x lt a k J2mE Vo 1px F Gel kzx where What s the general solution h 1 2a 71 2a 71 1 2 so Tv1CC1 Gk ek 1T smsza 39l with kza ZmaVor 1 Am 16EV0EV0 1J 4EV0EV0 1j hi VO Example 6 4 An 9 is incident on a rectangular barrier V0 10 eV and thickness at l8gtlt10 10 m 18 A Evaluate T and R as functions of the energy E of the incident e39 2rm12V0 2x9x103931x10x16x103919x 8x103910 9 h2 1068 Note that for E gt V0 T 1 whenever kza 11717 a result of destructive interference between re ections at x 0 and at x a This is reminiscent of the Ramsauer e ect in which 9 of certain energies of a few eV pass through a nobel gas solid as if the solid wasn t there T l and of scattering of n with certain energies in the MeV range from nuclei 7 this is the size resonance IzVU Finally we note that the timeindependent Schrodinger equation is of the same form as that governing classical wave motion For EM waves of frequency v passing through a medium with an index of refraction M 2 m w C Z M w0 where I is either the electric field E or the magnetic eld B Comparing to the timeinde endent 2 2 Schrodinger equation 1Pquot TrlE Vx 0 we nd M05 7mlE V00 so the behavior of an optical system with Mx should be identical to that of a mechanical system with Vx related to Mx One example is the coating of lenses to obtain very high light transmission Another example is in the imaginary Mx encountered in total step internal re ection which shows how we get frustrated total internal re ection when the layer of air wedged between the two solids is very thin Detailed treatment shows that Mx along ABC is real on AB but imaginary on BC There are EM waves in BC but their amplitude decreases exponentially with distance into air They are also standing waves so their Poynting vector S 0 so the light is totally re ected But if we add another block of glass we nd propagation of the wave through the 2 glass which is frustrated total internal re ection Frustrated total internal re ection was rst observed by Newton ca 1700 ngt0 And you can even see this with water waves Examples of Barrier penetration by particles e tunneling through a thin lm of SiOZ A1203 or CuOZ eg from one wire to another Oxide layers are typically 20 7 30 A thick so without tunneling we would constantly face problems of connections becoming disconnected or emission from ratemitting nuclei Historically a s withE 88 MeV emitted from 212Po and scattered from 238U showedthat Vr ZZeZ4rreor for distances r 2 r 3x103914 m where Vr 88 MeV But Z38U also emits 085 42 MeV So classically or emission should not occur In 1928 the problem was solved by Gamow Condon and Gurney Recall that forE lt V0 amp kza gtgt 1 we obtained T 16EVO1 rEVoexpr 219a expl LIME 7mm 1 vm chhc anew M urea r Garnow Condon and Gurney then approxrmated 7 c the coulomb potential as a sequence of adjacent rectangular barriers of height Vr amp width Ar d In the limit Ar 6 0 they foun T N exp 2V IZMiVO 7Eh21r Now T is the probability that in one trial the a will penetrate the barrier If the a is moving at speed v inside the nucleus the number of attempts per second it makes to penetrate the barrier is N N v2r so the adecay rate should be H V RNEexp 2 2m o E h2 r H233 mew mm Using v as the speed ofthe a after it escapes and r N 9 F 9gtlt103915 m yields an agreement between the calculated and observed R However R varies tremendously among nuclei 10 from 5gtlt10 18 sec 1 for 238U to 2gtlt106 sec 1 for 212Po This variation is due primarily to the E of the a The periodic inversion of the N atom in NH3 the oscillations of the N atom between the two minima l 71 3 occur at v 23786gtlt1010 Hz when the molecule is in m the ground state which is much lower than the vibrations uquot without barrier penetration Due to its relatively low but mZgL 4 J 1 5 U fundamental value this v1bration was used for atomzc clocks 1 rl MPV4 Atomic Clock The tunnel diode 7 will be discussed in Chapter 13 Physics 322 Aside on the Ammonia 1NH3 atomic clock How do we measure time We count something typically something with a xed period pendulum What do we typically call something with a xed periodfrequency An oscillator So what are the main elements of any clock l Oscillator 2 Counting mechanism accumulates the number of cycles 3 Mechanism to display count The stability of the oscillator is of course critical 1600 s 1700 s 1920 s 1940 s gt Water gt Pendulum gt Spring gt Quartz gt Atomic clock clock dr1yen watch clock clock 310 min 30 sday 10 sday 3 usyr noncloudy Let s look in some detail at the Ammonia clock we ll cover more modern atomic clocks when we discuss masers next semester 9 Q What does the potential look like If we solve Schrodinger s Eqn what do we expect 1p1 and 1p2 to look like 106 Assuming the molecule is in a superposition of its two lowest energy states what is the time dependent solution iEZIh w fuck W w2xe What does ltXgt look like Quiz oscillates VOL 0 moat wlxeriElth w2xeiiEZtfz What s the oscillation frequency x vE2E1 h no So how to do this in practice hint first appeared in the 1940 s During WWII the evolution of radar required the development of very sophisticated microwave circuitry ie the ability to control and measure microwave with high precision Apply microwave beam near frequency v to Ammonia gas tune v What happens when v E2 7 Elh Apply Feedback circuit to v tuning to maintain resonance Ammonia molecules are used as an ultrasensitive regulator of the frequency of the microwave frequency transmitter What well known device is this the basis for MASER 7 Microwave Ampli cation through Stimulated Emission of Radiation Atomic clocks are so accurate that minute adjustments must be made periodically to the length of the year to keep the calendar exactly synchronized with the Earth39s rotation which has a tendency to slow down There have been 17 adjustments made since 1972 addding a total of 20 seconds to the calendar In 1997 the northern hemisphere39s summer was longer than usual by one second An extra second was added to the world39s time at precisely 23 hours 59 minutes and 60 seconds on 30 June 1997 The adjustment was called for by the International Earth Rotation Service in Paris which monitors the difference between Earth time and atomic time The Square Well potential V V0 xlta2 orxgta2 Vx 0 a2ltxlta2 What do we expect classically If E lt V0 then classically i the particle will be bound to the well ifE 2 V0 particle is free Many systems can be approximated with a square well potential ie many closely space ions in a line The general solution of Schr dinger s equation inside the well is J 2mE h Alternatively the two independent solutions can be written as 1px B coska and 1px A sinkpc 1px A sinklx B cosij 1px Aeik x Be ik x where k1 1px Cek De39k 1px Feb Ge39W Outside the well for E lt V0 we have What can we immediately say aboutD and F Both zero so the general solution is 9 0 4m One wan Tnme zone m hue Many closely Suaccu mm m m AAMAAAAMAAAAAA x lt a2 2mV0 E xgt a2 H h As usual the relations between A B C and G are determined by the continuity of 1p and dwdx atx iaZ So we now have 4 equations and only 4 unknowns 7 this is a problem Why We need to leave one constant free for amplitude and so we can satisfy the normalization condition To solve this we treat the total energy as an additional constant that can be adjusted as necessary where we will find thatE can only take on certain values The general solution to the squarewell potential is quite complex and involves a transcendental equation 7 see Appendix H three bound eigenvalues three bound eigenfunctions V0 E1 0 4 it u 2 O a 2 u2 o n2 Note alternating parity w Vexy large v Mx m cos WinMs moderate height x F 27m MW I 2 o n2 90 n cos n MIMI0x in nite height AaZ 0 402 The In nite Square Well potential to Vb x lt a2 or x gt aZ a2 lt x lt 02 Vx 0 For 702 lt x lt 02 the general solution is still 1Isz ruz 0 u2 1px Asinkx Bcoskx where k h What are the boundary conditions 7 1px 0 at x 1112 AsinIm2 Bcoska2 0 Asinka2 BcosIm2 0 adding We nd ZBcoska2 0 subtracting We nd ZAsinka2 0 Since We require thatA or B be different from zero We get two classes of solutions Class OneA 0 coska2 0 Class TWoB 0 sinka2 0 1px Bcoskx 1px Asinkx Jul27r237r257r2 ImZmr kn mra n odd kn mra n even O p 7 2712quot 4 7 2m 2m 2mg2 E n consider E1 This is the lowest energy 7 the particle cannot have zero energy 7 basically due to the uncertainty principle Equivalently there must be zeropoint energy because there must be zeropoint motion WU a7 O rz2 Note alternating parity again p hzkj nzhznz 2m 2m 2mg2 V0 W E NZ uZ o W a2 compare classical dashed with QM 111 The Simple Harmonic Oscillator potential This case is so important because the local motion about any stable equilibrium point x0 can be approximated by simple harmonic motion To show this we show that the potential energy can be approximated as that of a SHO dV 1 sz Vx Vxo El n x x03 dxz LX x x02 Since the potential has a minimum at x0 the 1st derivative of Vx is zero at x0 and we can choose Vxo 0 so we get 1 dZV z 1 2 look magi 1x2 J 7 X Cxx familiar Quantum mechanically Planck already predicted that Vm En nhv n 0 l 2 3 but the solution to Schrodinger s equation yields En n 2hv n 0 l 2 3 where V Cm1227r is the classical frequency see Appendix 1 Therefore E0 12hv is the zeropoint energy Was Planck wrong 7 Yes although only the energy differences were important However in 1914 he published a speculation based on entropy g x x cons1derat1ons that the zero pomt energy should be Irv2 HOW to draw energy levels 7 What should the wavefuntions look like VIE I Cm 14 With u x J Quantum Number Eigcnfunctions 0 99 ul c quot l v 3 l lII 7 luc quot w 7 2 12 ll le le quot 4 w 3 lil 13u7 lusty quot 7 m 4 m l43 7 1213 0 Jam quot 5 95 451514 mm 415w Once again we see that the number of nodes n t m Note that 114x contain two terms k a polynomial responsible for the oscillations of 114x A a term expu22 responsible for the decay of 114x in the classically forbidden region Also note that the parity of 114x switches from even for even n to odd for odd n Name or Pllysi Polcnliul and Prnbzlbilily slgm unl Syuum Example Tmle Lnsrgm Dcnslly Feature Zero Promn in r E WW Resulls ust polunllul bcum l rum V or other Lytlnlum h 7 syxlems Slcp 39onduclmn w l cnslrulmu polcnliul clecm n new jb 14mv ulexcludcd lcnclgy erlucc M l x rcglOn below 0m mom 0 0 Slcp lmlron w l Punile rcllccr polcmlul lrylng to lion an energy a Ape pulmqu ubme mp nudem discommuil Burrler l l Tunnclmg pulenuul Qing m 1 energy mum 1 WW below mp Culmunh 1 burner Burner Flcclmn my No rcllccuon pmcnliul tclmgl39rnm i 7 3 ccnain Cncl gy ncgzmvcly l39 l cncrgm above lop lollicrl mom l 1 Vm 71 U n D U que Ncullou m hnergy square lmund m j T 9 Elm quunllmtmn all IlllClCHs X pulcnuul D n 0 n nl39Inile Molecule W Appruximuuon squurc smcuy l l m llnilc well con qu E q u 03am square well lmtclllial 10 lm l o a o ei K Fm Simple Almn ol Acmpmnl lmrmouic vlbluling 39 l energ oscillator dlnlomlt r39 pmcmlul molecule WW x39 J J Name of Syxlcm cho pmcnlml Step polcnliu energy below lop energy below Lop Ph mu Example Proton in bcum mm Cychxlmn Conduclmn eleclrnn ncur wrfACc 0139 Polcmim and Pmmbnny Tom Enmgles onsll E WW vm erl jE 1 WW x o q w w Signi cant Fculm39c Rcsuhs med for other Pcnstrauun uf cxcludcd Parlile rc ccr polemml d sconunuil Tunneling Pmbahlhly Slgm canl Potential and Name of Phy cal swam Fxnmple Tom Fncrgms Dcnslly Earner LL N0 re ccuou potcnlim 39 7 if ar I i cncvgy v 1y l39 l encrgm ubuvs mp ionicd nmm 1 Wu 4A x U u n Fmila Nounon WU Energy square bound n Wq qmxlIztm1 wH nucleus 7 4 V g gt x pomMia u a o a nlimlc Molecule VW Apploximuuon squm a stric y 1 1 to nite well con ned E q WM iqualcweH potential in box u a 0 a 7 7 Wu Simple Atom m empmm harmonic Ibldlinquot 39 mar monum dmmnm A potcmlu mulcculu WW 1 x Chapter 2 Outline Photoelectric effect Einstein s interpretation Photons and Xray production Bremsstrahlung braking radiation Compton effect Waveparticle duality for EM radiation Pair production and annihilation Antimatter Positronium Dirac theory Cross sections for photon absorption and scattering Assignment 2 Chapter 2 due Sept 9th Questions 6 21 Problems 4 5 8 9 11 18 29 Black body radiation 87212 87212 IN V dV 2 8 V dV 2 dv mm c3 U c3 eXphVkT 1 Rayleigh Jeans law Planck39s law s I 0 2000 4000 6000 9 nm Note that this plot is a function of 7c instead of v Planck s Blackbody spectrum The Photoelectric effect 1887 Hertz notices that the discharge between two electrodes works better when uv light falls on one of the electrodes Further studies revealed that negative particles were being emitted from a clean surface when it was 1900 exposed to light Lenard de ects these negative particles through a magnetic field and finds a charge to mass ratio the same as for J J Thompson s cathode rays aka electrons T Electron current I 0 Voltage Vgt VV0 Quartz Glass envelope window ll5l I A ncident f light 39 v Lenard s lt gt 39 Apparatus o l 3 Polarity reversing switch IIII Lenard notes that for positive V the current quickly reaches a plateau For negative V the current drops and goes to zero at V V0 V0 is referred to as the stopping potential What would we expect from a classical theory 1 Kinetic energy of the photoelectrons should depend on the intensity power density of the light higher intensity gt higher electric field gt larger acceleration of the photoelectrons The emission of the photoelectrons should occur for all wavelengths provided intensity of the light is sufficient Emission of photoelectrons should be delayed with respect to incident light in order for electrons to accumulate enough energy to perform work necessary to leave the solid Example 21 A potassium plate is placed 1 m from a feeble light source whose power is 1 W 1 joulesec Assume that an ejected photoelectron may collect its energy from a circular area of the plate whose radius r is say one atomic radius r z 1 x 103910 m The energy re quired to remove an electron through the potassium surface is about 21 eV 34 x 10 19 joule One electron volt 1 eV 160 x 1039 19 joule is the energy gained by an electron of charge 160 x 103919 coul in falling through a potential drop of 1 V How long would it take for such a target to absorb this much energy from the light source Assume the light energy to be spread uniformly over the wave front At Im from the source the power density is R 1Js 1W 474ml 1257m2 008W m2 Thus power radiated onto the target area is R1rr225x103921 W To emit a photoelectron one needs to accumulate energy larger than workfunction of 21 eV This takes time t W 21eV 21162gtlt10 19J 34 gtlt10 J T z 2 1111 Ran 25 gtlt10 W 25 gtlt10 W 251x10 21W S 3 n If we increase the light intensity do we expect V0 to change i M Bri ht Ii ht What about the current for V gt 0 will it change if we increase the light intensity o 1390 1395 7 How fast is the fastest electron mVZ K 6V0 What do we expect for different frequencies Intensity L is constant V gt Do we expect this pattern to continue i A for arbitrarily low frequencies f2 gt 1 7 WM There is a cutoff frequency v0 391 below whi Ch the photoelectric I f V effect doefs n0t occur 5l 0 5 10 15 V 39V02 V01 The photoelectric effect displays 3 inconsistencies with classical physics Classically the amplitude of the electric eld E should increase with the intensity of light and so should the force eE on an 6 so Kmax should increase with intensity but it doesn t The photoelectric effect should exist at all V but instead there is a cutoff frequency V0 below which it does not exist There should be a time delay between the absorption of light amp the observed photocurrent but none is observed Now back to our story Lenard s results were essentially unexplained until 1905 Einstein proposes theory of light quanta based Planck s idea of energy quantization He reasoned that inside Planck s cavity in going from a state nhv to a state nlhv an amount of energy hv would be released He then generalized that all light consists of bundles of energy called photons each of energy Applying this idea to the photoelectric effect he claimed the kinetic energy of an emitted electron should be K hv w where w is the work required to remove an electron from the metal hf Energy A quotquotz And for the least tlghtly bound electron we have hf o l 0 w Kmax I Filled W018 often referred to as the work functlon eIectron l states I From th1s we see that we can determ1ne the cutoff j frequency which is u Distance Inside i I Outside metal metal Kmax 0 3 hvo w0 Surface Note that the relationship between Kmax and V is linear and recall that Kmax 6V0 therefore we can write l9l4 enter Robert Millikan N l H Stopping potential V Frequency IONsec What is the slope of this line he According to Millikan this result gave proof independent of the facts of Blackbody radiation of the correctness of the quantum theory and Nobel prizes for Einsteinl921 and Millikanl923 Example 22 3 Deduce the work function for sodium from the figure E g 2 Kmax V0hV WO i 9quot 3951 So for V0 0 we have V0 V0 2 56 X1014 HZ I39IIJIIIIIIII W0 hVo 4 39 8 12 663 X 10 34 J8 X 56 X 1014 HZ I Fre uency 1014sec 37 x1019 J gtlt1eV16 x1019 J I 23 eV I I l 23 V gt1 Or for v 0 wO 6V0 To continue we now ask if the Vvs v plot will be the same for all photoelectric emitters 1L stripe same Review The Photoelectric effect Hertz Lenard Light shining on certain metal surfaces causes e s to be emitted Glass envelope Quartz window A lncident 395 o 5 1390 1395 7 uvo o Polarity reversing switch Lenard s M l fastest electron 2 Apparatus There is a cutoff frequency V0 below which the photoelectric effect does not occur Review The photoelectric effect displays 3 inconsistencies with classical physics 0 Classically the amplitude of the electric field E should increase with the intensity of light and so should the force eE on an 639 so Kmax should increase with intensity but it doesn t The photoelectric effect should exist at all v but instead there is a cutoff frequency v0 below which it does not exist 0 There should be a time delay between the absorption of light amp the observed photocurrent but none is observed Einstein proposes theory of light quanta based Planck s energy quantization idea E hv K hv w where w is the work required to remove an electron from the metal KmX O gt hvo w0 Review Relationship between Kmax and V is linear and recall that Kmax 6V0 h eV0hV gt Voz V 6 N w H Stopping potential V 8 Frequency 1014sec According to Millikan this result gave proof independent of the facts of Blackbody radiation of the correctness of the quantum theory Applications Very wide from high energy physics photomultipliers to solid state physics Angle Resolved Photoemission Spectroscopy ARPES to chemistry measurements of oxydation levels and chemical composition Technological photocell detection of light playback of sound in early movies etc Can you think of more examples Photoemission spectroscopy analyzer lt detector aorb sample N01 mulizsd imensily Numvuhlad innmil Typical pho roemission spec 7 T3omlt 7 pm m mum wkan 7275 7270 7265 rzen Energy 6V Numnlitnl intensity kw Trum from Bi2212 Bi 41 52 amp 41 72 SrEdZZ deSZ Theta3 deg bv500 eV Bi51 1251 32 conduction band valence band 200 Energy eV 100 O Valence and conduc rion bands simples r example HHHW HH H I Au5d poy Au 150m3 3 E 100 valence band E E a 397 93 50 E conduc non band nVV VVV WVVV VVVV VVVV VVV M W WWWW WWWW W is 74 Energy eV T100K T35OK mm um mm mm 01 00 01 02 Xrays On November 8 1895 R6ntgen discovered that mysterious radiation is emitted from a vacuum tube when cathode rays electrons strike an object This radiation could expose photographic plate in a closed drawer and caused certain materials to emit light Nobel prize in 1901 Later it was shown that this radiation has the same electromagnetic nature as more common Visible light but of higher frequency Evacuated tube Filament 540 athode Wilhelm Conrad R ntgen 1845 1923 Picture of Mrs R ntgen hand What causes this radiation What are its properties Energy hVEk Ek39 If the electron after collision comes to full stop Ek 0 hlmax Ek Or in terms of 9 gm ZE Ek K Xray spectral intensity w 5 5 E E 2 20 keV o 1 l D 02 00 0 08 Wavelength A 10 Review of special theorv of relativity Search for Aether or medium in which light propagates led Michaelson and Morley to perform very nice experiment and disproving it s existence in 1887 Halfsilvered mirror Light source 391 Observer Mirror 1 lt Aelher drift l7 In 1905 Einstein formulated Special Theory of Relativity built on following postulate The laws of electromagnetic phenomena as well as the laws of mechanics are the same in all inertial frames of reference despite the fact that these frames move with respect to each other Consequently all inertial frames are completely equivalent for all phenomena This of course required modifying either Maxwell s equations or Galilean transformation Einstein Chose the latter This required that C const in any reference frame Many properties people took for granted will never be the same Considsr following setup l d l a a t O UM inS NEE 21 C zzd2 my At 2 C 273 392 2 EN 72Al 2At Solving for At At A t Similarly ifd 11 d d39 V1 M2 C2 11 14202 Galilean transformation fx m y39 y z39 Z lquot 1 Adding velocities Vvu Lorentz transformation x39 7x ut y39 y ZZ t39 Kt lzx C l yVl LfCZ vu v l4 vuc Dynamics mass momantum rest energy 01 The Compton effect 1923 Arthur Holly Compton shoots a beam of xrays at fixed 7 at a graphite target and measures the intensity of the scattered xrays as a function of wavelength and angle Defining slit B S Calc1te ragg It crystal spectrometer y Shutter X ray tube lohmzaglon Mo target C am er He observes a shift in the wavelength of the scattered x ray This cannot be understood in terms of classical EM waves Compton along with Debye proposes that the incoming xray beam is actually a collection of photons each of energy E hv think billiard balls 3v mm mm Consider our photon as a particle of energy E and momentum p What is the total relativistic energy for such a particle in terms of its rest mass m0 and velocity v 2 mac 2 2 E 2 m6 mOC K see Appendix A xl vZc2 Since v c and since E 2 IN is not zero we must have that m0 0 l And from the expression E2 02192 m0022 we must have that p Ec hVc hxl 339 V Momentum conservatlon ghoton 11p1 I I E3533 Electron 1 I x p0 plcost9 pcosgp I o o vxvv a J o p1s1n49 ps1ng0 Ylectron 1p 0 o squar1ng these and add1ng Before After we have p0 p1COS 2 pzcoszgo plzsin26 pzsinzgo 1902 pl2 2p0plcost9 192 Conservation of total relativistic energy requires that E0 E1 K or in terms of 0 and p1 EO E1 Cp0 p1K But the square of the electron energy is given by E2 K mocz2 62192 mocz2 or K2 2m0c2K 62192 p2 262 2m0K But since Cp0 p1 K p2 70 p12 277106070 p1 Putting this into our squared conservation of momentum equation we nd 1902 pl2 2p0plcos6 p0 pl2 2mocp0 p1 p0p1C086 190191 17106090 I91 1 cos6 moc1p1 1190 1 056 mod lh Ai 2 10 h l cos6 DC where 2C a hmoc 243gtlt1012 m 00243 A Note that Al depends only on 9 not on 7t Compton wavelength antirparticles 7 1929 7 Paul Dirac is trying to come up with a relativistic form of the Schrodinger equation The equation he comes up with works well but has energy solutions of E i chZ moczZ What does a negative energy mean 7 Instead of simply throwing away the negative energy solutions Dirac pursued the possibility that perhaps they had some physical meaning

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