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by: Josiane Blick


Josiane Blick
GPA 3.72

William Baird

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William Baird
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This 79 page Class Notes was uploaded by Josiane Blick on Saturday October 3, 2015. The Class Notes belongs to PHYS 3802 at Armstrong Atlantic State University taught by William Baird in Fall. Since its upload, it has received 27 views. For similar materials see /class/217873/phys-3802-armstrong-atlantic-state-university in Physics 2 at Armstrong Atlantic State University.




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Date Created: 10/03/15
PHYS 3802 Notes follow and parts taken from sources in Bibliography The Step Potential A problem related to the particle in a finite square well involves a particle incident on a step This is what you would get ifyou removed one ofthe walls in the nite square well case We can write the potential as Vx 0 x lt 0 Vx V0 x gt 0 This potential will have no bound states but we can still have a particle incident from the left with an energy either greater than or less than V0 Classically we would see that there is a 100 chance of re ection if EltVo and a 100 chance oftransmission if EgtVo Where there is no potential region I we get the simple Schrodinger equation ll 2 w x k1 w x and in region II where the potential is V0 we get ll 2 w x W x where k lsz k l2mE V0 Our solutions are then 39k 39k l 1x Ae 1 Be 1 139ka 139ka ll Hx Ce De We will assume the particle comes in from the left moving towards the right Notice that that does not mean that we can set B0 since it will represent particles re ected from the barrier We can however set D 0 since it would mean particles coming from oo and we can restrict the problem to ignore those Using the condition that w10y20 and y 10y 20 where they meet we can write PHYS 3802 A B C kA kB kZC Solving in terms ofA gives us k k A 2 k B C A klk2 k1k2 The re ection coefficient R is defined as the square ofthe ratio of B left moving in region I to A right moving in region I 2 B Z k1k2 W k k2 Since there are no bound states the probability oftransmission T must be just 1R This can also be written as k2 4k1k2 T 2 2 k1 A 161 k2 Notice that there is a re ection probability R720 even if EgtVo which is not at all classical lf Elt V0 there is penetration into the step with an exponential decay just as was the case with the walls of the nite square well potential Barrier Potential lfwe replace the step with a barrier meaning the potential V0 extends over only a nite part of the X axis we can write Vx0 xlt 0xgta VxVO 0ltxlta As always we solve the Schrodinger equation in each region I II and III This will give solutions ofthe form PHYS 3802 I Ix Aeik1x Be39iklx xlt 0 IHxCe39 De 0ltxlta I HIx Feiklx Ge39iklx x gt a and we have k lsz a 2mV0 E 1 h h We can t set D O in this case since there could be re ection from either the front or the back of the barrier meaning A B C and D could all be nonzero We can however set GO if we again assume particles incident from oo only Repeating the same process as in the case of the step potential ie requiring continuity of the wave function and its rst derivative at each side of the barrier we get a transmission coef cient of 2 1 sinhza a W2 42 1 E V0 V0 which reduces in the limiting case or a gtgt1 to T16 1 5 e4 V0 V0 showing an exponential dieoff ofthe transmission probability with barrier thickness Expectation Values The probabilistic nature ofquantum mechanics means we can only ask about average values for interesting quantities the exact value at any particular time is in general unknown and unknowable This average or expectation value written as ltXgt for PHYS 3802 position ofan observable such as the particle s position X is calculated by finding the overlap integral lt xgt J P xtx l xtdx ln stationary states where there is no time dependence this reduces slightly to lt xgt Djill xxI xdx 7m The same approach works for nding the expectation value of any function of X We can find this value for the particle in a square well nite or in nite without any real work since X is an odd function and the product of v and y for either odd sine or even cosine wave functions will be even we get the integral of an odd function times an even function This again gives an odd function and its integral over a region symmetric about zero will be zero This is not unexpected the potential inside the square well is zero so no point is preferred by the particle and the average position is therefore in the middle of the well where x0 If you want the expectation value of momentum you have to write it in operator form 8 39h p lax It will operate on the wave function to its right and the result is multiplied by the w on the left before integrating Again using the square well solutions we can nd the expectation value of p as lt pgt ihlr xl r xdx Since the derivative of sine is proportional to cosine and the derivative of cosine is proportional to sine we will again end up with an even function multiplied by an odd one so that integrating from oo to co will give zero Again this is expected in the symmetric well since there should be no reason forthe particle s momentum to be more positive than negative or vice versa PHYS 3802 We can get less trivial results ifwe look at different functions of the position or momentum For example we can look at the expectation value of X2 in the infinite square well which does not have to be zero For the lowest energy state we get the integral below lt x2gt JW1xx2W1xdx 7m Our ground state wave function is 71717 X W X Sll 1 L After performing the integration we are left with L2 L2 3 2n2 7 2 We could calculate ltp2gt and we would get 2 2 2 lt 2 gt h n it p 39 T Look back at the energies for a particle in an infinite square well and you ll see that we could now write them as 2 Eltp gt 2m a result that we might have guessed The Simple Harmonic Oscillator Now that we have the Schrodinger equation to nd the wave function associated with a potential we can use it on one of the most important potentials in quantum mechanics the simple harmonic oscillator SHO As you may have seen in 2211 the motion ofa mass on a spring or a pendulum which has only been displaced by a small angle is simple harmonic motion The analogous problem in 2212 is a circuit containing only an inductor and capacitor The charge on one ofthe capacitor s plates will oscillate back and forth in just PHYS 3802 the same way as a mass on a spring oscillates It turns out that many interesting phenomena can be modeled as harmonic oscillators The potential for a SHO is 1 Vx Ema 2 x2 Whenever the oscillator is displaced from equilibrium a potential ofthis shape will give a restoring force that tends to bring the oscillator back into equilibrium The mass on a spring can be shown to have a total energy composed ofa part like the potential above plus the ordinary kinetic energy E ma 2x2 lmv2 Your book calculates the classical probability of finding the mass at a point X as PCxdXoc g v There is no classical restriction on the particle being totally at rest E0 Things will not be quite that simple quantum mechanically We can write the Schrodinger equation as l W gm xE39 x This is not a trivial equation to solve It takes a group of special functions known as Hermite polynomials There are many special functions whose primary use to us is that they solve a particular differential equation These polynomials fall in that group and we can look them up in tables or use a computer to nd them Because quantum mechanics can be formulated in terms of matrices there are some techniques from linear algebra that will be useful later Some terminology will be helpful now The wave functions that solve the Schrodinger equation can be viewed as eigenvectors and the energies associated with these wave functions are eigenvalues The eigenvectors also called eigenfunctions of the SH0 can be written as PHYS 3802 11 quotx Ct 6W HM The r H rrnrt ootynornrats have the generat torrn Hnx Axquot Bxquot z Cxquot 4 where the A B c are constants dependmg on n and therr srgns atternate M n are even so Wm Wm That s easy enough to remember The nrst 3 Wave mnctrons are shown oetow red ground state my green wx otue vim Nouce thatthere are n orossrngs of the x axts fontquotx The energtes assoctated thh these funcuons can be W tten as Enlhm 2 o o Stncetf r rt PHYS 3802 momentum zero if it has no energy and its position zero for a SHO with no energy perfectly This 12 h 0 term is known as the zero point energy Because the Hermite polynomials are orthonormal the integral of vmxyx from oo to co will be zero unless mn Think about what happens when a harmonic oscillator or something sort oflike one like an electron in an atom moves from one level to another energy in the form ofa photon will be absorbed or emitted to make the transition possible We can represent the light as a plane wave meaning it will have a form like em Forthe case where the wavelength of light is large compared to the size of the system called the dipole approximation which is the case when looking at visible light and the atom we can take the expansion ofthe exponential em 1 ikr k 2 z k3x3 6 and keep only the term which is rst order in kr In one dimension this becomes kX What we are looking for next can be considered to be an overlap integral We want to find the overlap between two or more functions This is very similar to the dot product which calculates something like the overlap between two vectors Orthogonal vectors like ones pointing East and North would have zero overlap as would our orthogonal Hermite polynomials When the two Hermite polynomials are of different orders n values we would ordinarily get no overlap If we sandwich some power ofX like XP between our two Hermite polynomials HX and Hmx we will get zero overlap unless n and m are separated by p units In the dipole approximation we sandwich X from the ikX term in between the two Hermite polynomials but they have to be exactly one order different from each other for this to give a nonzero result That nonzero result means that there is a nite probability for there to be a transition from one state represented by HX to another represented by Hmx where n m i 1 This is known as a dipole transition between the two states and it s by far the most common way for a system to move between two states Higher order transitions where n and m are more than one unit apart are called quadrupole octopole etc transitions and they are My unlikely to occur on Earth but are sometimes seen in high vacuum environments like outer space Dirac Notation Expectation values are very important in quantum mechanics and we can write them in a simpler form using the Dirac braket notation A ket is written as a wave function frequently just shortened to y sandwiched between a vertical line and a rightpointing arrow We write this as ygt The complex conjugate of this wave function is known as a bra and it points the other way lty In the harmonic oscillator case for example there PHYS 3802 are different wave functions for each different energy level n so we might write something like ygt or lty If we put two ofthese things together bra on the left ket on the right we get a bracket which represents a kind of dotproduct For example ltIr n w m gt W nlxlr mlxldx We can shorten this even further by getting rid of the w part altogether and writing lt n mgt W nlxll mlxldx where ltn and mgt represent the nth and mth energy eigenstates lfour eigenstates are orthonormal the expression above will give 1 if nm and zero othenNise This form allows expectation values to be written simply as well If we want the expectation value ofthe position operator x in state yn we can just write ltnI2Ingt lw zlxlxl nlxldx This is very closely related to a transition probability lfwe are looking for the chance that the state ngt could lead to the state ltm through the action ofthe operator x we would write lt m I ngt W lelxr nlxldx One ofthe most interesting operators used in the harmonic oscillator problem is the annihilation operator a We can de ne it as What will this operator do when it acts on a state of de nite energy ngt Using an expression from htt alileo h svir iniaeduclasses751mf1ifa02Sim leHarmonicOscillatorhtm we can write the harmonic oscillator wave functions as PHYS 3802 quotW a mm x2 2 n e h Usmg Matnernatrca We operate on one of tne Wave mnctrons say Wm wrtn tne annrnrtatron operator a o get H 3 hbarz e 12 hbarm x2 or 4m2 x4 J 2 hbarz W4 wnrcn ooks nornote We notrce though tnattnrs rs exacuy equat to 5m Mx KS 1 M51quot if an MEI emswmq Usmg the new notauon We can W te a much shorter form of the same resu t a 5 gt xg 4 gt m generat tor any state ngt We Wm get a n gt E n7 1 gt a trng n a 0 gt 0 Tne creatron operator a pronounced ardagger j rs tne Hermman contugate of tne a t nt m ail 2h mm utwe re not h cornptex conjugate so tnat We get PHYS 3802 As you rhrght guess rt we operate wrth thrs oh the state 5 we get sorhethrhg whrch rs a a h yery corhohcate w w 715 lobar an hharzmxz be so hba m2 x uz 8 m3 x5 as 2 x an user hm u m ahq agarh reduces to sorhethrhg srrhbte 2g xwl5l T axwt ml 61 rullsimplifyl R th the srrhote form we get a39ngtxn1n1gt Thrs rsthe reason these operators are caHed creatroh aho ahhrhrtatroh operators they erther create or ahhrhrtate ohe quantum ot ehergy th erther case the cohstahttactor rhuttrotyrhg the that state must the square root otthe targer ot the two states ertherthe rhrtrat or the that n er E nation in 3D hm Hr book ooks at the brobterh or a 3D rhrrhrte botehtrat WeH whrch must ah unbreakabte box A corhbtete set or waye ruhctrohs can be wrrtteh h the torrh v mm W MW AW 3 z Smce we haye the same bouhqary cohqrtrohs here as we had h the D rhrrhrte square WeH that rs the waye ruhctroh must be zero at the boundartea we wrh get the same who or Sotuttonsy gryrhg us V x yz AsinquB sinkzstin klz ere ength we er get ah Wh with expresston torthe ehergy otthe pamde tn the box We PHYS 3802 2 2 2 2 2 hit n1 n2 n3 Enn2n3 1 2m Li L L Notice that there is only one ground state where n1n2n31 but if the box is a cube so that L1L2L3 there will be three states with energies equal to h2 2 E 6 711712113 These will arise when two of the hi are equal to 1 and the other is equal to two which can happen three ways When different states have the same energy the states are known as degenerate Stretching the sides of the box by different amounts so that all the L are different would bring us back to the more complicated expression for E so that 112 121 and 211 would all have different energies This is known as lifting the degeneracy The Hydrogen Atom A single electron moving around the nucleus of an atom is another 3D problem When more electrons are present there is no exact closedform solution although the numerical approximations are very good If we want the atom to be neutral overall that leaves us with the study of hydrogen The most important part of the interaction between the electron and the nucleus is the electrostatic attraction between the Z positive charges in the nucleus and the single negative charge of the electron The potential looks the same here as it did in 2212 Zke2 Zke2 7 lx2y2z2 When the potential has spherical symmetry obvious in the rst expression above it makes sense to rewrite the Schrodinger equation in spherical polar coordinates We will also have to account for the fact that in a classical situation the electron and the nucleus both move The electron will move much more than the nucleus but to make the math easier we can define a reduced mass that we will use instead ofthe electron s mass By effectively changing coordinates to a system where the electron does all ofthe moving we don t need a term for the kinetic energy of the nucleus leayazl PHYS 3802 lfa mass m and a mass m are in orbit around one another we can de ne the reduced mass as mlmz 7711 m2 We ll use this u in place ofthe m in the time independent Schrodinger equation Writing out the full equation including expanding the Lapacian operator in the kinetic term gives us 2 a 2 Zk 2 h la Vz Wj t 1 a 51110 alj 1 a W e Iquot E 21 rz r 8r rzsin lt90 rzsinz Ml W W where v is a function of all three variables As a reminder the connection between spherical polar and Cartesian coordinates is r lxzir y2 22 x rSin0 C0s 6 C0S1i and y rSin0 Sin r 2 Tan1 Z rC0s6 y When we moved from the timedependent Schrodinger equation to the time independent versions we assumed that the wave function which depended on two variables X and t could be separated into two wave functions each ofwhich only depends on one variable We will make a similar assumption here and write vira6allerlfl6lglll lfwe substitute this form ofv into the Schrodinger equation and then divide by Rrf9gl we get E 2 2 h 1 irzder 1 ising 1 dg 2p r2 Rrdr dr 110st d0 d0 g sm20 d 2 We rewrite this separating the r dependent terms from the 9 and 1 dependent ones giving PHYS 3802 1 d 0df6 1 dzgw f0sma sm d0 g sin20 d 2 We were able to specify that V is really only a function of r V Vr since we re dealing with the Coulomb potential which is spherically symmetric The left side ofthe equation depends only on r while the right side depends only on 9 and l Each side must therefore be equal to the same constant We will call that constant 1 and rst solve the right side ofthe equation using it Again we rearrange things to put odependent terms on one side and 9dependent ones on the other We are left with 1 d2 sin0 d 1s1n20 gll all 0 d0 Again each side must separately equal a constant not the same constant as before but related to it which we will this time call m2 This makes the odependent equation particularly easy to solve since we have 0125 2 7m 1 gill 61152 The solution is quickly veri ed to be gl eim i sin0 d0 dfl As for the 9 terms the solution is much more involved but again is useful enough that the group of functions satisfying this equation have a special name the associated Legendre functions These functions depend on both constants e and m and can be written l Slnl9 m d m021 ZCZTSG 0050 1 Because these fm9 are so often found associated with the gl function in problems just like this the two combined are referred to as the spherical harmonics PHYS 3802 Yzml al M0 gmll The spherical harmonics are a complete set of functions covering a sphere What this means is that any pattern on the sphere like the boundaries ofthe US on a globe can be represented by a suitable combination of spherical harmonics just as any onedimensional function subject to some reasonable constraints can be represented by a suitable combination of sine waves with the correct amplitudes frequencies and phases Anqular quot39 f The allowed values ofangular momentum for a particle moving in a central spherically symmetric potential like the Coulomb potential are quantized It will be easierto show this ifwe rewrite the Schrodinger equation using the angular momentum L de ned as L rpsinA rp where A is the angle between the position vector r and the momentum vector p and p is the tangential part ofthe momentum Your book uses this to nd that p2p3 L2 2y 2y Zjlr2 This will give a Schrodinger equation that looks just a little different from the ordinary radial TISE 2 2 2y 2y r2 Vr Rr ERr We can make it look exactly the same by rede ning the potential to absorb the part containing the angular momentum The potential is then known as the effective potential Ve r L2 Ve r Vr Z rz Finding the differential operators that represent the parts of this new Schrodinger equation is slightly ugly so we just quote the results that PHYS3802 1 8 8 2 h2 2 prop r28rr er and 1 a a 1 82 L2 2 722 0 sm 86 sin26 M2 0 sine a 6 Notice that these two expressions are operators meaning they will act on the wave function to their right when they are actually used The very convenient thing for us about writing the TISE this way is the result of operating on a spherical harmonic with the L2 operator LipYzmle 6W lthYzmle i That will be much easierto evaluate than the long chain of derivatives and trig functions If the value of L2 is 1 W the length of the angular momentum vector L itself must be iii zz1h z0123 Similar work will reveal that the operator for L2 is 3 Lz0p which when operating on the spherical harmonics will give Lzop 3 2 3 so that the projection ofangular momentum along the z axis is just mh Because m is limited to values between and 4 we can see that the angular momentum vector L can never point exactly in the z direction since its length is at least slightly longer than This is because there is an uncertainty relation between the angular momentum components and having L2 L would mean LX Ly O with no uncertainty The reason the z axis is singled out in this treatment is that we used the coordinates 9 and l and 9 is de ned in terms ofthe z axis We will see later that in the event a magnetic PHYS 3802 field is applied we will orient our coordinate system so that it is along the z axis For this reason m is sometimes called the magnetic quantum number The Radial Equation Going back to when we rst broke the TISE into radial and angular parts remember that we set each side equal to 1 Using that and the Coulomb form forthe potential we can rewrite the radial part as h2 a 2aRltrgt Zprzar 8r kZe2 Wm 2 r 2 r Rr ERr Solving this differential equation gives us eigenvalues ofthe energy ofthe form 262 h it ZZEI E z 2n2 n2 n This time n known as the principal quantum number is still an integer but is now restricted to positive numbers n0 is not allowed The constant E1 is the familiar 136 eV binding energy ofthe electron in the ground state of a hydrogen atom I is restricted to lie between 0 and n1 and m ranges from to 1 This restriction explains the filling of energy shells in the periodic table when we include the concept of electron spin which we ll do shortly The eigenfunctions corresponding to the energy eigenvalues can be written in the form of special functions either the Laguerre polynomials or hypergeometric functions Again this is another class of special functions whose importance to us is that they solve a differential equation that describes something physically interesting The wave functions depend on both quantum numbers n and 4 as well as the radial coordinate r We can write them as Rn r An eiraon r nlra0 where Qquot represents the associated Laguerre polynomial This is derived from the nl h Laguerre polynomial via U1 d1 MAX nl E A few sample radial wave functions are shown in your book For example PHYS 3802 2 fr R10r 39 a0 J3 R30V 2 1 1 zrz 5 0 3 3a 3610 2703 4 2 32 2 r 9 mo 81 30a3 6102 Electric dipole transitions the most common between levels require that A i1 and A m0 or 1 Looking more closely at these wave functions we have at this point W nlmr76 Jq Cnlm RnlrYlm6 4 including the normalization constant Because the energy depends only on n and there are multiple values n2 ifyou count them all of I and m possible for a given n we have many degenerate energy levels This degeneracy can be lifted if there is a magnetic field present since the energy will then depend on m and 1 Finding the normalization constants Cnlm requires using the normalization condition in the form ofthe integral below 21 W V rzsin drda 61 1 000 Your book calculates this normalization condition for the ground state where n1 and I and m0 The angular part is easy since Yoo9 Yoo9 14 7 and R10 has no angular dependence This means that the integrals over 9 and i give 4 it exactly canceling out the spherical harmonics The Rm given earlier are already normalized as you can verify ifyou calculate i earL10 earL10 r2 dr 3 a0 0 which is equal to one Plots of the radial wave functions are also interesting R10 is a smoothly decaying exponential with a maximum at the origin meaning the electron is most likely to be found near or inside the nucleus This is not completely unexpected we earlier PHYS 3802 aiready as far in as it can be Aisor because 0 he eiectron has no anguiarmomemum in of wwiforri 7 rr I P00 na4aiue I Greer I4 Rea Bohrradii SQ n in me pnoton Many constant strearn ottnese photon wnrcn WiH be seen as a spectrai irne Under nrgner rnannin an more iines ans was known as tne ne structure ortne spectrum Tne expianation tor tnrs was tnattne eiectron acts as it it n intrinsic angular momentum r e an anguiar rnornenturntnat is arunoarnentai property ortne eiectron iust as its rnass and cnarge are 5 given sv17z Where 5 V2 for 5H eiectrons everywhere PHYS 3802 Notice that there is no correspondence limit in this case spin is a completely quantum mechanical effect with no classical analog In that sense the name spin is unfortunate since it seems that the electron now hastwo kinds of angular momentum orbital and spin just like the Earth has two kinds of angular momentum orbital rotation about the Sun and rotation about its own axis The problem is that the electron is a point particle meaning it has no physical size so the idea of it spinning classically makes no sense Nevertheless spin still acts like an angular momentum so the name has remained With a value ofs xed at 12 the corresponding projection ofs on the z axis is called ms and is restricted to either 12 or 2 Our earlier m is usually written as mto avoid confusion between the two The complete list ofquantum numbers necessary to specify the electron s position in an atom is now n I m s and ms Of course since 3 is the same for all electrons it is omitted One ofthe reasons the name spin has stuck is that there is a magnetic moment associated with spin just as there is one associated with the electron s orbital angular momentum since moving charges create a magnetic eld Following your book we can calculate the magnetic moment ofa current loop in this case the electron moving around in its orbit to be V 2l i u IA qZ rrrr q J We combine this with our earlier expression for angular momentum and get eh 2m 6 2 E 1 The projection of the magnetic moment along the z axis is then given by eh ml 2m 9 2 where m is the magnetic quantum number and me is the electron s mass These expressions can be written in a more general form if a fundamental unit of magnetic moment is de ned This is known as the Bohr magneton pa and is equal to PHYS 3802 e h U B 579 ueVTesla 2 m 8 The more general forms are W1lgLB zz39mIgL B The gL term is known as the gyromagnetic ratio or typically the gfactor This can be thought of as a measure of the effectiveness of a particular angular momentum at establishing a magnetic moment For orbital angular momentum the gfactor is called gL and is equal to one For spin angular momentum though the gfactor is gs and is approximately equal to 2 This means that the magnetic moment arising from the electron s spin is described by I SS1gSIuB U34gsluB qumsgsIuB Because the electron is moving around the nucleus it sees a circulating electric charge and therefore feels a magnetic eld The interaction ofthat nuclear eld with the electron s spin magnetic moment which only has two possible orientations causes what used to be one energy level in our simpler model to split into two levels The source of the splitting is just the interaction energy ofa magnetic dipole in a magnetic eld the direction of the eld is typically taken to be the zaxis Uz yZB The presence of this extra quantum number m5 i 12 means the complete electronic wave function has one more term besides the n I and m we already specify Since there are only two possible values for m they are typically referred to as up and down SpinOrbit Interaction The spin angular momentum and orbital angular momentum combine to form the electron s total angular momentum usually represented by J as a vector andjas a quantum number We can connect this new angular momentum with a quantum number PHYS 3802 7 jj 1 h J2 mj h The possible values ofthe new quantum numbersjand mj are j 6 5 0r jzl sl m j j1j 1j The difference injvalues between energy states which have the same n and l causes the level splitting mentioned in the previous discussion ofthe ne structure of spectral lines Because the electron interacts magnetically with the nucleus what would othenNise be a single energy level is split into multiple closely spaced energy levels The example in your book discusses the 2P levels of hydrogen The values of n and l are 2 and 1 for all electrons in the 2P shell but the possible values ofj are 12 or 32 The j32 level has an energy about 45 ueV greater than the j 12 level Your book calculates the magnetic field felt by the electron in the 2P level to be approximately 04 T lfthe atom is placed in an external magnetic eld the levels will split into 2j1 sublevels and the effect is known as the Zeeman effect This is useful in astronomy when trying to find the magnetic eld at the surface ofa distant star Symmetric and Antisymmetric Wave Functions When an atom has more than one electron an important restriction emerges no two electrons in an atom can have all quantum numbers exactly equal This fact known as the Pauli exclusion principle is actually true for all fermions which we ll discuss later If not for this effect all the electrons in a large atom and in small ones would eventually radiate down to the ground state where the excitation energy would be enormous approximately 136 eV Z The Pauli exclusion principle is enforced by requiring that the wave functions of two electrons in an atom are antisymmetric if the labels are exchanged This is a consequence ofthe fact that all electrons are identical particles Classically we can label them and keep track of which electron goes where in a given interaction but quantum mechanics says we only know what we measure For example if you have electron A and electron B going into a region and both coming out at some other time and you only measure the initial and nal states of the pair you have no idea which isA and which is B afterthe interaction This means that the probability density has to be unchanged if you swap the labels on A and B Since the probability density is the absolute square of the wave function there are two possibilities for the behavior ofthe wave function when labels are swapped the wave 22 PHYS 3802 function can be unchanged known as a symmetric wave function or the wave function can change its sign known as an antisymmetric wave function You can think of this as being similar to the fact that the equation x2 1 is satisfied by both x1 and x1 Another similarity is that the final wave function which is a product of in general an angular part a radial part and a spin part will have a symmetry determined by multiplication where each antisymmetric term is assigned a 1 and each symmetric term a 1 Therefore a wave function with two symmetric parts and one antisymmetric part would be 1 1 1 1 which is itself antisymmetric The Periodic Table The addition of spin means we can fit twice as many electrons in a given shell or subshell For example look at the periodic table below from httpwwwphysicsohio stateedulvw h sicsbi ertable PERiODIG TABLE OF THE ELEMENTS nd BIL 391 mmmmmmmmmmmmmmmmmmmmmmmmmmmmm E311mmmmmmmrainEmnmummmunmmmmmmmmnuumm 13311mmmmmmmmmaammmmmmummumuamammumam n T maquot I l 2 l V l iln39 um 12 m 11 a 4 a quotup 1 VA Inn IIII r quotr 7 r m 1 m 20 21 22 23 N U I n I Iquot I m In In 1 u t m 1quot Gas 50 11 V All NIM39U Wlwli Elli N W 8quotquot Mill llll llll m quotquot 39quot u 38 M 41 I A I a a N h Mum I I VIA VI 2 GT W quot u 3 quotIn M02 ml Wham 7 mquot 14 2 un m m 39 n m Sr Y 3945s I IIMVSL 36 WW sumn mm mm m m n w m u u m 1 VII quot 25 7 w L quotIn W n v 3 To nunn 39 u n MILWuASc5cmuA5c m hhm l mmmmam mwmmm PAPERTECH mm mm 2mw mmm Notice that the top row contains only hydrogen and helium ln hydrogen s lowest energy state it will have one electron in the n 1 state For this electron ifn 1 0 and m 0 On the opposite side of the table we have helium which has two electrons They can both have n 1 0 and m 0 meaning they are both in the ground state if they have 23 PHYS 3802 different values of ms they can t have different values ofs since they re both electrons and all electrons have s 12 The electrons in helium s ground state will then have the same quantum numbers for everything except ms and one will have m5 12 while the other has m5 2 This means helium has no more room for electrons in the n 1 level and we can then say that its outer shell is filled This makes it extremely unlikely that helium will try to combine with any other element since having a lled shell is the goal of any atom The next row starts with lithium which has the lled n 1 shell and one electron in the n 2 shell This shell has subshells with l O and l 1 In those subshells m can be either 0 I 0 or 1 0 or 1 I 1 That gives 4 possible combinations of and m which would mean that the second shell would be lled by the time we get to element number 6 However the fact that we now have the spin magnetic quantum number ms means we can double the available space for electrons and the second shell doesn t fill up until we reach element number 10 neon another element that generally refuses to combine with anything else because ofits lled shell Statistical Physics As we have seen before if we have a large number of noninteracting and distinguishable particles their occupation of various energy levels is determined by the Boltzmann distribution function fE below Ae EkT The actual number of particles with energy E is the product ofthe factor fE above and another function gE which counts the degeneracy ofthe energy state lftwo different sets of quantum numbers represent states that have the same energy those states are said to be degenerate lfwe ignore the spinorbit interaction responsible for fine structure and we have no external eld the electron in the hydrogen atom has an energy that depends only on n and so the levels with the same value ofn but differing values of m andor ms would be degenerate Maxwell Molecular Speed Distribution The speeds of molecules in a gas follow the Boltzmann distribution which was discovered earlier by Maxwell Your book writes the Maxwell normalized velocity distribution as 32 m mv v v 2kT Fanvyavz m and the related Maxwell distribution of molecular speeds as PHYS 3802 32 nv 47f N m v2 e39mvzZH 27f kT One ofthe differences between these two functions is that the first is a symmetric Gaussian centered on v0 and the second which describes speed is asymmetric The reason for this is that there is a lowest speed v 0 and there is no limit on the upper speed not strictly true relativistically but that s a minor point here Because ofthe asymmetry the most probable value of molecular speed is lower than the average value of molecular speed This is one reason why economists rarely talk about average incomes ifyou have 9 unemployed people in a room with one person who makes 1000000 a year the average income is 100000 but that doesn t really tell the whole story Another interesting distribution in your book is Maxwell s distribution of kinetic energy nEdE ew dE 7f Calculating the energy from this is done by treating it like an expectation value where the formula above takes the place of yy Just as the probability density has to be normalized we also have to divide the integral by the total number of particles N Integrating from 0 to co gives us an average energy of 32 kT Each degree of freedom essentially a direction where movement can be noticed carries 12 kT of energy per particle For a monatomic gas the only degrees of freedom are movement in the three directions of space so we end up with 32 kT per atom This can be measured by finding the heat capacity ofthe material As heat is added that energy is distributed among all the different degrees of freedom More degrees of freedom mean a higher heat capacity We can write the monatomic gas case as AEjmQ 2 gnRATanVAT 2 CVER lfa gas is diatomic there are additional degrees of freedom It is assumed that rotation along the axis is not allowed but that still leaves rotation around two other directions This would leave us with CV 52 R As shown by the table in your book on p325 the noble gases which are monatomic have measured values of CVR of 1 5 while at least some diatomic gases have CVR around 25 PHYS 3802 Considering that the atoms in the diatomic gas molecule could also oscillate like two masses on a spring we should have two more degrees of freedom one for the spring potential energy and one for the kinetic energy associated with the vibration This predicts CVR 35 but measurements don t support that for most diatomic gases The problem is that the vibrational degrees of freedom are not excited at room temperature As is shown by the graph in your book the translational modes accounting for 32 kT are present down to the lowest temperatures where the gas is still a gas and not a liquid or solid At higher temperatures the rotational modes can be excited and they begin to contribute to the speci c heat The vibrational modes are typically not seen until temperatures reach several hundred K or more Solids are assumed to be particles connected by springs in all three dimensions meaning we should expect a contribution of kT from each direction for a total of3 kT not 32 kT As temperatures rise this becomes accurate for all solids One failure of this Boltzmann based model is that metals and insulators should not have the same values of CVR since metals have free electrons roaming around their lattice that can be considered to be an electron gas This gas should add another 32 kT meaning we should measure CVR 3 for insulators but 332 45 for conductors The fact that this doesn t happen points to the need for a new statistical model Quantum Statistics Quantum mechanics requires us to replace the Boltzmann distribution for distinguishable particles with one oftwo other kinds of statistics either FermiDirac statistics or Bose Einstein statistics All three formulas are shown together below 1 1 1 fglEl W fpplEl W fgglEl W The e factor is present for normalization in all cases and the formulas for the three distributions look quite similar For EgtgtkT the resulting distributions actually are similar for the same value of or As is shown in your book at lower values of E we nd that fFD lt fB lt fBE This is a re ection ofthe fact that fermions particles of halfinteger spin which are described by fFD obey the Pauli exclusion principle and therefore are limited to one per state Bosons particles of integer spin described by fBE on the other hand preferentially clump together in the same state if possible Examples ofthis extreme behavior in the boson case would be lasers superconductivity and superfluidity In the fermion case we notice that the structure ofthe periodic table and the fact that two material objects can t occupy the same space at the same time illustrate these statistics PHYS 3802 Boltzmann statistics can still be useful Even though they were meant to describe distinguishable particles and we know all electrons are identical what really determines our ability to distinguish one electron from another is their separation relative to the size of their wave packet which is approximately the size of their de Broglie wavelength Your book states this as X ltlt ltdgt where ltdgt is the average separation of the particles Using the reasonable assumption that ltdgt VN 3 for a volume V filled with N particles and our earlier formulas for X and the particle s energy as a function of temperature we can write the condition for applicability of Boltzmann statistics as h V 3 h3 N ltlt 0r 32 ltlt1 43ka N 3ka V As an example can the free conduction electrons in copper be described by Boltzmann statistics at room temperature Using T300 K and NV 85 X 1028 kgm3 from the hyperphysics web site we get approximately 20000 which is definitely not much less than one We need FermiDirac statistics in this case Your book calculates this factor for helium atoms in the atmosphere very low density and it turns out that Boltzmann statistics are ne for this case Density of States To turn the statistical distribution functions into a measure ofthe number of particles with any given energy E we need to nd the number of states N available to a particle with that energy and then take the derivative of N with respect to E For the infinite square well in 3 D we have 22 71 2 2 2E2 2 2 nlnzn3 2mL2 111 112 n3 0 n1n2 113 The niare all positive so this is the equation of 18 ofa sphere The number of states is E 32 if N 6 E0 giving a density of states gE 32 dN 2n 2m V E g dE hs PHYS 3802 Your book goes on to indicate that electrons with spin 12 would have twice this density of states since there are two allowed spin states for each energy The importance of the gE factor is that it is part ofthe normalization which we can write for Boltzmann statistics as nglEldE lgglElfglEldE iggle dB 0 0 N 0 BoseEinstein Condensation When 4He the most common isotope is cooled to within a couple of degrees K above absolute zero it makes a transition to a state known as superfluidity In this state there is essentially no viscosity internal friction as all of the helium particles start to act as one Even though 4He consists oftwo protons two neutrons and two electrons all of which are spin 12 fermions it acts like a boson since the spins are all paired For bosons there is no occupancy limit on the ground state so all 4He try to get there Super uids display a variety of nonclassical behaviors due to their lack of viscosity A more surprising phenomenon occurs when 3He is cooled to within milliKelvins of absolute zero Although 3He has only one neutron and therefore can t be a boson pairs of 3He atoms can act as bosons and create super uid 3He The explanation of this following your book begins with the normalization condition 2717 m Ix N J0 13 2ka32JO de writing X in place of EkT The key difference in the case of super uid helium is that we shouldn t really be writing an integral for what is a sum over discrete states We can usually get away with it because the states are so incredibly closely spaced that we only introduce a small error by using the integral approximation The difference here is that the equation above ignores the ground state E0 since gE is proportional to E1 2 meaning it would be zero in the ground state When working with fermions ignoring the ground state in a macroscopic system means ignoring two particles out of 10 Because there is no limit on the occupancy ofthe ground state for bosons it begins to ll up at low enough temperatures and a reasonable fraction of the particles can be found there Your book calculates the critical temperature for liquid helium where the integral equation above is no longertrue to be about 3 K Nearthat point we replace the expression above with PHYS 3802 J eax1 27rV N N0 W dx 2kal32 j where go 1 N 0 ea eEOkT 1 ea 1 We get the interesting result that the number of particles in the ground state No is about T 32 NOzN 1 as opposed to the value it would have for fermions two The Photon Gas Planck s law for photons which we used earlier in the semester can be derived from the previous material Using 1 EkT e e 1 fgglE and taking into account that the number of particles photons is not something that is conserved in this case the normalization process that previously determined e now tells us that we must have oc0 This is required since photons can appear and disappear without restrictions placed on their numbers The density of states turns out to be The energy density is then PHYS 3802 or rewriting E as hf we get ulfldf mdf 3 C Heat Capacity amp Specific Heat Returning to the question ofthe specific heats ofdiatomic vs monatomic gases we note that Einstein modeled a solid as a collection of masses which could oscillate in all three directions of space He assumed quantized energies ofthe form E nhf and used the Boltzmann distribution to nd the average energy as m h f lt E gt J W 0 e This gives the classical expression ltEgt kT when the temperature is high enough to allow the expansion ofthe exponential to be terminated at 1hfkT For a solid containing 1 mole of a substance the 3D oscillators would have a total energy of 3N hf 3NAlt Egth e 1 and a heat capacity CV of 2 hfkT dU h 6 CV 3NAk f 017 kT ehWT 12 The Boltzmann distribution is obviously not going to be valid at all energy scales since the particles involved are not distinguishable The important temperature in this case is known as the Einstein temperature TE hfk and the energy distribution can be rewritten as fBEnAenhfkTAeI1TET Small changes in the number ofquanta n will produce small changes in fB if TgtgtTE and the energy can be regarded as a continuous rather than quantized variable meaning classical Boltzmann statistics will work PHYS 3802 Harder solids have higher values of TE than softer ones meaning they will only have hear capacities near the classically predicted 3R at higher temperatures The solids that have CV near 3R at room temperature are therefore the softer ones like lead and gold CV rises from zero to 3R for all solids in approximately the same way but at different temperatures To explain the problem with the speci c heats of gases ie the speci c heat is lower than would be predicted for a molecule either monatomic or diatomic that can rotate in 3 dimensions we de ne temperature scales for vibrations and rotations For vibrations we have TV hfk for vibrations of frequency fwhich should be the minimum temperature for vibrational energy to exist A molecule s rotational energy can be written as L2 if E 6 1 R 21 Sn 21 yielding a temperature TR 2 TR hT 8n 1k below which we should see no rotation Because the molecule s moment ofinertia is on the bottom ofthe expression a smaller moment of inertia means a higher critical temperature ie no rotations unless the temperature is above TR Notice that for rotation around the axis joining the two atoms in a diatomic molecule the moment of inertia will be tiny even compared to its moment of inertia in the other two directions Essentially that means that it would take extremely high temperatures to excite those rotational modes so high that the diatomic molecule would be disassociated Fermion Gases As we ve already mentioned another unexpected specific heat result involves the lack of a difference between a conducting solid and an insulating solid The conducting solid is modeled as having many free electrons traveling through the lattice of positive ions and this electron gas was predicted to add three degrees of freedom 32 R to the metal s heat capacity We can rewrite the familiar FermiDirac distribution in the form below ifwe make the substitution or EFkT where E is known as the Fermi energy PHYS 3802 Notes follow and parts taken from sources in Bibliography Ionic Bonding When two atoms form a molecule the binding is usually somewhere between ionic and covalent If one of the atoms loses an electron to the other the donor atom will have an overall positive charge and the atom with the extra electron will have an overall negative charge When one of the atoms from group I of the periodic table which has only one electron in its outer shell comes in contact with one from group Vll which is only one electron away from having a lled outer shell this electron transfer is likely to happen The important numbers here are the ionization energy of the donor atom and the electron affinity energy released when an electron is captured of the acceptor atom Because electron af nities are typically smaller than ionization energies the formation ofthe molecule would not be energetically favorable if not for the Coulomb attraction between the ions The energy ofthis attraction is ke2r and it allows the binding to occur There is also a term involving the exclusion principle which keeps the ions from getting arbitrarily close as the Coulomb term would like Although the extra electron can move to the new atom relatively easily as the ions get closer together electrons closer to the nuclei would begin to overlap in violation ofthe exclusion principle To avoid that some of the electrons would have to move to higher energy levels This energy is typically modeled as having the form 1rquot where n may be relatively large approximately 10 for Na and F as calculated in your book The energy of attraction in a crystal will be more complicated than the simple ke2r for two ions There will also be attractive and repulsive forces from other ions all over the crystal The sum total of these attractive and repulsive forces for a particular ionic crystal structure is represented by the Madelung constant or and the energy per ion pair is ockeZro where r0 is the equilibrium separation Covalent Bonds The bonding in the H2 molecule is not at all ionic Your book models it as the limiting case oftwo square wells moving closer and closer to one another with a single electron equally likely to be found in either well Because the wells are identical the probability function must be symmetric as seen from the point between the wells As we ve seen in other cases a symmetric probability requires either a symmetric wave function ys or an antisymmetric one VA The amplitude of the wave function will be largest in the center of each well with the expected exponential dieoff outside the well The antisymmetric wave function requires that w O at the midpoint between the wells while the symmetric wave function will be the sum ofthe two positive exponentially decaying terms and is therefore greater than zero at the midpoint PHYS 3802 Moving the wells closer to one another would eventually cause vs to look like the ground state wave function of a single well and VA will look like the rst excited state ofthat single well Therefore the state vs will also have a lower energy than the state VA in the double well case just as the ground state is lower than the rst excited state in the singlewell case Although the Coulomb potential around a proton is more complicated than that ofthe nite square well the idea is the same We nd that the symmetric wave function has a lower total energy than the antisymmetric function which makes sense on physical grounds when we look at the value of v12 everywhere For yA we see that the probability of finding the electron between the protons is zero but that s exactly where we should put it electromagnetically to attract both protons to one another The vs case has a large probability of nding the electron between the two protons and each is attracted to the electron s negative charge The Hamiltonian in the case of the H2 molecule ie the left side ofthe Schrodinger equation is ke2 i ii r1 r2 r0 H 2 pop 0 2m where r0 is the distance between the protons and r1 are the distances from the electron to each proton Detailed analysis of the solution to the Schrodinger equation in this case will con rm that the VA wave function will not lead to a bound molecule since the electron is effectively never found between the two positive charges but the vs wave function will each proton is attracted to the electron s location between them thereby effectively attracting them to each other An outline of this solution in terms ofenergies is presented in your book but the basic idea is that in the limit where the distance between the protons approaches zero an electron with the vs wave function should see itself as orbiting a 22 nucleus and therefore feel a binding energy ofapproximately 136 eV 22 as if it were a single electron orbiting a helium nucleus For the VA wave function the repulsive force between the approaching protons is always greaterthan the attraction of either one to the electron and there is no minimum of the curve ofenergy vs proton separation meaning no bound state for yA In the vs state the protons are 0106 nm apart and the electron s energy is 163 eV meaning it is 27 eV more tightly bound to the pair of protons than it would be to a single proton For a neutral hydrogen molecule H2 we can analyze this by realizing that for interatomic separations which are large we should have a total energy of 2 136 eV As the atoms approach each other the electron wave functions begin to overlap and the space part of the wave functions will be either W or WA For a symmetric space part ofthe wave PHYS 3802 function the electron spins are in opposite directions and the total spin is zero making this a singlet state For the VA space wave function the spins are in the same direction and add to a total value of 1 forming a triplet state since the magnetic quantum number m can be 1 0 or 1 Because of the attraction between the electrons and protons the symmetric state Vs is the bound state and the antisymmetric state is known as the antibonding orbital The electrons in the symmetric state are called sbonded since each is in an 3 state In this state the nuclei are 0074 nm apart and the binding energy is 317 eV which is of course lower than the 272 eV for two isolated H atoms As in the case of the H2 ion there is no minimum for VA since one of the electrons would have to go to an orbit of higher n Sharing electrons is known as covalent bonding since these outer electrons are also called valence electrons This is the bonding which holds together diatomic molecules where the two atoms are the same also called homopolar or homonuclear This makes sense because there would be no net energy bene t if one oxygen atom grabbed an electron from another oxygen atom so they could bond ionically For other compounds bonding is commonly a mixture of both ionic and covalent types To quantify this the size of the ionic bond whether it exists or not would be calculated by assuming it would result in e charge and e charge separated by re the distance between the centers of the atoms The actual dipole moment of the material is compared to this theoretical maximum to determine the fraction of the bond that is ionic The remainder is covalent Other Kinds of Bonds Other bonding mechanisms involve the interaction of dipole moments If an atom has a permanent electric dipole moment like water it is known as polar and the force between two permanent electric dipoles can be shown to be 1h The key to seeing this is to remember that the eld ofa point charge drops off as 1r2 and the eld ofa dipole isthe difference ie related to the derivative between two point charges so it decreases as 1r3 The force between the dipole and a point charge would then be proportional to 1r3 but since another dipole is a combination of opposite point charges that reduces the force to 1h This is a relatively weak force since it drops off dramatically with distance but it is very important in determining the character of the interaction between water and other materials as well as the phase transitions for a given polar substance Even ifa molecule doesn t have a permanent dipole moment there will always be a fluctuating dipole moment due to the constant motion ofthe electrons At any given instant they will be distributed around the nucleus according to the probabilities predicted by their wave functions it is extremely unlikely for these electrons to arrange themselves so that they exactly cancel each other s electric field at a particular point in space occupied PHYS 3802 by another atom lfthey don t there will be a dipole eld acting on the second atom and it will tend to polarize it drag the electrons one way and the nucleus the other way The dipole moment induced by an electric eld in a linear medium is given by 1320517 where or is the polarizability of the atom or molecule lfthe electric eld acting on the atom to be polarized is a dipole eld that means it drops off as 1r3 so the induced dipole moment also depends on 1r3 The energy of interaction between these dipoles is proportional to pE which gives 1r6 Since the force is the derivative ofthe potential with respect to r the force will be proportional to 1r7 These forces are known variously as van der Waals forces London forces andor dispersion forces These are the only forces acting between noble gas atoms to cause them to condense into liquids or solids as the temperature is reduced It s clear that there cannot be covalent bonds between them since they all have lled outer shells or ionic bonds two identical atoms can t form a covalent bond Diatomic Molecular Spectra Molecules have spectra similar to those seen in atoms ie due to excitation ofthe electrons from one state to another with the addition of rotational and vibrational energy transitions as well In general electronic transitions will still be in the eV range vibrational transitions will be somewhere around 01 eV and rotational transitions will be less than about 10393 eV The molecule s rotational KE can be written as 1h2 21 E 1E0 in terms ofthe molecule s characteristic rotational energy Em A molecule without a permanent dipole moment can only make an electricdipole related transition between rotational levels ifthere is also a vibrational or electronic transition at the same time If there is a permanent dipole moment the electric dipole selection rules require that A X 1 so the change in energy must be 1 1 AE lfthis energy is known the masses ofthe constituent molecules can be used to nd the average interatomic separation lfthe separation between them is re we can use the reduced mass u m1m2m1m2 to get PHYS 3802 Zuroz It is common to use unified mass units to describe these masses where 1 u 16605 X 103927 kg For typical values of l the energies associated with a rotational transition are far below kT at room temperature meaning it is easy for molecules to be excited to X gt 0 states Vibrational Energy Levels For the vibrational energy levels ofa diatomic molecule we model the molecule as two atoms connected by a spring This is the de nition of a simple harmonic oscillator The energy levels should then be spaced according to Ev12hf v 012 where the frequency fis 1 K er ii for reduced mass u and spring constant K The vibrational selection rule is that Avir1 meaning the frequency of the emitted photon is just fas well At room temperature molecules are commonly in the v0 state meaning they can absorb incident radiation and make a transition to a higher state For a diatomic molecule the vibrational transition must be accompanied by a rotational transition to conserve angular momentum The possible energy differences for these transitions are of the form AE hf 2 1E0r 0r hf ME For the rst formula the transition to a higher value we can have 0 12 but for the second lower K we are restricted to X gt0 since there is no 1 level This means there is a gap which is 4 E0r in size centered on the vibrational frequency hf this would correspond to a pure vibrational transition which doesn t happen in a diatomic molecule Finding the population of the various rotational levels can be done using Boltzmann statistics and will also depend on the number ofdifferent degenerate states which is 2 1 in this case PHYS 3802 hf zz1EO kT nE 25 1e This population determines the relative strengths intensities of absorption lines As increases and there are more ways to achieve the energy E we see the intensity increase Of course at larger values of X the increase in 2 1 is more than offset by the exponential Boltzmann factor and n drops off again Your book nds the value of X that maximizes n by calculating dnd and setting it equal to zero This gives X m 3 at room temperature meaning the X 3 absorption line is the most intense The assumption that the rotational energy is simply X X 1 Eor begins to fail for larger because the separation between atoms tends to increase as the molecule spins faster This changes the spacing of the peaks representing rotational transitions Scattering Absorption Emission The interaction ofa photon with an atomic electron can take many forms In Rayleigh scattering the photon loses no energy during scattering but the probability of scattering varies as f4 which is the reason the sky is blue high frequency more heavily scattered and the sunsetssunrises are red most ofthe blue has been scattered out ofthe Sun s white light leaving a reddish color In Compton scattering a photon usually ofhigh energy hits a free electron or an electron bound to an atom with an energy that is small compared to the photon s energy resulting in a free electron and a photon which has lost energy Raman scattering occurs when a polarizable molecule interacts with a photon and the molecule s rotational or vibrational mode changes This results in the photon leaving with an energy that could be greater if the molecule was in an excited state before the interaction and is in a lower state aftenNards or less molecule initially in a lower state is elevated to a higher state than its initial energy lfan electron is in a lower energy state or the ground state there is a chance it will interact with incoming photons and absorb energy The probability of this happening depends on the energy density of the incoming photon stream and the Einstein absorption coefficient written as B12 ie transition from level 1 to level 2 Once the atom has been excited it can return to the ground state by either spontaneous emission described by the Einstein coef cient of spontaneous emission A21 or by stimulated emission The rate of stimulated emission depends on the energy density of incident radiation just as absorption did When an atom deexcites by stimulated emission the emitted photon will have the same phase and direction of travel as the incident photon Spontaneous emission can be viewed as emission which is stimulated by uctuations in the vacuum of space around the atom PHYS 3802 The rates of emission and absorption must of course depend on the number of atoms that can emit atoms in state 2 and the number that can absorb atoms in state 1 We already have a measure ofthe ratio ofthese two numbers in the form ofthe Boltzmann statistical factor N 2 2 e hfkT N1 ifthe energy difference between the states is hf In equilibrium we will have the same number of atoms emitting N2 gt N1 transition as absorbing N1 gt N2 transition so we can write N1 312 N2A21 321 Your book solves the equation above for uf and uses the fact that we already have an expression for uf in the form of Planck s law Setting the two equal gives us 32 z 87 hf3 1 71312 1 cs kehfkT 1 N2 321 where substitution ofthe previouslyderived Boltzmann factor for N1N2 on the left leaves us with i8 hf3 3 21 C B12 2 321 and Analysis of this gives us very interesting results First since A21 is proportional to B21 f as the energy difference between levels increases spontaneous emission becomes more important You can also write A21 ehfkT 1 321 and note that for hf ltlt kT stimulated emission is typically much more important than spontaneous emission PHYS 3802 Lasers and Masers The basic principle ofthe laser an acronym for light amplification by stimulated emission of radiation and the maser microwaves instead of light is that stimulated emission which produces coherent radiation is more probable than spontaneous emission which produces incoherent light Stimulated emission means we will have a ux of photons with energies equal to the gap between levels 1 and 2 Under normal circumstances the number of atoms in level 1 will be larger than the number in level 2 due to the Boltzmann factor The fact that B12 B21 means that it will therefore be more likely that a photon stimulates a transition from level 1 to level 2 than vice versa there are so many more atoms in the lower state a photon is more likely to interact with one of them and be absorbed than it is to nd an atom in level 2 and cause it to emit The rst necessary ingredient for a laser is then to create a population inversion where there are more atoms in the higher state than the lower one This inversion can be created by an intense blast of ordinary light Typically this light will raise the atoms in the ground state which we ve called level 1 above level 2 The atoms can deexcite without emitting a photon by exchanging energy with the crystal bringing them to level 2 lfthe initial ash of light was intense enough this will invert the population and put more atoms in level 2 than level 1 As soon as one of these excited atoms spontaneously emits a photon that photon will trigger the release of more photons like it and the chain reaction will continue The laser cavity has mirrors on both ends to trap the photons and force them to make multiple trips across the cavity and stimulate the release of more photons Of course one mirror hasto be less than perfect so that some light can escape In a HeliumNeon laser the ones in the lab about the size ofa loaf of bread helium atoms are excited to levels 25 and 2p with energies 1972 eV and 2061 eV relative to the ground state These energies are very close to the energies of the 3s and 25 levels of neon Collisions between the helium and neon atoms transfer the energy from the excited helium atoms to the neon atoms where they will decay when stimulated to a level that is 1870 eV above neon s ground state This will result in the emission of a photon with 16328 nm The laser will also produce coherent light with 7v1100 nm when an electron in neon s level 1983 eV above ground falls to the same 1870 eV level The key is that by using multiple energy levels it is easier to create a population inversion since there were no or almost no atoms in the 1870 eV state in the first place Solid State Physics While solids can be amorphous like glass various crystalline forms are more commonly found when the material is cooled slowly from the liquid state These crystals each of which is made of a regularly repeating fundamental shape known as a unit cell can take one of14 forms lfthe atoms in the crystal are bound to each other ionically the overall PHYS 3802 effect ofthe alternately attracting and repelling neighbors is summed up by the Madelung constant mentioned previously Different crystal forms will have different values ofthis constant As was shown earlier with ionic bonding the Pauli exclusion principle forces the ions to repel each other at close enough distances and this repulsion can be modeled as A rep rn for a total potential energy of ke2 A ke2 r 1 r U a n a 0 0 r r r0 r n r where the last equality results from minimizing the left side ofthe formula with respect to r to find the minimum separation r0 Experimental measurement of re 028 nm shows that n is about 9 in the case of ordinary table salt NaCl Covalent bonding diamond for example provides another means for solids to form Metallic bonding metals occurs when the outer valence electrons in atoms become free to roam the lattice It s another case ofelectron sharing but instead of a couple of atoms sharing an electron the whole metal shares these valence electrons The electrons act as they would in a lowtemperature plasma being bound to the material as a whole but not to any particular atom The potential wells of each atom in the lattice begin to overlap with one another and the net effect is to restrict the electrons to a smaller region This lowers their potential energy since they are now closer to the positive ion cores but it also increases their kinetic energy according to the uncertainty principle The total energy must of course be lower forthe bond to be stable and it is The free electrons are responsible for many ofthe properties we associate with metals Classical Conduction Theory Treating the electron gas in a metal with Boltzmann statistics will obviously fail at some point we ve seen that the predictions of an additional 3 degrees of freedom and corresponding increase in speci c heat for a metal as compared to an insulator are wrong Starting with Boltzmann statistics however we can predict that the average speed ofthe electrons traveling through the metal at temperature Tshould be PHYS 3802 8kT Em ltVgt Ifyou recall the velocities we predicted for gas atoms at room temperature 1000 ms notice that these velocities will be much larger since the electron s mass m in the denominator above will be about 110000 the size of even helium s mass The square root means that speeds will be 100 times larger or around 105 ms This is not the speed at which electrons travel long distances though just as air molecules don t cross a room in a fraction of a second Collisions act to randomize the direction of motion of a particular particle air molecule or electron on a very short time scale lfthere is an electric eld present this will bias the direction of the electron and tend to guide it towards the source ofthe eld The speed ofthis macroscopic motion is described by the drift velocity vd which can be connected to the current I and crosssectional area A ofthe conductor by I neA Vd where the electron number density and charge are n and e The speed ofthis net progress is very much slower than the average speed found above typically millimeters per second Note that neither ofthese speeds is the rate at which electricity travels since the signal to move is transmitted to the electrons at a speed closer to c Your book calculates other values related to collisions between the electrons and the underlying lattice such as the mean free path 7v which is lltvgtl 1na7rr2 where 1 is known as the relaxation time time between collisions and n5 is the number density of ions with radius r These quantities can be connected to the resisitivity p and its inverse the conductivity 5 by pmltvgt neZl neZl mltvgt This classical picture ofdistinguishable electrons bouncing off of xed ions correctly predicts that the resistance is independent ofthe applied field which is a restatement of Ohm s law but the values predicted for p are not very good Also experiments confirm the 2212 formula stating resistance is a linear function oftemperature but the expression above has p increasing with temperature through ltvgt as T1 2 10 PHYS 3802 We already know that the electron gas doesn t give an extra 32 R contribution to the speci c heat as predicted by Boltzmann statistics Just as FermiDirac statistics were needed to explain that problem they are needed again here FreeElectron Gas in Metals The study ofthe free electron gas is made easier by starting with the onedimensional in nite square well lfthis could be physically realized it would essentially be a chain of N atoms which is L long The allowed energies can be written as n2 E E1 represents the ground state energy and at O K each level will be occupied by two electrons from the ground state up to the state containing the Nth electron The energy of this electron is the Fermi energy and we can write it as N 2 h2 N 2 h2 N 2 EF 2 Z E1 2 2 m L 2 32 m L While the linear density ofelectrons NL doesn t seem to make much sense in the real 3D world we can substitute the cube root ofthe actual number density ofelectrons to nd this value For copper this gives a Fermi energy of 1 8 eV which is about 70 times the room temperature value of kT The average energy of these electrons can be written as 1N2 lt Egt E 2n2E Nn1 Calculating the integral ratherthan the sum itself rememberthat this is OK here since we re dealing with Fermions gives us 2 N2 2 3 ltEgt J2n2dnz h LN E F N 0 8mL2 3N 3 which is still far largerthan 12 kT The FermiDirac distribution function isjust a step function at O K with all states below EF occupied f 1 and all states above EF empty f0 Your book shows how to get the density of states gE in this problem and calculates it to be E1391 2 E391 2 giving nE the number of electrons with energy E as nE gE24E 4 for Elt E lEE For EgtEF of course nE 0 PHYS 3802 Extending this to three dimensions we get LL M22 8m r 8m V V m 2 h2 0WWW 2 h2 3 F with the last step following for T0 We can invert this and solve for E as h2 3N W3 F2m S V To nd the average energy we need to nd the number of electrons at a given energy nE nE 8 m w 5 2 h2 2 E2 giving an average energy of lt Egt iTEnEdE 3E 39 N 39 5 F 0 As we saw before with the FermiDirac distribution at temperatures where kTltltEF only a small fraction ofelectrons are able to move to higher states This is because only those electrons within kT ofthe boundary between filled and empty states at E could get the energy from a collision that would promote them to one ofthe empty states Because of this connection it is sometimes convenient to talk about a Fermi temperature TF EFk PHYS 3802 At the end or the 9 century physrcs was consroereo to be atrnost oornptetety sotyeo oy sorne As rt turned out there were seyerat rnoonsrstenores whroh eyentuahy ted to two reyotutrons m physrcs the theory or retatryrty and the quantum theory One or hght gryen ott y tater 39 39 F nHaH then trayet away trorn rt through a vacuum rt necessary The tarnps that keep tooo Warm n a restaurant or catetena Work on thrs pnncrpte As the temperature rncreases the peak or y 39 aw H s yery rrnnt wayetength rs 1m T 00029mK 0 0029 rn K rs known as Wrens constant Thrs teHs usthat sornethrng at 310 K you raorates rnore photons wrth a wayetength or 0 0029 rn msm x 9 4 x10 rn than any 01 er m h r 530 x10 rn green hght That s about Where the Sun s surtace rs Notrce that we grant m r a hurnan body as tor a targe coueotron or hydrogen and hehurn T0 ayery good approxtmauom rt doesn t rnatter A 400 K turnp or goto raorates the sarne who and number or photons as a 400 K pork chop We caH thrs approxtmate oehayror blackbody radia ion p p Here s the shape or a few orachoooy curves vaeYVWImAZ Wavelengvhm 75 gtltl quot 1 xl quot 125 gtlt1 quot15gtlt ln ms xl quot PHYS 3802 Notice that the peak of the curve moves to shorter wavelengths as the temperature increases but also notice that the area under the curve goes up dramatically The law describing this increase is called Stefan s law and we can write the power heat transferred per second emitted by this blackbody as P d Qz a As T4 dT 5 is called the StefanBoltzmann constant and is equal to 567 X 10398 Wm2 K4 A is the area ofthe object a large stove eye will radiate more power at a given temperature than a small one s is called the emissivity of the material ranges from O to 1 and describes how close to a blackbody it really is with s 1 meaning a true blackbody and T is the absolute temperature ofthe object Notice that temperature enters to the fourth power going from 300 K to 600 K doubling the temperature increases the power output by 24 16 times We may be more interested in the net gain or loss of energy due to radiation In that case we have to consider the surroundings as well A 300 K block of material will radiate much more energy than it absorbs if we put it in a 100 K cryogenic cooler but it will absorb much more energy than it radiates if we put it into a 500 K oven The net power is then where Tenvwould be the temperature ofthe environment This formula gives us an idea of why we call a blackbody perfect absorber a perfect emitter as well It must radiate a temperaturedependent amount of energy regardless ofits surroundings However ifit s going to be in equilibrium with those surroundings same temperature it must be absorbing just as much as it s radiating For that reason a good absorber must be a good emitter The problem here is that this behavior does not agree with the othenNise successful theory of statistical mechanics which says that each different mode of oscillation ofthe electromagnetic eld in this case carries an energy of kT where k is Boltzmann s constant and T is the absolute temperature Finding the number of modes allowed in the classical cavity requires a bit of algebra and a little calculus but it predicts that the number of modes per volume the density of modes at a given wavelength X is We combine these two ideas and predict an energy density at a given X of PHYS 3802 87f kT ull gt 4 To nd the total energy density we need to integrate over all wavelengths w 87 kT 014d Ul Unfortunately this goes to in nity as we move to shorter wavelengths If everything above absolute zero which is everything had an in nite energy density energy would not be a useful physical concept Because moving from the visible to shorter wavelengths is a move towards the ultraviolet this problem was known as the ultraviolet catastrophe Obviously something is wrong with one ofthe initial assumptions and it turned out to be that the energy ofa mode is always kT lfwe use MaxwellBoltzmann statistics which just assume that the vibrational modes inside the cavity are noninteracting and distinguishable we predict that more energetic modes are populated according to Ae EkT where fE represents the probability of having an energy E and A is a normalization constant A plot ofthis function looks like this 7 1 39239 40 753 an 100 where the top line blue has a temperature 10 times that ofthe middle line red which has a temperature 10 times that ofthe bottom line green What this is showing is that the chance ofa highenergy state being occupied at low temperature is very small and that chance increases as the temperature goes up PHYS 3802 When the temperature is low only the lower states are expected to be occupied Of course the size of low depends on the energy needed to move to the next higher state For example we can look at the chance that roomtemperature thermal uctuations will move a hydrogen atom from its ground state to the first excited state To do this we just compare fE2 to fE1 The energies involved are E1 136 eV and E2 34 eV At room temperature 300 K kT is about 0025 eV Although the A factors are different for the two states they are of the same order of magnitude so we will ignore the difference This gives fE2fE1z e lEz Ellsz 10 In other words room temperature is very cold ifwe re talking about exciting electronic states of hydrogen atoms as there is essentially no chance that could happen On the other hand the rotational energy states of molecules can be separated by less than 1 millielectron Volt Finding the ratio of the occupancy ofthe rst excited rotational state to that ofthe ground state would give us a fraction not much smaller than one in other words the two levels would be almost equally populated and movement back and forth between them would be easy at room temperature Back to the blackbody problem the average energy ofan oscillator described by Maxwell Boltzmann statistics would just be the energy of a state weighted by its probability of occupancy integrated over all energies 00 m E j EfEdE j EAe39EkT dE 0 0 Instead of the continuous energy distribution fE used above Planck suggested that only discrete energies multiples of hv were allowed That changes the integral above into so i2 E EnAe EnkT 0 n 0 n Evaluating these sums gives us hf E ehfkT 1 We then multiply by the earlier gure for the number of modes per volume not to get PHYS 3802 8727 he 1 Wm 5 This function agrees with the observed blackbody spectrum at both large and small wavelengths For large v we can rewrite the portion of the expression in parentheses as 7v kThc to rst order in X We then get 87f kT wr as predicted by the RayleighJeans distribution which works ne at large 7v For small 7v UOv 9 0 avoiding the ultraviolet catastrophe Photons Shortly after Planck made his ndings known Einstein realized that if the radiating oscillators can only have certain discrete values of energy they must be radiating discrete values of energy ifyou can only have 1 bills in your pockets you can t give someone a quarterl Einstein called these discrete pieces of light energy photons One ofthe early experiments where light seemed to be behaving as a particle or photon is known as the photoelectric effect The basic idea here is that when light shines on certain metals the photons hit electrons in the metal hard enough to eject them from the rest ofthe metal entirely Einstein said that this happened because the photons were carrying an energy equal to Ezhf which should look familiar We can calculate the energy of a redlight photon as an exercise For red light fis about 4 x 1014 Hz so h fgives us 265 x 103919J Obviously the Joule is an inconveniently large unit of energy forthis kind of situation so we sometimes use the electron Volt eV We ve seen this before when talking about the amount an electron s energy changes when moving through a potential difference of one volt and it s 16 x103919 J On this scale our red photon has an energy of about 17 eV Blue light at about 79 x1014 Hz would have an energy ofaround 33 eV The big idea here is that there is a connection between the frequency of light and its energy Before this people would have assumed that you could use something like a dimmer switch to turn the intensity of a light down as low as you d care to without limit Running a blue light bulb off ofthis dimmer switch would allow you to produce for example a constant dim stream of blue light at a rate of1 eV per second for example PHYS 3802 The connection between energy and frequency says you can t do this if you want blue light 79 X 1014 Hz anyway the smallest piece you can have is 327 eV Anything smaller won t be blue light You won t see a constant stream of light in this case but instead you ll see a discrete lump of blue light every few seconds with nothing in between This is very much like the situation with atoms the smallest piece of Uranium you can possibly have has a mass of about 4 X 103925 kg No one can give you 2 X 103925 kg of Uranium because it just doesn t eXist You can have 2 X 103925 kg of Hydrogen because the basic pieces atoms are smaller Similarly you can have 1 eV ofinfrared radiation or microwave radiation etc but you can t have a 1 eV visiblelight photon Xray photon etc Getting back to the photoelectric effect it was observed that certain colors of light near the blue end ofthe spectrum could trigger this electron ejection but other colors near the red end couldn t The intensity of the light number of photons per second wouldn t allow red photons to eject electrons no matter how bright the light bulb Changing the intensity of a blue light bulb would change the current number of electrons per second however A wave theory of light couldn t eXplain this A slowly vibrating wave should over a longer time be able to deliver the energy needed to free an electron just as well as a more quickly vibrating wave could do in a shorter time Also no time lag was ever observed between turning on the light and catching the rst ejected electrons lf waves were shaking them loose it should take some time to get the rst ones moving The photon theory is able to eXplain all of this First while free electrons in a metal may not be bound to particular atoms they are certainly bound to the overall body of the metal lfthey were totally free you could pick up a piece of steel and shake electrons out of it It takes a nonzero amount ofenergy to remove an electron from a metal The exact amount depends on the metal but it is called the work function For potassium it s about 23 eV while for platinum it s closerto 64 eV What will happen ifa red light photon hits a potassium atom Since 17 eV lt 23 eV it won t be able to eject it The atom involved will either ignore the photon altogether or absorb and very quickly nanoseconds reradiate it Increasing the intensity of the red light just means more atoms are doing the same thing What iftwo photons hit the atom at about the same time Then you could have an ejection but the problem is that no light source from the early 2Oh century was capable of this kind of intensity The reradiation happens so quickly that only powerful lasers can drop enough photons on the surface quickly enough For our purposes red light won t give us any photoelectrons What about violet light Since 33 eV gt 23 eV electrons will be ejected In fact there will be energy left over after the ejection That extra energy goes into the kinetic energy of the electron so it leaves with a greater speed With greenish photons having energies just slightly above 23 eV we would get photoelectrons but they d be moving very slowly since there is very little energy remaining after freeing them The formula is hf KEmaX W0 Where We is the work function for the particular metal hf is the energy of the incoming photon and KE X is the maXimum kinetic energy an electron can leave with More PHYS 3802 intense violet light gives more photoelectrons ejected per second which means a higher current Digital cameras use this effect to give us pictures without lm In another popular application we can collect these energetic electrons and use them to power an electric circuit a battery after all isjust a source of moving electrons and we ll have a solar cell Xrays As recognized by Einstein the photoelectric effect can work in reverse When electrons moving at high speeds approach a nucleus they tend to be decelerated by the Coulomb interaction attractionrepulsion of unlikelike charges Accelerated charges emit electromagnetic radiation The probability that an electron will emit radiation in this situation is known as the cross section symbol 0 for Xray production If we were trying to calculate the chance oftwo spheres colliding one ofthe important things to know would be the crosssectional area of each it s much easier to hit one basketball with another than it is to hit one BB with another We can derive not easily the formula below 2 4 2 ZZ 82 137 me2 0C In this formula 2 is the number of elementary charges on the projectile Z isthe number in the target nucleus and m is the mass of the projectile This kind of radiation is known as Bremsstrahlung which is German for braking radiation The Bremsstrahlung spectrum is continuous meaning that if we have enough of these electrons interacting with nuclei we will get photons of all wavelengths in a range The limit to this range is reached when the electron gives up all of its kinetic energy in a single interaction For example ifan electron is accelerated through a potential difference of 120000 volts so that it has a kinetic energy of 120 keV it could produce a single Xray photon with that much energy or multiple photons with smaller energies It can t of course give off photons with energies higher than 120 keV This gives the short wavelength cutoff to the Bremsstrahlung spectrum PHYS 3802 Ecllpselll XRay Spectrum Ag Target so kV Au Kw Au Kb ta 2 Ener kew n the mage above taken from mtg HWWW amgtek Comechgse 3 mg We can See the X a N er approxwmate y 22 2 Rev and 25 o heV The rest otthe hm rs the cohthuous spectrurn p keVr Because the detector rs hot pertect and has a hhrte energy resolution We don t see the razorrsharp cutoff We rmgm otherwrse expect oh the other end of the Spectrum hotrce that there is a sharp cutoff at a Mme Tess maer kev Thrswash t predrcted rh ourprevrous ahaTysrs out rt s because of ahotherpractrcaT cohsrderatroh the May detector rs oehrhd a Beryururn wrhdoh T rs a 0W5 a vacuum to be for or course a m srnah arnouht of shrerrhg for the detector The owes trenergy photons Wm not be aoTe to penetrate the Wmdow and reach the detector out that s the Compton Scatte g The photoeTectrtc effect dtscusses the consequences of a photon of reasohaoTy mgh e ectron nStead m ms case the photon cannot be absorbed by the eTectroh Pamdes Ween thew the one More swmp y PHYS 3802 momentum in the reaction below y equot gt equot not allowed This doesn t mean the two particles can t interact though What we will get in this case will be ye39 gt ye39 where the two particles will exchange energy and momentum during the interaction This process is known as Compton scattering Because the only two particles involved in Compton scattering are the photon and the electron and because the energymomentum relationship of each is well known we can relate the change in the photon s direction to its initial and nal energies The derivation ofthe result is not overly dif cult but it does involve some reasonably tedious algebra so we will just quote it directly Al LU cos0 m c e The change in the photon s wavelength which can give us the energy change very quickly is M On dimensional grounds therefore hme 0 must be a wavelength as well and it is known as the Compton wavelength of the electron which is 243 X 103912 m This tells us that a small change in the direction of the photon necessarily means a small change in its energy Conversely the largest possible change in the photon s energy occurs when the scattering angle is equal to 180 This is known as backscattering since the photon heads back along its initial direction For example ifa 100 keV photon is backscattered its energy after scattering will be 71893 keV The plot below shows the energy of a scattered 100 keV photon as a function ofangle PHYS 3802 Energ kevl Compton Scattering of I00 keV photon 100000 95000 90000 85000 50000 25 50 75 100 25 Rutherford amp the Nuclear Model Them tp outar H v eah quot w and was RHOWH as the oturh puddmg rhooet h t the eteotrohs were my pomt partrotes One of of expe me t H v 19M Ruthertoro red at partrotes hehurh huoter erhtteo H v sorhe went through Because the sheet was so thtn each apartrote oouto ohw ehoouhtera r 00 trrhes as an rhasswe as an eteotroh the ohw soho thrhgto scatterfrom as tar as anyone knew Kum Mord deteotrohs of srzes that were hot oohsrstehtwrth oohsrohs wth eteotrohs Ruthertoro verytmy vomme thh a radtus ofabouHO15 rh at the oehterotthe atom ratherthah oerhg spread out aH over tt Thts oresehteo sehous prooterhs tora hurhoerotreasohs Ftrst the Coutorho reputsroh of At the Wet ddth naH around to 1 rh the atom around the Sum wth ahgutarrhorhehturh preventmg the oouaose there was a orttereht prooterh Apartrote oah ohw move W a orrote uhoerthe m uence ota toroe whroh wrt PHYS 3802 cause a centripetal acceleration The problem in the case of the electron is that an accelerated charge like one moving in a circle must radiate energy Calculations indicated that electrons moving in circular orbits around the large positive nucleus would radiate all their excess energy away and collapse into the nucleus in much less than one second which obviously did not happen Rutherford was able to estimate the size ofthe nucleus by rst calculating what kind of deflections are predicted if the or particles scatter from very dense concentrations of charge and mass at the center ofthe atom The derivation of the Rutherford scattering formula based on Coulombic repulsion can be found at httphyperphysicsphy astrgsueduhbasehframehtm and gives 2 NnLZZk2e4 4r2 KEa Zsin40 2 where Ni is the number of or particles incident n is the atomic density atomsvol ofthe target L is the target thickness Z is the target s atomic number k and e are the Coulomb constant and electron charge KE is the kinetic energy ofthe on particle r is the target detector separation and 9 is the scattering angle Rutherford wanted to investigate with higher on particle energies to see ifthere was a deviation from this behavior Because he was limited to less than 8 MeV for the or particle s KE he had to work from the other direction and change the strength ofthe repulsive force by substituting aluminum Z 13 for the gold foil Z 79 Rutherford s estimate of 103914 m for the radius ofthe aluminum nucleus was certainly the right order of magnitude and very different from the 103910 m radius ofthe atom itself Spectra Any theory of the atom would needs to explain the fact that light emitted from isolated atoms like those in a relatively thin gas is only emitted at certain discrete wavelengths rather than across a continuum The atomic model developed by Bohr did this nicely In the Bohr model the electrons move in circular orbits without emitting radiation and spiraling in to the nucleus The allowed orbits are restricted to those having discrete values ofangular momentum speci cally an angular momentum equal to n h where n is an integer greater than zero and h is Planck s constant divided by 2 7c lfthe centripetal force holding the electron in the atom is to be supplied by the Coulomb attraction between the electron and the nucleus we get PHYS 3802 mev2 kZe2 r r2 The v is the electron s velocity in its circular orbit and Z represents the total number of charges equal in magnitude to that on the electron We can solve this for rand use the fact that Lmvr E 27 Combining these ideas we can write 22 nh r 2 2 47 m k Z 6 and then since the energy of a circular orbit is wellknown to be 2 E lm kg 2 r we can write the energy as E kZe2 27r2mk2e4 Z 2 2r h2 n2 We will get slightly more accurate answers if we recognize the fact that the nucleus is not in nitely massive and it too has kinetic energy lts momentum will be the same as the electron s but its KE will be much less since we can write KE p22 m and the nuclear mass will be huge compared to the electronic mass Making this correction amounts to replacing the m in the above equation with something known as the reduced mass u de ned as mM H where m represents the electron s mass as before and M is the nuclear mass 12 PHYS 3802 For the hydrogen atom with Z 1 we calculate the energy for an electron in level n to be 1356eV i2 71 Again the restriction on n as being an integer greater than 0 still applies When an electron moves from a higher level n2 to a lower one m it will emit one or sometimes more than one photons with a total energy equal to E2 E1 This formula explains that energy levels aren t evenly spaced The factor of 1n2 tells us that the levels get closer together as the electron gets closer to being freed For example to jump from the ground state to the 1st excited state n 2 we need to supply 1 1 Ef Ei 136eV 2 2 1 2 102eV but if we want to move from the 48th excited state n 49 to the 49th n 50 we only need EfE136eV 224x10394eV 50 49 It s occasionally helpful to think ofthe electron moving between energy levels as being like a ball on a staircase though the stairs are spaced strangely Moving from a lower step to a higher one takes an investment of energy the move back down the steps releases energy Once the ball reachesthe bottom of the steps ground state no lower drops are possible lfwe want to know how much energy it will take to remove the electron entirely from the atom that also depends on the starting state of the electron The nal state can be considered to be n oo so our formula isjust 1 136eV n2 n2 Ef Ei 136eV 0 Thistells us about the line spacing in the hydrogen spectrum To put this in the same form as the Lyman series Balmer series etc we recognize that 17 f c We can then write PHYS 22m PTugg 1 1 1 R 2 2 11 hh Hhe na stem unhe amm Tsthem excited s1a e me na ate Tsthe 2 W2 ng he aahheysehes h seeehg EXENEd 513 2 rh amen enhe Paseheh sehes ohee hTs dscuv sma 2 e empmeaummmaegTseeveyeg by amev aseheh L mah 2 W212 nuw uhgeysmeg ahg EXp amEd The hhespmgueeg by e EXENEd gasammsmvm h 39s39n spemummnhe gas STmHav y MHEMHvambuWchmuvsMmeWHO S passeg Waugh a eeeh uwrpvessuve gas canamwaw engthswm he ahsmheg bythe gas heywmvequ TehTy EV r amated bmm yahgemgheehehswmmhg ah zhsnr Iinn spemvum The eehheeheh hemeeh EVWSS DH absuv Mum and maehhegyspeeha can he seehheTew The Sun a gehse in mm gas pmguees a b ackbudyspemmmr au eeTmswnhh a vangE Semeehe uukmgthmugh a phsm gheeuy anhe Sun pmmA Wm see mam vambuw ahg Wave engms umswdemewswb e hunhars hm mpu am hgh1 nuw Semeehe Teehhg anhe Sumhmugh a Wm eeeT gasth hehee eenam hnes stsThg ThTs 14 PHYS 3802 is because light matching one of the energylevel transitions ofthe atoms in the gas gets absorbed by the gas and used to move the electrons to a higher state The electrons will very quickly drop back down but they ll emit the photons matching the missing parts of the rainbow in random directions Only a tiny percentage of them will be emitted in the original direction so the person at point B will see very dark lines when comparing them to the full spectrum nearby This is the absorption spectrum and it s really all we can see of the Sun since it has its own atmosphere around it Someone at point C who wouldn t be seeing anything if not for the gas will see some of the scattered photons emitted by the gas Since they are comparing those photons to the darkness beside them the lines look bright in comparison and they get an emission spectrum The Fine Structure Constant A number which comes up frequently in atomic physics and later is known as the fine structure constant usually represented by or As it turns out it is a measure ofthe relative strength of the electromagnetic force It is a dimensionless combination of Planck s constant the electronic charge the speed of light and the Coulomb constant Your book has a method for finding this based on the speed ofthe electron in the ground state of hydrogen as a fraction ofthe speed of light This can be written as vke2 1 1 0 c hc 137 Of course the problem with the derivation in the book is that it is based on the semi classical Bohr model ofthe atom where the electron is moving in a circular or elliptical orbit The name comes from the fact that there were small details in the arrangement of spectral lines ie fine structure ofthe spectrum that Sommerfeld thought might be due to the effects of special relativity acting on the reasonably highspeed electron Relativistic effects can be expected to be oforder vc2 or 12 which was the right size for the observed structure The cause ofthe ne structure is actually not something that can be represented classically so this is a coincidence or is still a useful quantity though since the energy of a hydrogen atom Equot and the Bohr radius ao can be written in terms of it as 2 2 mc a h En 2 a0 2 n mca The FranckHertz Experiment One ofthe important early con rmations ofthe prediction ofquantum theory came in the form ofthe FranckHertz effect In this experiment the current due to a beam ofelectrons moving across a potential difference in a lowpressure tube is measured as function of PHYS 3802 voltage In a resistor we would expect to see a linear relationship via Ohm s law In this experiment the lowpressure gas in the tube causes fluctuations in the current as the voltage is increased As shown in your book on p175 there are peaks and valleys in the graph of current vs voltage The current increases with voltage until it peaks at about 49 volts when the lowpressure gas is mercury followed by a steep dropoff in current as the voltage increases until it begins to climb again As the voltage keeps rising another peak is found at 98 V 2 X 49 V and at other multiples of this voltage The explanation of the effect is that electrons moving with less than 49 eV of kinetic energy are unable to excite the mercury atoms from their ground state to the first excited state and therefore pass through without dif culty Once their KE reaches the 49 eV level some ofthem will collide with the outer electrons in mercury atoms and excite them meaning the projectile electrons will lose 49 eV in the process At voltages only slightly above 49 V the electrons won t be able to climb the slight voltage barrier in the middle of the tube so the current drops off As the accelerating voltage increases the peak at 98 V represents an electron which undergoes two collisions with mercury atoms Further proof ofthis explanation is found in the radiation given off by the FranckHertz tube lfthe accelerating voltage is less than 49 V there are no inelastic collisions between the atoms and projectile electrons so no energy is lost and therefore no photons are radiated After crossing the 49 V barrier when electrons start to lose energy to atomic electrons the excited electrons quickly return to their ground states releasing photons with 49 eV of energy This corresponds to a wavelength of 253 nm which is the wavelength of light the tube begins to emit when the accelerating voltage exceeds 49 V de Broglie Waves The idea that light could be both a particle and a wave was still young when Louis de Broglie hypothesized that it could be applied to matter as well Not only was an electron a particle but it would also be a wave In other words it would have both particlelike and wavelike properties depending on the experimental situation lfwe look at the wavelength X of a photon we see that it can be related to the photon s momentum p by de Broglie assumed that the same relation would work for particles He also realized that the formula above would allow an interesting physical interpretation ofthe quantization condition for the electron s angular momentum in hydrogen Since it was PHYS 3802 and the electron s wavelength could be written as hmv we could write ml 27 r In other words an integral number ofelectron wavelengths would fit around the circular orbit of the Bohr model This is just the condition for a standing wave which is a satisfying result since the electron which was now assumed to be nonradiating was in a stationary state It is easy to see that macroscopic objects and even many microscopic ones will have wavelengths much smaller than their actual physical size making it essentially impossible to observe any ofthe normal wavelike effects interference diffraction etc For a baseball m 015 kg moving at 30 ms we would get a wavelength of hmv 14 X 103934 m far smaller than the size ofa proton and not far from the Planck length the smallest length that can be said to make any sense under current physical theories On the other hand if we restrict ourselves to small masses and low velocities we can get more reasonable answers An electron accelerated through a potential difference of 100 V would have a momentum of 2 100eV16x 103917J p54X103924kgms m which corresponds to a wavelength of0123 nm This is in the neighborhood ofthe spacing between planes ofa crystal Just as crystals had been used as diffraction gratings for Xrays with similar wavelengths it turns out that they can function the same way for electrons The crystal a 3D structure can be thought of as a collection of planes 2D structures that hold the atoms making it up The atoms are arranged in a repeating and consistent manner like the designs on wallpaper One difference between diffraction from a crystal and diffraction of visible light is in the angle convention Instead of being measured from the normal the angle 9 is measured from the crystal planes as shown below PHYS 3802 d plane spacing 9 measured he Ifwe want to measure from the normal we could call that angle or and it will be equal to 90 9 The angle of incidence ocwill equal the angle of re ection or so that the initial and final directions are 20c apart While d is the spacing between planes the spacing between atoms making up the plane can be written as d D Sin or so that our scattering condition becomes ml 2 Dsin2x 2 Dsinq The information about the lattice spacing for a particular crystal could be found from well established Xray diffraction so the equation above could be solved for the wavelength of the electron and compared to the value predicted by de Broglie s formula since the potential difference crossed by the electron was under the experimenter s control The equation above is valid as long as there are planes in the crystal at an angle of o2 with the surface this is not guaranteed to be the case for any random choice of i of course since there are a nite number of planes in the crystal Similar experiments have revealed the wave nature of other larger particles such as neutrons atoms and molecules Wave Packets f matter can behave like a wave we should be able to treat it like a wave mathematically The wave equation in one dimension is from introductory physics The solution to this equation can be written in terms oftrigonometric functions or imaginary exponentials Your book uses harmonic waves ofthe form yxl yo coskx a I yo coszl bc v1 PHYS 3802 This is one form for the equation ofa wave traveling to the right towards x with maximum amplitude yu wavelength 1 39equency VI and phase velocity v 139 L 39 L 39 39 39 39 all space the x axis at all values of time quotquot 39 39 39 39 39 wuv u n s which exist only in some region of space for some duration oftime This requires the combination of quot3939 39 139quot 39 39 quot of two waves The gure below is a plot of two waves of the same amplitude but the green wave s wave number is 10 larger than that ofthe red wave This means that the waves move in and out ofphase as time passes at the origin both waves are a maximum but at about 15 on the x axis the waves are 180 out of phase 1 ll l1 quotl n l 4175 i The result of adding these two waves together is shown below Two periodic patterns are 39 quot quot 39 39 equal the red r r amp green quot 39 equal rx u A g u PHYS 3802 This is characteristic of adding two trig functions since we can write 51 51 2 2 Coslxji Carly 2Cos 2 2 C05 Your book expands the mquot expression into yw 2yncosArv A a JcosUrrr m where the bars indicate average quantities and the deltas represent differences between es In this form we can see thatt e igher 39equency wave moves att e phase velocity vp and the lowerfrequency wave moves at the group velocity viI where vpw k vngwAk We can connect these two velocities with dvp dk vg vpk When the phase velocity depends on wavelength dvpldk is not 0 the material is called dispersive Water glass etc are all dispersive for electromagnetic waves which is the 20 PHYS 3802 reason prisms and water drops split white sunlight into a rainbow lfthe medium is nondispersive a pulse will keep its shape as it travels since the different frequencies it is composed of are all traveling at the same speed For a particle like an electron the phase velocity which can be rewritten as Ep would be p2m v2 ifwe assume vltltc The group velocity though would be dEdp which is pm v Therefore the group velocity is equal to the electron s actual velocity Classical Uncertainty Even in classical physics there is an uncertainty associated with measurements ofa periodic phenomenon such as the oscillations in a wave Think about what you do when you exercise and measure your pulse You can count heartbeats for 10 seconds and then multiply by 6 to get your pulse rate but the possible answers are things like 60 66 72 etc You can t get a pulse rate of 65 this way You could count for 30 seconds and multiply by two but you ll only get even numbers in this case You will get a better estimate by measuring longer You could count for a full minute but what if your real pulse rate is 674 beats per minute You will still have the potential for error in the determination ofthe pulse rate as long as you count for a nite time We can see this uncertainty when we look at Fourier analysis Fourier Analysis There are two equivalent ways to describe any waveform We can talk about it as the amplitude of something speaker position electric eld location of molecules in rocks during an earthquake etc as time changes the time domain or we can think of it as a collection of sine or cosine waves ofdifferent amplitudes and different frequencies the frequency domain lfwe look for the connection between these two ideas the Fourier transform takes us from the time domain to the frequency domain and the inverse Fourier transform goes in the other direction lfwe were to plot the position ofone of the tines in a recentlystruck tuning fork as a function oftime we would get a sine wave Because ofthe uncertainty principle we could only be completely certain of the frequency ofthis wave if we had an in nite number of cycles to count To understand this imagine that you re taking your pulse while exercising lfyou count it for 10 seconds you know that you need to multiply your answer by six to get your true pulse You can also quickly see that you will never get an answer like 63 beats per minute since that s not a multiple of six lfyou wanted to narrow this uncertainty in the pulse down to something smaller you could count for thirty seconds and double your answer Of course you re limited to even numbers forthe pulse in this case You could extend it to one minute and get to the nearest whole number but what if your heart doesn t happen to beat exactly x times in a minute You need to count for two minutes and then divide by two etc You can see that reducing the uncertainty in your pulse frequency to zero would mean counting for an in nite time You can see the two plots below time domain on the left frequency domain on the right PHYS 3802 u an mu an 2m Theta Frequency If a signal is narrow in one domain it must be wide in the other domain just as a particle with a small uncertainty in position has to have a large uncertainty in momentum In a similar way if you want to produce an instantaneous displacement of a speaker cone or your eardrum a huge range of frequencies will really be present The graphs below are constructed from plotting a simple function from 1r to 1 It can be written as fxj Cosi x For example fx1 is just Cosx fx2 Cosx CosZx fx3 Cosx Cos2x Cos3x etc In the frequency domain this function will just look like spikes What do we get in the time domain Look below PHYS 3802 As you can see adding more frequencies making the signal broad in the frequenc domain bunches it up in the time domain To get a perfect spike which is infinitely tall and infinitesimally wide called a delta function you would add an infinity of these frequency terms so the frequency domain plot would look like a single horizontal line at some nonzero height Being able to represent any waveform either way means that if you could have millions really an infinity of people each with a tuning fork of a different frequency and you could control how hard and at what instant each person hits his or her fork you could exactly reproduce the Gettysburg address in Lincoln s voice or any other sound you can imagine Other properties of the Fourier transform that can affect the imaging process include aliasing where a frequency higher than the sampling frequency f is incorrectly identified as being a different frequency lower than f as shown below Imagine that the green vertical lines represent the instants in time when samples are taken The red line is the signal were trying to sample but notice that everywhere the red line intersects a green line it also intersects the m line which is a lower frequency signal There are an infinite number of sine waves that have the same value as the red line at each sampling interval green line We will always assume that the lowest frequency wave that fits the data is the signal actually being detected because it s the most reasonable assumption It is up to the people designingusing the equipment to know its limitations and to not exceed them PHYS 3802 ll ll The highest frequency we can reliably measure when we sample at a frequency fS is just twice that frequency This is known as the Nyquist condition and it s the reason CDs are sampled at a rate of 441 kHz even though the upper limit of human hearing is about 20 kHz We can t accurately detect 20 kHz sound waves if we don t sample at 40 kHz or better Aliasing can be seen when someone on TV wears clothes with very thin and closelyspaced stripes they tend to look like they re moving or made of different solid colors because the spacing is smallerthan the resolution of the TV signal The classical uncertainty relations tell us that wave packet width amp wave number uncertainties are related y A k A x z 1 and the frequencytime uncertainty relation is AwAtzl The Uncertain Princi le In quantum mechanics the uncertainty principle changes its importance by removing the idea of determinism Determinism basically says that if you knew the positions and 24 PHYS 3802 momenta ofall objects in the universe exactly even forjust an instant you could then calculate their positions and momenta at any later time The outcome of every coin ip lottery drawing football game or stock purchase after that would be known in advance This was largely a philosophical argument since nding the exact position and velocity of even one particle had not been done when it was proposed Still there was nothing to prevent this in principle it would just take better and better measuring instruments which could be expected to be developed through technological progress In fact it was later discovered that this could not be done even for one particle and even in principle Heisenberg s Uncertainty Principle says that you or anyone else cannot know the exact position and momentum ofany one particle at the same time It further restricts the product ofthe uncertainty in momentum and the uncertainty in position as shown below h A AxZ p 2 Therefore the better you know a particle s position the lower AX the less you know about its momentum You can think ofthis as follows if we want to measure the position of an electron how do we do that At the most fundamental level we look at it which means we have to bounce photons off of it In doing that though we transfer some momentum to the electron lfwe want to measure the electron s position to within one nanometer for example we need to use light with a wavelength of less than one nanometer What momentum will the light have We can use our previous results to see that h 139 llt10399m s0 gt f p l p 10399m Getting a better measurement of position means using light with a smaller wavelength but that means each photon has a high momentum Some unknown fraction of that momentum will be transferred to the electron so measuring positional information causes us to alter and therefore lose momentum information This is not a problem in our macroscopic world but it can be one on the microscopic scale lfan electron moving at 1000 ms has a momentum ofabout 103927 kg ms an uncertainty of 103924 kg ms is ridiculously large Of course an uncertainty in momentum will evolve into an uncertainty in position if the uncertainty in a particle s velocity Apm is 10 ms its uncertainty in position is growing by 10 m each second If we localize a particle very well measure it precisely so that its AX is small the uncertain amount of momentum transferred to it may completely remove it from our viewing area In this case about all we can say is that a particle was at this position earlier PHYS 3802 There is a similar uncertainty relation for energy An accurate determination ofthe energy of a state depends on how long it is observed Basically we are allowed to violate conservation of energy l my brie y and borrow some energy from the vacuum as long as we return it very quickly The product ofthe energy uncertainty and the time uncertainty satis es the same inequality as above AEAIEL 47 It s strange and probably a little disturbing to see that one of the things so important in physics conservation ofenergy seems to be disappearing Before you get too worried do a sample calculation if you only want to borrow some energy for a picosecond 103912 seconds how much can you have Plugging 103912 seconds in for At in the formula above gives 53 X 103923 J or about 33 X 10394 eV Not exactly a huge amount ofenergyl One ofthe consequences ofthe uncertainty principle is that certain features of classical mechanics are radically altered For example ifwe imagine a ball bouncing inside a drinking glass it is very straightforward to nd out whether it will escape or not lfthe kinetic potential energy ofthe ball is less than the potential energy the ball would have at the rim ofthe glass it absolutely won t get out The uncertainty principle allows the ball to borrow a small amount ofenergy from the empty space around it and return that energy once the ball has gone overthe side Since the time that energy can be borrowed is inversely proportional to the amount borrowed and Planck s constant is so small we won t see this happen with a ball amp drinking glass The probability of penetrating a barrier drops very quickly as the barrier s thickness increases Also if the barrier is very large compared to the particle s energy the probability of penetration drops rapidly In the smoke detector example the process can be modeled by assuming the on particle istrapped inside the nucleus like a particle in a well The on particle bounces back and forth within the nucleus until by random chance it grabs a small amount of energy from the vacuum climbs the side of the well and appears outside the con nes ofthe nucleus Once it s on the outside the repulsive force between the two protons of the on particle and 93 remaining ones in the nucleus cause the on particle to rocket away and be detected by circuitry in the smoke detector For this reason the energy ofthe escaping or tends to increase as the Z of the nucleus increases Something interesting to notice isthat the energy borrowed really is returned to the vacuum We can calculate the energy an alpha particle needs to have to escape its con nement by the nucleus and we find that the emitted alpha particle has a lower energy than this lfthe borrowed energy didn t have to be returned and therefore conservation of energy could be violated on the large scale the emitted alphas would be ejected much more quickly than they really are PHYS 3802 This principle also explains the reason that atoms are stable The electron can t radiate photons and fall into the nucleus because that would make its position more certain since the nucleus is smaller we d know where it is to within 103914 m or so instead of 103910 m which would mean its momentum would be more uncertain making it potentially larger lncreased momentum means increased kinetic energy A smaller value of position on average would give a more negative potential energy but a more positive kinetic energy The atom s size is the result of a balancing act between confining an electron to a tiny volume which the Coulomb force would like to do and keeping its momentum and kinetic energy from increasing wildly The Schrodinger Equation The requirement of quantized values of the electron s angular momentum can be arrived at by another means If we consider an electron with momentum p to be a wave with a wavelength X hp the idea of light as a particle gained support around the same time that the electron was postulated by de Broglie to have wavelike characteristics and propose that an integral number ofwavelengths must fit around the electron s orbit 2 7 r in size we would get nh nh 27f r gt mvr p 27E which is the same condition on rthat was seen before lfthe electron can be considered to be a wave there must be a wave equation describing how the electron moves This equation is known as the Schrodinger equation The wave which is expected to be in general a function of position and time is therefore known as the wave function and is usually represented as yxt As a trial wave function we can draw on what we will see will be many parallels between quantum theory and optics and try a plane wave of the form ikx a t V xatl 06 A classical wave either a physical wave in a string or an electromagnetic wave ofthis form would obey a wave equation that was second order in both time and space something like mm Bx v2 812 PHYS 3802 lfwe substitute the solution for y mentioned above into this equation we get 2 a a 22 23 k V V which we can write as o ck for electromagnetic waves Multiplying by Planck s constant on both sides ofthis equation allows us to write E p c which we know is true for photons For particles with mass the relation between E p and m is different This was used as a starting point for a relativistic wave equation by Dirac Schrodinger tried to work something like this out but was unable to do so Instead he used the Newtonian relationship between energy and momentum for an electron which isjust 2 1LVE 2m The equation Schrodinger eventually produced is in one dimension 2 2 iii g nanvdmzm 2m dx dl You can verify that the plane wave solution for y reproduces the Newtonian energy momentum equation Because the Schrodinger equation contains the first derivative with respect to time but the second derivative with respect to position it looks very similar to a diffusion equation which speci es how two materials mix a drop of milk in a glass of water for example The weird thing about the Schrodinger equation is that it is a diffusion equation in imaginary time notice the on the right side What is diffusing in this case The wave function itself which describes the electron s positionmomentum An imaginary wave function can t directly represent the electron s position or momentum since those are measurable and therefore real values The wave function itselftells us the probability amplitude and the absolute square of the wave function is therefore the real and nonnegative probability density Since this is proportional to the chance of finding the electron in any given region we can say that this expresses what we saw earlier with the uncertainty principle after we locate the electron its position gradually gets less certain with time due to Ap as the wave function spreads out or diffuses through space 80 that this probability is physically reasonable we require that the total chance of nding the electron anywhere in space must equal one We ll see later that this provides a PHYS 3802 normalization condition that sets the overall scale ofthe wave function We can write this condition again in one dimension as It dx1 00 This means that the wave function has to approach zero at large values ofX fast enough to make the integral on the left hand side nite Solving the Schro39dinder quot The Schrodinger equation is usually given in either time dependent or time independent form The example above is the time dependent case the potential is a function of position as well as time and the wave function is also The time derivative on the right is the signature oftime dependence As a first step we ll restrict ourselves to potentials and wave functions that are independent oftime Most potentials you ve seen in other physics classes would t this description and the wave functions that are time independent are known as eigenstates ofthe system or stationary states Under these conditions we can separate the solution to the Schrodinger equation into two terms one of which depends only on position and one which depends only on time as shown below rmmwmwn Substitute this into the Schrodinger equation and divide by W to get 2 hchggnm wme 2myx dx MI dl Notice that the left hand side ofthe equation is only a function ofX and the right hand side is only a function oft The only way these two sides can be equal is ifthey are both equal to a constant which will be E the energy of the particle The energy is a constant in this case because time independence is deeply connected to the conservation of energy In the same way translational invariance is deeply connected to the conservation of momentum ifyou re interested in learning more about this it s called Noether39s theorem First we ll solve the right hand side by writing it as PHYS 3802 LdW im hm dr Egt z e We can then rewrite the left hand side as i dzr x M W wool x El x which is known as the Time Independent Schrodinger Equation TISE There are a few conditions on the wave function which we can understand on physical grounds 1 yx has to satisfy the Schrodinger equation 2 yx must be continuous ifit were not the momentum associated with the particle which is proportional to deX would be in nite and so must its rst derivative if deX were not continuous its derivative would be in nite and the kinetic energy ofthe particle is proportional to this derivative 3 So that we get physically measurable results y and deX must be nite and singlevalued and nally 4 yx 9 O as X9 oo fast enough to allow the normalization integral to be nite The Infinite Square Well One ofthe simplest problems we can solve using the TISE isthat ofa particle inside an in nitely deep potential well This is essentially a region with walls that are in nitely high and therefore con ne the particle The potential inside the well is zero everywhere meaning that there are no forces acting on it We then have a free particle restricted to a certain region in space and we can examine what this does to its energy An unrestricted free particle one that can be anywhere in space can have any energy just as you would expect from classical physics We can describe the potential as Vx 0 0 lt x lt L Vxoo xlt0andxgtL The wave function is therefore forced to be zero outside ofthe well since there can be zero probability that the particle leaves the well We need a wave function that satis es this requirement as well as the ones mentioned above continuous finite etc Inside the well we have a freeparticle Schrodinger equation since there is no potential there we can use the de nition of kto rewrite it as PHYS 3802 dzr x 2mE 012w x W h2 lx gt dxz k x The solutions to this equation are the two trig functions and any linear combination of them which we can write as l1x Asinkx l2xBcoskx Requiring the wave function to vanish at X 0 means that the constant B multiplying the cosine function must be zero Our conditions then reduce to 1L AsinkLz o This requires k to satisfy mr k n123 This is the rst appearance of quantization meaning all possible wave numbers are m allowed This is in contrast to the situation with a truly free particle not restricted to any part of the axis which can have m value of k The expression for energy in terms ofk means that we have hz 2 E 2n2 n123 2 m L Quantizing k means the energy is also quantized Notice that the lowest energy state is equal except for the factor of 72 to what would be predicted for the kinetic energy ofa particle of mass m with uncertainty in position equal to L2 We can now use the normalization condition to nd the value of the constant A in the wave function We integrate the absolute square ofthe wave function y over the entire x axis but since the wave function is zero outside of the well that integral from oo to oo reduces to the integral over the well itself mrxdle L JAZ sin2 0 PHYS 3802 The integral on the left is A2 Lsin2mr k2 4717 However the second term in parentheses disappears so we have the much simpler A2 212gt A 2 L leaving a nal wave function of lltxgtgsin x The classical probability density for this particle which has a slightly different meaning classically the particle s position has no uncertainty and can be precisely located at any time This probability density can be thought of more as the chance of nding the particle in a given region ofthe well at a random time would be uniform across the well The particle s speed is constant all over the well only changing at the edge ofthe well when it is re ected The chance of nding the particle in a part ofthe well dX in length would be dxL Because quantum mechanics at large values ofthe quantum number n should reduce to the classical result we can look at what happens to y2 at large n The sinusoidal term will oscillate so rapidly that we can replace it by its average which isjust12 The probability density v2 would then reduce to 2L12 1L which is the classical result Your book looks at the interesting result obtained when we rejoin the separate wave functions yx and lt back into i xt We then get LP nxt sinknxe39 t which can be rewritten in the more useful form PHYS 3802 P quotxjf iJZelknxwnt eikx cont 2 L This looks like a lefttraveling wave and an equalamplitude righttraveling wave together they produce a standing wave This means the particle s wave function in the infinite well looks like the waves on a string anchored down at points 0 and L where an integral number of halfwavelengths exist The Finite Square Well Moving a little closer to reality we take the in nite square well and replace the potential at the walls by some finite value of potential We can center the well at the origin for convenience Vx0 lxllt a2 VxV0 lxlgta2 Notice that the Schrodinger equation looks the same inside the nite well as it did in the in nite case L19 m x where k is still equal to W2 m E h The different part is that now the wave function does not have to be zero outside the well Instead the Schrodinger equation outside the well becomes 3 le Ell x a 2w x Because V0 gt E outside the well 12 gt 0 Although the wave function is not forced to zero at the well boundary we still have the fundamental condition that the wave function and its first derivative are continuous everywhere and that includes the boundary We have to match the inside solutions to the outside solutions We still have two possible solutions for or corresponding to positive and negative roots ofthe equation above The equation will be solved outside the well call the left side region and the right side region III by functions ofthe form PHYS 3802 l Ix Ae39 Be39 xlt a2 ll HIx Feax Ge39 xgt a2 Since the limit ofthe wave function has to be zero as you approach too we require that the constants B and F equal zero as those terms grow in size as we move away from the well Inside the well region II we have V x Csinkx Dcoskx At the left edge xa2 continuity ofthe wave function and continuity of the rst derivative ofthe wave function also evaluated at xa2 gives Ae39M2 Csin ka2 Dcos ka2 and 1 X Ae39 W2 kCCOs ka2 kDSin ka2 2 On the right side at xa2 we have Ge39 2Csinka2Dcoska2 and 3 a Ge39MZ kCC0ska2 kDSinka2 4 We can take equation 3 and divide it by equation 1 and then equation 4 divided by the negative of equation 2 to get 2 CSmka2 DCoska2 A CSmka2 DCoska2 A CCoska2 DSmka2 G39 DSz nka2 CCoska2 Multiplying these two equations together gives PHYS 3802 DCOska2 CSinka2 C C0ska2 DSz39nka2 1 DC C2 D2C0ska2Sinka2 Verify that this equation will be solved ifeither C or D but not both is set equal to zero We will therefore get two classes of solutions inside the well either 00 and our wave function will be a cosine or D O and our wave function will be a sine The C0 case will then require that G A and Ae39 W2 Dcos Ira2 and X Ae39 W2 kDSz39n Ira2 Dividing the second equation by the rst gives 0 kTanka2 Both sides ofthis equation can be written as functions of E but we can t solve for E algebraically This is known as a transcendental equation and we can solve it numerically using a calculator or computer or graphically To see how we rewrite it as 0 T anka 2 k and plot each side as a function of the energy E Wherever the curves cross we have a bound state at that energy In the D0 case we get AG since the wave function all three parts must be odd Notice that the probability amplitude which is proportional to the square ofthese coef cients will be equal on both sides of the symmetric well On the left side ofthe well we get Ae39 W2 Csin Ira2 and or Ae39 W2 kCcos Ira2 which we combine in the same way to get 0 kcotka2 PHYS 3802 or tanka2 5239 The number of bound states in a particular well depends on the depth and width ofthe well and can be determined by calculating i 2W0 7th Rounding this value upto the next whole number will give you the total number of bound states in the well The values of the normalization constants A and C or D depending on the particular wave function are determined by enforcing the normalization condition which says so JV Nx V Nxdx1 700 for each wave function VN associated with each bound state energy EN When we know the energy we will also know both or and k so the boundary conditions will just give us a way to write C or D as a multiple ofA The normalization integral above then sets the value of A and we re done


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