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Quantum Mechanics I

by: Quinn Larkin

Quantum Mechanics I PHY 851

Quinn Larkin
GPA 3.67


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This 75 page Class Notes was uploaded by Quinn Larkin on Saturday September 19, 2015. The Class Notes belongs to PHY 851 at Michigan State University taught by Staff in Fall. Since its upload, it has received 12 views. For similar materials see /class/207632/phy-851-michigan-state-university in Physics 2 at Michigan State University.

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Date Created: 09/19/15
Lecture 27 February 13 2001 Tensor Operators rotation operators and the WignerEckart theorem Often physics involves calculating transition elements of the form I I k Z m Tq 1772 1 where the labels in the kets and bras denote the angular momentum of the initial and final states and T is an operator that transforms like spherical harmonics Ym under rotation Such operators are known irreducible tensor operators Example are H Lz which transforms Ym D L which transforms Yb O3 7 2 which transforms Yogo 9 a or pfr which transform Ym erl Operators such 32 can be written a sum over gs and the coefficients can be found by viewing the expressions for the Clearly one could expand any analytic function of 3 y and z in terms of irreducible tensor operators The term irreducible refers to the fact that under rotations the operators mix only amongst irreducible subsets k 71 a k k RaTq R dmqu 2 The rotations mix only the 2k 1 operators with the same C but different 1 The term irreducible refers to the fact that rotations can mix in any of the different 1 components The rotation matrices dmt can be expressed in terms of the overlaps dime ltm Rltagtz mgt WWWquot391 mgt lt3 The d matrices have a variety of properties many of which are related to those of the spherical harmonics For instance the orthogonal property is 011 J2 arm mz 071139 m 2 ii 1392 dudma u dmgw 211 1 4 Finding the d matrix elements Finding the d matrices a function of a can be painful but luckily canned routines exist in standard computer languages If one needs to calculate them it is best to consider the rotation 04 in terms of Euler angles 2 inQhveiinleheiiJ2 h Lecture 27 February 13 2001 where the y axis is the y axis after an initial rotation around the original 2 axis by an angle a and z is the new 2 axis formed after the rotation of 9 about the y axis Thus P u n Z P uanP uyanPUZmn 6 Munn Z E uy0n8 uzun8uymn Substituting these into the expression for R allows one to write the rotation matrices without mentioning primed axes R 915 Z 8 HmnE uyanE uzwn 8 One is now in a position to find the elements Simple expressions follow for when either m or m is zero or when some of the angles are zero but otherwise it can be bit tedious to find the elements The elements can be written in terms of the Yams The Wigner Eckart theorem Since T rotates like an angular momentum state one can define the state 8 is 3 2 M E 2063417nikjj 7n gtT8jm 9 mm which must rotate an object with angular momentum j and projection m Here we have written the Clebsch Gordan coefficients with the convention E1524 mEEI 2 mhmz Furthermore a matrix element formed with this state would be zero unless the angular momentum of the bra state are identical lt5 kajaj wig kajaj 773 Z 5m m 3quot339f3 3 kajaj 10 This conclusion follows from the fact that the matrix element should be invariant under equal rotations of both the bra and the ket which requires both matrix elements to behave identically under rotation and therefore have the same angular momenta Furthermore the matrix element should be independent of m Since rotations mix values of m the function f can not depend on m This result can be proven rigorously using orthogonality relations of the rotation operators d See Baym or Messiah Expanding the state 3523 mj m EMUquot mquot qui3J mgtltkiji I mikJJ39 m39 5m m 5jquot3 f8 3 kidquot 11 mm Multiplying both sides by Clebsch Gordan coefficients and using the orthogonality relations 2 3399quot mquotTjg3 3 mltk 3 q mik 3 3quot m ltkjj m kj 1 m 12 qmm m lt6 k 3 m Tj 8 2 m gt 13 WM m ikj 1 m gtf3 3 k 3 j 14 This is the statement of the W igner Eckart theorem It is usually expressed in the following form 3 j EETquot EE3jgt 5 lt6 kjj 7n iTq i3jmgt m 1 m kjqmgt Lecture 25 February 7 2001 Isospin The angular momentum algebra developed last semester has a broad range of applications many of which have little or nothing to do with rotating systems but instead deal with symmetries in some other space One such example is isospin In nuclear and particle physics isospin represents a mirror of the spin mechanics used to discuss spin 12 particles In quark language the up quark is a I 12Iz 12 particle while the down quark is a I 12 Iz 12 particle In nuclear physics the proton and neutron form an isodoublet with the proton being the Iz 12 particle The strong interaction is invariant to rotations in isospin space This means that the symmetry is analagous to the rotational symmetries discussed with angular momentum For instance composite particles those made of several quarks have good isospin which is determined by adding the isospin of several particles Example The p meson Consider the pHquot mesons which form an isotriplet I 1 and can decay into two pions Pions 7WDquot also form an isotriplet Find the decay branches for each of the three p gt 37r decays First look at the N or p The only branchings for each decay are p gt 7r7r0p gt 7r 7r0 However the p0 might go into one of two branches p0 gt 7r7r or 7r07r0 To solve for the relative strengths of the two branches nd how to add two pion isospins to form a I 1 I3 0 state Start by writing the I 2 I3 2 state i1 213 2 mm then use the lowering operator to nd the I 2 I3 1 state 1 231 i oy Ei a ogt By orthogonality 1 1 I 1112 1 i o r 7527 By applying the lowering operator 14110 0 1 1411 141 1 I 113 0 fin 7W 14110 0 1 7r7r 2 1 1 I 1IZ 0 7w 7W i 7r7r Lecture 20 February 7 2001 Scattering Central Potentials Partial Waves and Phase Shifts If potentials have spherial symmetry one can solve the scattering problem through the con sideration of spherical waves rather than plane waves This effectively reduces the problem to the solution of onedimensional Schrodinger equations where each partial wave is char acterized by a specific angular momentum i and requires a different centrifugal potential First we consider the kinetic part of the Hamiltonian The Schrodinger equation can be written in either Cartesian coordinates or spherical coordinates 2m 832 33112 832 2m 7quot 872 T T2 sin2 6 132 T2 sin 6 86 5m 86 222 1 a2 L2 WT 27mquot2 2 If the potential spherically symmetric angular momentum is a good quantum number and one can write solutions corresponding to a specific 5 and m Hagar Ewan 3 with the wave function being written a product of a radial part and an angular part 1quot mea 1247 4 The radial wave function is a solution of the equation 7amp2 1 82 2 1 39EEW m 7 Remember that the radial wave function depends only on i and not m since the centrifugal potential is determined by i only RAT WHEAT 513W 5 If the potential is zero the solutions RAT are referred to jg Plane wave solutions can be expanded in terms of the spherical solutions through 6 r 2 225 1i jgkrPgcos6 6 where cos E 12 f This is known the partial wave expansion Note that the expansion are in terms of the Legendre polynomials 47T PgltCOS 6 V mnmuw 13 All angular functions can be expanded in terms of Yams and since Eik r ammo does not depend on 239 it is no surprise that the expansion contains only Vanna terms The partial wave expansion can be derived by combining orthogonality relations of the Legendre polynomials and spherical Bessel functions with the Rodriquez formula 1 d 32 1V HWZWYd Lecture 20 February 7 2001 The solutions jg are normalized as 4 i 6m Gag 17 H 3 l lit gt00 21x and are known spherical Bessel functions They are a linear combination of outgoing and incoming waves each of which is a solution to the Schrodinger equation The relative phase between the incoming and outging waves is chosen so that the solution goes to zero at a 0 In fact a 12039 z 8 As an exercise one can verify the small 3 expansion by applying Schrodinger s equation with the centrifugal potential There also exist a class of solutions which are orthogonal to the spherical Bessel functions but do not satisfy the boundary conditions at zero These are known spherical Neumann functions and have the opposite relative phase between the incoming and outgoing parts while being quite divergent at the origin wng Examples of a few spherical functions for low E are 4 sin a cosa 1037 a 7 71037 sin 3 cos a 10 cos a sin a 1137 3 x 7 15 T 11 4 3 1 3 3 1 3 07201 Sll lCC COSCC cosa 731113 12 Both jg and w are real The spherical Hankel functions are de ned in terms of jg and w iy1 hME WWW 7WMw m if 1 kWE Wszjwww as 3 Here kg and h behave outgoing and incoming waves respectively Scattering Phase Shifts By making combinations jgii g one nds solutions that correspond to incoming or outgoing waves When adding a potential of nite range there still exists solutions which look like thmquot or h kr for 7quot beyond the range of the potential but have modi ed forms at small 7quot Just in the case with no potential one can nd a linear combination of the incoming and outgoing solutions which goes to zero at 7quot 0 However the relative phase between the incoming and outgoing phase will be adjusted by a phase 62W due to the existence of the potential The large a k7quot behavior is then Rxgtm gmMm mm 15 2 Lecture 20 February 7 2001 where 6 is known the phase shift Here a is any distance large enough such that the potential is zero We de ne the overall phase of Hg such that the incoming phase has the same phase does the incoming part of jg If one scatters a plane wave off a potential one can consider the solution to be the original plane wave expanded in terms of partial waves plus the correction due to the interaction mu 22 1i RgkrPgcos9 16 a 225 1z 12kr gum Pglcos 17 where the choice of phases in the definition of Hg allows the incoming waves to be identical to those of the solution with no potential Expanding the answer at large T one obtains 2kquot I lg 22 L mm c Z j2 1g 0 12ikTPgcos9 18 A ikr all 225 newquot sin 6 Pgcos 9 19 g Only the latter term contributes to scattering the plane wave continues to travel forward after the wave packet leaves the region of the scatterer One defines a quantity ll the scattering amplitude and gives it dimension of length ll E 2218l6 Sln6gPgCOS9 20 f 82761quot mm a 7fsz 21 Note that f no longer depends on T It can also be related to the differential cross section To see this we first relate the differential cross section to the flux of particles per solid angle 0 d0 LN 17 dlldt 22 The flux per unit area can be found by multiplying the square of the wave function in Eq 3 by the velocity and dividing by the volume dN 3 Jul M1th V r 23 Comparing the two equations above allows one to see that ll is directly related to the differential cross section 1 a m Julquot 24 One can see that the differential cross section at a given energy is determined solely by the phase shifts Mk Lecture 20 February 7 2001 Integrating over if to obtain the whole cross section eliminates the cross terms involving dif ferent is resulting from squaring f due to the orthogonality of the Legendre Polynomials In fact one obtains 47f k2 039 225 1 sin2 56 25 3 Example Hard Sphere Scattering Consider a hard sphere of radius 1 Find the contribution to the cross section from s and p wave scattering a function of the momentum The i 0 case is simple the solutions for ingoing and outgoing waves in the region 7quot gt a are the Hankel functions which must go to zero at 7quot a 1 m a 12001 5 9 0ka mm o 26 Plugging in the expressions for fag one obtains emquot sinka icoska sinka icoska 0 27 which gives the i 0 phase shifts 50 ka 28 The contribution to the cross section from the i 0 partial waves is then Mr k2 sin2ka 29 09 Note that k gt 0 the cross section approaches 47ra2 four times the expected geometric cross section Calculating the contribution for the i 1 partial waves is a bit difficult In this case the incoming and outgoing waves are R101 9215mm mm o 30 This yields the following expression for 51 when requiring that R U 82151 coska iM isinka Sinam coska iCOSUm isinka 0 ka ka ka 31 Solving this for 51 coska sin t39 5 2 1 sinka coskaka 3 l The contribution to the cross section is 12 a k sin2 51 Lecture 11 February 7 2001 1 Spherical Harmonics The kinetic energy term in Schrodinger s wave equation may be written 2 2 2 2 2 HK2 h22 7 128 h 1 a 18sin92 1 2m 872 2mr2 sin298i v2 sing 39 Furthermore the components of angular momentum may be written in terms of angular derivatives L iha 2 L ih sin 13 cot 9 cos Ly ih cos 13 cot 9 sin Li magi ii cot 98 5 Using the relation L2 L LLisz one can see that the V2 term for the kinetic energy may be written 7L2 7amp2 82 2 8 L2 7v2 7 7 W 2m 2m 872 7quot 87quot 2mr2 6 If the potential is spherically symmetric the Hamiltonian commutes with all the components of L since they only involve angular derivatives and each component of L commutes with L2 Thus we may write the solution an eigenstate of L and L2 denoted by i and m W1quot mummy 7 Schrodinger s equation may then be written as 52 32 2 8 v 2 EMMY w 5 Wm 5 il 1 emquot 8 Our immediate goal is to understand the angular functions ngw 139 which are eigenstates of L and L2 and are refered to spherical harmonics In terms of bras and kets Y mga 7n 9 which implies the normalization dadcosmmw a 1 10 Given the requirement that the raising operator see de nition above working on Y gives zero one can write the expression for Yaw 1 yidga 058i Sing 97 1 Lecture 34 March 22 2001 Cooper Pairs Pairing of electrons is responsible for the phenomena known superconductivity The connection between forming pairs and reducing the conductivity to zero is actually rather tenuous The formed pairs which have net negligible momentum can coalesce and move a coherent unit through the material One could imagine resistance coming from several sources 1 As we will see a gap energy will be associated with the pairs In order for the moving pairs to deexcite one must excite the pairs by at least the gap energy If one imagines a current moving through a circular loop at a temperature near zero there is not enough thermal energy available to break a pair D One could imagine a drag force acting on the condensate due to moving the condensate through the liquid of noncondensate particles The drag force would then behave deg 0 712 P x U3 1 where u is the condensate velocity Since the power goes U3 and the resistance is de ned P 12R 12 0 U2 2 one sees that since the power from drag contributes at a higher power of u drag can be neglected with respect to it s contribution to the resistance if u is small Since the coalesced pairs move a unit they can carry a large current with a very small velocity Our goal in this section will not be to understand the correlated structure of the pairs but only to show that pairs exist a lowerenergy solution than a Fermi gas In order for electrons to pair there must exist an attractive interaction between two electrons with momenta near the Fermi surface The idea of two electrons interacting attractively is most peculiar since electrons have the same charge The excuse for assuming an attractive interaction is that the electrons interact with one another via the lattice An electron might interact with an ion which might then interact with the other electron Intuitively one might expect such an interaction to provide an effective polarizability that reduces the electric repulsion but does not reverse it However electrons are not static in a medium and the movement of the electrons seemingly miraculously results in an effectively attractive interaction For our purposes we will follow the work of Bardeen Cooper and Schraefer BCS and assume a simpli ed interaction between electrons U I I 5k1k2k lk 2 kf lt k1 k2 k1 k2 lt kw otherwise k1k2EVEk 1k 2 3 where v0 sets the scale of the interaction V is the volume and k is close to kf Thus the model assumes the attractive interaction is con ned to particles within the neighborhood of the Fermi surface Lecture 34 March 22 2001 Since total momentum is a good quantum number eigenstates can have the form K paired Z akIK k l K2 k kt K2 k 4 k where the primed sum is over all relative momenta k E k l 2 such that both k l and kg are inside the region between 19 and 1 Solving for the eigenstates k1 k2 H K paired k1 k2 E K paired 00 I 7 k39 akIK kl k2lkK C v Summing both sides over k I 00 I 1 I akK 7 ak39 0 g lE k1 k2 Now one can see how the simplifying assumption that the matrix element UM was indepen dent of k and 19 in the subspace simpli es the problem It allows the sum over amplitudes to be canceled from both sides of the equation above and result in 1 1 I 1 f 7 f 8 14 l k kl 6k 1 I 1 i i 9 VgE k ok 1 I 1 V E hZKZle thZm where the last step involved writing the energy as a center of mass energy plus an energy of relative motion The solutions to the above equation can be found graphically To illustrate the solutions we consider the function 1 1 ltIgt v E 27 K VE I39Lz1K2m I zzlcz4mg 11 and graph the function to see for what energies E the function ltIgtE equals 1v0 Lecture 34 March 22 2001 Every time E passes an energy h2K24m h2k2m I changes from 00 to 00 Thus for every value of k there exists a solution with E lt h2K24m h2k2m where I 1v0 The vertical dashed lines in the gure represent the energies h2K24m h2k2m which would be the eigenenergies if 120 were to be zero The intersections of the blue curves with the horizontal dashed line represent the solutions If K 0 the rst value of k that enters the primed sum is k kf Thus there exists a solution with energy E lt 26 even though the solution formed from momentum states above kf Thus the paired is energetically favorable compared to being a momentum state at the Fermi surface Note that 220 is increased the solution for the energy becomes lower The Gap The expression for I in Eq 11 can be integrated analytically for the case with K 0 if one approximates the sum over states Z gt Mex d k 12 k f This amounts to assuming the density of states pkk is constant in the region of integration The expression for I then becomes u 1 MB pm fq mm 13 p2 f E26a EE r 72 log 26 14 where k 7R f y 2 it 1 pm 2W lt 0 Solving for E 2 a 26f82p ml E g20 Uol 1 Lecture 34 March 22 2001 The difference between this energy and 26f is known the gap energy A 26 E A E 44 m 2 a 6f 82pltfvol 139 18 One can do a similar calculation for K 95 0 but the binding energy would be smaller since the interaction links fewer pairs both electrons in the pair with momenta ki K 2 J k must lie within the shell in momentum space One can see that the energy required to break the pair is 2A because in order to break the pair one must move the electrons outside the Fermi sea that is increase their energy beyond 26f Wave Functions The relative wave function can be found by Fourier transforming the wave function in coor dinate space For K 0 the relative wave functions seen from Eq 6 in momentum and coordinate space are 1 K 0 k 39 K 0 7 lt 72gt m gt E 26k 19 17kt E 39 N likig ltmgt d E 26k 20 47r 39ka sinkr 7 k k7 r M d E 26k 21 This integral results in rather obtuse combinations of cosineintegral and sineintegral func tions One can expand the result and see that for large 7 cos k r cos k r v 22 N T2 As a homework problem you are to show that this is squareintegrable but that T2 0 Since T2 2 gt0 one should use a different criteria for finding the effective size One such criteria is to find the square of the wave function at zero if M71 2 W 0V2 23 One can define an effective radius that corresponds to a sphere that would give the equivalent density at T 0 1 E 311 r 0 i2 24 47ng 3 The resulting effective radius is a complicated function of the reduced mass the Fermi energy a and the gap A For larger gaps one finds smaller radii Lecture 17 February 7 2001 TimeDependent Interactions Many problems in quantum problems involve timedependent interactions Obvious exam ples include spin magnetic resonance problems where the interaction explicitly varies in time A less obvious example is a scattering problem where the incoming wave packet slowly enters the region where it feels the potential then leaves This problem is treated by considering asymptotic momentum states with a potential that slowly turns off and on with time rather than with a xed potential with wave packets Thus nearly all perturbative scattering eg Feynmann diagrams treatments are application of timedependent perturbatin theory even though the potential is not actually varying with time The TwoState Problem an Exactly Solvable Model Consider two states 1 and Q interacting through the potential H0 hw1 11 hw2 22 tag saga W YCOSW ox Y sinwt0y EV1 Y m This is the form of an interaction with timedependent magnetic eld or written a matrix H H0 1150 805 Bg 13m coswt g Simtam One can make the substitution to write the new Hamiltonian Hw122 75m ya t w122 where tam E E1 E22 Writing the evolution in the interaction picture for the components and 1amp2 8 w r Y v am z 2mm wig trma 6 w xv z 212313205 zgc tgw1 Now by making the substituion E mtQvl elm2w Lecture 17 February 7 2001 one can derive the evolutions 6 w m w gym Limmgwm an 8 w w 1v a WZLflgMWv no The problem now looks like a Hamiltonian without a time dependent interaction HI RCA12 w 2 I my 14 This is the same problem as we worked out for neutrino oscillations The evolution operator becomes 8 33quot cos 2i In sin 2t W12 w Q 4 a 0 cos9azsin90y 2 tan9 7 M s12 If the state begins polarized in the z direction the maximum probability of becoming a spin down state is 2 2 v I V 7l maxElEc HtETE2 15 gtquotWWWMWM Note the resonant condition can 2 w Also note that 7 plays the role of the half width in the Lorentzian The Lorentzian is a common in time dependent systems The resonant form clearly displays that driving the system at the resonant frequency w tan results in the greatest chance for ipping the spin NMR works on very similar principles only in this case the time dependent eld usually oscillates only in one plane ie t Bg By coswt 16 This is a bit harder to work out as compared to our example but the resonant conditions remain the same The Interaction Representation Picture Before one can begin time dependent perturbation theory we need to consider the interaction representation which is an alternative to the Schrodinger and Heisenberg representations Summarizing the representations W term 2W 0m 17 wi 05 Heisenberg states are xed in time 18 w eiH wtgts 19 2 Lecture 17 February 7 2001 where the Hamiltonian is divided into an understood part Hg and a perturbation V Note that if V 0 that the interaction states become equal to the Heisenberg states Since matrix elements ABC must not depend on the representation operators in the Heisenberg and Interaction representations are de ned as Ana intEASE hn 20 At EiIIot ASE lgt 21 mtn where we have assumed that H is not time dependents otherwise 6 would be replaced with an evolution operator Solving from the evolution of a state in the interaction representation r M0 Huizl tgtz 6 quot quot 39Hu Vliwths 2 dt mayW 23 The subscript I is omitted in most literature and the knowledge of which representation is being used is left to the astute reader If the potential has an explicit time dependence the explicit time dependence must be absorbed into V t V50 Z mntimgtltnig 24 mm then V becomes Vat e lt quotquot ngt 5mntimgtltns 25 where the eigenstates of Hg are fawn TimeDependent Perturbation Theory Consider the evolution operator in the interaction picture de ned by W UIUgtulitl Uul 26 Using the equations of motion 1 may to WWI7U in 27 with the boundary conditions 17t t9 t9 One can rewrite the differential equation a differential equation 39 t Utt 1 dt vt Ut t 29 to i t i 2 t t39 1 7 may 7 dt dt V1t Vgt 30 7a to h to to Tl t t39 re1 dt dt X dtlnxIt xIt xItn 7a to to to 3 Lecture 17 February 7 2001 This is known the Dyson series named after Freeman J Dyson Keeping only the rst term in the expansion in V amounts to rstorder perturbation theory Also note the the operator U is related to the Schrodinger evolution operator by US E iHotDrIEiHot 32 which means that transition probabilities which go the square of matrix elements Emil30 are the same in both representations since the states are eigenstates of Hg To second order perturbation theory the matrix element becomes NEUUJOEO 7 dt ltn Vst i2gte ltEquot E 33 to 2 t t o o dt dt n V5t Emmil5U ine EquotEquot39 e wm Eie il h m t0 t0 Higher orders are simple to write down it just takes space Note that the this expression accomplished by noting that ltn wlttgt mgt ltnvSlttgtmgte ltEquotEmgt u lt35 Fermi s Golden Rule We will now derive one of the most important expressions derived in the class Fermi s Golden Rule for the transion rate from the state to the state We will derive the same expression twice to demonstrate how robust the derivation is First we consider the case where we turn on the perturbation slowly nilgaff antiquotm 7 gt 0 36 where the times being considered are less than or near zero meaning the interaction turned on slowly from t 00 To rst order I t l l niUAa ltgt0iigt Vm x riff EVE quotW 37 Vm F F trnt 1 an 3911 I En E mn d Thus the probability goes as EV P t PM 39 2 n En EiV flingf In obtain Fermi s Golden Rule one does not calculate iltn U i but instead calculates ddt n U i 2 to obtain transition rates rather than transition probabilities 1 2mm 7P quott 40 it quot En E37fn1 l Lecture 12 February 7 2001 Separating out relative and centerofmass coordinates Consider the Schrodinger equation for two particles interacting through a potential lrg r1 The kinetic energy term a a H 7 2 7 2 1 K 27an1 27ngv2 needs to be rewritten such that the derivates are with respect to centerofmass and relative coordinates 7R1 1391 TRQI39Q E R 2 r r2 r1 m1mQ Using these de nitions one can show that that the kinetic energy becomes 7amp2 7amp2 H V 7 2 7 2 K QMVR 2 r 3 where M E m 7732 and u E 7n17R27R1 mg The wave function may then be written as a product of centerofmass and relative coordi nates 1Rr E K39Rgrelr with the overall energy being a sum of the eigenenergy of the relative wave function plus 52K22M Note that if one of the masses is much larger than the other that the reduced mass 1 equals the smaller of the two masses If both masses are equal the reduced mass is half the mass of either of the two individual masses For our purposes we will solve problems such as the hydrogen atom assuming the potential is xed For the real case one need only replace the mass with the reduced mass to include the effect that the source of the potential is itself mobile Separating the center of mass coordinates is convenient whenever one has a potential that is a function of r1 r2 only Writing Schrodinger s equation 72 72 1sz neRan wageRan Ee K39R gtr a One can factor out c K39R after operating with Viit and get a simple equation for the relative wave function cfr 72 v3 gtltrgtvltrgt ltrgt Evian 6 52K H 2M TE E m Example Threedimensional Harmonic Oscillator Cartesian Basis This factorization works whenever the potential is a function of r and not B One especially easy example is the threedimensional harmonic oscillator In this case the problem is even Lecture 12 February 7 2001 more factorizable 1 1 1 Vrrg 5km r22 ETYLUJQTQ 5k 112 y z 8 Since this potential can be written the sum of an IIJdependent a ydependent and a zdependent piece 90 IIIWNWML 9 one can separate the Schrodinger equation for the threedimensional relative wave function into three onedimensional Schrodinger equations 7a 2 1 2 128 114 Q E 39 11 2 y 2 11 Q l W91 l h8239 1k E39 12 2M gask2 z QM any Here E Er Ey E By multiplying the rst equation by g yg39as the second equation by with and the third by 91 then adding the three equations one nds that they provide a solution to the threedimensional equation Thus one is able to take a sixdimensional equation factor out the centerofmass motion and recognize the factorizability and reduce the problem to three trivial onedimensional equations of motion The energies are Er 711 127zw Ey my 127zw E 71 127zw 13 where a ku The total energy is thus the sum E 71r my 71 14 These solutions are known the Cartesianbasis solutions to the harmonic oscillator Since the harmonic oscillator has a spherically symmetric solution there also exist eigenstates of good angular momentum which can be expressed in terms of Yams One convenient aspect of harmonic oscillator potentials is that they can be solved form the N body case where N particles of m interact through mutual harmonic oscillator potentials Consider the potential 1 lr14 A ArN 5k ri r1 15 iltj This potential is independent of the centerofmass coordinates That is a translation of all coordinates does not affect the potential One can therefore write the solution a product of a centerofmass wave function and a wave function that depends on N 1 relative coordinates Lecture 12 February 7 2001 The above problem would be intractable if not for a trick which is unique to the harmonic oscillator One adds a ctitious potential that depends only on the center of coordinate V gk r1 r my N2kchmg2 16 The energies are thus equal to Em Ere n 32 WNkm 17 where Ere is the energy of relative motion which is our ultimate goal Here the term n 3231m replaces the usual kinetic energy of the centerofmass The trick comes in noticing that the sum of the original potential and the ctious potential turns out to be the same that of N independent oscillators 1 Vw V V 51 r r3 18 Thus the energies are Em m 12 quotIAN 3N2h1Nkm 19 The ground state energy we are interested in is thus Emmi 0 minus the energy of the center of mass 3 2h Nkm The wave function of the ground state can be written the product of all the ground state wave functions divided by the wave function of the centerofmass Solving the Radial Wave Equation for Spherically Symmetric Cases Similar factorization ideas are applied any time one deals with a spherically symmetric potential In that case one can write the wave function a product of gr and mw 05 and reduce the problem to a onedimensional problem of the radial coordinate In that case the onedimensional Schrodinger equation for gr becomes 712 2 2 WM 1 E 87 7 WV WT WU EWU 20 Although this is a onedimensional differential equation it is not a onedimensional Schrodinger equation due to the extra derivative term 2 r8 The Schrodinger form 0r can be regained by de ning E r r 21 The wave equation for 71 looks like a 1d Schrodinger equation with a centri gul potential 7amp2 82 1325 1 r lr Eur 22 The boundary condition on u is that it must go to zero at the origin so that d is nite at the origin Lecture 12 February 7 2001 The wave equation is particularly simple for s waves 5 U such solutions reduce to simple 1d problems with an in nite potential when 7quot lt 0 Considering nonzero 5 introduces a divergent potential at the origin In the neighborhood of r 0 the solution looks like either umquot 7i 1 07quot 07 2 A A the regular solution 23 or wr 7 1 Oquot 07 2 A A the irregular solution 24 Clearly only the regular solution satis es the boundary conditions Even when a Coulomb potential is added which is also divergent at the origin only less so the behavior at the origin can be expanded shown If the potential is more strongly divergent then there is a problem When the potential is zero the solutions are known spherical bessel functions The solution that behaves regularly at the origin is referred to jgkr while the irregular solution is referred to TlgUCT39 For the low is 4 sin 3 cos a v W T g quot00 lt29 4 sin 3 cos 3 cos a sin a aw 32 a mm 26 12W 3 333 3 3 x73 COSQ 5723111 3 27 sinat 3coscc 712kr 28 By taking a linear combination of jg and 71g one can nd a solution which behaves like an outgoing wave at large 7quot hgkr jgkringkr 29 Umquot N 1 3 I m asr gtoo 30 As an example we consider 5 1 W m 1 3 31 Examples 1 Solve for lowest 5 0 and i 1 states of in nite spherical well of radius R Note that for the i 1 solutions a transcendental expression will remain 2 Outline how one would solve for the boundstate of an i 1 solution for a well of nite depth Lecture 12 February 7 2001 Spherical Harmonic Oscillator Spherical Basis The spherical harmonic oscillator can be approached either through cartesian coordinates described in above or in a spherical basis First we review the Cartesian solutions Since the potential can be written lr 116 116w g2 22 32 2 2 i the solutions factorize into L39 y and 2 dependent functions W1quot nyy m2 33 Each piece is a solution of the 1 dimensional Schrodinger s equation and the total energy is E N327zw N E 72fr ny 713 w ku The N 0 and N 1 eigenstates have the form gamnymoltrgt WM 35 an1nnorgt 36gt anon1norgt garW 37gt annonlr ze WM 38 39 By looking at the form of the Ygams one can see that the N 0 state has a i Cum 0 angular dependence and can therefore be written a spherically symmetric function 8 722 13 multiplied by Yoto The solution for the 71fr my 0 na 1 state can be written a product of Ym and the radial function rc T2i2 3gt By taking linear combinations g nl1tnynl0 1 g n gtnyltan one finds solutions which can be written the same radial wave function multiplied by Hi1 In spherical coordinates the labels 711 my and 713 are replaced by N i and m Mapping the solutions for higher N is a bit tricky For N 2 2 one can count the states from a Cartesian perspective One needs to know the number of to get three integers to add to N First the number of ways d to get two integers to add to N J is dJN Z NJ 1 40 n10N Requiring that a third integer adds to N gives the total degeneracy dN Y Y N 1 N 2 Z dJNJ 41 N0N Thus there is one to get N 0 3 to get N 1 six to get N 2 etc To determine which i multiplets combine to create the dN Iartesian solutions with a given energy N 3 2hw consider two pieces of evidence First creating the cartesian state with 5 Lecture 12 February 7 2001 n3 N nfr my 0 is represented by the spherical state YgN m therefore there must be at least one i N multiplet Secondly all the multiplets for N 2 must have corresponding multiplets which are generatedy by operating on those states with the spherically symmetric operator a We can now determine the multiplets by dUV 2N 1 dN 2 any others 42 Solving for the number of others one nds there are no others Hence the excitation N 2 states have the same states but with one more multiplet of i N 2 Since one knows that for N 0 there is one i 0 state and for N 1 there is one i 1 multiplet one can quickly nd all the multiplets for any N As an example the N 5 states are covered by one i 5 multiplet one i 3 multiplet and one i 1 multiplet Note that all states with even N have even parity and all states with odd N have odd parity The Hydrogen Atom There are three standard problems of spherically symmetric potentials where the solutions are analytic the ini nite well the harmonic oscillator and the Coulomb potential Here we consider the case where the potential is attractive V0quot i 43 One may rewrite the Schrodinger equation a 5 1 2122 872 T2 107quot qu k2wlr lt44 where the Bohr radius is de ned 10 E hymn2 and k2 2uEh2 For large T the potential and centri gal terms are negligible and the wave function must behave 23quotquot multiplied by terms that vary more slowly in 7quot The solutions to the Schrodinger equation can be written in terms of associated Laguerre plynomials 12 at 2 371 5 1 39 27quot g 27quot y R L 1 that L2 1 4 Mir 7 nag 2mm l3 P W10 quot1 nag a For a given 5 there are many solutions labeled by the integer n gt i The eigenenergies can be written simply 232 1 E 7 n 2 46 If larger charges are used the above expressions are modi ed by scaling 10 by 1Z1Z2 Writing a few solutions 2 R150 723 7 0390 Lecture 12 February 7 2001 1 I 1 7 2 i g quot 2 10 1220 MW a0 48 1 I 1 R 74 7 g quot 2 10 251 2 auxZ The degeneracy where energies with different i have the same energies seems accidental but is related to the similar degeneracy in the harmonic oscillator Both degeneracies can be explained by considering the Lenz vector which commutes with the Hamiltonian for the special case of the Coulomb potential 1 e2 REpXL pr 7r 50 This operator is Hermitian and if the commutation between p and L is ignored is identical to the classical expression for the Lenz vector i m K ER 51 which is a little odd since one is de ning the square root of an operator However since we are considering only eigenstates with negative energies this is not too sick One can also de ne a scaled operator One can show that the components of K and the components of the angular momentum L obey simple commutation relations Kb ih ijkLk Kn ih iijk These commutation relations are reminiscent of angular momentum commutation relations and in fact if one de nes two new operators L K E Z 53 L K N E 7 54 2 one can see that M and N obey the same commutation relations L ML Mj iheijk lk A2 NJ iheijka ML NJ 0 55 Since M and N are linear combination of the L and R they commute with the Hamiltonian and since they commute with one another they may simultaneously be de ned Furthermore some algebra reveals that the Hamiltonian may be written as me me H 396 2K L2 h 22M2 2N h a 1 Lecture 5 February 7 2001 1 Wave Packets A plane wave can be written l 1 I Well y 1 which is normalized in the volume V but is not really an eigenstate of the momentum operator due to the sharp cutoff at the boundaries A more physical description is a wave packet which is described by not just a momentum it represents a spread of momenta described by a function g dp ix 1 w f gltp k lt2 For our purposes we will consider 9 to be of Gaussian shape p k k 929 aoxp 4A 3 As an exercise one should check that this resulting wave packet is properly normalized by a mlRE 4 By inspection one sees that the width of the wave packet in momentum space is A To understand the spatial shape of the packet one can see that p p 1 2A 2A2 1 2 2 z 7 i 39 WU m h 0X1 7Z2 2 2 0 which is also of Gaussian form with the spread being hQA Thus the product of the spread in momentum space multiplied by the spread in coordinate space is 15 2 If the packet had been described with x x0 instead of x the packet would be centered at 1 0 instead of the origin If we consider the wave packet at arbitrary times 129 mm H v r wxtie quot mquot k 6 A Qm 929 We expect the packet to move in time The packet has contributions from all momenta which make differential additions with a wide variety of phases At every time t there is a point 1 for which the phases are constant in the region near k At this point the differential contributions add inphase and the wave function in coordinate space is a maximum To nd that point we take a derivative of the phase factor with respect to p at k and require it to be Zero iEpt ipx p k 0 7 Lecture 5 February 7 2001 Given that dEdp v even for relativistic particles one sees that vt 8 which is not too surprising Re ecting a wave packet and time delays As an example where wave packets are involved we will consider a packet as described above incident on a potential barrier 0 gt 0 I gt in lt0 m We also assume that the packet is narrow and that V0 is larger than Ek The incoming wave packet can have the form 27m 2 F t m amp winmt Tdpe quot 1 E 1 90716 10 Such a packet should have a reflection with the reflected wave packet looking like 2m zfiout t T fdim lEpt p EETmpyP The negative sign is chosen so that in the limit of an infinite potential 61 0 The reflected packet has the same amplitude but with a phase factor that might be momentum dependent The factor of two in the phase is a convention which we will encounter again when we discuss scattering theory Looking for the point where the phase is stationary one finds the expression d quott 276 12 r L dp p Thus the packet has a lag in space or a delay in time which may be expressed as m 4151M At 2hi3E 13 H dp p dE This time delay is relative to the case where V0 00 The Uncertainty Principle When two operators do not commute one can not usually specify states which are eigenstates of both operators An example of such operators are momentum and position Proof Consider two operators AA and AB AAA W ngBB W w on Now we consider states and Lecture 37 April 2 2001 The Dirac Equation The Dirac is another example of a relativistic wave equation However the Dirac equation differs in that it describes spin 12 particles whereas the KleinGordon equation works for spinless particles Dirac motivated by nding a linear equation that consistend with the relativistic constraint H 2 P202 M204 By linear Dirac looking for and equation that linear in the derivatives In order to accomplish this Dirac needed to use a matrix equation H 01ch aych ach3 8mc2 1 where a and 8 are matrices Given that one needs to satisfy our knowledge of relativity H2 p202 M204 2 there are constraints on the matrices ahaj 261562 1 11 8 0 Twobytwo matrices are insufficient one can only come up with three anticommuting matrices the Pauli matrices but one needs four Threebythree matrices are also insuf ficient but fourbyfour matrices are sufficient The following fourbyfour matrices satisfy the equations 10 0 0 05 010 0 50 3 00 1 0 4 00 0 1 Actually there exist an infinite number of choices one can transform the matrices by a unitary tranformation U ZUl 5 and find a new choice of matrices The choice above is known the Dirac representation which is convenient for massive particles For highly relativistic particles it is sometimes convenient to employ the chiral representation 0 539 0 0 0 5 3 1 0 We will confine our discussions to the Dirac representation i QOQ QOOH GOP 0 CE V It should be stressed that the four components of the wave function are not a relativistic four vector They correspond to spinup spindown and the two corresponding antiparticle solutions Like any wave function the components of the wave functions mean nothing physically Only expectations have a physical meaning By changing the arbitrarily chosen representation the solutions are changed Conserved Quantities Lecture 37 April 2 2001 The particle current hmns mwwum nosw nwwmo U is conserved One can demonstrate the conservation of the current by taking the four diver gence 9739quot and by applying the Dirac equation 9 me mi 65 61 7 I 8 8 gt mc quot th ca 4 WT z7wl 9 to see that a ajod39v j 13 10 Now we consider conservation of angular momentum Unlike the Schrodinger equation the Dirac Hamiltonian does not commute with the orbital angular momentum Hf gtlt p 2725 X 1 11 One can also de ne a spin operator 3 iii2 where i 2139 eijkaiaj 12 gt c 0 2 0 13gt The spin is also not conserved lip Basil Zflajagil 14 7 i7lpx Z 4 Jeikllaiaako ll 10 jkl ihpj 2 Z eijkakeikjak jk 2 2 iRqukajpk jk Thus the combination j FX 13 is conserved lE U as Solutions for free particles Analogoust to the KleinGordon equation the Dirac equation will have both positiveenergy and negativeenergy solutions The positiveenergy solutions with a given momentum 15 are 2 Lecture 37 April 2 2001 referred to 715 while the negativeenergy solutions are referred to The label 8 refers to the spin Ema x3m02u 19gt Epvltm row p mew m 20 The momentum labels on the negativeenergy solutions were labeled with the opposite mo menta since they would correspond to the destruction of antiparticles with momentum rather than the creation of particles of momentum 5 i To nd the solutions rst nd the solutions for p 0 The solutions are then 1 wp0 3 21 U U wp0 5 22 U U Mp0gt E 23 U U viltpogt 3 24 1 Note that the solutions v are labeled with a spin index opposite to the eigenvalue of D Again this is because the solution will correspond to the destruction of an antiparticle It is to check that these solutions are eigenstates of the Dirac equation with eigenvalues zme2 Finding solutions for nonzero momentum can be accomplished by multiplying the zero momentum solutions by the operator me2f 62 A e 3 E This results in a solution to the Dirac equation because E i A e 8me2 E 07 A e 8me2 0 25 Thus we de ne the solution to be E li Cl 8me2 u p u p 0 26 54 2me2me2 E A which will also be a solution to the Dirac equation E i A e 8me2u 0 27 3 Lecture 37 April 2 2001 The square root in the denominator chosen to result in the normalization Plus27 Epm 28 The choice of normalization is motivated by the fact that ulu turns out to be related to the zeroth component of the current which must transform like the zeroth part of a four vector E Cm As an example we nd where is along the z axis For instance E m0 0 p20 0 1 1 0 E m0 0 p30 0 u pz 29 A 2m02Ep mg p30 0 Ep MC 0 0 0 p30 0 E1 m0 0 E m0 1 0 30 2mc2 Ep m0 1750 Thus one can see that at nonzero momentum the positiveenergy solutions have a mixture of upper top two and lower bottom two components Nonrelativistic limit and the 9 factor Interaction with the electromagnetic eld can be accommodated by replacing I with EliC and maat with maat 61 where I is the electric potential Using the Dirac representation one can write the four components in terms of two two component vectors 0 and X where d and X refer to the upperlower components 31 One can now write the Dirac equation two equations for d and X s E E 2 mat Civ CA 0Xcltlgtmc 3 32 rag aV EA ltIgt 2 33 l at c i C 00 23 m0 X In the nonrelativistic limit the lower components will be small and one can make the replacement ith mc2X which results in a simple substition for X using Eq 2 7a 23 A 2mc X m 0 3V EA mp 30 4 Lecture 37 April 2 2001 Substituting this expression into Eq 32 2 8 1 7L 8 837827 iV iA c39r39 d 8 mczdx 36 8t39 2mz c gt i39 If the gradient and vector potential operator commuted with one another one would use the anticommutation relations to obtain the usual kinetic energy piece in the Schrodinger equation from squaring the p eAc term and employing the anticommutation relation However taking account of such terms yields an extra piece 8 1 h e 2 2 h I I H mad d 22 mo d goio aAg 4483lt 3t 3 V SA 28k m M t i Ea mm 2m 1 c 39 1 39 2mc 39 Since hi 2 25 the last term explains why the 9 factor of the electron is 2 Previously this had been inserted by hand However the Dirac equation requires that the 9 factor be exactly two Note that the 9 factor of the proton and neutron are not equal to two because they are composite particles The only difference between the expression above and the usual Schrodinger equation is in the additional term meg However this merely adds a constant to the energy as long is conserved and does not affect any observable in the nonrelativistic limit The spinorbit interaction One can make a rigorous expansion of Eqs 32 and by iterating Eq 33 1 h 22 1 8 2 X 39O39tf ma mo 22 X 39 1 h 22 1 8 2 7 2mc V CA 0d 27RC2 m8t 7m 81 40 1 h 22 1 8 2 7 m ma mo 81 X 41 One can repeat the iterative substitution and nd an expression for X to arbitrary power in lm Once a satisfactory level for X has been found it may be substituted into Eq 32 to obtain a wave equation for d If one pursues the expansion one step beyond what done previously with the magnetic eld to nd the 9 factor of the electron one nds with a substantial amount of work that the approximate Hamiltonian for the upper components 1 has the extra terms If eh H 74 f 6 8m2 4m2620 Exp m The rst term is merely the nextorder expansion of E VpQCg M264 while the second term is the spinorbit term To show that the second term is the spinorbit term we note 0 Lecture 37 April 2 2001 that if X 43 for a radial electric eld E Finally it is noted that one may go through the same exercise Without any electromagnetic eld but instead with a position dependent mr One then nds that the spinorbit term looks identical except that E gt 8mrv 87quot This has the opposite contribution for an attractive interaction This is important for un derstanding nuclear physics Where the spinorbit interaction is surprisingly large It can be explained by an attractive scalar interaction like a position dependent and a repulsive vector interaction similar to a Hydrogen atom but with opposite sign The interactions cancel each other out to a large degree far the binding energies are concerned but the contributions from the spinorbit terms add together 44 The Dirac equation is one of the great triumphs of twentieth century physics Motivated by aeshtetic considerations several previously adhoc assumptions fall out naturally the 9 factor particleantiparticle symmetry the spinorbit coupling etc When combined with the relativistic coupling to the quantum electromagnetic eld incredibly accurate calculations can be made of g 2 using perturbation theory But this is the material for another course More notation iv matrices To more clearly demonstrate the covariant nature of the equations it is common to de ne the Dirac y matrices yo E 8 E 362 45 The three spacelike y matrices are antiHermitian While y is Hermitian The convenient covariant behavior of the y matrices comes from the property W V 29 46 Where 9quot is the metric tensor 1 U U U U 1 U 0 1w V n 9 0 0 1 0 4 U U U 1 This property means that the y matrices transform like four vectors A very common notation is 26 E 27 48 which means that the Dirac equation multiplied by Yo can be expressed 16 49 This has the attraction that 16 is a scalar 6 Lecture 9 February 7 2001 1 Propagators Green s Functions and Integral Equa tions A propagator is merely an incarnation of an evolution operator and is often confused with being the same a Green s function which it sometimes is A propagator is de ned H t t Kxtx t E x exp x 1 a If the eigenstates a of H are known the propagator may be written Kxt x t 26quot 39ltx agtltaix gt 2 0 Since K is the evolution operator one sees that determining the propagator is equivalent to solving the Schrodinger equation Mm faxzt wxzt 3 In fact the propagator is a solution of Schrodinger s equation for t gt t hgvg 39 z z x a z z 7 7 v i 7 t 2m xltxgt Altxtxtgt mmmxm gt U while being zero for t lt t and equal to 63x x when t tg We now consider the simple case of a free particle in one dimension In that case the eigenstates of the Hamiltonian are momentum states and I lt5 7 I yz z 7 ROCK t x t 27m 7 2m m at 2 6 V 2mm t exp 2m t l The integral in the last step performed by completing the square Note that the phase in the exponential looks 12mv2t7L where the velocity is given by ActAt A similar expression can easily be found in higher dimension To see that this form approaches Mat 3 t gt 0 one can check 1 that datKattat t 1 2 that at small time differences the phase oscillates quickly except when a at Adding a potential makes nding the propagator much more difficult One can see that if the propagator is expressed 1 39 Kc tgat t KQJ t at t dt dat Kgat t 3 WW 56 t Kc t at t 7 1 Lecture 32 February 28 2001 Slater Determinants One to write antisymmetric wave functions is through a determinant Imagine three identical particles with states described by the wave functions at 0543 and 05430 The total wave function 123 can be written a gt w 13T13T g 0532 3 2 c5 i 3 4553 ZC E c33 By expanding the determinant one can see that the overall wave function is antisymmetric Wan crusts ax1 bx2 cxs ax1 cx2 bxs bx1 ax2 cxs bx1 cx2 axs cx1 ax2 bxs cx1 bx2 axs 2 Angular and spin wave functions The overall wave function for a manyelectron state described by L S and J can be written the sum over products of angular and spin wave functions L S J 17 Z L S J AMLQML A15 L AMLXS A15 JULst The angularspin wave functions EL AMLXS A15 must rst be written in terms of products of singular particle wave functions summed over with the help of ClebschGordan technology As an example we consider a twoelectron state with orbital spin wave functions in a shell of angular momentum i WL S J A39HJQng7n57n lt9Q m5m 5 LSJM Z ltLSJMJMLMSQQ LMLltm5mggsMS 5 MLMS Z LSJMJ MLMS 6 MLMS 39 Z MLMamam2gtY2m9Y2m9 Z 1212SMgWL mg 7n5 mgmgimg Z LSJAIJ AILi7l15gtltLAILETRg nD AWILAWImm 1212Sng m5mQHMAQDQMAQ39 The expression would be even more complicated if there were three electrons in the shell Example Constructing the 30 state in Carbon Lecture 32 February 28 2001 The two electrons in the p shell of Carbon can be in any number of con gurations Using the expression above PL0S0J0MJ0Qg 9 quot359 mg 8 Z ltLSJ1MJ 17UL17USltZ 1Z 1L OHM Unnam mum vlt1212S 01115 2 11mg TRISH21 QHQQWEUI Z Z 21 1L 2 COMM U mgm Yg1ml 921m29 9 mum 1 0m5l20mg l2 0m5 l20mgl2 mmmplm Ywm 319 mama39s 10 390m5120m 12 0m5 l20mgl2 Note that the wave function is symmetric with respect to interchange of Q with Q and antisymmetric with respect to interchange of mg and mg making the overall wave function antisymmetric Permutation symmetry When coupling two particles together the permutation symmetry goes 1quot for the spatial part while the interchange of spins is symmetricantisymmetric for S 10 For this reason one can notice that for two electrons in the 2p shell carbon the orbital states must be L 0 or L 2 if the the spin state is S 0 and must be L 1 if S 1 Thus the possible states in carbon are 30 Dg 3P0 3P1 and 3P2 loupling a higher number of particles together can lead to mixed symmetries In this case the angular wave functions may be neither symmetric or antisymmetric while the spin wave functions might be mixed well However the overall wave function needs to be antisymmetric For example the flavor and spin wave functions of the three quarks that constitute a proton or neutron are in states of mixed symmetry This will be discussed in detail later Zeeman Effect A particle in a magnetic eld feels the interaction 88 Hmag L 233 11 27m where the magnetic eld is assumed to point in the z direction We wish to calculate the change in energy for an atom in a state of good J M J due to the interaction AEmm 2 ltL 3 Alma 2321 s J M 12 AIJ l LSJAIJ SZ LSLiMJ 13 Lecture 32 February 28 2001 The challenge in calculating the splitting comes from finding 33 We expect this to be proportional to AM it is the only label available The Wigner Eckart theorem and Clebsch Gordan technology come to the rescue L S JMJ Sz LS JMJ LSJ S LSJ 14 L S JMJJ3LSM1J L S J HLSJ39 This follows by applying the Wigner Eckart theorem to both 33 and J3 and noticing that the same Clebsch Gordan coefficients appear in both terms Thus if we can find the ratio of the reduced matrix elements we will have ful lled our mission To do this we first step aside to perform a proof Consider a vector operator A JlMgA A JngM Z JlMgAmtJm JlM 15 71 Sm gl Z JlMgAmrJillquotJAIquot Jm JAI 16 ilgm gl7JSIlI SJ Note that the inserted states were only those within the same J M multiplet This is valid since Jm does not mix different multplets Now by applying the Wigner Eckart theorem one can write the matrix element A A J JMEA JEJMgt ngMXJEEAEEJXJEEJEEJ 17 f0 M Z 18 ilgm gl7JSIlIquotSJ 1 A l l l Hquot 1 1JJ Mm M 1JJ M m M 2J1 9 Since I is determined solely by ClebschGordan coefficients one can see that JMgAAJgJM J J 20 JMgJ JgJM JIIJggJ Now that our proof is finished we can see that L SJMJ JAS L S JMJ 21 LSJMJ JAJgLSLMJ JJ1SS1 LL1 22 2JJ 1 where we have taken advantage of the fact that 1 J S JJ1 SS1 LL1 23 Finally we are able to insert our result for the ratio of the reduced matrix elements into our expression for the splitting to get AB g hBWJ 24 27m J 1 SS l LL 1 g 1Jltltgt ltgt 5 2JJ 1 Lecture 7 February 7 2001 1 Interactions of a Charged Particle with the Electro magnetic Field Electric and magnetic elds are determined by the vector potential A and scalar potential Q 8A The Hamiltonian for a charge 23 For a negative charge 23 lt 0 is written in terms of the vector and scalar potentials 1 E V E BVgtltA 1 1 EA2 Hquot RP d l i 2 AA 2A2 21 3 27m p 0p p c Q Solving for the equations of motion 5 dx hr H d7 m a gmm wmm wM m wmm wmm 352W EAjCXM EAjC 11239 EAjCl z lpj EAjC Pi EAjCl mz m 12 61420 773 As an exercise one can solve for the force and nd Edij Bxdj 8 d2 7 7V E dtz dt dt Note that since dxdi includes a derivative that the two similar terms in the brackets above can be different As shown in a previous homework assignment the density 0x i X7 il x7 i 9 and current density ih 23 2 39 57 J J J Apt 1 Jmogman2wmwmy m satisfy the continuity equation a i V 39 0 11 at 7 Lecture 7 February 7 2001 Also from the previous homework remember that gauge transformations 1 A t A gtAVr xt gym 7L 12 0 8t yield a new Hamiltonian whose solutions are identical to the old Hamiltonians with an extra phase added 7C8Ax t XJ gt0xp m l wx t lt13 Circular Motion in 3 Magnetic Field A constant magnetic eld in the z direction can be described with the vector potentials A 0 A0 0 A4 2 p13 14 which has a nice symmetry as the vector potential winds around the z axis or through a gauge transformation Ag 2 Bx Ax 11A 2 U 15 This appears to violate the symmetry but yields the identical magnetic eld The advantage is that it is easy to solve The wave function can be written in the form y z WWW 16 because both P9 and P2 commute with the Hamiltonian The differential equation for a then turns out to be 2872 l xv 2m 8102 L 2mc2 where E19 2 E Ra QM and 9 E Mayo813 18 But this is the harmonic oscillator Hamiltonian with w BB 19 7730 the usual expression for a classical particle The solutions thus look like a particle whose w position is centered about 909 which is deter mined by ky and whose position in the y and 2 directions are simply uniform Furthermore it seems odd that a particle with circular motion would be an eigenstate of P9 The solution to the paradox is that if P is constant that means mvy eBayc constant 20 Lecture 19 February 7 2001 Propagators Timeordered Evolution Operators A propagator also known as a Green s function is nothing more than the evolution operator with a theta function tacked on Gm Whom 1 W E MMnail 2 The Theta function is tacked on because of the timeorderings involved in the expression for the evolution operator in the interaction representation Rembering that the evolution operator in the interaction representation is Udtt0 6mmL Lo hC MU Lo L 1 dt1em L1Vem 1 4 Lo 7 2 L L1 1 7 dtle HOLIle m dtgiHot2Vequot 24 5 a 0 0 one can write the propagator Gt t0 e i gt 39Ut8t t0 C iHoL Lo 8t to AA 10 VV 8 jm dt WW Wag t1Vequot 1 gt et1 t0 oc 2 OC dtl C iHoL L1het t1 rei oL1 L2 etl t2vrei oL2 Lohet2 to l oc i i i mr a immHm tnmu m 7 2 06 3 dtldt2gt t1Vgt1 t2Vgt2 t0 73ltgtc 3LjMMbWmwmMWWWWW mw This expression has a nearly identical form to that for the evolution operator the only appar ent bene t being the incorporation of the 8 functions which allows one to more efficiently express the relations The real bene ts to this formalism comes when one considers the Fourier transform of G The Fourier Transform of the Propagator G Let us rst consider the Fourier transform of g the propagator without V Assuming the states m and mgt are eigenstates of H0 gmvxw dtemtgnmt 6mmdtG w iEnxLVn met 11 i6 7 nm a n gt 12 w EmfzHin39 1 9 Lecture 19 February 7 2001 The in nitesimal 7 gt 0 represents an extremely slow exponential decay at large time differences By using Cauchy s theorem one can integrate awful270 over all to to see that the propagator gt is recovered In fact if one ipped the sign of 7 to 0 the result would look the same except with 6 t The pole in tells us at what energy the particle propagates In this case one sees that the pole where the propagator blows up is at w Enfz in Note that if 7 were replaced with a nite value FQh that the square of the propagator would go 8 H thus Ffz would be associated with the exponential decay constant The simple case Where V is diagonal If V and g are both diagonal in the same basis one can forego the matrix notation and consider the states oneatatime One can then write the expression above for G in a recursive form 1 7 Gmm 9mm l fgmm WV1n1nG1n1n 1 l Zhgmm WMmm i 15 w meh i Thus the only difference between the full propagator C3 and the original propagator g is that the energy of the pole is shifted by V exactly as one would expect from knowing the eigenvalues of H H0 V The general case Where V has offdiagonal elements One can not do the same trick when V has offdiagonal elements However by inspection of the expansion of G one sees that one could accomplish the same simpli cation by replacing V with the diagonal matrix T de ned by 2 v 1 7 7 1 7 7 7 7km 5 Wm Xlkryz rW rm lquot39kzquotyz z llrg ga a l b m 16 Here the primed sums refer to sums over all states except m Thus T absorbs all the offdiagonal terms in the expansion of G i Gmm l i 19 w 17 To second order in the potential 72m can be separated into real and imaginary parts 7 1 72mm lrni ml39m 20 V l gt v v 7rz6hw Exmvm 21 2 Lecture 19 February 7 2001 One can now view the expression for mmw for w Em to see how the propagator is affected to second order y if V7 717 Em M 2 Em 3quot 2 7 Z Lni l 1n6E1ll E37 Here we have used L 7mm EM 7 a 0 25 w EhFi w Eh b 9 Thus a pole of the propagator is adjusted by the interaction in such a way that the real part of the pole moves by an amount consistent with stationary state perturbation theory while the imaginary part is consistent with Fermi7s golden rule Resonant Scattering Here we consider the problem where an initial state C scatters to a nal state 16 through a resonant channel R For simplicity we consider the matrix element of the momentum state C with the resonant state R to be 0416 REVEk 26 lt i i gt W t gt We will assume that one has performed all the necessary integrations to nd the matrix element and that we call it X aside from the 1 VV due to the normalization of the plane wave To simplify our discussion we will assume that a has no dependence with respect to the direction of k One could calculate the rate of decay of the resonance using Fermi7s golden rule and obtain 27r30 2 H FR XV AMER 21 k2 y y WE 28 where pkER is the density of states of the outgoing particle An example of resonant scattering could be a photon scattering off an atom Here k labels the momentum of the photon while R would refer to a speci c excited state of the atom that could be attained due to the interaction with the photon To simplify our derivation we will assume that a has no dependence with respect to k It was shown in the previous lecture that the cross section for scattering could be written to second order 27rV i 0 Z i77e39k 20Ek Eh 29 k Uh 3 Lecture 19 February 7 2001 mi 30 Elmk i z V i V 7 6 7 kmw Em Iin 31 One could have extended the treatment for scattering to find that by writing the propagator of the intermediate state in the form derived in the previous section the answer could have been written I Z E39k lk39k Z lk39mem39lw m39b l mm39 juggk By resonant scattering we mean that the only interaction is between the resonant state R and the momentum states The matrix element then becomes 33 1 7 f C to f 7Tc39k a lk39li 11ml lkawEk Z 1 r 1 CHEW w Elifa iFRQ39 3 where F is the decay rate of the resonance The cross section is no longer infinite when E 2 ER due to the factor F R appearing in the denominator Plugging the expression for T into the expression for the cross section one obtains 27rV 3043 1 l l E 1 E 3 0 m V2 Ec ER2222F45l k k l 0 27rVEaE4 1 I l L E 36 W W Ec E12a2r4p l l k2 4 1 37 7041 WW Ec ER2 him4 39 2 2 MM 38 E E E2 MR22 The last line is known the BreitWigner form for scattering through a resonance Note that the cross section is determined by two numbers the width of the resonance and the Energy of the resonance If the resonance has spin SR then the effect is multiplied by the number of degenerate states through which one might scatter If the incoming particles have spins 91 and 92 the effect is correspondingly reduced due to the fact that many of the states of the resonance would not be reached with particular combinations of 91 and 92 The BreitVVigner form for resonant scattering is then 0 231 1 47F 2311232 1 k2 E E2 MR22 39 Lecture 10 February 7 2001 1 Angular Momentum BakerCampbellHausdorff Lemma Before we go further let us step back and derive the Baker lampbellHausdor 39 relation 8AB EAEBgiCa for the case the commuting A and B gives an operator C that commutes with both A and B eg the unit matrix C E A B 2 To see that this is true consider the expansion 8AB Z N H 23 7 3 Expanding A Bquot gives all terms with all orderings of the 77 operators For example for n 5 one of the terms is ABBAB We wish to move all the A operators to the left which requires commuting them past the B operators Everytime an A operator moves past a B operator one must add a term where the BA pair is replaced by C BA Using the binomial theorem one can then write ABquot AW 77 AHBH n 7 4 AHBH n 7 Z 7 7 7 77 HFquot 77 iljl C n iljl 1W C n M 20 AHBH w C n Vellijw n nj where M is the average number of to pick 5 AB pairs from an order n term under the constraint that the B operators in the pair were initially to the right of the A terms This number is simply the number of such independent pairs times 12 to account for the fact that only half the time does a given pair start off with the BA ordering 7 1 ii 1wi EHjj le EH y M lt gt lt 4 gt lt gt a 2 l One can then factor the exponentials in the expression above to get Eq 7 The rotation group Without going into group theory we consider the rotation group which consists of unitary operators 12w gem 1 6 To understand why this a rotation consider the case where i is along the z axis One can then see that d 8 8 39 L 39 ref 7 47 3 zhxay mg l Lecture 10 February 7 2001 which means that V I n a 2 82 n EngaTvHQ 1 39 39 8 f a 9 What de nes the rotation group is that the effect of subsequent rotations of i and 3quot result in a single rotation i Pmi 6PM a Pmi y 10 Since the different components of E do not commute it is nontrivial to nd the equivalent single rotation i given i and 3 Given the de nition E F X 1 it is straightforward to nd the commutation relations L5 Our goal in this section is to discuss the requirement of using different operators Sm 3y and S3 to generate rotations not in coordinate space but in a discrete vector space meaning that can be expressed matrices The important requirement for S to be considered a rotation is that u u u Ems ugsns 5 Ems 7 12 whqre the same results from a given i and 3 would have resulted from using E instead of S We wish to demonstrate that if the components of 339 obey the same commutation of the components of L the rotations will be identical To see this we divide the two rotations into N smaller rotations with N gt 00 Pins fths a ms gN 8 Ems 6N Ems 6N8239TLS 5N8239h8 aN Ems Nv 13 To nd the equivalent single rotation one must expand each exponential then commute them in a manner done for the BakerCampbellHausdor 39 lemma earlier There are of order N 2 such commutations For our purposes we wish to consider the inner two exponentials V V 732 2 732 msgm thaIV 4 r r 4 r quot r r 23 c 1LhS39a3ll WS39a3Wl339a3w30 4 q 4 Y J32 4 4 H 7 1 expz RS a 3 39LQNJS 13 51 19 Since there are of order N2 such commutations we must perform to nd i we may throw all terms of order 1 N 3 or higher One can then see that the nal rotation is determined by knowing the commutation relations i e if the commutation relations for the components for g are identical to the commutation relations for E that the equivalent angle 77 will be the same in both cases Lecture 10 February 7 2001 In group theory the rotation matrices expih are the group elements and the components of S are referred to generators of the group The simplest example of three operators Sm Sly and S which generate rotations is the 2 X 2 representation Sm E 2035 3 E 20 S E 20 16 Many Lagrangians in physics have rotational symmetry which would suggest that the an gular momentum is conserved This can be seen by commuting the rotation operator for an in nitesimal rotation with the Hamiltonian If LZ commutes with the Hamiltonian then the eigenstates of H can be simultaneously chosen eigenstates of L2 However if the Hamiltonian has terms such L 39 S neither L nor S commutes with the Hamiltonian However in that case the operator J E L S does commute We perform this an example in class D Matrices This is for the most point an exercise in notation Rather than expressing rotations as a function of am my and 04 one can express the rotation a function of the three Euler angles which represent subsequent rotations about the z y and the new z axes N iJza in8 iJn H Dmma 3 exp h exp h exp in 1 The label 5 refers to dimension of the matrices used to represent the rotations eg for two component matrices i 12 while m and m refer to the components of the matrices 3 m g i For instance for i 12 N 7312 exp 204 exp L39 exp L32 7 8711 le2 cos32 e il l ll2 sin32 ell lMV2 sin 82 ellmw2 cos82 39 More Angular Momentum The angular momentum operator LZ ih88 commutes with the Hamiltonian if the Hamiltonian is invariant to rotations about the z axis Further more if the Hamiltonian is invariant to rotations about any axis all three components of E commute with H One may then de ne states which are simultaneously eigenstates of H and L2 or LC or L2 But one may not necessarily nd states which are eigenstates of Lac Ly an d L since these operators do not commute with one another However the operator L2 Li L L is spherically symmetric and thus commutes with any of the three components of E One may there for de ne eigenstates of a spherically symmetric Hamiltonian that are also eigenstates of L2 and L2 We de ne the eigenvalues in terms of m and 5 L55 m mh m L2 m 5 1h2 m 18 3 Lecture 35 April 3 2001 Landau Levels and the Integral Quantum Hall Effect Consider particles of mass m and charge 6 moving in a two dimensional world the x y plane A magnetic eld is present B B The Hamiltonian for such particles is H p eAcgt2lt2mgt 1 If one chooses as a vector potential A 2 one obtains the desired magnetic eld Since there is no 3 dependence in the Hamiltonian solutions can be chosen as eigenstates of the operator Py with eigenvalue py One can then rewrite the Hamiltonian P3 pg wenc 7 H 2m 2m 3 Where py is simply a number and Pr is an operator Since py is a number one can consider the second term an offset harmonic oscillator potential P2 1 1702 H 7w 2L 4 2m27W T EB l EB so mg 9 Thus the Hamiltonian looks like a one dimensional Harmonic oscillator with a frequency equal to the Larmor frequency The harmonic oscillator is centered at a pyccB Note that the eigen energies 71 12fzw do not depend on py Thus there are many solutions with different py that have identical energies The energy levels are referred to as Landau levels Each level has a degeneracy equal to the number of values of py for which there exist solutions The density of such states is dN L 7y 6 dpy 27m where Ly is the size of the sample in the y direction The limits on py are determined by the a dimension Since py is related to the offset of the center of the Harmonic oscillator in the a direction pyc 0 lt FB lt Lx 7 Thus the number of such states is BL L 23 V V 27rhc 8 Assume the sample has a number of free electrons per unit area n If the number of free electrons exactly ts an integer number of Landau levels the levels will be exactly lled and the conductance will be relatively small Thus varying the magnetic eld one sees conductance minima for 27mm m 7 integers 9 23 Lecture 13 February 7 2001 Adding Angular Momentum In a spherically symmetric potential the orbital angular momentum L commutes with the Hamiltonian and i and 7m are good quantum numbers 1 and ma are the eigenvalues of L2 and La respectively Many particles also have intrinsic spin even those particles which are currently considered point particles such electrons and photons Thus in addition to the orbital quantum numbers two more quantum numbers may be used to describe the eigenstates of a single particle in the potential 8 and Mg which describe the magnitude and projection of the spin angular momentum Often a term exists in the Hamiltonian which couples the two types of spin 1750 aL A S 1 which is known the spinorbit term This term originates from relativistic considerations which we will see later in the course Since the term is written a rotational scalar and does not involve an externa eld which would explicitly break the rotational symmetry we expect that the overall angular momentum remains conserved Indeed one can see that the total anuglar momentum J E L S 2 commutes with the spinorbit term even though neither L or S commute with H50 Fur thermore the total orbital and total spin angular momentum also commute with Hm Thus there are two new quantum numbers j and mi which replace 7m and m5 good quantum numbers while ma and m5 are no longer good quantum numbers A clearer insight into the spinorbit term can be attained by rewriting it explicitly in terms of j i and s LS2 L2SQ2LAS 3 1 L48 5J2 L2 S2 4 which means that the spinorbit term may be rewritten in terms of the eigenenstates of the j m j basis m2 H50 7 jj1 1 ss1 5 The coupling of spins is a common occurence in all branches of physics In nuclear physics the spinorbit term is surprisingly large and is responsible for the basic scheme for nuclear shell structure In describing hadron spectroscopy a spinspin interaction is largely responsible for the difference of the spin 3 2 delta baryon and the spin 12 proton which are comprised of quarks of the same flavor The coupling of angular momentum in physics is thus really a study of changing bases from the Wig m5 basis to the j mj basis When changing bases the number of states involved is 2812 1 can be determined by considering the number of combinations of 7m and 7m However changing to the j mj basis only mixes states with identical mj 7ng m5 Since the states in a multiplet described by j must be complete running from j to j we see that the maximum value ofj is j s 6 Lecture 13 February 7 2001 Since there is only one state with 7m i and m5 8 there is only one state with mj 5 3 and thus only one j multiplet with j i 8 Counting the number of pairs of 7m and m5 that add up to a speci c value of mj and realizing that every j multiplet must be complete lets one see that the values of j involved are jmax 8 7 jmin These are known as the triangle relations as they can be considered as constraints involved in adding vectors One can not add two vectors of lengths i and s and obtain a vector of length j outside this range Furthermore if s lt i the number ofj multiplets is 23 1 while the average j of the multiplets is i which means that the number of states is the 28 12 1 as mentioned before Clebsch Gordan Coef cients Changing from the mum5 basis to the jmj basis is described by the overlap of matrix elements 5 8j mjf 8 mg mg 9 Such matrix elements are known as Clebsch Gordan coefficients and are refered to through a variety of confusing notations such as CE 3 nu ma j m and nearly every other possible permutation of the arguments Sometimes the coee cients are labeled by superscripts and subscripts and sometimes they look like matrix elements jmj 87Rg7R5 The notation is remarkably confusing given that the only purpose of not advantage compared to writing down the matrix element is that the labels 5 and s are not written down twice To emphasize that the coefficients are matrix elements and that the two indices 5 and 8 remain unchanged in the transformation we will refer to the coefficients in terms of the matrix elements in this lecture Finding the matrix elements is straightforward given the algebras for raising and lower ing angular momentum First remember that the matrix elements are all proportional to 6mbmzmV Now since there is only one multiplet with mj i s that matrix element is simple to write down 58j smj 8 sm m5s1 10 To generate the coefficients involving the asame j i s but reduced mj one can use the lowering operators 4 1 4 87nj 89197njgt 10 1 mjmj 1 1 H L s x 1 mi 12 10 1 m mv 1 Lecture 13 February 7 2001 Applying this to the case where mj j i 8 one generates an expression for the matrix elements with mj i s 1 4 1 s is7nj s 1 H L 3 smg Am 513 JO 1 mjmj 1 711 1 m mj 1 v 1 TIETIM 1 sm i 1m5 s 15 xss 1 mgm5 1 3 7m 5 m5 s 11 One can now read off39 the Clebsch Gordan coefficients For instance 14 1 1 58j smj 8 1 sm 1m5sw 17 JJ1gtmimi 1l Finding the Clebsch Gordan coefficient for j i s is a bit trickier By knowing that the j i s 1 states are orthogonal to the j i 8 states allows one to write down the j i 8 1 states down by inspection For instance sj s 1mj s 1 18 1 v 1 7ng7ng 1 sm 5 m5 s 1X19 711 1 mjmj 1 88 1 m5m5 1 sm i 1m5 20 Of course one could multiply the states by any arbitrary phase and the coefficients would work well The convention is that the coefficient j1j2jj j1jg 7m j17n2 j j is real and positive Example Find j 32jg 1j 32m 32ij1 32 2 17m 32m2 0 First using J J12 J21 ij 52m 52 m1 32mQ 1 5272 5232J 52 m 32 3252 3212 7m 129m2 1gt ij 52m 32 gum 12mQ 1 iml 32m2 0 Now since the state ij 32 m 32 is orthogonal to the state above 2 ij 32m 32 iml 327n2 0 giml 12mQ 1 24 3 Lecture 13 February 7 2001 Another Example A neutron and proton are each in an 3 wave of a nuclear potential The two particles feel a spin spin interaction V135 asp Sn and also feel a magnetic eld of strength B V uB s MB Squot 26 In terms of oz up 11 and B nd the four energy eigenvalues This problem is made difficult by the fact that Vb is diagonal in the mp mquot basis while V55 is diagonal in the jm basis We will omit the 3p 12 3 12 labels in the bras and kets to save space m2 4 4 2 11 1 33 1 33 1 6772mm 6mnmBh np un in janliviisij n Sid quotMm mp mudbung m n To proceed further one must choose a basis We choose the j m basis with the following eigenvalues 1 U U U 1 U j1m 1 0 1 1m 1 0 1 1m U 1 j Um U U U 29 In this basis V55 is diagonal 0134 0 U U 0 334 0 0 55 0 0 ah24 0 30 U U U 3ah24 while writing Vb requires rst rewriting each of the states in the mquot mp basis j1m1 mp12m12 31 1m 1 mp 12m 12 32 1 1m 0 gmp 12m 12 mp 12m 12 1 Um0 7 mp12mn 12 mp 12m12 If the problem involved higher angular momentum one would have to go through the pro cedure of the previous exercise utilizing raising and lowering operators to write the j m states in terms of the 7m7m basis Lecture 3 February 7 2001 1 Coordinate and momentum space One of the most famous equations in physics is Schrodinger s wave equation 1 a R2 82 8t 2m 812 Rather than thinking of a wave function it is more revealing to recognize it overlap of the state with the state Mr xiv gt 2 The only difference between a label in coordinate space and a label that denotes a discrete variable such spin is that since the x label is continuous the normalization has to be changed 133119 Vll Wlxl 1 fix 51 1 3 This infers that has dimensions of inverse length to the onehalf power The completenes relation becomes lt xgtltx gtdx W lt4 Next we wish to show that the Schrodinger equation is merely the continuum limit of a matrix equation where 01331 will represent at 1 11613261quot To do this we write the second derivative 92 9 x6x2 x 6x2 Ml SEED a I 6a2 a a6a Hm 6 51139gt0 012 Thus by making the substitution M gt I 71 H w a n F 1 39 39lx one may rewrite Schrodinger s equation in terms of discrete vectors i715 Vxz 8 923391 2m61 2391 139 The normalizations are 2 in 1 lt9 1 f 1 lt10 For the discrete case one may also write the Hamiltonian matrix 22 2 1 VX1171 0 U 7i 1 2 1 U V1 U 11 2m0x2 Lecture 3 February 7 2001 Here the potential term is diagonal While the kinetic term is barely band diagonal Momentum states A momentum state is no more than a linear combination of coordinatespace states ewnix 12 With this de nition of the state p the normalization becomes MWMWg US and the inverse transformation is 1 Erzgid39Wm M exgt m pa 22gt lt gt Expressed a completeness relation dp 7 39E r r 1 mwmow WW Changing the problem to 71 dimensions only affects the expressions here by changing 27m to 27quot The label 1 here refers to a continuum of momentum states If a system is con ned then no eigenstate of the momentum operator really exists In that case discrete states are possible Sometimes the label p is used for discrete states in which case the normalizations are different It is the duty of the watchful reader to accurately interpret the notation The momentum and position operators The momentum and position operators can be expressed as X 16 P 5WWS UU From the de nition of X one can see that WW Memo 18 One can also perform a similar operation with the momentum operator mm dimmpwm lt19 27m However the customary to view the momentum operator is not in momentum space but a derivative in coordinate space Expanding in terms of coordinate space states one can see that d 139 a 2 279214 fdxdy me W pegw lt20 2 Lecture 15 February 7 2001 Approximation Methods I In this lecture we study 3 kinds of approximation methods 1 WKB Method 2 Variational Method 3 Sudden and Adiabatic Approximations In the next lectures we cover stationarystate and timedependent perturbation theory WKB Approximation The W KB VVentzel KramerBrillouin approximation is a useful method for estimating wave f quot and quot p1 L LT for smooth potentials or for potentials with only a few discontinuities The W KB approximation for a wave function can be written Aatcw ltl Aatc w lt 1 x mewh 2 where the lower limit of the integral is absorbed by the arbitrary phases in A and A The function is de ned by W W W 2 ME W 3 and can be thought of the momentum of a classical particle with energy E at position To see the accuracy of the approximation one must apply the Schrodinger equation to the assumed form for Here we use only the A term H Ewen 4 2 flat E won 4 E we mewW 2pltxgt xaltxgt magma 22 7L2 82 it 7 7 o M 2m 832AT E The rst term disappears from our choice of while the second term will vanish if we choose AW 0lt p 2 7 The last term does not vanish but is neglected in the limit that h is small it is proportional to 152 One can understand the accuracy of the approximation by taking the second derivative of ACC in the last term and comparing it to the other terms One then sees that what is important is whether characteristic lengths of the the potential 5 are much longer than WWquot Physically one may understand the 17 dependence of A by realizing that the WKB approximation has no reflection associated with it thus conservation of ux requires p A 2 to be constant 6 Lecture 15 February 7 2001 To estimate a bound state wave function one sets the phase to be zero at one turning point point where 0 then solves for at the other turning point By nding an energy for which the phase changes by 7r one then has a solution For one may then combine such a solution with the A solution to return to the original turning point incurring a net phase change of 27r Perhaps the most common use of the W KB approximation is in estimating tunneling proba bilities In this case the wave functions have exponentially growing and decaying amplitudes In fact one usually ignores the a dependence of the amplitude and merely states that the tunneling probability form going from turning point a to turning point i is Par exp g f dx127nIat E 8 with the factor of two coming from squaring the amplitude As an example we estimate the ground state energy of a particle of m in the one dimensional harmonic oscillator potential ICC mw2x2 9 In this case one fourth of 27m must result from integrating from zero to the turning point mr x2m a 1 2 2 1 2 2 7 V 7 V 1 2 h 0 elm27m a 27m 3 0 where the turning point 1 depends on the unknown ground state energy E mw2a22 Solving the equation above for 1 gives 2717 2 7 11 a 7m E 717w 12 which is correct aside from an extra fats2 Variational Theory Variational calculations are used to calculate ground state wave functions by using the simple fact that all states must have energies greater or equal to that of the ground state Thus by writing a state in terms of some parameters 1 and minimizing E with respect to the parameters 1 one knows that the energy can never fall below the true ground state energy The minimization procedure therefore can be used to estimate the ground state energy and ground state wave functions As an example we consider the potential 82 V0quot 7 13 r We suppose that we were lucky and guessed the hydrogen atom wave functions for the trial form 2 W7quot megaquot 14 Lecture 15 February 7 2001 where a is the variational parameter One would then minimize the expectation of H with respect to a 7amp2 e2 H 2ma2 10 The minimization ii9a 0 yields 7Z2 1 may 16 which in this case gives the exact wave function and ground state energy but only due to the fortunate choice for the form of the wave function In general one would obtain an approximate wave function with on overestimate of the ground state energy Variational calculations are popular in a variety of manybody applications where the inter actions and manybody wave f quot can be quot x l c Sudden Approximation The sudden approximation can be used for calculating transition probabilities for cases where the Hamiltonian changes rapidly between two times t and t2 In fact there is no real technique involved in the approximation but merely an understanding that the reaction so quick that one can approximate the transition amplitude by a simple overlap lta Ut27t1i8gt 18 17 where g is usually an eigenstate of the Hamiltonian before t1 and is usually an eigenstate of the Hamiltonian after t2 When a potential is changed slowly the probability remains assigned to the same state This is because for each differential change in the potential a differential change 6 is induced in the wave function However if the changes occur at much different times the differential amplitudes contribute with uncorrelated phases and the net change in the probability goes Z 6 2 0 Thus if a particle is in the ground state and the well changes slowly it remains in the ground state afterward This also implies that entropy is not generated hence the term adiabatic Expanding Well Example The most common example used to illustrate the sudden approximation is the case of the expanding well Here we consider an inifinte square well confining particles to the region 0 lt a lt a which suddenly expands at time t 0 to allow particles to occupy the region 0 lt a lt 2a Assuming a particle in the ground state of the old well 0 What is the probability of being in the ground state of the new well 0 Vhat is the probability of being in the state n of the new well Here x sin Twat2a o What is the expectation of the energy H after the expansion of the well 3 Lecture 2 February 7 2001 1 Evolution in time States evolve in time EW UU t zl tgta 1 where is a unitary matrix known the evolution operator The unitarity is necessary to preserve probability throughout time In order for to be unitary the state must evolve gvmwH wmt lt3 or equivalently 01M iHtUtat a 3 where H t is a Hermitian matrix referred to the Hamiltonian Either of these equations could be referred to Schroedinger s equation One can see that the restriction that H is Hermitian is necessary to ensure unitarity by integrating forward in time by a small amount At Ut Att Ut t iHtAtUtt 4 The product only gives unity if H is Hermitian Thus Schrodinger s equation is merely a statement that states must evolve in such a to preserve probability 11 Twocomponent problems Many problems in quantum mechanics can be reduced to twostate problems Aside from the spinup spindown problem the twokaon problem see Baym the solar neutrino problem along with many other examples are really simple manifestations of the twostate problem The twostate problem is especially nice because all twobytwo matrices can be written in terms of the unit matrix and the three sigma matrices 042 4331Wm c These matrices are Hermitian traceless and obey simple commutation relations 03 O39j 2 2160760 In fact we will see later the matrices 62 obey the same commutation rules angular momentum Additionally the square of any 0 matrix is unity awf1 a The evolution of states under a Hamiltonian H6a s Lecture 2 February 7 2001 11 Twocomponent problems is especially simple In this case the evolution operator is U E z Hth 1 new 73 i tc m2 2 new m3 3 v 10 cos8t E 73 sin8t 11 Example Problem A spinupalong the zaxis particle is placed in an environment at t 0 where it interacts with a magnetic eld pointed along the 3 axis H 3 12 Find the probability of being in the up state as a function of time Solution W rawWU 13 cos8t isin8tcrx I 14 t Cisin8t 39 19 Thus the probability of being in the up state is cos2 t Another Example Two species of neutrinos the 22 and 2 have masses mu and m7 The Hamiltonian that describes these masses could be written 2 m c 0 H 1 0 0 711702 gt 6 Now we consider an extra term added to the Hamiltonian that mixes the two flavors of neutrinos 0 102 H Hmixac2 0 First let us nd the energy of two new states We do this by rst writing the Hamiltonian in terms of the sigma matrices 1 1 H m m7c21 m m7c202 102 18 1 m m7c21 1302539 19 Here 8 is the ma nitude of the two terms that multipr si ma matrices g e g 3 0 2 1 m7 mul22 20 Lecture 2 February 7 2001 11 Twocomponent problems and 73 is a unit vector pointing in the direction 73 cos9 rsin9 21 sin9 3 22 3 Finding the eigenvalues is simply a matter of rotating the 0 matrices such that 73 is in the z direction The two energies are then 1 2 2 Ei m m7c 1 1 Jo 23 Heisenberg and Schrodinger representations Usually one wishes to caculate expectations of operators eg dxAB A 4 40312 where the states evolve a function of time but the states are considered independent of time Con sidering an evolution operator U lmh one can express the time development of the expectation of AB A A AC in either of two equivalent representations WWW C EMW 0107243 0071210 24 050 UlAU UTBU AAAD TCU 25 The upper line Eq 7 known the Schrodinger representation with the states evolving while the operators are xed In the Heisenberg representation the time development of the operators can be written a differential equation where the rate of change of the operator is given by the commutation of the Hamiltonian with A AHt E UAstUt 26 MT gb tt 4stUtat l 27 mt mm AslUt t gt W 3450 W t 28 Here the subscripts S and H refer to Schrodinger and Heisenberg representations respect fully If there is no explicit time dependence in Ag then any operator that commutes with the Hamiltonian represents a constant of the motion The most obvious such operator is H itself Thus if the Hamiltonian has no explicit time dependence the expectation of H the energy is a constant of the motion The spinprecession example from above could also have been written a differential equa tion In that case the time development of 03 and 0y could be written as gem i3UT0wozU2 oyt 29 gm i b llowoylbquot43040 so or equivalently a at 4 ozt 31 3 Lecture 31 February 26 2001 Multi electron atoms The ThomasFermi approximation A very crude to find the density of fermions in an external well is the ThomasFermi approximation This approximation is semiclassical and assumes that the local density is a function of the local potential only 231k3 v 7 77 6W2 f 2m lr f The density would extend only in the region where lr lt 6 Given the potential and f it is straight forward to find the density For a finite system with a fixed particle number one would have to adjust Cf to get the correct total number of particles The ThomasFermi approximation is crude and gives rather nonsensical results for atoms but is more reasonable for estimating the electron density near surfaces As a general rule it is valid for large systems where the potential changes slowly Hartree Fock The principle dif culty in handling multielectron atoms comes from the fact that the elec trons interact with one another If it were not for the mutual interactions of the electrons one could treat the electrons independently Of course this would yield horribly unphysical results For instance one could place an infinite number of electrons into the the hydrogen atom where in reality no more than two electrons can be bound to hydrogen The Hartree approximation is the first method one might consider for treating such atoms One then solves the Schrodinger equation with the potential 82 Wt ZTB 2 d3r ltr gt lt2 I m 1 r If there are n electrons one must solve n coupled equations One can go one step further and solve the Hartree Fock equations h V 410 d3rhri ltr gt l r ir 633I ltrgt iltr gt1 war 3 y 2m 7quot to find the wavefunctions One can then find the energy by calculating Again this requires solving coupled differential equations but the Fock term also makes the equations nonlocal which brings along an added computational dif culty The difference of the energy found by solving the Hartree Fock equations and the Hartree equations is the exchange energy Since the Hartree Fock wave functions force the electrons to from one another due to antisymmetrization the repulsive Coulomb interacton between electrons is weakened which means that the exchange energy is negative The Hartree Fock equations were derived with the assumption that the solution is a product state As that is a variational assumption the true ground state energies are about 1 eV Lecture 31 February 26 2001 lower than Hartree Fock solutions This difference is refered to correlation energy It can be calculated perturbatively by considering the mixing of oneparticleonehole and two particle two hole states into the wave functions This is done perturbatively by considering the nelectron Hartree Fock ground state the ground state Then one considers excited states being those nparticle nhole states formed by considering 71 of the singleparticle solutions of the Hartree Fock ground state to be replaced by 72 singleparticle solutions from the set of singleparticle states left unfilled in the Hartree Fock ground state The periodic table Singleelectron wave functions are labeled by 71 i s j and mj In an atom with only one electron the energy depends principally on 72 states with same 72 but different i 0 1 A 4 471 1 are degenerate aside from the spinorbit interaction Of course 3 always equals 12 and j can be either 5 12 or i 12 The existence of the other electrons destroys the accidental degeneracy of the hydrogen atOm and allows states With different i and the same 71 to have signi cantly different energy This is due to the screening of the positive charge The accidental degeneracy allowed states with fewer nodes in the radial wave function but larger angular momentum to have the same energy states with smaller angular momentum but more nodes in the radial wave function By screening the charge an advantage is created for states that have a relatively greater probability of being near the origin Since radial wave functions behave 7quot near the origin a state with a lower 5 but the same 72 will have lower energy due to screening This difference can be large enough at times to allow a state to move lower than states of higher 71 but higher angular momentum For instance in some cases the 48 states can move below the 3d states and similarly the 4d and 53 shells compete well The 53 shell is always well below the 4f shell When shells compete eg the 4s and 3d electrons the choice of orbitals is nontrivial In these cases the configuration can vary from one element to the next and in fact solutions might contain a mixture of configurations Chemical properties are determined largely by the outermost electrons When shells are filled the elements are less reactive The rare gases also know as inert gases all have filled p shells with the exception of Helium The set of orbitals with a specific 71 and i is known a shell The degeneracy of a shell is 45 2 Electronic configurations are labeled by the shells and the filling eg 1822822p3 with the superscript labeling the number of electrons in the shell Con guration splitting Different configurations are generally split by a few electron volts Within a configuration the splitting is complicated Although there are 45 2 singleparticle orbitals the number of ways to arrange several electrons among these levels can be rather large Ignoring the spinorbit interaction the total angular momentum L and total spin S of the electrons commute with the Hamiltonian The 2L 123 1 states of an L3 multiplet are then further split by the spinorbit interaction Thus the spinorbit interaction invalidates AMI and A15 good quantum numbers and replaces them with J and AMJ Lecture 30 February 23 2001 Interacting Fermi systems the Hartree Fock approximation Twopoint interactions The simplest sort of interaction is a twopoint interaction H2 vaaga 1 a 393 Such an interaction allows particles to behave independently To demonstrate what is meant by independent consider the evolution of the product state t lmha uEaLEU 2 Vimhaggmh Vimhaggmn E gmnaj gmn Em Thus the evolution in time simply adjusts the creation operators to timedependent quan tities amt E NIthaLEz FIth minim 11 am E z Htn Hyatt E z Htn Z waagu 8 A A A A C71 0 V V V V Therefore a twopoint interaction leaves a product state a product state and the evolution can be reduced to the evolution of independent creation operators Unfortunately physics is not usually so simple If all interactions were of the twopoint type all problems could be solved by considering the independent motion of independent operators and there would be no need for manybody theory Interactions of the type H d3 wxwx Vxxyxxyx 8 are of the twopoint form and can be solved by considering a single particle in the system then creating the manyparticle system a product of many singleparticle solutions Fourpoint interactions An interaction between particles separated by r r1 r2 is written classically 1 39 5 d3md3r2pr1pr2Vru r2 9 The factor of 12 corrects for double counting since the integral is without a r lt r2 quali er Written in terms of eld operators this interaction becomes 1 39 Hint Edsrld iTQlrl 1392215 T139119T 10 1 Lecture 30 February 23 2001 For the moment we will omit spin indices and assume only particles of a speci c spin are involved One peculiar aspect of the Hamiltonian is that the two 111 s are on the left while the two 111s are on the right whereas the product of densities would suggest a product of Illllls However such a product would result in an energy for a oneparticle state the particle would interact with itself Fourpoint interactions or threepoint interactions eg the electromagnetic interaction J A are in nitely more complicated than twopoint interactions For one if one wishes to calculate the evolution of Illa one needs to commute the Hamiltonian with However commuting a fourpoint function which is a product of four creation destruction operators with a single creation operator results in a product of three creation destruction operators This destroys the independentparticle nature of the evolution In fact the eigenstates of a Hamiltonian with fourpoint terms is not a product state but is instead a complicated linear combinations of product states In fact such problems are in general not solvable forcing one to resort to approximations One such approximation is the HartreeFock approximation Example correlations in a Fermi gas Consider a onedimensional gas of fermions moving in a large region of length L 1 Find the correlation function relating the ratio of the probability of nding two parti cles separated by 7quot to the probability of nding the two particles at arbitrary locations To do this problem we rst write down the twoparticle probability P231g32 lt 1 l31 1 32 1 32 1 1M5 11 gt H am 12 Mk Commuting the eld operators and towards the ket yields EH91 1 Eikynz f I kg aha k1 k Z Z aklak2 kg ah 91 Pik2m2 22 z i quot2 H hawk VT Vi kltk k k1k k2 Here the sign i depends on whether lk1lk2 appear in the sameopposite order as alk1alkg in the product of creation operators all W4 One can nd a similar expression for The overlap of the two states is zero unless Cl and k2 are the same in the expressions for the bra and ket This yields 1 ai l lm Pl2 1 2 1 1 3gt E p 1 C05 92 900 531 10 k1k2ltk Taking the ratio of the twoparticle density to the square of the one particle density gives the correlation function 2k1k2ltk1 COS k2 k1372 2k1k2ltk gr E 32 an 16 Lecture 30 February 23 2001 The sums can be changed into integrals W 1 EIEZ 17 dkeikr I f kfltkltkf 18 f Icfdcdcf dk s1nkfr 19 kfr 2 The particles now feel a mutual interaction lar 20 Find the exchange contribution to the energy per unit length due to the interaction in rstorder perturbation theory In rstorder perturbation theory the correction to the energy is H ifdmldm2lt iwmlIIar2Par21m1waxan a2 21 L drlt 1 01 rxprxpogapT 22 where we have used the fact that the two particle density depends only on 711 272 By using the results of 1 one can nd the energy per length 712 fie d7quot grVT 23 Egg00 dr 11672 sin2kfr 24 2 1 1 1 5km 1 s1n2kfa 20 where n is the number of particles per unit length Note that the rst term describes the potential energy one would expect from particles with a uniform density interacting through the potential larl This is called the direct term while the remainder is referred to the exchange term Hartree Fock The HartreeFock approximation yields an expression for the ground state of a manyparticle system It is basically a manifestation of the variational approximation One assumes a form of the wavefunction then solves for the parameters that minimize the energy The assumed form of the wave function is W a a a0gt that is one con nes the search to product states 26 Lecture 30 February 23 2001 Next one writes down the expectation of the Hamiltonian l 7amp2 39 39 3 72 3r39l 39 H 9114 lt0 d 7 11 ltrgt mv Mr d w my mirm lt20 1 O lt 5 d37 1d37 2l 1 2 Til l lTil l lr2 1 r2 1 ril gta 28 where the spin indices have been omitted temporarily One can now write down the energy in terms of the wave functions of the occupied states oz w v 3 ms 52 2 r lt gtsz gtgt 2 f d mm 37v Mr 29 a m 1 331 I I gteltl l 5 2 f d rd 7 meow mama i ourwmmr mm 141 The second term in the twobody interaction where the i refers to bosonsfermions is known as the exchange term with the name coming from the exchange of the oz and 0 indices The next step in the variational procedure is to minimize H gt with respect to changes in the wave functions subject to the constraint that each wave function is properly normalized One accounts for the constraint by multiplying the constraint by a Lagrange multiplier A then adding it to the function one wishes to minimize L lt EHE gt A7 d37 ltrgt yltrgt 0 so 507 i i 7 More correctly one would vary both the real and imaginary parts of by which is equivalent to varying either 7 or The resulting expression is 722 Ame v2 Um Mr 31 d37 Vr r Z39rl a rl vr i rl 7rl a r The two terms involving Vr Iquot are known as the Hartree and Fock terms respectively The Hartree term looks like a potential felt by the particles due to the presence of the other particles Thus the Hartree equation can be solved by selfconsistently nding the solutions to the effective potential l flrlartregr E d37JVr rl rl a rl The Hartree equations are often solved iteratively One guesses at the wave functions finds the Hartreepotential solves the Schrodinger equation for the singleparticle wave functions then iterates the procedure until the wavefunctions converge The Fock term presents a different challenge since Linr can not be factored out of the differential equation The Fock term is nonlocal in that gin1quot appears in its place 4 Lecture 8 February 7 2001 1 Path Integrals and the Aharanov Bohm Effect 1 1 Path Integrals One to perform Quantum Mechanics is through path integrals Path integrals are usually a rather inconvenient to go but sometimes come in handy The name refers to the fact that a sum over all intermediate states can be thought of a path To get a better idea we consider the matrix element 4m lie mar w Z w ixv1gtltav1iw x2gtltx2 e39i 5 ix1gtltx1ie39mwi NunVii We 22 MM Here 6t tN and the approximation becomes exact in the limit of large N If the states 33 correspond to positions in coordinate space arranged in a mesh of size 63 the succesive points in the path 3 and 3341 are constrained to be neighbors since H is local That means that each term in the sum can be thought of a continuous trajectory where any at each step in time the trajectory either remains at the same position or moves by 633 This motivates the name path integral though one might more accurately state that one sums over all trajectories rather than over all paths The classical limit of quantum mechanics comes from the constraint of choosing the trajectory for which the phase becomes xed with respect to small variations of the trajectory We will not pursue this further In principle all matrix elements can be considered in this fashion Lattice gauge the ory which provides a powerful tool for numerically calculating the structure of the non perturbative QCD vacuum is built upon exactly such concepts Path integral techniques are also often used in statistical mechanics by making the analagous decomposition 8 13 Z almaNil 63H 8396 3041x041 ie39a m 3 lt 39 1gtltlt N 1 5042x0423 12 Free particles and the AharanOVBohm Effect When we studied the Hamiltonian for a free particle from the perspective that the space described by discrete points in space separated by 63 we that the kinetic term of the Hamiltonian mixed states from neighboring sites a iiHii 1 7 4 lt gt 2m6332 while the potential term well the remainder of the kinetic term purely diagonal If it were not for the offdiagonal piece probability would not spread over space with time The interaction with an electromagnetic field A also contributes an offdiagnoal piece due to the presence of the term iefzmcA O V The offdiagonal term to the Hamiltonian is 1 1 Z wfgu iH6txN1ltxNgmgxgxxggu iH6ta1lta11 iH6ti g Lecture 29 February 19 2001 Symmetrization and antisymmetrization for fermions Consider a multi particle state 11052 where the ordering 061062 signi es that the rst particle is measure to be in the state m the second in 12 etc The permutation operator 7 is de ned by 7912lima2a3 EOQgOtuOtsg Piauaaasg 1 Any multiparticle state gtshould have a good symmetry with respect to permutation since the particles are identical This means that all states of the system should be eigenstates of the permutation operator 7 79iji gt i1i gt 2 Only ii are possible eigenvalues of the permutation operator since 792 1 In order for the state to be an eigenstate of all permutations the N particle state must be either totally symmetric or totally antisymmetric 1 a WPQBEIWPaMMWL lt3gt where up is the number of pairwise permutations required to make the desired permutation For fermions no two states eg a and 8 can be identical Fermi creation and destruction operators Writing states as a sum over various permutations is a rather clumsy to consider the Fermi nature of the particles as it requires assuming there is a rst particle second particle and so on The algebra of anticommuting creation and destruction operators offers a more natural means to incorporate antisymmetrization With this formalism a state can be noted simply by their labels with no mention of permutations Matrix elements are then calculated according to the algebra of the creation and destruction operators which account for the symmetrization The operators obey the algebra am 1 3 606 4 at am 0 a Any time two creation operators or two destruction operators with the same index are next to one another the result is zero aaaa 0 This enforces the Pauli exclusion principle Similar to the Bose example the vacuum is annihilated by the destruction operator VV 1030 0 6 The commutation rules written above assumed the states 16 A were orthogonal If the indices refer to states which are from different different bases the algebra becomes am 4 gamut a ltajgt lt7 1 Lecture 29 February 19 2001 We are now in a position to calculate arbitrary matrix elements for N particle states izjzkzn39 iam gify Z 1quot ltiiagtltji3gtltkiivgt O O O perms of 139j A 9 vv As an example we consider twoparticle matrix elements 104 WY fa If i and j refered to positions a and y the result would read 3 yia 3 3a 53y ay 35 which signi es that the two particles can not be at the same position Note that the overlap of an N particle matrix element would yield N terms Thus it seems little has been gained using eld operators rather than writing symmetrizedantisymmetrized wave functions which also have N terms However when writing the bra and ket with all permutations both the bra and ket have N terms with the extra N being cancelled by the in the normalization of the wave functions It should also be noted that all the same results are valid for bosons except that the 1s disappear Fermionic eld operators and the density operator The eld operators and create and destroy a particle in the state They obey the anticommutation relations 1 x1 y 53xY 10 The eld operators commute with other operators 1 xa2 ax 11 where x x a is the single particle wave function of a particle in the state 04 Thus the eld operators are no different than other creation and destruction operators except that their dimension is length 372 and their anticommutation relations are expressed in terms of Dirac deltas instead of Kronecker deltas The density operator is MK PX PX 12 The expectation of the density operator in the state Ea 8 O O O is 0057 O O O E Px 1 xiau3z O O O Whammy O Ol Px PxO O Oaialgailim 13 Z 2x nx 14 Rea3Oquot Lecture 4 February 7 2001 1 Potential Problems in One Dimension Consider Schrodinger s wave equation 7amp2 82 8 77II WImi 7II i 1 2m8ac2 miLu L39 8i L39 If Iw i is an eigenstate of the Hamiltonian the solution becomes me i amt mm 2 and the time derivative i88t can be replaced by E in Schrodinger s wave equation For the most part we will consider the steadystate case where we look for eigenstates Aside Before we nd solutions we regress to consider the notation and vocabulary associated with Schrodinger s equation One often refers to the Hamiltonian in this context Hum E19310 Hz w 6 VL39 psiw 3 39 2m 8102 This notation is actually incorrect as it does not represent the fact that the Hamiltonian is an operator More correctly the left hand side of the above equation should be replaced by ELM gt ltwiHiz2gtv 4 However this clumsiness usually does not cause problems and we will not go out of our way to avoid it To nd solutions for problems where the potential is continuous it is sufficient to nd solutions to the wave equation that have the correct behavior at L39 gt 2100 If the potential is discontinuous at certain points boundary conditions must be enforce at every point where a discontinuity is formed If the discontinuity is nite not a delta function the boundary conditions at the disconti nuity are that 1 7990 is continuous 2 8 8w Ms is continuous If the rst derivative were discontinuous at y the second derivative would be in nite at that point To understand the boundary conditions we integrate Schrodinger s equation from at 0 L b 122 8 8 1b 7 iz wE 77J x 2 dwE Vw ux 5 m A 4 864 1 lt M H Lecture 4 February 7 2001 We now consider a discontinuity at y and perform the integral over an in nitesimal range centered about y If the potential is nite one obtains 7L2 9 9 7 iJ39a 7quotIE 0 6 2m 81 lye amid 4 e thus demonstrating the requirement that the derivative be continuous However if the potential is a delta function integrating the righthand side of Eq 2quot over the in nitesimal range can yield a nite value Thus if m 5m y 7 the boundary condition becomes 1amp2 9 9 7 74 7 7 2m await 83w x my 8 For the remainder of this lecture we merely consider a large number of potential examples Square Well Example Consider the potential 00 1 lt 0 VI 2 V0 0 lt 1 lt a 9 0 1 gt a Solve for the binding energy and wave function of the lowest energy state Solution Assume the energy is negative In the region of the well solutions of Schrodinger s equation are sines and cosines with wave number k 2mVg B 122 where B is the binding energy positive number The BC at the origin is that the wave function must go to zero due to the in nite potential Thus only the sine piece remains 1I sink1 10 We have chosen an arbitrary normalization constant of unity In the second region exponentially growingdecaying solutions work with a decay constant of q 2mB7L2 The exponentially growing piece can be thrown out we wish to nd a solution where some probability is near the origin 11I 2 AK 11 Thus far we have neglected the two BCs at 1 a which will be suf cient to determine the two unknowns the binding energy B and the normalization factor A Writing the two BCs sinka 46 12 kcoska qu q 13 Lecture 4 February 7 2001 Dividing the two BCs eliminates A and gives the relation k tanUm 7 14 I This is a transcendental equation for B which can be solved graphically or on the computer One can see that the solution disappears if the depth V0 is too low or if the the width 1 is too narrow At this point the solution has binding energy zero and from the previous equation one can see that tanUm 00 or ka 7r2 with k 12mV0h2 One physical to understand the constraint is that the wave function must turn over and have a negative slope at a a in order to match to an exponentially falling solution Thus ka must be larger than 7r2 It is worthwhile to note that if a potential remains below zero everywhere that there is always at least one bound state In this case the potential is in nite for a lt 0 Delta Function Example Consider the potential V0quot 353g 3 gt 0 15 Find the binding energy of the bound state Solution Assume the existence of a bound state of binding energy B The solutions are exponentials with decay constant k 12mB 152 and the requirement that they go to zero at a too gives L I OXpki a gt 0 CXpUCCC a lt 0 16 Plugging them into the BC at a 0 Eq gives 15 7 8 139quot m I Thus the binding energy is Tray2amp2 Barrier Penetration Consider a plane wave incident on a positive barrier 0 a lt 0 r gt l 5 V0 3 gt 0 18 Find the reflection and transmission probabilities for a particle with energy E which is greater than V0 Solution Solutions must be of the form f c e k UrAeW kzwQmEW 19 WW 86quot q 2mEVu z2 20 3 Lecture 33 March 13 2001 Molecules and the adiabatic approximation The force felt between atoms largely derives from the distortion of the electronic wave func tions due to the proximity of a second potential Since atoms move slowly with respect to the electronic motion electronic wave functions largely adjust in such a to remain in the ground state for a given atomic separation This refered to the adiabatic or BornOppenheimer approximation Examples where the adiabatic approximation were applied included a particle in a slowly expanding box If a box were slowly expanded or contracted a particle would remain in the ground state of the box Of course this implies that the energy of the particle is not kept constant For the example of an expanding box the ground state energy falls the volume expands and the lost energy appears in the kinetic energy of the piston In fact that energy is equal to the work PdV done by the expansion In the case of atoms the gain or loss of the electronic energy appears a loss or gain of the heavy ion s kinetic energy The electronic energy calculated a function of the ion s separation 7quot serves a potential for the atoms In a previous problem set we worked out the energy felt by two electrons in harmonic oscillator potentials situated far apart such that they felt a dipoledipole interaction Calculating the correction to the energy it found that the energy went lrs Example A neutral atom and an ion separated far apart For this example we consider an electron and a positive ion where we assume the states available to the electron are described those of a harmonic oscillator with frequency An ion of charge ZE is placed a distance R from the atom The perturbative potential between the ion and the atom is V Z62 d3rpriR1ri 1 w Z62d37quotpr 2 Zia2 zop 3 One can then calculate the energy to second order perturbation theory Z264 nz1 2 g0 AE R4 4 7284 2R4mw2 5 This energy serves the potential between the atom and the ion Note that if one were to consider an induced dipole moment proportion to the electric eld multiplied by the electric eld which falls off 1122 one would have expected the 1124 behavior Example H2 ion Consider two protons separated by a distance R with the addition of a single electron In the adiabatic approximation the binding energy of the electron 12 serves the potential 1 Lecture 33 March 13 2001 between the protons when combined with the protonproton interaction At large distances the binding energy is one Rydberg while one would expect an increase in the magnitude of the binding energy R is shortened At very short distances one expects the interaction to become repulsive when one penetrates the electronic cloud and the interaction is dominated by the Coulomb interaction between the protons Calculating the binding energy a function of R can be crudely accomplished with a variational calculation assuming a trivial form of the wave function 1M1quot Ci W40quot i erll 6 where 1214 and 1331 are the bound state wave functions of an electron to each of the protons mm marl28 RA 7 The factor Ci is merely a normalization constant 1 C 8 i QiQSUB l l 3amp1 dwlnrwm lt9 R R2 ratHo 1 70 a 10 This can be considered a variatonal calculation with zero variational parameters This integral SUB is most easily calculated in elliptic coordinates where the three compo nents of 139 are replaced by Er R4EEr RBE E 11 R lt gt Er R4E Er RBE 5 h 12 L R l l a E arctanyat 13 Thus u is the scaled sum of the distances to the two protons v is the scaled difference of the two distances and a is the usual azimuthal angle with respect to the axis de ned by the two protons Some J acobian manipulations would reveal fdsr r fl du 11 dv fug dq agwg v2fr 14 Once one has made a transformation into these coordinates the integral becomes rather trivial since Wm ma a quotW 15 If you have not worked with elliptic coordinates before you can understand the name el liptical since the set of points with a xed sum of distances r RAE if R33 de nes an ellipse Lecture 33 March 13 2001 One is now in the position to calculate the expectation of the energy AgHgA BgHgB i 2AHB AEHEA i AEHEB H lt221gtiltswgt 1 Where e 92 2 92 3 ABA 61 E wrmd r 18 2 q SE 1 e m D 19 and 2 2 ltAHBgt BE 3 f 20 e2 e2 R The term 61 is merely the electronic binding energy of one electron With a proton 10 Rydbergs The integrals were calculated With the help of the tranformation into elliptic coordinates The potential between the atoms is effectively VAR HE e1 22 When plotted against R the resulting curves looks like VR 176 ev L u 13 Angstroms Lecture 28 February 16 2001 Fermions When a single particle is in a system it does not matter that it is a Fermion or a boson How ever when multiple particles are present the behavior is entirely different The difference between fermions and bosons can be explained by a variety of perspectives 1 When considering populations of singleparticle levels no more than one fermion of a given type and spin can be assigned to a particular level while an arbitrary number of bosons can be placed in a given level This constraint on fermions is known the Pauli exclusion principle D For identical fermions of the same spin the wave function 32 must be com pletely antisymmetric while the wave function for bosons must be symmetric C Creation and destruction operators for fermions obey anticommutation relations while those for bosons obey commutation relations Examples of fermions are electrons quarks and neutrinos Examples of bosons are photons and gluons Composite particles made of an odd number of fermions eg a proton which is made of three quarks are also fermions Composite particles made of an even number of fermions are bosons eg a 12C atom which is made of 6 protons 6 neutrons and 6 electrons All bosons have integral spin while all fermions have half integral spin The relation of statistics to spin is a profound consequence of time reversal property in quantum eld theories In this lecture we will consider some of the consequences of the Pauli exclusion principle and leave the discussion of eld operators and wave functions for the next chapters Fermi gases Consider a large number N of fermions of spin 8 and m in a box of volume V If the fermions are placed in the lowest levels consistent with the Pauli exclusion principle the highest singleparticle energy is known as the Fermi energy 6 and the momentum of that state is known the Fermi momentum pf The density is a function of the Fermi momentum Y V Km 3 1v 231W dp 1 N 1 n 7281ng 2 Note that the Fermi momentum is directly dependent on the density and does not depend on the particle s Of course the Fermi energy does depend on the mass 6 One example where the Fermi energy plays a pivotal role is in neutron stars Due to beta decays neutrons can change into protons through the emission of an electron and neutrino n gtpci9 pc gtnv 3 Since neutrinos can exit the star due to their neglibible masses and small crossections they need not be considered with respect to conservation laws However baryon number neutrons Lecture 28 February 16 2001 plus protons and electric charge must be conserved In fact the electric charge density must be zero Thus beta decays and inverse beta decays can proceed long the following constraints are met nnlnp n1 4 VV ne up 5 Beta decays will proceed until the energy is minimized for the given value of the baryon density 713 Neutrons are more massive than protons by an amount mu mph2 13 MeV which more than accounts for the electron s of 0511 MeV When the energy is minimized changing a neutron to a proton plus electron must leave the energy unchanged if the particles are added and removed from the top of the Fermi surfaces Thus 6 13 MeV Eji 651 0511MeV 6 Combining this equation with the two constraints allows one to nd all three Fermi momenta in terms of 713 At low density the extra 13 MeV favors protons over neutrons Therefore in normal stars protons are much more numerous than neutrons At high density the electron Fermi momen tum plays the pivotal role Since electrons and protons have the same density they have the same Fermi momentum but the electron s Fermi energy is much higher due to it s lighter mass sfzxm2ip3e mzpi2m 7 Thus at high density the system must try to reduce the electron density which in turn requires a reduction in the proton density and results in a large excess of neutrons This is certainly the case for the interior of neutron stars where densities are of the order 01 baryons per cubic fm The resulting electron Fermi energy is on the order of 10 MeV is the resulting neutron Fermi energy Thus the electrons are extremely relativistic while the protons and neutrons remain nonrelativistic Filling the harmonic oscillator with fermions As a second example we consider the lling of harmonic oscillator energy levels This simple picture provides a surprising amount of insight into the structure of light nuclei First we must understand the degeneracy the singleparticle levels in a 3d harmonic oscil lator with an energy of N 3 2hw Considering the problem in a Cartesion basis there are NJ 1 to arrange nfr my NJ Then the number of to arrange n1 my and Hz to add up to N is dl V 1VL1W 8 Thus there is one combination of nmny and Hz to get N 0 three combinations to get N 1 six to get N 2 etc Next we wish to calculate the number of states of a speci c N with a given orbital angular momentum i This will come in handy when we consider the spinorbit coupling on top of Lecture 28 February 16 2001 the harmonic oscillator structure The solution will be that the states with excitation N are accounted for with one i N multiplet one i N 2 multiplet one i N 4 multiplet etc To prove that this is the case we make an inductive proof assuming it is true for N 2 For every state with excitation N 2 there is a state with the same angular momentum with excitation N which is reached by applying the operator which is a rotational scalar There are N 1N2 such states Furthermore there must be at least one i N multiplet because the state transforms like part of a i N multiplet But one i N multiplet with degeneracy 2N 1 and the N 1N states accounted for by counting the states with excitation N 2 completely account for all the states with excitation N N 1 N 2 N 1 N Y X 211 9 2 2 Thus increasing the excitation by 21 adds one more multiplet with i N As an example for N 4 there are multiplets with i 4 2 0 with degeneracy 9 5 1 15 5 62 If no spin orbit terms were present adding neutrons to a harmonic oscillator would lead to shell closures with neutron numbers equal to 2 8 20 40 and 70 meaning that 40 neutrons exactly lls the N 0 1 2 3 shells Adding the spin orbit term adjusts the singleparticle energies by an amount Es Wei8 gnu11 1 s11 10 For every value of i there are two values of j j i J 12 Thus one nds the energy levels by rst labeling the states by N and 5 Then splitting each level into it s two values of j and nding its energy EN32hw jj1 1 ss1 11 The shell structure changes due to the spinorbit interaction and the numbers of neutrons required to reach a condition where there is a large gap are called magic numbers In nuclei the magic numbers are 2 8 20 28 50 82 and 126 Nuclei are considered doubly magic if both the neutron and proton numbers are magic numbers Examples are 4He 160 gCa 48Ca ggNi 48Ni 78Ni 80891 and 38171 The magic numbers of nuclei were only explained by assuming an anomalously large spinorbit coupling The reason this large coupling surprising and will be explained later in the semester when we discuss the Dirac equation Zeropoint surface energy for Fermions It costs energy to divide a piece of metal into two pieces The associated surface energy energy per surface area has a component deriving from the penalty associated with the kinetic energies of the particles To understand the source of this energy we rst consider the onedimensional case Consider a box of length L which is divided into two boxes each of length L 2 The initial energy levels are given by 2 2 Enh27 kLmr n123 12 3


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