New User Special Price Expires in

Let's log you in.

Sign in with Facebook


Don't have a StudySoup account? Create one here!


Create a StudySoup account

Be part of our community, it's free to join!

Sign up with Facebook


Create your account
By creating an account you agree to StudySoup's terms and conditions and privacy policy

Already have a StudySoup account? Login here


by: Garett Kovacek


Garett Kovacek
GPA 3.95


Almost Ready


These notes were just uploaded, and will be ready to view shortly.

Purchase these notes here, or revisit this page.

Either way, we'll remind you when they're ready :)

Preview These Notes for FREE

Get a free preview of these Notes, just enter your email below.

Unlock Preview
Unlock Preview

Preview these materials now for free

Why put in your email? Get access to more of this material and other relevant free materials for your school

View Preview

About this Document

Class Notes
25 ?




Popular in Course

Popular in Physics 2

This 43 page Class Notes was uploaded by Garett Kovacek on Thursday September 17, 2015. The Class Notes belongs to PHY 5645 at Florida State University taught by Staff in Fall. Since its upload, it has received 39 views. For similar materials see /class/205531/phy-5645-florida-state-university in Physics 2 at Florida State University.




Report this Material


What is Karma?


Karma is the currency of StudySoup.

You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 09/17/15
PHY 5645 Summary of Lecture 19 Central potential scattering in 3D Recall that m7ww mm W w D where the Greens function satis es v2 k2 our r 6r 7 1M and the solution is chosen such that the second term in Eql corresponds to an outgoing wave Then 1 eiklr7r l G 77 kltr7rgt 47Tlr7r l and in the asymptotic limit of r 7 00 Mr i m 1 fT 6 TIRED 22w We 7 m2 dgr e k vltrgtwkltr gt T eikr T w9m where 09415 and the angles 6 and b are the angles between the vector de ning the detector f39 and k the vector de ning the incoming wave For central potentials7 ie if Vr llrl7 then fkt97 ie the scattering amplitude does not depend on the azimuthal angle 15 To determine fkt9 we need to nd the solution of the Schrodinger equation m A I 27Th2d3r671krr Vrwkr 2 hZ h k2 7V2Vlrlw m w 7 w nmlt6 gt l 7 2 12 62 h2ll 1 2 1sz 7 l 2m 2 VltTgtgt WU 7 2m WU For Vr with a nite range d7 we have shown that for 7quot gt d we have lt h 62 UM h2k2 7 2m 2 2m W T and the solution is a combination of spherical Bessel and Neumann functions ulr sinkr 7 lg B coskr 7 lg r kr l Aljla quot Blnlkr Al kr Letting 7 tan 6 we can rewrite the above expression up to a normalization constant as um sinkr 7 lg 61 r kr Now since we are seeking the scattering amplitude with azimuthal symmetry7 we can write the solution of the Schrodinger equation as a superposition of YMFO spherical harmonics only w7imwwwWW l0 sinkr 7 lg 61 00 alkPlcos l9 0 M 3 where the Legendre polynomials are 1dl Hw7 mtnkRwnawaaw7 w7mm We x the coef cients alk by equating the expression 2 to the expression 3 e sinkr 7 lg 61 kr 6ikcos9 fkw ialk31cos l9 7 which must hold at large 7 and where we chose the coordinates by letting the incident wave propagate along Z direction Note that due to an entirely separate argument 0 sin kr 7 lg elkTCOSQZ2l1ZlHCOS6 lt 2 10 kr sin 77 lg f 6W sinkr 7 lg 61 2l1z39lPlcos0 ialkPlcos 19 kr M8 H o H 0 We x the coef cients alk by matching the incoming spherical waves on both sides of the above equation Note that this does not involve fk6 since the scattering amplitude controls the outgoing spherical wave Since sinx em 7 e 2z we have mug 211 lei5l Therefore 1 fk6 E 22l 1e sl sin SlPlcos 9 10 Note that 61 is a function of k and therefore a function of the incident energy If Mk is known we can reconstruct the entire scattering amplitude and consequently the differential cross section The phase shifts must be determined from the solution of the Schrodinger equation Physically7 we expect 61 lt 0 for repulsive potentials and 61 gt 0 for attractive potentials Also7 if lk gtgt d7 then the classical impact parameter is much larger than the range of the potential and in this case we expect 61 to be small The differential scattering cross section is d 1 W 2 i lfk6l2 7 Z2l1e slsin61Plcos0 d9 k 10 By integrating 3 over the solid angle we obtain the total scattering cross section 47139 Utot E l 1 sin2 61 M8 1 H o which follows from the orthogonality of the Legendre polynorninals 1 2 d P P 5 1 95 M 95 2l1 Finally note that since 1311 1 for all l7 we have amt 43mf0 This relationship is known as the optical theorern PHY 5645 Summary of Lecture 9 Feynman path integral formulation of Quantum Mechanics We showed last time that since the Hamiltonian 2 10 1 2 H 7 7 2m l 2 does not depend on time7 the evolution operator is Ut 0 54W 1 We were interested in nding its projection on the position eigenstates mm 095 ltzle ml20gt 2 since physically this tells us the probability amplitude that if the particle is at a position 20 at the beginning t 07 then it will end up at a position x at some later time t Feynman7s method is Ut 07 le imlx le im e im e im ei m lm 3 where the product appears Nitimes and NAt t As At a 0 we can write A r 2 r 6 Am exp 7At2pimgt exp 7Atgz2gt 0At2 where the last term means that the error is of order At27 and therefore in the limit of At a 0 we have mm 095 mi m ltzle m gtlt gtlt flammgco 4 i lt At p2 Ats 2 X X At p2 Ats 2gt gt 7 2exp m2m exp m 22 exp m2m exp m 22 20 Now7 we inserted these resolutions of unity A 00 dp 1 7 5 700 2Wlpgtltpl ltgt 1 dxxz 6 So A 00 00 d d d Ut 07 L dxldzg dmN1L g 2 At p Ats 2 ltzl exp legtltleexp gt lxN1gtltxN1l gtlt At p2 Ats 2 x mm exp mm exp mm w 00 00d d d d ldigdN1 700 27139 27139 27139 At 712102 At H ltlPNgt exp ltleN71gteXp ltE V1gt ltN71le71gt gtlt At p AtH 2 gtlt ltx1lp1gtexp ltp1lz0gtexp 00 00d d d d ldigdN1 700 27139 27139 27139 i m At 7121 7i m At K i z e W exp 2m 6 W N 1 exp 6 W 1 N 1 gtlt At p At H X tplml if 1P110 7 2 e expltm 2m 6 expltm 2x0 00 00 d d d d ldigdN1 foo 700 27139 27139 27139 2 2 eipmm w ll exp ZION exp eipN 1 N 1 mN 2 gtlt z m z At p2 At H 1171m1 mo 771 if 2 gtlt e expltm 2m expltm 2x0 7 Now we performed the integral over the p s which can be done for any potential because it amounts to gaussian integration only We have evaluated such integrals in an earlier lecture and we found that 00 d 39 A 52 m 39 ac39iac39ii2 e mw mj lkilf pg 7 e 700 27139 2mht which is nothing else but the free particle propagator in 1D A m 0 1Atmmnm2 Ut0z LmdzldzgdN1e 2 A02 2N4 gtlt 39 Af rm729 m i r1tac 7w 2 a iAt m 2 At 7 1 o 7 2 X 675 2 A02 272 tea 2 A02 2 8 This is again just a string of Gaussian integrals and we could in principle just evaluate it In order to illustrate the physics however we chose to interpret this as Feynman did in terms of path integrals We considered a path xt which was de ned through a series of points W H 7N717N727 7170 H lt95j1 951gt2 t jAt At Then 2 t t Hm 6 39 PHY 5645 Summary of Lecture 11 WKB approximation In this lecture we studied the semiclassical approximation of Wenzel7 Kramers and Brillouin WKB7 which is a technique for solving typically a 1D Schrodinger equation As mentioned before7 exact solutions are rare and not generic enough to study the variety of quantum phenomena Much can therefore be learned from nding a controlled approximate methods We motivated the approximation by noting that for a 1D Schrodinger equation 2 12 62 7 11496 VWWW BMW 1 a smoothly varying potential will introduce smooth variations to the wavelength The key idea is that the potential should vary over lengthscales much larger than a wavelength We would expect that a 77 local wavelength77 can be de ned as 2 h 2 71 AW L 4 We 2mE 7 Vz 1996 x2mE W90 The approximate wavefunction can then be written in terms of the phase accumulated from 0 to x as and a local momentum as ii an 1 dm we wltz0gte We gt where i corresponds to the rightleft moving wave We then introduced formalism which allowed us to develop this intuitive picture in a more rigorous manner In the limit of h a 07 the wavelength Ms also approaches zero In such case any potential appears smooth So what we need is an asymptotic expansion of the solutions of the Schrodinger equation in h We built such solutions by writing 206 WW 19296 wze 7 g 2 h hz 0 which is the equation that must be satis ed by Ms if ww 61quot is to satisfy the Schrodinger equation Since we did not assume anything about reality of x the transformation is exact Next7 we expanded Ms in powers of h ie z boa mm h2 2z and substituted into the equation for x lt 6x rugs1m mas2m Z l gm h fz 222 gm pzm 7 12 l 7 1 7 12 If this equation is satis ed for each power of h then we built an exact solution of the Schrodiger equation The hierarchy of the equations in increasing powers of h is 45 19296 7 hz 7 0 2 2 i g 0 1 g 0 0 3 0050 4 where we have written down the equations of the order 7172 and h 1 and omitted equations of higher order In the WKB approximation we keep only these two equations The solutions follow immediately7 since the equations are of rst order only 2 2 2 7g gltzgt altzgtg gltzgt 07 1ltzgt 1ltz0gt 1n 0 gt 00 j Apd 1996 and so the WKB wavefunctions are 10950 ii f MIMI Ibi c wi co 6 h 0 lt gt lt gt M We mentioned that the probability density associated with ww behaves as px 1 This dependence on classical velocity is familiar from classical mechanics Think of a particle con ned to move in some region in space The probability to nd it between z and z dx is proportional to the time it spends there 7 dx dx WW lpl39 Which is exactly what the WKB wavefunction describes dt dx i 73xd ldtl What is the region of applicability of this approximation From the de ning differential equation for Ms must have 206 WW 2 12 2 1 Clearly7 the above conditions are NOT satis ed when Vzmmmg E because the momentum at such a turning point vanishes pzmmmg 0 The WKB approximation breaks down in the vicinity of such turning points7 as is clear from the divergence of the wavefunctions Exactly how close to the turning point can we not trust the WKB approximation Near xmmmg we have p2z 2mE 7 2mVztmmng 7 72mVmmmmgx 7 zmmmg 1 1 7 gtgt27T dA 19196 dyc gtgt 2 1996 h gt and so the condition 2 p 96 h 19196 gtgt PHY 5645 Summary of Lecture 1 0 Con ict with reality and instability of matter we estimated the rate p at which a classical atom within the 77planetary model radiates energy We did this based on dimensional analysis p pw7 60713 a fer0wc 1 where the last replacement comes from looking at the power radiated far away from the dipole Since the units of p are energytime the only combination of er07w70 with the correct units is N 7 2 So as the dipole loses energy by radiating7 the radius of the orbit decreases7 which only increases the rate of emission Therefore7 classically the matter as we know it should collapse 0 The double slit experiment with electrons bullets and waves and Heisenberg uncertainty principle AxAp 2 For thorough discussion of the double slit experiment see Feynman Lectures Vol 37 Quantum Mechanics 0 Probability amplitudes and the Schrodinger equation We introduced the probability amplitudes 1 and stated that Quantum mechanics pro Vides mathematical description of the laws obeyed by the probability amplitudes ln non relativistic quantum mechanics this law is the Schrodinger equation 3 A mama Wm where 7jr7 t is a complex function which represents the probability amplitude and 72 is the Hamiltonian operator For a particle in an external scalar potential Vr we have A2 A P 7 V H 2m r where the momentum operator is For general Vr the solution to the Schrodinger equation is not known We will study some of the problems with soluble potentials and ecctract some of the generalfeatures which hold even for potentials which cannot be solved eacactly o librtl2 as the probability amplitude We showed that the Schrodinger equation o 712 2 maxim 7 7v no wont implies 6 2 h i 7 giant v i WWW 7 vi maxim 7 0 which is a form of a continuity equation This in turn implies that unless there is an in ux or an out uX of the particles we get 6 a dgrwmi 0 which means that the Schrodinger equation conserves probability This also de nes the probability density current as 3 i mem i ltwltntgtgtiltntgtgt o The probability amplitudes form a linear vector space called the Hilbert space The 77vectors of this in nite dimensional space will be denoted by 77ket77 MM and the basis will look something like Wokl 1gt7l 2gt7l jgtwd What this means is that in the space of functions which belong to the Hilbert space any function can be written as a linear combination of the basis functions ie n0 The scalar product sometimes also called the inner product is de ned as d3r rtirt lt lwgt where the last term represents the Dirac 77bra77 77ket77 notation o Operators Hermitian adjoint and Hermitian operators Loosely the action of an operator A on a wavefunction libgt generally produces another wavefunction AW W PHY 5645 Summary of Lecture 10 Feynman path integral calculation of the 1D harmonic oscillator propagator We showed last time that for the Hamiltonian H g WZ the evolution operator can be written as Ut 0 eifm 1 And the propagator 0lx0gt is Ut x 0 m m D zt g dt i2t 12t 2 z mo To evaluate this propagator we decided to shift the variables by writing the path as 965 96015 W37 where the classical path 0107 solves the Euler Lagrange equations of motion for a trajectory which at t 0 starts at 0 and at t end at x The path yt is a deviation from the classical trajectory Since the classical trajectory passes through the starting and ending space time point the deviation must vanish at the endpoints ie y0 yt 0 Then for the exponent we have dt t 7 gm dt g rem 9W 7 g ltsz aw lt3 0t dt rim nextmm m e 3 gm mommm W lt4 Now lets analyze the rst cross term t t t t t cit220mm 2i01tyt 7 mammm 7 7 malemm O t0 O O because y0 yt 0 Therefore we found that t dt Tim 7 595 7 5 0 2 2 t t t dt gem 7 39cm 7 dt mom 7 meme W dt gm 7 gm 0 0 0 But note that mth 7 mcl t 0 because its just the Newton7s equation of motion along the classical path Therefore we found that Otdt gay 7 gym 7 6 0t dt gzim 7 3m 0t dt gm 7 gym And so the propagator is yt0 002 07 6Sclmtmo0 f dt92tly2tx y 00 where t Sm o cw Exam 7 Exam 0 2 2 The important point is that the factor t 0 r K Mum K y D mg fotdt y2t gteay2ltt gt y00 does not depend on either the starting or the ending positions ie its independent of 0 and x Therefore it is only a function of time t mass m and the spring constant K Thus Ut z 0 x At m me czltwgttswogt0gt 8 In the limit of H 0 we already know the solution since this must reduce to the free particle propagator So m t 0 7 4 m V2m ht and so we only need to nd out the H dependence of the path integral yt0 Dyt5 f dt 2ltt gt7y2ltt gt y00 Everything else reduces to a classical problernl Next we calculated the classical action and found that letting H mwz Scl t 0 0 z2 953 coswt 7 2zz0gt 2 sinwt Finally we evaluated the path integral and found that yt0 i t mZ w 2 mu D t Ef 1 9 quot9 tl 9 y00 le 0 2 2 27Tz39hsinwt This concluded our path integral calculation of the 1D harmonic oscillator propagator Our nal result was ltlUt0l0gt m exp z2 x3 coswt 7 2xx0gt 10 PHY 5645 Summary of Lecture 16 Hydrogen atom The Schrodinger equation for two particles interacting via a central potential Vlr1 7 rgl is z h v tr r pipivqr in mm 139 at 7 17 2 2m1 2m2 1 2 7 17 2 The time dependence separates therefore 7tr1r2 e Etw l 1392 and 2 2 E771quot171quot2 L g VlI391 I392lgt 770139171392 27711 27712 The two particles are assumed to have distinct masses m1 and m2 Note that this is a 6 dimensional differential equation Now introduce the center of mass and relative coordinates mlrl m21392 R I39 r1 7 r2 7711 7712 In these variables the Schrodinger equation is 2 12 62 7 12 62 43MEE7ZE WMWE EWE where M m1 m2 is the total mass and M is the reduced mass Since the potential is independent of the center of mass variable the wavefunction separates as 7Rr fR7I39 and fR eiK39R ie the cm wavefunction is a planewave We can just take K 0 and restore it at the end if we wish simply by shifting the nal energy by h2k22M So 712 62 m m 7rE7I l012 Now since the potential is spherically symmetric we can chose the eigenstates of the resulting Hamiltonian to be simultaneous eigenstates of L2 and L i ur wmmmw l and 12121ltl1gt 2M6r2 2 r2 V0 ur Eur For attractive Coulomb potential Vr ZEZT we expect bound states at E lt 0 in addition to the scattering states at E gt 0 Now focus on the bound states De ne a dimensionless variable 2ME P T and given the asymptotics we derived in previous lectures let umvaw where wp must satisfy dZw dw 7 2l 1 7 2l 1 0 dp2lt pgtdpltpo HM and iZeZ 2 P07 h Expand wp in power series wp a0a1pagp2 k1kak12l1k1ak172kakp072l1ak 0 therefore 1H1 2kl17p0 a k1k2l2 This power series must terminate at some nite power otherwise as k a 00 aft g wpe2p aspaoo Therefore p02Nl12n where N012 then Zez 2 Z2 4 Z2 4 ZZR 7 i2n EJ 6 7 6 7 9 h 2712712 2on2 n2 where the Bohr radius is a 72 and Rydberg is 136eV Note that the energy depends only on the combination N l i n 1meansN0l0 n 2meansN0l10rN1l0 etc The total degeneracy not including spin is therefore 712 Now 1H1 7 w 0 F 7N2l22 a k1k2l2 p 1 1 p where the con uent hypergeometric function is a z aa 1 22 F 1 W 77 1 1017672 cllcc1 2X1L And nally the solution is 2Hrle quot n l wnzmn 07 1F17n l 1 21 2 2mY1m6 3 2l1l 27101 7171 where H x72uEh Lastly we discussed the nodes of the radial wavefunction in the context of the oscillation theorem and found that each radial wavefunction has n 7 l 7 1 nodes between 0 and 00 and states with angular momentum higher that l 0 have an additional node at the origin LZ39 4W olt 7439 fu zcra L L L 7 3 lA j T 1 3927 141 31dquot E43007 i 39 2 I 174C V011 N b 1 e Maw7A p 74V ymfi 4 42 9 Z 4 1 fL 4104396 e 14w 2 W 4 9 V4 m m M MW 3 MV 7 WE MW We 4w 4 m gym MM A 13944 14 m 4c Wmam rot47w 2 09 22 7 9 70 22010252 WL WM N mquot 7717 74 Ar w gjanj 139 M M 44 79441 m39f W 1 4 1 3914 mg haW I39LL Wm WM W W I 7 m m dam 44 M 74k QMj39 7 m M paw mow swam 47 94 wk qu39m MW Wm M 1 W 4 MZW 1 7quot w r39 74 5 ampVaw5 1x M vvvm w Ul39laa m 39 MI39C ZL 6 L9 we m bud Iv Ooidb 739 6644 m J 5 9144 AV Xvi abv dw 2 Mi a Maw kW aw fy1 CMvWJ 0Hquot 249 ng 39fQ vaw 1 Mvle Eda 21de 2 W 274 97 3 2144 1 94 93964 9 fn g Wvf d XI 31 fwu39 406 3 WM AW W 6 1W74 J L 7 2 f e ggq Ziiwz 393 MW y7 T 14 2 1 JI W 0 L W 6 W I 2 w 49 quot IW WW 3 939 MW Jx gaffer Wm H A M L L gtZ V 59 l 7V Wyvmv r M V a w 27 W 1W7 m 397VU394M I I 34739 I W 071 0L 9Z 0 J V e e 5722 g n e LIEC y he 72f gj s 9quot 439A 7216 7amp6 Z c Lq o 4 M 351quot r39IL I 31 32h uquot ris i 115i A 7924 Hmixmi 7 mm v2 4 3 ELZ3 L3 3 2 17 97 1171 Z 72 ue fob24501 aMX39M i 74499quot 1 6 2 2 2 p39 77 mm quot02 7er 2 36 222 quotW W M A 4L 444494 is ndW 1 a 939 fi 3 32 2Z Zaz 73 f7 g7 252 2 sz Amy401474 szh g7 q 51w VANWW2 g 3 z39AZ V 7 1 4 3 1 9 b 27 4 24 A 54773 54 57351 56 M 9 14 f ah W4 110 N h I LL I Z 41114Ml 7z4h142 t wang7 0 90 01 IL Z an 111 g A an LL30 0 l ga k ifgp 0 LI0 39 7 r LL Z 41m n1 gut QltL mL j 2 n1 fIA 1 4 4 UL If 39L a M1 4 dr Fm J2 4 0 ccj L 0L4L ark I CU cnI 2 w lt1 PM 4 1 42 7 AZ 7A 7 C a 1 A414Cr ZJ J fbm o 74M Ploj i k a cuj EMC a d Fcrwj 7 z rc 392 a g z 39 gtgt7 We 441 W war12 l394 7quot 74414 21 IIAA AL JCe f w W744 4 39ampL 4 zli F g g E 6 f171 t gt 4L9 71444101 firm Mimic 139 Za39 m AJZmA r 97 h39m Mo 1739 c a T A zlaMALU Z m9 39ff T 4 39 A 14V r V rU a 1M AQm E 74 IK Q 346 Z f7 lfl NW1 E 4 r 0 u I39umMmp t fr 7r 7 a 6 Z f77u 7 41 magma 7739 L 1 M L a 1 2 C 79 d V M m L 7 7 a m A 7A L z F M 747 M I39KMW mm39 w 39 E I Iw 16 1 11 839 L A m 4m H g 7m 22 M 7717 73 2 72 NM 39 A 7 7l39l1 7 p7 I L L fmzzi FHA PHY 5645 Summary of Lecture 6 1D Harmonic oscillator and coherent states oHarmonic oscillator We considered a harmonic potential in 1 spatial dimension ie V k2 where k is the 77spring constant The Schrodinger eigenvalue equation is then 2 12 d2 1 777 7k 2 E l mm 2 x we we We pointed out that the potential is invariant under parity ie z a ic and since it is con ning7 we expect based on the non degeneracy theorem that we proved that the eigenstates are non degenerate and simultaneous eigenstates of parity they will be either even or odd under inversion Note that classically7 771211272 7km and the natural frequency of the oscillator is w So 71 d2 mu h 777 7 2 1 H lemwdz22hl Now de ne d 7 sz h d 2 7 2h 277m dx A r h i T 0 2h z 277m dx 3 The Hamiltonian can now be rewritten as A A 1 H ha ala This interesting because we could easily show that Then7 let 7 old and assume that we have found an eigenstate of 7 ie Next we showed that lngt n71amplngt 4 mum nlampllngt 5 This means that dlngt is an unnormalized eigenstate of 7 with eigenvalue n 7 l7 and dllngt is an unnormalized eigenstate of 7 with eigenvalue n 1 So lngt nln71gt 6mm anln1gt lt A TCTJ VV We found the normalization by requiring that 1 Then 00 O VVVVV n ltnldldlngt ltn71l f nlnilgt 62 since we can always choose n real Sirnilarly7 AAA HHHAA DH K3 HHH UIHgtOO i H g g 3 V AAAAAAA HH NO VVVVVVV H Q So dlngt mnew lt20 dllngt V711ln1gt 21 The d and ampl are often called ladder operators7 or destruction and creation operators respectively Next we argued that n has to be an integer otherwise we could lower the energy inde nitely The state whose 72 eigenvalue is 0 is the ground state and it is annihilated by 37 ie 3 0 0 A generic state is constructed by multiple application of the creation operator Wln n W30 n 0172 22 So we can think of an excited state ln 2 of the oscillator as containing 71 7quanta77 of energy ha in addition to the zero point rnotion energy We also showed that the wavefunctions in the coordinate representation can be found from the coordinate representation of the ground state wavefunction 20 Now7 0 2640 lt fxil22wgt 20 23 x 920 5 24 We noted that this wavefunction is even under z 2 is as expected by the theorems we proved before We generated the next excited state by applying ampl and we found that wax ltz1gt2 woltzgt Notice that this state is odd under x gt 7x We can generate all the higher energy state by the repeated application of ll to nd that W where is the Herrnite polynomial of order n d 2 H 71 27 4 z lt gt e me We brie y mentioned coherent states lozgt which are eigenstates of the destruction operator lm alagt and we found that up to a normalization factor it is lozgt 2200 PHY 5645 Summary of Lecture 12 WKB approximation continued derivation of the Bohr Sommerfeld quantization conditions In the previous lecture we studied the semiclassical approximation of Wenzel Kramers and Brillouin WKB and mentioned that near a turning point awning the solution 10950 ii f pmdm Ibi c wi co 6 h 0 lt gt lt gt W breaks down because pztmmng 0 We found that the exact condition for how close to the turning point the above wavefunctions remain accurate is h z 7 39 gtgt turnmgl Q mV mmmg l which means that as h a 0 the region where WKB breaks down shrinks Nevertheless for arbitrarily small but nite h the wavefunction diverges exactly at the turning point In analysis such perturbation is called singular In this lecture we studied a method to handle this case The method is called asymptotic matching and consists of nding an exact solution of the Schrodinger equation in the vicinity of the turning point We argued that this can always be done because the potential is a simple power law near gummy typically linear and we can solve the Schrodinger equation exactly for the linear potential Thus for one turning point on the right 721mwltxgt vltzgtwltzgt Ewe lt1 Vatmm39ng V tmm39ng 96 7 turninQDwa 2 i V tmm39ng 96 7 tummgw 0 3 Next we shifted and rescaled the coordinate as 1 2mV tu39rnin g 5 Ty W mmmg we found that the differential equation becomes 62 7 7 0 4 6222 w We argued that even though the equation is second order there is only one physical solution This follows from the non degeneracy theorem we proved previously for the 1D potentials which demand that the wavefunction vanishes at either ioo We constructed the solution to this equation by Fourier transforming or equivalently by rewriting it in the momentum representation We found that if we write the solution as we dqewwq then the Fourier coef cients satisfy 2 3 gs 1qu wq0 7bqconstgtltea where the solution in term of Airy function follows immediately since the equation is of rst degree Therefore DltP where 5 000 cos 15 q3gt We next noted that for nite z 7 awning 5 a ioo as h a 0 where the sign depends on which direction from the turning point we move So for a right turning point where the potential slopes upwards the sign is for the z to the right of gummy and 7 for the one to the left Therefore to match the exact solutions near the turning point to the approximate WKB solutions away from it we must work to the same degree in smallness of h For this we need asymptotic expansions of the Airy functions i 1 iggg 13130 5 2 gie 5 i 1 2 g 7139 gamma a 7 smgltiggt2 lt6 The order of limits that is taking h to 0 BEFORE taking the z to awning is the crucial point of the argument Remember the WKB expansion is organized in powers of h The exact solution that we found for the linearized potential near the turning point is valid in a region near xmmmg which INCLUDES part of the region where WKB approximation is valid In other words the two regions have a nite overlap and if h a 0 before z a xmmmg the two regions will always have a nite overlap We also noted that as h a 0 the exact solution in the vicinity of xmmmg is 1 an lim a feigf umiwlpwwml wmwfummg 2 l 1 wumm lim 7m a 7 t haven3 s O lt 7 sin wmwammg pz h w 4 This solution must match the WKB solution which is B 73 x 1 dm 6 h f M l Mummy for z gt mmmg 9 WW A 1 w mum IbWKBW Sin EA t gPWVw f 04gt for lt mmmg 10 mm wWKB The two solutions match if 2B C C A7 and Oz E The nal solution to the problem of one right turning point can now be written as A 1 w mum 7 Sin 7 t g pd 7 for lt ztummy and turnmg 7 gtgt 1i pltzgt h w 4 2lmvltzmmggti 1 1 1 5 2mV xt g A q 7 urnin lbw lt2thxmmmggt 2 12 t 9 7 2Vztu39rnin for lx 7 Lyman l lt g g V tu39rning A 77 w 1 dm W96 76 h LWW39W W l 7 for z gt zmmmg and z 7 zmmmg gtgt 1 2 2lmVzturning l g In the case of two turning points at 1 and 2 we found A 7 I I g wWKBz 67fw1dw Wm for z lt zl amp 1 7 z gt a 1 12 WW leV 9 1l g WKBz i sin p d 04gt for z gt 1 amp lzlg 7 xl gtgt 13 1095 h M QlWV12l Z wWKBz Leifw2dmllpwgtl for z gt 2 amp z 7 2 gt h al 14 V lp9 l leV WN 3 and from the exact solution near the turning points D 1 x1 lim 7m 7m x MW 15 Wm 21 WW D 1 1 7T 1 7 d 7 16 1331 W h pm z 4 lt gt E 1 12 7139 lirn Ms sin 7 pz d 7 17 1 OW h m 4 E 7quot M w lirn Ms 76 5 702 p 18 Wm 2 WM 19 Matching the coef cients we found AQ BD 513 05 2 4 2 and importantly 1 m 1 m B sin h A1 p d Esin 2 p d 20 1 w 1 w Esin 110xdx g A12 p d 21 7 1 m 7T 1 m2 77gt i E31nlthm1pd 4 h M pz d 2 22 B E71m1 71 2pz dx 7giim7r 771717727737 We thus derived the Bohr Sommerfeld quantization conditions 777 m 1 faxWM 2 210zdz 27TH lt71 n 0 1 2 1 PHY 5645 Summary of Lecture 17 L811le vector and the degeneracy of the Hydrogen atom bound states The extraordinary degeneracy of the hydrogen atom Kepler problem is connected to the fact that the classical orbits do not precess The vector which points from the origin along the semimajor axis of the orbit is a constant of motion Quantum mechanically this vector operator called Lenz7s vector is R 71 XL Lx 62 Lenz 2mp p TI which is written in an explicitly Hermitian form A direct calculation shows that p2 62 RL5mH 0 where H 7 7 7 2m 7 As we saw before the fact that the angular momentum L is a constant of motion ie commutes with the Hamiltonian with central forces lead to 2l 1 fold degeneracy of the energy levels Therefore we expect that the additional conservation law associated with RLW will lead to additional degeneracy Now RLenz L L RLenz 0 Moreover 2 4 2 2 2 It is an unusual feature of the hydrogen atom problem that the Hamiltonian can be written in terms of other constants of motion ie in terms if Rim and L2 Since RLW is a vector under rotations RLenzuy Liz thEcLlARLBTLZA also RLenzuy RLmzy 7 EMVALA It is convenient to absorb the factor 47 into the de nition of a new vector 7m K 7 R enz V 2H L This vector operator is Hermitian when acting o the negative energy states which are the ones we7re interested in Then KM Kzl mewALA KM Ly ihEMVAKA I M Lvl mewALA AAA VPWM From Eq1 we also have 7 64m 7 K2 L2 23239 Which means that if we can nd the eigenvalues and degeneracies of K2 and L27 we have solved the bound states problem To decouple the above algebra7 lets de ne 1 M LK N7LK then MMle ihEMVAMA leNzl iheublANA MMNV 0 Notice that both M and N obey commutation relations of the angular momentum In terms of M and N the Hamiltonian is AAA ICTJCT VVV 77164 7 72 2M2 2N2 232 Since M and N obey angular momentum commutation relations7 we know all about their eigenstates and eigenvalues We can nd simultaneous eigenstates of M27 MZ7 N2 and NZ which we denote by lM7N7M7Vgt where leM7N7M7Vgt hZMMl 1llM7N7M7Vgt NZlMNLVgt h2NN1lMNLVgt 9 leMNLVgt hulMNLVgt 10 NleNLVgt hylMNLVgt 11 where M can take on values 0 l 1 3 as can N In addition7 7M77M1M 71jl7azn7d V 7N77N1N71N Because7 ELEML07 we haveKL0and M2 i N2 0 So the only states which are physical are therefore the ones with M N The eigenstates of H are lM N711 Vgt and HlMN7M7Vgt h2lMN7M7Vgt ZlMN7M7Vgt 64m 7 64m 24h2MJ1 1 2h22A 1 where 2M 1 17 27 37 which is the principal quantum number n For a xed value of M N there are 2M 1 values of u for each of which there are 2M 1 values of V Therefore the total degeneracy not including spin is 2M 12 n2 Thus7 the additional symmetry associated with the conservation of the Lenz vector gives the complete account of the nzifold degeneracy of the hydrogen atom bound state spectrum Note7 that lM N7 Vgt are eigenstates of Lz7 however7 they are not eigenstates of L2 PHY 5645 Summary of Lecture 3 Motion in 1D Recall that for the scalar potential the Schrodinger equation in 1D is 612 7 7 12 62 0 In nite square well 1 Va 0 if lt 2 Va 00 if m 2 3 So for 2 g the wavefunction vanishes We an think of this as a boundary condition on the solution For lt 7Lt e mx and ha 8 satis es im m ewe lt4 with the boundary condition Mia2 0 The wavefunctions are 2 2n 17rx 5 7 5 we iacos a gt ltgt we sin lt6 712 2 Silt gt 71172 2m 1 We noted that o The ground state has non zero energy This is an example of quantum zero point motion 0 The ground state wavefunction is node less lt vanishes only due to the boundary condi tions The rst excited state has one node7 the second has two nodes etc o The wavefunctions are orthogonal and the eigenenergies are l dzxmmm dxxfnzxnz a NW N o The wavefunctions are either even or odd under inversion x a ix oGeneral properties of eigenfunctions When there exist operators commuting with with the Hamiltonian lf 0 then the eigenstates of H can always be chosen to be simultaneously eigenstates of O as well In the above example the parity operator commutes with H and the wavefunctions could be chosen even or odd under z a 7x We proved this theorem in the case of non degenerate eigenvalues and stated that ifthe spectrum is nifold degenerate there are always niproper linear combination of the degenerate eigenstates such that the resulting rotated degenerate basis is both orthogonal and diagonalizes As an example we showed that the eigenstates of the free particle are 6 and 6 While these eigenstates are not simultaneous eigenstates of parity there are two linear com binations namely lteikm i 6 which are orthogonal and which are i1 eigenstates of parity oSquare well V V5 for 2 12 and Vz 0 for lt 12 The eigen wavefunctions has the form 9x Ale M Ble ik z lt 7 7 a XW Azelk 3257mm lt i 8 9x Age m Bge ik z gt g 9 The eigenenergy 8 relates k1 to kg to k3 as g 7 W V W 7 mg 07 2m 2m 2m So we can choose k1 kg The solutions of the Schrodinger equation must be everywhere continuous and differen tiable therefore lim 9x lim 9x 10 mai mai lim Xz lim Xz 11 mai mai The boundary conditions at the left step lead to Ale ik1 Bleik1 Ageiik2 Bgeik2 Z klAleiik1 7 k1B16ik1 Z kgAgeiik2 7 iszzCik2 while at the right step they are A26ik2 B267ik1 A36ik1 B367ik3 iszink2 7 Z kngeiik2 iklAgeik1 7Z k1B367ik1 The eigenenergy S can be either smaller or larger than Vb For 8 lt Vb we will have bound states while for S gt 0 we will have scattering states see online supplementary material Lets study 8 lt Vb then i k1 g 2m Eilb Em That means that A1 B3 07 otherwise the wavefunction diverges at ioo Our set of 4 equations is then B167H Azeiik Bzeik 16 KB157H Zyszzeiik Zyszzeik 17 A25ik2 Bge ik1 A367H 18 z kgAgeik2 7 z knge ik KA357H 19 We used symmetry under inversion parity to reduce this system of equations We can take the solution to be either even or odd under z a 7x Even A2 B2 and B1 A3 B167H 2A2 cos 162 20 KB1 7H 2k2A2 sin Dividing the two equations gives us k H kgtan 22 2m 2mE 1 2ma2E FUbiE hz tan 2 hz gt 23 or7 writing the energy in the units of V5 the last equation becomes 1EiEt 1E Vo voan2avo where Vb a h22ma239 A2 7B2 and B1 7A3 gt a a B167 sin KB167H 22k2AZCOS PHY 5645 Summary of Lecture 8 Quantum dynamics We considered the time evolution of quantum states for a general time dependent Hamiltonian The basic question is Given an initial state W02 of the system how is the state at t gt to determined from lwt0gt7 Thus we must solve 3 A ZhEWW Htl tgt For a time independent Hamiltonian we already know the solution iwm megt lmportantly we mentioned that for a time dependent Hamiltonian the solution is NOT obtained by simply exponentiating the time integral of the Hamiltonian ie i t 7 todtHt immee gtww because in general Ht1 Ht2 31 0 Rather we showed that the result is iwmTGf mhwww where T is the time ordering operator For two operators T orders them in a chronological order the earliest to the left V H V TltAt1i t2 4 EW for tlgtz t2At1 for t2gtt1 Um D V We de ned an evolution operator 20212 T 67 0 dt Ht gt which takes a state at to ie WWO and evolves it according to the laws of quantum mechanics to the state at t ie lwtgt Ut t0l7t0gt We also showed that the evolution operator is unitary ie U 1tt0 UTtt0 This follows from the conservation of probability ie if lt7Zt0l7t0gt 1 then WWWW 1 lt follows from the Schrodinger equation that atUtt0 EHtUttO oSchrodinger and Heisenberg picture We mentioned that the representation in which the states evolve in time is called Schrodinger picture That is7 if we start in a particular state at to at some later time it generally becomes a different state In the Heisenberg picture7 the states are xed in time7 but it is the operators which evolve To see that the two pictures are equivalent we looked at a time dependence of an expectation value of an operator A A15 r t which for clarity is written as an explicit function of momentum and position operators as well as any possible explicit time dependence that it may have Then ltwtlA15ftlwtgt ltwlttogtlwltntogtAltmtgtUltmogtwlttogtgt lt3 In the Heisenberg picture7 the operator Ap r It depends on time not only through the explicit dependence7 but also from the time evolution operator Thus7 in the Heisenberg picture AHQ Ult7t0A157f tUt7to As a simple example we discussed the momentum operator7 p7 which is time independent in the Schrodinger picture7 but depends on time in the Heisenberg picture 15Ht Ultvt0f0t7t0 At t to the operators and the state in the Heisenberg and the Schrodinger picture coincide The above time evolution of operators can be written as a differential equations We considered the time derivative of AHt 21M Ultt0AprtUtt0gt 4 301on Ana 2 Wm Mm game mtvto Ultt0AprtUtt0 5 Ultt0lf1tgt ApftUtt0 mm gape a Utt0 0t7toA157ftHtUt7to lt6 Ultt0HAprtlgt 002 7 gm HAHltpftgtgtH lt8 The last equation is also known as the Ehrenfest theorem Next7 we considered a couple of examples First we showed that the Heisenberg equation of motion for time independent Hamiltonian is 412 dt m Second we showed that for the 1D harmonic oscillator 21H dt m inf dt H The solution to the last set of equations is A A 130 i t 0 t 7 t M cosw mw s1nw 13t 130 coswt 7 mw O sinwt where w Hm


Buy Material

Are you sure you want to buy this material for

25 Karma

Buy Material

BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.


You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

Why people love StudySoup

Steve Martinelli UC Los Angeles

"There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

Kyle Maynard Purdue

"When you're taking detailed notes and trying to help everyone else out in the class, it really helps you learn and understand the I made $280 on my first study guide!"

Jim McGreen Ohio University

"Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."


"Their 'Elite Notetakers' are making over $1,200/month in sales by creating high quality content that helps their classmates in a time of need."

Become an Elite Notetaker and start selling your notes online!

Refund Policy


All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email


StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here:

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.