ψ n (x) ψ n (x ) e ien (t t )/ exp x 2 + (x ) 2) cos ωt 2xx ]] 2 sin ωt t t iε/ω 2πi (sin ωt iεcosωt)

Transcription

1 Problem 1 The propagator K(x, t; x t ) x, t x, t ca be rewritte i terms of the eergy eigestates ψ (with eergy E ) as follows: K(x, t; x, t ) ψ (x) ψ (x ) e ie (t t )/ Suppose we perform a path itegral to obtai the SHO propagator: K(x, t; x, 0) πi si ωt imω ( exp x + (x ) ) cos ωt xx si ωt where we set t 0 WLOG. Now we wat to extract iformatio about the eigestates ad their eergies. a) We set xx 0 to obtai: ψ (0) e ie t/ πi si ωt We perform Fourier aalysis o the RHS to determie the eigeeergies. We eed a prescriptio for goig aroud the brach cut sigularity of u located at u 0. The correct prescriptio is to add a small egative imagiary piece to the time: t t iε/ω where ε is a ifiitesimal dimesioless regulator. Thus, si ωt si (ω t iε) cosh ε si ωt i sih ε cos ωt si ωt iεcosωt, ad we fid: ψ (0) e ie t/ εe/(ω) To see that this is the correct prescriptio, check the case t 0: ψ (0) e εe/(ω) πi (si ωt iεcosωt) π ε The LHS is a coverget, regulator depedet sum, ad is therefore a positive real umber, as is the RHS. Havig regulated the propagator, we see that K(0, t; 0, 0) has a period 4π/ω, rather tha the aive expectatio π/ω. This is because crosses the u brach cut (covetioally alog the egative 1 ε cos ωt+i si ωt real axis) oce each period π/ω (at times t br (π + π)/ω), pickig up a extra mius sig; thus, K(0, t; 0, 0) satisfies K(0, t +π/ω; 0, 0) K(0, t; 0, 0) 1. 1

2 We write K(0, t; 0, 0) as a Fourier series: K(0, t; 0, 0) c e iωt/ Thus, c ω 4π/ω K(0, t; 0, 0) e iωt/ dt 4π 0 The atiperiodicity of K(0, t; 0, 0) over the iterval π/ω implies that the c vaish for eve. I priciple, all we have to do to extract the other coefficiets (for odd) is to perform the above itegral. However, this is somewhat difficult to do i practice. Cosider the followig shortcut: istead of addig a small regulator term, suppose we add a large egative imagiary part to t: t t i α/ω where α >0. We fid: ψ (0) e ie t/ αe/(ω) π (e α+iωt e α iωt ) Suppose we take the limit α. The RHS vaishes, whereas the LHS oly vaishes if all the E are positive. This must therefore be the case. Now take α to be large but fiite, ad expad the RHS i a power series i e α (we set t 0 for simplicity; it s easy to see that the e iωt s always lie up whe the e α s do): ψ (0) e αe/(ω) e α/ mω (1 ) e α 1/ π e α/ mω 1+ 1 π e α e 4α + We read off the eigeeergies: E ω/, 5 ω/, 9ω/, ad the coefficiets: ψ 0 (0) mω π ψ (0) 1 mω π ψ 4 (0) 3 mω 8 π 1. This property is obvious whe we obtai the propagator from the eergy eigestates, as the eigeeergies are E ( + 1/ ) ω, where the + 1/ cotributes the crucial mius sig.

3 Lookig at Sakurai eq..6.17, We see that these coefficiets should take the geeral form: ψ (0) 1! H (0) mω π It s easy to verify that the first few work out as expected. The odd eergy levels do ot appear sice their wavefuctios are odd fuctios of x, so that ψ 1 (0) ψ 3 (0) 0. Why is this procedure valid? Suppose the regulated sum ψ (0) e ie t/ εe/(ω) is coverget for ε ifiitesimal. Icreasig ε ca oly make it more coverget. Thus, takig α, we obtai the aalytic cotiuatio of the propagator to t iα/ω for α large. Sice the RHS is aalytic, we must obtai the same result upo aalytically cotiuig it to t iα/ω, allowig us to match coefficiets i a regime where the expasio i e α is valid. b) We geeralize the procedure of the previous sectio to the case xx 0. The propagator becomes: K(x, t; x, 0) πi si ωt exp π (e iωt e iωt ) e iωt π (1 e iωt ) ( ) x cosωt 1 i si ωt exp x exp x ( e iωt +e iωt e iωt e iωt ) ( ) 1 e iωt + e iωt 1 e iωt We perform the same procedure as above, replacig t iα/ω ad expadig i powers of a e α. The term i paretheses iside the expoet becomes: Thus, defiig γ x, we fid: 1 a +a 1 a (1 a) (1 a)(1 + a) 1 a 1+a 1 a 1 +a ψ (x) a (E/ω) a 1/ (1 a ) 1/ π aγ e γ exp 1+a a 1/ ( ) 1 1 a γ e γ π! a a exp 0 1 a 1 + a a 1/ +γ a 3/ + 1 π (γ 1) a 5/ + e γ a0 3

4 Thus, we fid the spectrum E ( + 1/) ω, ad the modulus squared of the correspodig wavefuctios: ψ 0 (x) exp m ωx π ψ 1 (x) x exp x π ψ (x) 1 ( ) x 1 exp m ωx π These are easily checked agaist Sakurai eq..6.17: ψ (x) 1! ( mω H π ) x exp x The ext step If we assume that the wavefuctios are real, the the above procedure is sufficiet to recover them from the propagator. Obviously, this is ot always a good assumptio. To circumvet it, we let x ad x take arbitrary values. Writig this out, we fid: K(x, t; x, 0) π (e iωt e iωt ) exp mω (e iωt e iωt ) 1 ( x + (x ) )( e iωt + e iωt) xx Replacig e iωt with a, ad defiiig γ mωx ψ (x) ψ (x ) a (E/ω) a 1/ 1 a π a 1/ 1 a π ad γ mωx exp 1 1 a exp as above, we obtai: 1 ( γ + (γ ) )( 1 +a ) aγγ γ + γ aγγ a ( γ + γ ) exp 1 a Expadig the RHS i a power series i a, we fid: ψ (x) ψ (x ) a (E/ω) a 1/ + γ γ a 3/ + 1 π (γ 1)(γ 1) + exp γ + γ Factorig the aswer, we read off the wavefuctios: ( ) 1/4 ψ 0 (x) exp 1 π γ ( ) 1/4 ψ 1 (x) γ exp 1 π γ ψ (x) 1 ( ) (γ 1/4 1) exp 1 π γ 4

5 where the overall phase is chose arbitrarily. Problem We wish to fid the mometum space propagator: where the state p, t satisfies: ad therefore takes the form: K(p, t; p, t ) p, t p, t pˆh(t) p, t p p, t p, t U (t) p ( ) iht exp p where p p, 0 is the Schrödiger picture mometum eigestate. Thus, for a free particle, with Hamiltoia H pˆ m, K(p, t; p, t ) p e ih (t t )/ p p i e m (t t )/ p p δ (d) (p p ) exp i p (t t ) m where d is the umber of dimesios i the problem (typically 1,, or 3). Problem 3 We wat to solve the differetial equatio ( m ) x E G(x, x, E) δ(x x ) where G(x, x, E) G(x x, E) is the oe-dimesioal Gree s fuctio. There are various tricks for obtaiig Gree s fuctios. For istace, oe ca Fourier trasform the above equatio to fid a algebraic equatio for G (p), which is easily solved. Cotour itegratio ca the be used to obtai G(x). However, i this case, a brute force approach is tractable. We set x 0 WLOG to obtai ( m ) x E G(x, E) δ(x) We first write dow the homogeous solutios (without the delta-fuctio source), which take the form: G H (x, E) Ae ike x +Be ike x 5

6 where k E me/. Sice the source is localized at x 0, we must have G(x, E) { Ae ik E x + Be ike x x <0 Ce ike x +De ike x x >0 for some coefficiets A, B, C ad D. To determie the relatioship betwee A, B ad C, D, we itegrate the defiig differetial equatio from x ε to xε for small ε: G(x, E) m x ε ε E ε ε G(x, E)dx 1 The cotributio from the secod term o the LHS is ifiitesimal, ad ca be dropped. We the fid or Cotiuity of G(x, E) at x 0 further requires The solutio is G(x, E) G(x, E) x xε x x ε m (ik E C ik E D) (ik E A ik E B) m A+B C + D C A B D i m k E To fix the remaiig ambiguities, we eed to specify boudary coditios. For potetial scatterig problems (E > 0), we may wat to exclude icomig waves. To do so, we must set AD 0. Thus, G(x, E) i m ike x e k E This is also a appropriate choice for boud state problems (E < 0), where k E iκ E for κ E me/, so that G(x, E) m κe x e κ E ad therefore G 0 as x ±. I either case, for E 0 we fid G(x, E) m x after discardig a diverget costat piece ad terms which vaish as E 0. This is easily see to satisfy the defiig equatio, sice which ca be verified by itegratig both sides. 1 d dx x δ(x) 6

7 Other choices of Gree s fuctio will differ by the homogeous solutios G H (x, E) discussed above. Solutio by cotour itegratio For completeess, we also preset the cotour itegratio method. To begi with, we Fourier trasform. Defie 1 G (p, E) dx e ipx/ G(x, E) π We start with the defiig differetial equatio ( m ) x E G(x, E) δ(x) Multiplyig by 1 π e ipx/ ad itegratig over x, we evetually fid ( ) p m E G (p, E) 1 π after itegratig by parts ad droppig the boudary terms. This is the promised algebraic equatio for G. Thus, we obtai the formal solutio G(x, E) where k E me/, as before. 1 π dp e ipx/ G (p, E) m π dp e ipx/ p k E The RHS ca be evaluated by cotour itegratio: for x > 0, we close the cotour i the upper half plae, ad for x < 0, we close it i the lower half plae. The itegrad has poles at p ± k E. For E > 0, the poles lie o the real axis, ad we must choose whether to go over or uder them. To reproduce the Gree s fuctio for potetial scatterig discussed above, we go uder the p k E pole ad over the p k E pole. Thus, for x > 0, the itegratio is couter-clockwise, ad we pick up the p k E pole givig G(x >0, E) i m k E e ike x For x < 0, the itegratio is clockwise, ad we pick up the p me pole, givig G(x < 0, E) i m k E e ike x Thus, G(x, E) i m ike x e k E as we foud previously. A differet prescriptio for goig aroud the poles will yield a differet Gree s fuctio, correspodig to differet boudary coditios. For E < 0, the poles lie at p ± iκ E, where κ E me/. The most atural choice of cotour is alog the real axis. Thus, for x > 0, we pick up the p iκ E pole, givig G(x > 0, E) m κ E e κe x 7

8 Similarly, for x <0, we pick up the p iκ E pole, givig G(x < 0, E) m κ E e κe x so that G(x, E) m κe x e κ E as we foud previously. Problem 4 Cosider the wavefuctio: ψ(x, t) exp is(x, t)/ If we allow S to be complex, the this asatz is geeral so log as ψ is ovaishig. We apply Schrödiger s equatio: Thus, S t eis/ i ψ t Cacellig a overall factor of e is/, we fid: Hψ m ψ + V (x) ψ(x, t) x ( i S m x 1 ( ) S ) e is/ + V (x)e is/ x i S m x V (x)+ 1 ( ) S S + m x t Suppose that S has a characteristic legthscale of spatial varatios L ad a characteristic ( ) magitude S 0. Therefore S S x 0/L ad S S x 0 /L. The S S term is much smaller tha the x x term so log as: or S 0. Thus, i the large S/ limit, we obtai: S 0 ml S 0 ml 0 V (x)+ 1 ( ) S S + m x t ad S may be take to be real. We recogize this as the classical Hamilto-Jacobi equatio for the Hamiltoia H p S +V (x), where p. m x Now cosider a free particle, with V 0. Thus, S t 1 ( ) S m x We assume the asatz: S W(x, α) α m t 8

9 for some costat α. Thus, α ( ) W x or W αx+s 0 where the arbitrary sig is absorbed ito the defiitio of α. Thus, S αx α m t +S 0 ad we fid the wavefuctio: i ψ(x, t) exp (p x Et) +iφ 0 where p α is the mometum, E p /m is the eergy, ad φ 0 S 0 / is a arbitrary phase. The solutio is exact i this case, sice S x vaishes idetically. Problem 5 A coheret state of the 1D SHO is defied by where a is the aihilatio operator ad λ C. a) Cosider the state We compute a λ λ λ λ e λ / e λa 0 a λ e λ / ae λa 0 e λ / a, e λa 0 λe λ / e λa 0 λ λ where we use a 0 0 ad a, e λa λ a, a e λa, as prove i problem 1 of PS #. Thus, λ is ideed a coheret state, with a eigevalue λ. We ow check that it is ormalized. We have: λ λ e λ 0 e λ a e λa 0 To simplify this expressio, we first derive a idetity. For operators A ad B such that A, A, B B, A, B 0, we already showed that e A e B e A+B e 1 A,B. Applyig this idetity twice, we fid: e A e B e A+B e 1 A,B ( e B e A e 1 B,A) e 1 A,B e B e A e A,B. Sice a is ot Hermitea, it ca have complex eigevalues. 9

10 (This is the idetity suggested i the prompt for part (e).) Applyig this idetity to e λ a e λa, we fid e λ a e λa e λ e λa e λ a Thus, so the coheret state is ideed ormalized. λ λ 0 e λa e λ a b) We have x mω ( a + a ), p i mω ( a a ) Thus, we readily compute (for the state λ): x To compute x ad p, we rewrite (λ +λ mω ), p i mω (λ λ ) x ( a + a ) mω ( a + aa + a a +a ) mω ( 1+a + a a +a ) mω ad similarly p mω ( a a ) mω ( a + aa + a a a ) mω ( 1 a +a a a ) That is, we place these operators i ormal order (all a s o the right, all a s o the left) which makes the computatio of the expectatio value easier. We ow fid: Thus, ad so x which saturates the boud. p mω ( 1+λ + λ + (λ ) ) mω ( 1 λ + λ (λ ) ) mω ( 1 + (λ+λ ) ) mω ( 1 (λ λ ) ) ( x) x x ( 1+(λ +λ ) (λ +λ ) ) mω mω ( p) p p mω ( 1 (λ λ ) + (λ λ ) ) mω ( x) ( p) 4 10

11 c) We wat to express the coheret state i terms of the ormalized eergy eigestates ( ) a 0! which is a eigeket of the umber operator N a a with eigevalue. From the defiitio of λ we are able to read off λ e λ / 0 λ ( a ) 0 f()! 0 where f() λ e λ /! The probability of measurig a eergy E ω ( + 1/) is the f() λ! e λ which is the Poisso distributio with expectatio value λ, so that E ω ( λ +1/ ) However, the expectatio value eed ot be the same as the most probable value (mode)! For istace, the expectatio value eed ot be a eigevalue, whereas the mode always is. I this case, oe ca either look up or check by had that the most probable eergy is 3 E (mode) ω ( λ + 1/ ) where x deotes the floor, i.e. the greatest iteger less tha or equal to x. d) We wish to compute e ipl/ 0 for l real. As before, the best approach is to write the operator i ormal orderig, at which poit all computatios become trivial. We have e ipl/ mω (a exp l ) a exp λ ( a a ) where mω λ l Applyig e A+B e A e B e 1 A,B, we obtai e ipl/ e λa e λa e λ / 3. For λ a iteger, both E ω ( λ + 1/ ) ad E ω ( λ 1/ ) have equal probabilities. 11

12 Thus e ipl/ 0 e λ / e λa e λa 0 e λ / e λa 0 λ as desired. e) There are several possible approaches to this problem. The simplest is to use the results of part (d), together with the kow groud-state wavefuctio: But the where l λ mω. Thus, ψ 0 (x) ( ) 1/4 exp 1 π mωx ψ λ (x) x λ x e ipl/ 0 x l 0 ψ 0 (x l) ( ( ) 1/4 ψ λ (x) exp 1 ) mω x λ π mω ( ) 1/4 e λ exp 1 mωx + mω λx π However, this procedure is oly valid for real λ, sice we implicitly assumed that l was real. To establish the aalogous result for complex λ, we follow the procedure suggested i the prompt. The defiig equatio a λ λ λ becomes a differetial equatio whe writte i terms of the wavefuctio. We have Thus, or where x 0 /mω. The solutio is for λ C. We compute the ormalizatio: ψ λ (x) dx N a mωx+ip mω mωxψ λ (x)+ x ψ λ(x) mω λψ λ (x) ( x x0 λ)ψ λ (x)+x 0 x ψ λ(x) 0 ψ λ (x) N exp 1 ( ) x x0 λ x 0 exp 1 ( ) x x0 Reλ x +(Im λ) dx N x 0 π e (Im λ) 0 1

13 Thus, N 1 λ) x 1/ e (Im 1/4 0 π +iφ where φ is a ukow phase factor, so that ψ λ (x) 1 λ) x 1/ e (Im +iφ exp 1 1/4 0 π x 0 1 λ) x 1/ e (Im λ +iφ exp 1/4 0 π 1 / x 1/ 0 π 1/4 e λ e λ / e i φ exp where we absorb a extra phase ito φ o the last lie. Note that x ( ) x x0 λ x + x λ 0 x 0 x λ x x 0 + x 0 λ λ e λ / λ / 0 e λ a e λa 0 e λ / λ / e λ λ 0 e λa e λ a 0 e λ / λ / e λ λ where we apply e A e B e B e A e A,B as i part (a). Thus, we fid ψ λ (x) ψ λ (x) dx 1 e λ / λ / e λ / λ / i e (φ λ φ λ ) x 0 π exp x x + ( λ + λ ) x 0 x 0 ( x exp 1 ( λ + λ )) 1 + x 0 e λ / λ / e λ / λ / e i (φ λ φ λ ) e e λ / λ / e i (φ λ φ λ ) e λ λ ( λ +λ ) / ( λ + λ ) Thus, we must have φ λ φ λ, so we may fix φ λ 0 (sice it is idepedet of λ). I et, we fid ψ λ (x) 1 / e λ / x 1/ 0 π 1/4 e λ exp x x + x λ 0 x 0 which matches that give i the problem assigmet after accoutig for the differet ormalizatio covetios. Now we wat to examie how the coheret state evolves with time. Sice λ 0 λ! e λ / we see that e iht/ λ 0 λ e iωt e λ /! 0 ( e iωt λ )! e λ / λe iωt Thus, ψ λ (x, t) ψ λ(t) (x) 13

14 where λ(t)λe iωt. We rewrite the time-depedet wavefuctio as ψ λ (x, t) 1 x 0 1/ π exp 1/4 1 ( ) x x x0 Re λ(t) + i (Im λ(t)) x 0 x 0 I this form, it is clear that ψ λ (x, t) is a wave packet of fixed width, which oscillates back ad forth with cetral positio x ce (t) x0 Re λ(t). The x-depedet phase (the secod term i the expoetial) represets a time varyig mometum p ce (t) Im λ(t). I particular, takig x 0 λ(t) λ 0 e iω(t t0) with λ 0 real, we fid x ce λ 0 cos ω (t t 0 ) mω p ce mω λ 0 si ω (t t 0 ) so the positio ad mometum of the wave packet oscillate 90 out of phase with p m ẋ, as oe would expect of a classical particle. Problem 6 The Hamiltoia for a charged particle i the absece of electric or magetic fields is: where V (x) is some uspecified potetial. H p m + V (x) Suppose we icorporate a vector potetial via the prescriptio p p e A. Thus, c H 1 ( p e ) e (p A + A p)+ m c c A Sice i geeral A A(x), A does ot commute with p. Suppose that the magetic field B is costat. Oe ca check that the vector potetial obeys A B. Thus, the Hamiltoia becomes: H 1 ( p e m c p m p m A 1 B x ) e (p (B x)+(b x) p) + B x + V (x) 4c e B (x p p x) + e 4mc e (B L)+ e mc where i the last lie we use x p p x, sice 8mc B x (B x) + V (x) 8mc B x (B x) + V (x) ε ijk (x j p k + p j x k ) ε ijk x j, p k 0 14

15 The secod term i the Hamiltoia is simply µ B, where µ e mc L is the magetic momet iduced by the orbital agular mometum. Now cosider the third term, ad write B Bẑ WLOG. We fid: e B 8mc x + y At first, the appearace of this term may seem slightly suprisig, as we bega with a setup which was (potetially) ivariat uder traslatios x x + x (assumig that V (x) is a costat), whereas this term is maifestly ot ivariat uder traslatios (it looks like a D simple harmoic oscillator potetial). However, upo writig A i the form A 1 B x we have made a gauge choice which is ot ivariat uder geeral traslatios x x + x, but oly uder z traslatios (ad rotatios i the x-y plae). The caoical mometum p ca p phys + e A is ot c gauge ivariat, hece the appearace of other terms i the Hamiltoia which are ot gauge ivariat. However, i cases where V (x) breaks x ad y traslatio symmetry itself, this gauge choice is rather atural. Problem 7 Start with the Lagragia for a free particle: L 1 m ẋ We add a iteractio term (for a particle of charge + e): L 1 m ẋ + e c ẋ A(x) The Euler-Lagrage equatios give: d L dt ẋ i L x i Thus, d m ẋ i + e dt c A i(x(t)) e c ẋ j i A j m ẍ i + e c ẋ j j A i ad so m ẍ i e c ẋ j ( i A j j A i ) e c ẋ j ε ijk ε klm l A m e c ε ijk ẋ j B k 15

16 or m d x dt e c ẋ B which is the Loretz force law for magetostatics. Startig with this Lagragia, oe ca obtai the Hamiltoia of the previous problem by the stadard procedure. The caoical mometum is: as we saw before. Thus, H p ẋ L p ca L ẋ m ẋ + e c A p phys+ e c A (m ẋ e c A ) ẋ 1 m ẋ + e c ẋ A(x) 1 m ẋ 1 m p phys 1 m p e c A I the above solutio, we assumed that A, ad hece B, was idepedet of time. If we allow B to be time-depedet, the E will be ovaishig i geeral, with E A φ. The A term which we t t omitted above will cotribute the additio ee term to the Loretz force law, as expected (we should also iclude eφ(x) i the Lagragia i this case.) 16