QM1 Problem Set 5 solutions Mike Saelim

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1 QM Problm St 5 solutions Mik Salim If you find any rrors with ths solutions, plas mail m at mjs496@cornll.du. Rcall from th last problm st that th Hamiltonian is givn by Π H m + φ x p A x m c + φ x p m mc [ p A x + A x p + mc A x + φ x Lt s solv for th probability flux first. With th Schrödingr quation, w don t hav to dal with bras and kts - just th wavfunction. t t ψ x ψ x ψ x ψ x + ψ x t t ψ x ψ x ψ x + ψ x t t Th tim drivativs of th wavfunctions ar givn by th Schrödingr quation: ψ t ψ t i h Hψ x [ h i h m ψ h mc i Aψ + A h i ψ + mc A ψ + φψ i h m ψ + [ Aψ mc + A ψ + [ i h mc A ψ + φψ i h m ψ + [ Aψ mc + A ψ + [ i h mc A ψ + φψ, whr ψ, ψ, A, φ, and H ar all c-numbr non-oprator functions of x. Plugging ths in, t i h m ψ ψ + [ Aψ mc ψ + A ψ ψ + [ i h mc A ψ + φ ψ + i h m ψ ψ + [ψ Aψ mc + ψ A ψ + [ i h mc A ψ + φ ψ. Notic that th i h trms on th right asily cancl, and th mc trms in th middl add up to two total drivativs which ar th sam. W can mak th scond drivativ trms on th lft into total drivativs if w add So now w hav and thus i h m ψ ψ + i h m ψ ψ. t h m Imψ ψ + mc A ψ t + j 0 whr j h m Imψ ψ mc A ψ.

2 W gt th continuity quation for our probability dnsity, with an xtra trm in th probability currnt proportional to th probability that th particl is at a spot and th EM fild strngth at that spot. Th coupling to th EM fild changs how th wavfunction volvs with tim. For th scond part of this problm, w ll b working solly with oprators. From th Hisnbrg quation of motion, dπ dt i h [ Π, H i h m [ Π, Π + [ Π, φ. Lt s work with th individual componnts hr. I ll work in Einstin notation, which is a common convntion for quations in physics: any rpatd indx in a trm is undrstood to b summd ovr. In ffct, I m just dropping th summation signs in front of th xprssions. Any indx that you only s on copy of in a trm is not summd ovr. Th scond commutator is prtty simpl: [Π i, φ [p i, φ i h i φ whr i x i. Th first commutator, howvr, is not as simpl as it looks. Th canonical commutation rlations involv x i and p i, and Π i contains both of thm. Whil [Π i, Π i 0, [Π i, Π j is not ncssarily 0. [Π i, Π j [p i, A j + [A i, p j i h c c ia j j A i i h c δ kiδ lj δ kj δ li k A l i h c ɛ mklɛ mij k A l i h c ɛ ijm A m i h c ɛ ijmb m whr ɛ ijk is th Lvi-Civita tnsor and I v usd th contractd psilon idntity and th dfinition of th cross product. If you ar not familiar with Einstin notation, this is a good opportunity to larn: i and j ar not contractd anywhr in any of this stuff how can thy b, sinc thy ar not summd ovr in th commutator? but th rst of th indics ar contractd and summd ovr. With th forsight w hav by looking at what th final answr is, lt s try to comput what dx i dt is for this Hamiltonian: dx i dt i h [x i, H i h p i m mc A i Π i m, [ x i, p jp j m mc p ka k + A k p k so, as som may hav xpctd, th position oprator volvs with tim according to th kinmatic momntum Π, just lik it did for th fr particl Hamiltonian.

3 Finally, w can put this all togthr: dπ i dt i h m Π j[π i, Π j + [Π i, Π j Π j + [Π i, φ ɛ ijm Π j B m + ɛ ijm B m Π j i φ mc Π j ɛ ijm c m B Π j m ɛ imj B m i φ m [ d x c dt B B d x i φ i dt i d Π dt [ d x c dt B B d x φ. dt Sakurai p.50, #37 W want to comput th phas shift producd by th tim volution of th nutron s wavfunction in th magntic fild. For simplicity, w ll only considr th componnt of th magntic fild prpndicular to th nutron s motion sinc that s th only part of th fild that affcts th nutron anyway, and w hav th frdom to lt that fild b in th x-dirction, whil th nutron travls in th z-dirction. Our tim-volution oprator is xp [ īh HT [ xp ī p z h m p z [ xp ī h µ B T m g B n m n c S x T Th first trm in th xponnt will b th sam for both arms, but th scond trm will b diffrnt bcaus it dpnds on th magntic fild. Bcaus th S x oprator rturns an ignvalu of h/ rgardlss of whthr th nutron is spin-up or spin-down, But, w also hav ī h µ BT ī h g hb n m n c T. T l v z l lm n p z /m n h/λ λlm n π h whr th nutrons travl through a magntic fild rgion of lngth l with d Brogli wavlngth λ. Thus, th nutron picks up a phas shift φ g nb λlm n m n c π h g nλl 4π hc B from th intraction with th magntic fild. Th diffrnc in th phas shifts btwn nutrons in magntic filds that corrspond to succssiv maxima in th counting rats must qual π, so φ g n λl 8π hc B π B 4π hc g n λl.. 3 SHO Partiton Function 3

4 a I hop you lik Gaussian intgrals. Thy ar awsom bcaus dx x not that sinix i sinhx and cosix coshx. π. Also { mω Z dx πi sin ωt xp mω x} i sin ωt t i hβ [cos ωt { } mω dx πi sin iβ hω xp mω i [cos iβ hω x sin iβ hω mω { dx xp mω } [coshβ hω x π sinhβ hω sinhβ hω To gt this into Gaussian intgral form, mak th transformation mω u coshβ hω x. sinhβ hω mω sinhβ hω Z π sinhβ hω mω coshβ hω coshβ hω β hω + β hω + β hω β hω β hω/ β hω βj+/ hω. j0 du u b It is prtty simpl to show that β ln Z i E ip i : β ln Z Z β i i E i βe i Z βe i i Thn, w can show that β ln Z quals th listd valu: E i P i Ē. Ē β hω/ β ln β hω [ β hω ln β hω β hω β hω + β hω hω + β hω 4.

5 Altrnativly, w can also show that th dfinition of Ē also quals th listd valu, though this taks a bit mor work: βn+/ hω β hω hω Ē βen E n n + Z n n0 { n n } hω β hω β hω + n β hω n0 n0 { hω β hω + β hω β hω n0 hω + hω β hω β hω d n β hω d β hω n0 hω + hω β hω β hω d d β hω β hω hω + hω β hω β hω hω + β hω. c For this part, it hlps to know that β T For high tmpraturs, θ E T trms: C N 0 Ēcrystal T β hω k B T and θ E T β hω. 3N 0 Ē β N 0 β T hωβ hω 3 hω β hω k B T hω β hω 3k B k B T β hω θe θ E/T 3k B T θe/t. β hω/ n } n β hω, so w can xpand th xponntials and cut off highr-ordr C 3k B θe T θ ET θ E T k B. For low tmpraturs,, so w can rwrit th fraction in trms of xp θ E /T and argu that th xponntial in th dnominator is much, much smallr than : θe θ E/T C 3k B T θe/t 3k θe B θe/t. T 4 Sakurai, p48, #5 a Alright. Lt s solv th Schrödingr quation for this wavfunction, pursuant to th boundary conditions: ψρ a, φ, z ψρ b, φ, z 0 h ψρ, φ, z Eψρ, φ, z ψρ, φ, 0 ψρ, φ, L 0 m ψρ, φ + π, z ψρ, φ, z 5

6 Sinc w ll b naturally oprating in cylindrical coordinats, w ll nd th Laplacian in cylindrical coordinats: ρ + ρ ρ φ + z. As with most of th tractabl partial diffrntial quations givn to physics studnts, this on is solvabl by sparation of variabls, so lt s lt ψρ, φ, z RρY φzz. This allows us to rwrit our diffrntial quation by introducing this dfinition of φ and dividing both sids by φ. In th nd, w gt ρ Rρ + Y φ Rρ ρ Y φ ρ φ + Zz Zz z + m E h 0 with boundary conditions Rρ a Rρ b 0 Z0 ZL 0 Y φ + π Y φ Not that, in this form, th Zz trm is th only on dpndnt on z and it is not dpndnt on any othr variabls. So, th trm must b a constant. With a littl forsight and th knowldg that whatvr w gt will hav to satisfy th boundary conditions at z 0 and z L, w ll lt this constant b Cz. Zz z Cz Zz This admits solutions Zz A cosc z z + B sinc z z. Applying th boundary conditions on Zz, w find that lπ Zz A z sin L z C z lπ L l,, 3,... whr our rstrictions on th possibl valus of l nsur that our solutions satisfy th boundary conditions, ar nontrivial, and ar uniqu. W rturn to our original quation and not that, if w multiply both sids by ρ, w again gt a trm that only dpnds on on variabl, φ, and is th only trm dpndnt on φ: ρ Rρ ρ Rρ + Y φ Y φ φ + ρ [ lπ + m E L h 0 Th diffrntial quation for Y φ is th sam, and w ll lt it b Y φ A φ imφ. Bcaus of th priodic boundary condition, m is constraind to b an intgr. Lt s go back to our original quation, xpand th first trm, and multiply both sids by Rρ. ρ Rρ + ρ Rρ [ lπ + {ρ L + m E h m } Rρ 0; l,, 3,... ; m Z This is almost in th form w nd to raliz Rρ as a solution to Bssl s diffrntial quation, but w nd to gt rid of th cofficint of ρ Rρ. Lt m E lπ ξ h L 6

7 and not that solving th diffrntial quation in trms of ξρ instad of ρ will lav th form of th first two trms untouchd. Now w hav ξρ Rξρ ξρ + ξρ Rξρ ξρ + [ξρ m Rξρ 0 which admits solutions Rξρ AJ m ξρ + BN m ξρ whr J m and N m ar Bssl functions of th first and scond kind, rspctivly, with ordr m. Applying our boundary conditions givs us a systm of quations AJ m ξρ a + BN m ξρ b 0 AJ m ξρ b + BN m ξρ b 0. with which w can solv for th allowd valus of ξ that mak th radial wavfunction bcom zro at both th innr and outr cylindr walls: J m ξρ b N m ξρ a J m ξρ a N m ξρ b 0. Lt us dsignat th nth valu of ξ that solvs this mth-ordr quation k mn. W can now writ out th full non-normalizd wavfunction: lπ ψ lmn ρ, φ, z A[N m k mn ρ J m k mn ρ J m k mn ρ N m k mn ρ imφ sin L z whr ρ ρ a or ρ b, l,, 3,..., m Z, and k mn is th nth root of th quation J m k mn ρ b N m k mn ρ a J m k mn ρ a N m k mn ρ b 0. Finally, w can gt th nrgy lvls from th dfinition of ξ k mn : m E lmn k mn h b Our Hamiltonian changs to th form H lπ [ E lmn h kmn + L m p m m c p A + A p + m c A. lπ. L So what is A, if all w know is B Bẑ for ρ < ρ a and zro lswhr? Th asist way I know to gt A involvs applying Stoks thorm to a flat, D circular disk in th ρ φ plan and cntrd at ρ 0. In th rgion ρ < ρ a, w lt th disk hav radius ρ < ρ a : A ds A dl S S B ds S S A dl B πρ A φ πρ A φ B ρ; ρ < ρ a. If w lt th othr componnts of A b zro, w find that w gt th corrct magntic fild insid ρ < ρ a. Of cours, this is gaug-dpndnt, and you can pick any gaug you want. Outsid of ρ a, th magntic fild drops to zro, so B πρ a A φ πρ A φ Bρ a ρ ; ρ ρ a. 7

8 Again ltting th othr componnts of A b zro, w rcovr zro magntic fild outsid ρ a and a continuous vctor potntial. Lt s put our answr for A into th Hamiltonian. With this choic of A, it turns out that A and th gradint oprator commut. H With th bnfit of forsight, lt p m m c A p + h + i hbρ a m m c m c A ρ ζ Bρ a hc, which is dimnsionlss. Thn, our Hamiltonian bcoms H h m [ i ζ ρ Rdoing th sam stps with this nw Hamiltonian, ρ Rρ + [ Y φ Rρ ρ Y φ ρ φ iζ φ + B ρ 4 a 8m c ρ φ ζ ρ Y φ φ. + Zz Zz z + m E h ζ ρ 0 with th sam boundary conditions as bfor. Th solution in th Zz sctor is th sam, but w nd up with a diffrnt diffrntial quation for Y φ: Y φ φ Y φ iζ φ + C φy φ 0. Knowing that w still nd a sinusoidal solution in φ that obys th priodic boundary condition, w can try a trial solution Y φ A φ imφ whr M Z. This givs us th quation Putting this back in, M + ζm + C φ 0 C φ M + ζm. Rρ ρ ρ Rρ ρ Rρ + ρ M + ζm lπ + m E + ρ Rρ {[ + L h lπ + m E L h ζ ρ 0 } ρ M ζ Rρ 0 W gt th sam quation w had bfor, with th sam formulas for th wavfunction and nrgy spctrum, xcpt that m M ζ Bρ a hc! This shifts th ordrs of th Bssl functions to diffrnt valus, which ar not ncssarily intgrs. Th spacing btwn th ordrs is still, howvr. c If switching on th magntic fild dosn t rsult in an nrgy chang in th ground stat, thn th Bssl ordrs M ζ must still b intgrs. But sinc M is an intgr, this mans that ζ must b an intgr: ζ Bρ a hc N Φ Bπρ a π hc N hc N; N Z 8

9 5 Chargd particl in a magntic fild a Classically, th chargd particl will oby th Lorntz forc law: F c v B. B A Bẑ, so mẍ B c ẏ mÿ B c ẋ m z 0. Idntifying ω 0 B mc, w can solv th first two quations by taking anothr drivativ:... ẍ ω 0 ẏ x ω 0 ÿ ω0ẋ... ÿ ω 0 ẋ y ω 0 ẍ ω0ẏ With som forsight, lt s choos th solutions so that our final answr nds up bing ẋt ω 0 A sinω 0 t + δ ẏt ω 0 A cosω 0 t + δ xt A cosω 0 t + δ + x 0 yt A sinω 0 t + δ + y 0 zt v 0z t + z 0. b W mak th canonical transformation to Q c B Π x c B P Π y p y B c x. p x + B c y W can show that this transformation is indd canonical by computing th commutator: [Q, P c B [Π x, Π y whr I hav usd [Π i, Π j i h c ɛ ijmb m from problm. Writing th Hamiltonian in trms of P and Q, H m c i h B c B i h B c Q + P + p z P m + B m mc Q + p z m. Th first two trms rprsnt a D simpl harmonic oscillator with ω 0 B mc, whil th last trm rprsnts a fr particl in th z-dirction. So, E n + hω 0 + h k m. You might b wondring how what w did is koshr. How did w go from a D simpl harmonic oscillator to a D simpl harmonic oscillator, just by rdfining our variabls? W sm to hav lost two dgrs of frdom. Wll, far not: w can dfin a scond st of canonical variabls that ar conjugat to Q and P, Q c p y + B B c x P p x B c y, 9

10 which satisfy [ Q, P i h, [Q, Q 0, and [P, P 0. Ths canonical variabls do not appar in th Hamiltonian bcaus thy rprsnt hlical motion opposing th dirction prfrrd by th magntic fild. c Expand! H [ p x + B m c y + p y B c x + p z p x m + B m x + p y mc m + B m y + p z mc p x ω0 x + p y m + m ω0 m + m ω0 H, m ω 0 L z y + p z m ω 0 L z Th basis for diagonalizing H ω 0, m will also diagonaliz H if [H ω 0 [ [ ω0 H, m ω0, H H, m { ω 0 ω 0 ω 0 0, ω 0 L z ω0 m [p x + p y, xp y yp x + m { [p x, xp y [p y, yp x + m m ω0 { i hp x p y + i hp y p x + m m ω0 m B mc xp y yp x, m, H 0: [x + y, xp y yp x } } y[x, p x + x[y, p y } i hyx + i hxy So, this is indd th cas. Th full Hamiltonian not only shars th sam stats as th Hamiltonian of a D SHO with angular frquncy Ω ω 0, but also th sam nrgy lvls and cration and annihilation oprators: mω a x x + ip x mω a x x ip x h mω h mω mω a y y + ip y mω a y y ip y h mω h mω with commutation rlations [a x, a x [a y, a y and 0 othrwis. Th stats ar { n x, n y, hk z }. W can also rwrit th angular momntum oprator in trms of ths laddr oprators, L z xp y yp x h mω h mω a x + a x i a y a y i ha x a y a y a x and w gt th Hamiltonian mω h i a y + a y H hω + a xa x + a ya y i hωa x a y a y a x + p z m. 0 h mω a x a x

11 Howvr, sinc th classical motion is circular and our ignstats of th Hamiltonian ar also dfinit stats of angular momntum L z, lt s rwrit th ntir systm in trms of clockwis and countrclockwis cration and annihilation oprators a + a x ia y a + a x + ia y a a x + ia y a a x ia y. Ths cration and annihilation oprators also satisfy th normal commutation rlations: [a +, a + [a, a and 0 othrwis. Just as th prvious st of laddr oprators cratd and annihilatd oscillations in th x- and y-dirctions, this st of laddr oprators crats and annihilats oscillations in th clockwis and countrclockwis dirctions. To put th Hamiltonian in this form, w ll nd th rvrs convrsion: a x a + + a a x a + + a a y i a + a a y i a + a so that a xa x + a ya y a + a + + a a n + + n a x a y a y a x ia + a + a a in + n, and L z hn + n Thus, w obtain th nrgy spctrum H hω + n + + n hωn + n + p z m hω 0 + n + p z m. En, k z hω 0 + n + h k z m with angular momnta m n + n. Not that th numbr of quanta for th + oscillator dosn t figur into this quation! This mans that vry nrgy lvl has infinit dgnracy. To writ it in th form indicatd in th prompt, not that { n + m m > 0 n + + n n + + m m < 0 so that p z H hω + n + + n hωl z + m E hω + k + m hωm + hk z m hω 0 + k + m m + h k z m

12 whr k 0. In this form, w s that for ach stat of dfinit angular momntum, thr is a towr of possibl nrgy lvls. For stats with angular momntum m > 0, th m-dpndnc disappars and you gt th normal towr of lvls starting with nrgy hω 0, but for stats with m < 0, th towr ffctivly starts at + m hω 0. This is bcaus stting m constrains th diffrnc in numbrs of quanta n + n, and n + and n must both b 0. You might b having a lot of troubl trying to visualiz what this all concptually mans. Thr s a rason for this! L z is th canonical angular momntum, not th physical angular momntum, analogous with how p is no longr th physical momntum upon th introduction of a vctor potntial. Th physical angular momntum is K z xπ y yπ x ; Π p A c L z B c x + y h + a a + a + a + a + a. Calculating th xpctation valu of this for a givn stat n +, n, k z, w gt th rlation E ω 0 K z + h k z m which is what w would xpct classically, givn that K z will b ngativ for B > 0.