ψ = i c i u i c i a i b i u i = i b 0 0 b 0 0
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1 Quantum Mechancs, Advanced Course FMFN/FYSN7 Solutons Sheet Soluton. Lets denote the two operators by  and ˆB, the set of egenstates by { u }, and the egenvalues as  u = a u and ˆB u = b u. Snce the set of egenstates s complete n the Hlbert space, we can expand every state ψ n the egenstates, ψ = c u Now we have  ˆB ψ =  ˆB c u = c  ˆB u = c Âb u c a b u = c ˆBa u = c ˆB u = ˆB c u = ˆB ψ  ˆB = ˆBÂ. Soluton. (a A and B are Hermtan f and only f A = A and B = B. Ths s true for A (t s dagonal and real valued, and b b B = b = (b b ( b T b = b b (b A s already dagonalzed, so we smply read ts egenvalues off the dagonal: λ = a and λ = a wth λ double degenerate. For B we fnd b µ det(b µi = µ b b µ = (b µ(µ b = (b µ (b + µ gvng the egenvalues µ = b and µ = b wth µ double degenerate. (c We have ab AB = BA = ab ab A, B] = AB BA =. (d We want to fnd common egenvectors of A and B. Note that the basc unt vectors e, e, e 3 are egenvectors of A (they are the natural egenvectors of every dagonal matrx but only e s an egenvector of B, so t s not certan that we only need to fnd the egenvectors of one matrx and then be done (ths s n fact because the egenvalue a of A s degenerate. Incdentally, f you fnd the egenvectors of B, they turn out to be egenvectors of A too. (If you don t remember how to fnd the egenvectors of a matrx, consult your books/notes from your course on basc lnear algebra! = B However, f you re not satsfed wth ths happy concdence of the egenvectors of B, the general method s as follows: ( Fnd the egenvalues and egenvectors of A. If all egenvalues of A are non-degenerate you are done. ( For each degenerate egenvalue of A, fnd the matrx representng B n that subspace. Meanng, f the egenvalue λ s k-degenerate wth egenvectors u, =,,..., k, fnd the k k matrx B λ = u Bu j].
2 Quantum Mechancs, Advanced Course FMFN/FYSN7 (3 Fnd the egenvectors of B λ, whch we denote by v = (v (v (v k ] T. (4 Then the vectors w = (v u + (v u + + (v k u k are egenvectors of B, and smultaneously egenvectors of A wth egenvalue λ. For further detals, you can check out pp. 4-4 n Cohen-Tannoudj et al, 5. Quantum Mechancs, Vol., Wley-VCH. Now, lets follow ths procedure wth A and B (whch ncdentally s pretty much the same as fndng the egenvectors of B, snce e, e, e 3 are egenvectors of A. We have already completed step (. The egenvalue λ s double degenerate wth egenvectors u = e = ] T and u = e 3 = ] T. The matrx element of B λ n row and column s u Bu = b. We fnd the other matrx elements n a smlar way and fnd B λ = ] b. b The (normalzed egenvectors turn out to to be v = ], v = ] (or those vectors multpled wth any complex constant you lke gvng w = ( u + u =, w = ( u + ( u =. The bass of common egenvectors of A and B s then {e, w, w }, that s,, Soluton.3 Note: The operator s the dentty operator, that s for every state φ one has φ = φ. It s lke the dentty matrx I n lnear algebra t leaves others alone, doesn t change a thng. Snce  s an observable, ts egenvalues are real numbers. Let α be an egenvalue of  and α a correspondng egenstate. Then,. α = α = Â3 α =  ( α =  α α =... = α 3 α α 3 =. The only real soluton to that equaton s α = so the only egenvalue of  s. Now, snce  s an observable, we can choose an orthonormal bass { u } for the state space consstng of the egenstates of  (hence,  u = u for all. Every state ψ can then be wrtten n terms of ths bass ψ = c u and we obtan  ψ =  c u = c  u = c u = ψ  =. Soluton.4 (a Let ψ be some arbtrary state. Snce { u } s a bass, there s a unque representaton of ψ n the bass ψ = c u. Then we fnd ( ( u u ψ = u u c j u j = j j u u u j c j = }{{} δ j c u = ψ
3 Quantum Mechancs, Advanced Course FMFN/FYSN7 and therefore u u =. (b Let ψ be an arbtrary state. Usng the closure relaton we have ψ = ψ = w α w α ψ dα Defnng we have c(α := w α ψ ψ = ψ = c(α w α dα Therefore we see that any ket ψ has an expanson n w α. To show that ths expanson s unque, we assume that we have two expansons ψ = c(α w α dα and ψ = c(α w α dα Subtractng we obtan = c(α c(α] w α dα Applyng w α on both sdes of the above equaton and usng the orthonormalzaton relaton, we obtan = c(α c(α] w α w α dα = c(α c(α]δ(α α dα = c(α c(α Therefore, for any α we have c(α = c(α, and consequently the expanson of any state ψ s unque. Hence, { w α } consttutes a bass for the state space. Note: In most of the followng exercses we use the short-hand notaton φ n = n for the nth egenstate of the harmonc oscllator. Soluton.5 Knetc energy: ˆT = ˆp m = ω ˆP = ω (â â ] = 4 ω((â â â ââ + â = 4 ω((â â â + â Denotng the nth egenstate by n, we now get the expectaton value of thet knetc energy ˆT n the nth egenstate of the harmonc oscllator n ˆT n = 4 ω n ((â â â + â n = 4 ω n (n + (n + n + n n n + n(n n ] = 4 ω ] (n + (n + n n + n n n n n + n(n n n = 4 ω (n + (n + δn,n+ nδ n,n δ n,n + ] n(n δn, n = 4 ω( n = ω(n + In a smlar manner we fnd the potental energy: Û = mωˆx = ω ˆX = ω (â + â ] = 4 ω((â + â â + ââ + â = 4 ω((â + â â + + â 3
4 Quantum Mechancs, Advanced Course FMFN/FYSN7 n Û n = 4 ω n ((â + â â + + â n = 4 ω n (n + (n + n + + n n + n + n(n n ] = 4 ω ] (n + (n + n n + + n n n + n n + n(n n n = 4 ω (n + (n + δn,n+ + nδ n,n + δ n,n + ] n(n δn, n = 4 ω(n = ω(n + Another (and shorter way to fnd the expectaton value of the potental energy s to remember that the Hamltonan s Ĥ = ˆT + Û and that the energy levels of the harmonc oscllator (whch we already know are the egenvalues of the Hamltonan,.e. ts expectaton value n each egenstate Therefore E n = n Ĥ n = ω(n +. n Û n = n Ĥ ˆT n = n Ĥ n n ˆT n = ω(n + ω(n + = ω(n +. Note that we have shown that for the harmonc oscllator n ˆT n = n Û n = n Ĥ n = E n. Ths s a specal property of the harmonc oscllator. Soluton.6 (a We start by fndng φ usng â on φ : (â φ = â φ = φ so φ = (â φ. Now, so Ĥ = ω(â â +, â φ n = n φ n, â φ n = n + φ n+ Ĥ φ = ω(â â + φ = ω(â φ + φ = ω( φ + φ = ω( + φ = 5 ω φ That s, φ s an egenket of the Hamltonan Ĥ wth egenvalue 5 ω. Here s another way: and therefore Ĥ = ω(â â + = ω( ˆN +, and ˆN φn = n φ n Ĥ φ = ω( ˆN + φ = ω( ˆN φ + φ = ω( φ + φ = ω( + φ = 5 ω φ And yet another (longer way, usng that â φ = null : Ĥ φ = ω(â â + â φ = ω (â ââ â + â φ = ω (â 3 â + â + â φ ( = ω (â 3 â φ + 5 â φ = 5 ω â φ = 5 ω φ 4
5 Quantum Mechancs, Advanced Course FMFN/FYSN7 where we used â ââ â = â (â â + â = â ââ + â = â (â â + + â = â 3 â + â n step (. Try to compare ths to the amount of work you d have to do, dong the same calulatons n poston space: From eq. (.6 n the compendum â ( ξ d dξ so φ = (â φ gves φ (ξ = ( ξ d dξ, φ (ξ = c e ξ /, c = ( mω /4 mω, ξ = π x c e ξ / = c ( e ξ / d dξ e ξ = c / d ( ξe ξ eξ dξ = c ( eξ / e ξ + 4ξ e ξ = c e ξ / ( ξ Now, the Hamltonan s We fnd H ξ = p m + mω x = m = ( ω ξ d dξ H ξ φ (ξ = ω ( ξ φ (ξ = c d ( dξ ξ e ξ / e ξ / = c d ( dξ d dx + mω x = mω m d dξ + mω d dξ φ (ξ = ω ( ξ φ (ξ φ (ξ ξ 3 e ξ / + 4ξe ξ / + ξe ξ / }{{} 5ξe ξ / mω ξ = c ( ξ 4 e ξ / 6ξ e ξ / 5ξ e ξ / + 5e ξ / = c ( ξ 4 e ξ / ξ e ξ / + 5e ξ / ( = ξ c ( ξ e ξ / e ξ / 5 c (ξ e ξ / + e ξ / = ξ φ (ξ 5φ (ξ and therefore H ξ φ (ξ = ω ( ξ φ (ξ φ (ξ = ω ( ξ φ (ξ ( ξ φ (ξ 5φ (ξ = 5 ωφ (ξ. Ths confrms that φ s ndeed an egenfuncton of the Hamltonan, wth egenvalue E = 5 ω = ω( + (correspondng to n =. But we had to calculate a lot more than compared to the algebrac method! (b We have Ĥ(â n = ω(â â + â n = ω((ââ + â n = ω(ââ â n = ω ( ââ â n â n = ω(ân n â n = ω(n â n so â n s an egenstate to Ĥ wth egenvalue ω(n. Another possblty s to argue that snce we know that n s an egenstate of Ĥ wth egenvalue ω((n + = ω(n and that â n n, then â n must also be an egenstate of Ĥ wth egenvalue ω(n. (The symbol means proportonal to ; that s f y x, then there exsts some non-zero constant c such that y = cx. 5
6 Quantum Mechancs, Advanced Course FMFN/FYSN7 Soluton.7 We have that ˆx = ˆx ˆx, ˆp = ˆp ˆp. In Ex.. we found that so Now, Û = mωˆx = ω(n + and ˆT = ˆp m = ω(n + ˆx = mω (n + and ˆp = mω(n + ˆx = n ˆx n n (â + â n = n ( n + n + + n n = ( n + n n + + n n n = ( n + δ n,n+ + nδ n,n = and smlarly so Fnally, we fnd ˆx ˆp = ˆx ˆx ˆp ˆp = ˆp = n ˆp n n (a a n = ˆx = ˆp = mω (n + mω(n + = (n +. Note that ˆx ˆp = (n + for all n, n accordance wth the Hesenberg uncertanty prncple. Soluton.8 For every tme-ndependent operator  we have that see eq. (.35 n the compendum. d dt  = Ĥ, Â] (a Now, Ĥ = ω(â â +, so Ĥ, â ] = Ĥâ â Ĥ = ω(â â + â ωâ (â â + = ω(â ââ â â â = ω(â (â â + â â â = ωâ Hence and thus Smlarly d dt â = ωâ = ω â â (t = A e ωt, where A = â (. Ĥ, â] = Ĥâ âĥ = ω(â â + â ωâ(â â + = ω(â ââ ââ â = ω(â ââ (â â + â = ωâ gvng d dt â = ωâ = ω â â (t = B e ωt, where B = â (. 6
7 Quantum Mechancs, Advanced Course FMFN/FYSN7 (b (Note: There s another way than the followng, that does not use â and â. There one uses that Ĥ, ˆp] ˆx and Ĥ, ˆx] ˆp, to derve a nd order dfferental equaton of the form d dt ˆp ˆp. The result s, of course, the same but the constants appear a bt more naturally. We know that ˆp = ˆP = (â â = (â â so usng the results n (a ( ˆp (t = A e ωt B e ωt, A = â (, B = â ( If we lke to have the constants n terms of ˆp ( nstead of â ( and â (, we do a lttle trck ( ˆp (t = A e ωt B e ωt ( = A e ωt B e ωt + ( B e ωt B e ωt + ( A e ωt A e ωt = ((A B eωt + e ωt + (A + B eωt e ωt = (A B cos(ωt (A + B sn(ωt From the calculatons above, we know that (A B = ˆp (. Notng that we fnd ˆx = mω ˆX = mω (â + â = mω (â + â (A + B = ( â ( + â ( = mω ˆx (. Fnally, pluggng thngs together, we arrve at ˆp (t = ˆp ( cos(ωt mω ˆx ( sn(ωt. Soluton.9 From Ex.. we know that ˆp = (â â + (â â so the matrx elements are n ˆp m = n (â â + (â â m = (m n + m (m + (m + m + ] m(m m = (m + δ n,m (m + (m + δ n,m+ ] m(m δ n,m Hence, the matrx has three dagonals wth non-zero elements (the central dagonal and the ones two steps off-center, whle the rest of the matrx elements are zero. 7
8 Quantum Mechancs, Advanced Course FMFN/FYSN7 Soluton. (a Applyng â on φ β â φ β = â C ] β n n! (â n β n φ = C n! â(â n φ We turn â(â n nto a normal ordered product (all â terms precede â usng ââ = â â+ repeatedly Remember that â φ =, so Now we fnd â φ β = C â(â n = (â â + (â n = â â(â n + (â n = C k= = â (â â + (â n + (â n = (â â(â n + (â n =... = (â n â + n(â n â(â n φ = ((â n â + n(â n φ = n(â n φ β n n! â(â n φ = C β β k (â k φ = β C k! β n n! n(â n φ = C k= β k k! (â k φ ] n= = β φ β. Thus we have shown that φ β s an egenstate of â wth egenvalue β. (b The tme-dependent Schrödnger equaton reads We have Ĥ = ω(â â + so we look at â â φ β = â (â φ β = â (β φ β = C = C k= β k k! k(â k φ φ β t β n+ n! = Ĥ φ β. (â n+ φ = C Pluggng thngs nto the Schrödnger equaton above we acheve ( d dt (C(tβn (t (â n φ = ω n! Comparng coeffcents we fnd C(tβ n (t n(â n φ + n! β β n (n! (â n φ β n+ (n +! (n + (â n+ φ C(tβ n (t (â n φ n! d dt (C(tβn (t = ω ( nc(tβ n (t + C(tβn (t C (tβ n (t + nc(tβ (tβ n (t = ω(n + C(tβn (t C (t C(t + (t nβ β(t = ω(n + ( β (t n β(t + ω = C (t C(t ω Ths equaton must hold for all n. Hence, both sdes must be zero, and we fnd whch gves C (t C(t = ω and β (t β(t = ω C(t = C(e ωt/ and β(t = β(e ωt. 8
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