PHYS 215C: Quantum Mechanics (Spring 2017) Problem Set 3 Solutions

Transcription

1 PHYS 5C: Quantum Mechancs Sprng 07 Problem Set 3 Solutons Prof. Matthew Fsher Solutons prepared by: Chatanya Murthy and James Sully June 4, 07 Please let me know f you encounter any typos n the solutons. Problem 0 a Let and label the two z-bass states of a spn-/ partcle. The quantum state wth equal ampltude to be spn-up and spn-down s ψ = +. The correspondng pure state densty matrx s ρ p = ψ ψ = On the other hand, the densty matrx descrbng a mxture of the two bass states wth equal probablty s ρ m = + We have ˆσ z p = Trˆσ z ρ p = ˆσ z + ˆσ z + ˆσ z + ˆσ z = = 0 3 ˆσ z m = Trˆσ z ρ m = ˆσ z + ˆσ z = = 0 4 but ˆσ x p = Trˆσ z ρ p = ˆσ x + ˆσ x + ˆσ x + ˆσ x = = 5 ˆσ x m = Trˆσ x ρ m = ˆσ x + ˆσ x = = 0 6 Thus, the two stuatons lead to the same result for ˆσ z but dfferent results for ˆσ x. Ths result s obvous f one notces that ψ s the state wth defnte spn-up along the x-axs. b Consder the densty matrx for a spn-/ partcle wth magnetc moment µ n a magnetc feld, wth Hamltonan Ĥ = gµ ˆσ 7

2 PHYS 5C: Quantum Mechancs Sprng 07 Problem Set 3 Solutons Page where g s a constant. The equaton of moton for the densty operator s d dt ˆρt = [ Ĥ, ˆρt ] = gµ j[ˆσ j, ˆρt ] 8 where a sum over j =,, 3 s mpled. Tracng both sdes wth ˆσ k yelds d dt Trˆσ k ρt = gµ j Tr ˆσ k[ˆσ j, ˆρt ] 9 Snce the trace s both lnear and cyclc, Tr ˆσ k[ˆσ j, ˆρ ] = Tr ˆσ kˆσ j ˆρ ˆσ k ˆρ ˆσ j = Tr ˆσ kˆσ j ˆρ ˆσ j ˆσ k ˆρ = Tr [ˆσ k, ˆσ j]ˆρ = ε kjl Tr ˆσ l ˆρ 0 where the last equalty follows from the commutaton relatons of the Paul matrces. Usng ths n??, d dt Trˆσ k ρt = gµ j ε kjl Tr ˆσ l ˆρt Swtchng to vector notaton and dentfyng the polarzaton vector P = Tr ˆσ ˆρt, ths gves whch s what we d expect classcally. dp dt = gµ P Problem 0 Consder a bpartte quantum system bult from a drect product Hlbert space of the two parts, H = H H. Let the dmenson of H be n and that of H be N n. The Schmdt decomposton states that, gven any quantum state Φ n the full Hlbert space, there always exst orthonormal sets, ψ, φ j wth, j =,,... n, such that Φ = n v ψ φ 3 where the v are nonnegatve real numbers satsfyng the normalzaton condton n = v =. = a For notatonal convenence, I ll wrte ψ φ nstead of ψ φ, and so on. Usng the above Schmdt form of Φ, t s easy to obtan expressons for the reduced densty matrces: ˆρ = Tr Φ Φ = n v vj ψ ψ j Tr φ φ j =,j= n v vj ψ ψ j δ j =,j= n v ψ = ψ 4 and smlarly ˆρ = Tr Φ Φ = n v vj Tr ψ ψ j φ φ j =,j= n,j= v v j δ j φ φ j = n = v φ φ 5 It s clear from these expressons that the reduced densty matrces are Hermtan wth egenvalues λ = v and assocated egenvectors ψ, φ.

3 PHYS 5C: Quantum Mechancs Sprng 07 Problem Set 3 Solutons Page 3 b Now consder two spn-/ partcles, labelled and, n a normalzed pure state Φ = straghtforward calculaton gves the reduced densty matrx for partcle : and that for partcle : ˆρ = ˆρ = The egenvalues and correspondng normalzed egenvectors of ˆρ are whle those of ˆρ are λ ± = 4 λ ± = 4 ±, ψ ± = ± ± [ ], φ ± = ± The egenvalues agree, as they should. Comparng wth the result of part a, we obtan the Schmdt decomposton of the state Φ as Φ = λ + ψ + φ + + λ ψ φ Problem 3 0 Consder a two-ste verson of the quantum Isng model n a transverse feld wth Hamltonan Ĥ = J ˆσ z ˆσ z hˆσ x hˆσ x where the two spn-/ operators are gven by Ŝµ j = ˆσµ j. convenent orthonormal bass of states whch spans the full Hlbert space for ths model conssts of a drect product of egenstates of ˆσ z denoted φ = ; φ = ; φ 3 = ; φ 4 =. 3 a To compute h αβ = φ α Ĥ φ β, we use the fact that Ĥ = J ˆσ z ˆσ z hˆσ + + ˆσ hˆσ+ + ˆσ 4 n terms of σ z rasng and lowerng operators. Then we can smply explctly calculate the matrx elements. Some basc algebra gves J 0 h h h αβ = 0 J h h h h J 0 5 h h 0 J

4 PHYS 5C: Quantum Mechancs Sprng 07 Problem Set 3 Solutons Page 4 b Usng your favourte method, you can fnd the lowest egenvalue s E 0 = J + h 6 wth egenvector Ψ 0 = φ, 7 where = α α α +, α = h h J + h J = E 0 + J 8 c and d Let Then and α = N α 9 α α α α ˆρ = Ψ 0 Ψ 0 = N α α α α α α 30 α α ˆρ = Tr ˆρ = N α + α / h α α = h + 4h / +J 4h +J = / h/e0 h/e 0 / 3 e We all know how to calculate the egenvalues of a matrx. The egenvalues are: λ ± = ± h E 0 = ± h/j h/j + 3 f The von Neumann b-partte entanglement entropy s defned as S = Tr ˆρ ln ˆρ = λ ln λ 33 We fnd where λ ± are defned n part e. S = λ + ln λ + λ ln λ 34

5 PHYS 5C: Quantum Mechancs Sprng 07 Problem Set 3 Solutons Page 5 g s h/j, we have that λ + and λ 0. Thus the entropy becomes S = 0 35 s h/j 0, we have that λ + / and λ /. Thus the entropy becomes S = / ln + / ln = ln 36 When h/j >>, the J term can be gnored and the Hamltonan factorzes; there s no couplng between the two stes. Thus, there s no entanglement. When h/j <<, the J term that couples the two stes domnates and there s maxmal entanglement. Problem 4 0 a The sx possble states wth two electrons are φ = ĉ ĉ 0 φ = ĉ ĉ 0 φ 3 = ĉ ĉ 0 φ 4 = ĉ ĉ 0 φ 5 = ĉ ĉ 0 φ 6 = ĉ ĉ 0 b The states that have total spn zero along the z-drecton are just the frst four states lsted n part a. c We now compute the matrx elements h j = φ Ĥ φ j of the -ste Hubbard Hamltonan Ĥ = t ĉ σĉσ + ĉ σĉσ + U ˆn ˆn + ˆn ˆn 37 σ=, n the spn zero sector. Usng the antcommutaton relatons for the fermonc operators gves 0 0 t t h = 0 0 t t t t U 0 38 t t 0 U Your matrx may look dfferent by permutaton of the bass states or by factors of f you defned some of your bass states wth a dfferent phase.

6 PHYS 5C: Quantum Mechancs Sprng 07 Problem Set 3 Solutons Page 6 d Usng your favourte method, fnd the egenvalues and egenvectors of the matrx n part c. The lowest egenvalue s E 0 = U 6t + U 39 The egenvector s gven by Ψ 0 = α α α +, α = 4t 6t + U U = t 40 E 0 We wll also express ths state as Ψ 0 = φ. e The ground-state densty matrx s gven by and the correspondng reduced densty matrx by ˆρ = Ψ 0 Ψ 0 4 ˆρ = Tr ˆρ 4 Let 0,,, denote respectvely the four states on ste no electron on ste, an up-spn electron present on ste, a down-spn electron present on ste, both an up and a down spn electron present on ste, and smlarly for ste. Notng that φ = δ,, φ = δ,, 0 φ = δ,3, φ = δ,4 0, 43 the trace gves ˆρ = f We see that the reduced densty matrx s already dagonal. The von Neumann entropy s then gven by S = λ ln λ = 4 ln 4 3 ln 3 45 snce = and 4 = 3. In the lmt U/t, we have that α U t =, so that S = ln 47 We could have predcted ths because n the U/t lmt, the term that domnates prefers to have one electron on each ste no hoppng, but couples the spns on the two stes n a maxmally entangled state.