851HW9_09Solutions - PHYS851 Quantum Mechanics I, Fall 2009...

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Unformatted text preview: PHYS851 Quantum Mechanics I, Fall 2009 HOMEWORK ASSIGNMENT 9: SOLUTIONS 1. The Parity Operator: [20 pts] Determine the matrix element x|Π|x′ and use it to simplify the identity Π = dx dx′ |x x|Π|x′ x′ |, then use this identity to compute Π2 , Π3 , and Πn . From these results find an expression for S (u) = exp[Πu] cosh u in the form f (u) + g(u)Π. 1 What is x|S (u)|ψ ? Express your answer in terms of ψeven (x) = 2 (ψ (x) + ψ (−x)) and ψodd (x) = 1 2 (ψ (x) − ψ (−x)). Compute x|S (0)|ψ , lim x|S (u)|ψ , and lim x|S (u)|ψ . u→∞ Answer: u→−∞ x|Π|x′ = x| − x′ = δ(x + x′ ) Π= Π2 = dxdx′ |x δ(x + x′ ) x′ | = dxdx′ |x −x|x′ −x′ | = dx|x −x| dxdx′ |x δ(−x − x′ ) −x′ | = dx|x x| = 1 So Π3 = Π2 · Π = Π, which generalizes to Πn = 1; n = even Π; n = odd Now we have 2 S (u) = = = = x|S (u)|ψ = 3 I + Πu + Π2 u + Π3 u + . . . eΠu 2 3! = cosh u cosh u 2 2 4 5 u + u + u + ... 1 + u + u + ... 2 4! 3! 5! +Π I cosh u cosh u cosh u + Π sinh u cosh u I + Π tanh u x|ψ + tanh u −x|ψ = ψ (x) + tanh uψ (−x) = ψeven (x) + ψodd (x) + tanh u (ψeven (−x) + ψodd (−x)) = (1 + tanh u)ψeven (x) + (1 − tanh u)ψodd (x) So that x|S (0)|ψ = ψeven (x) + ψodd (x) = ψ (x) lim x|S (u)|ψ = 2ψeven (x) u→∞ lim u→−∞ x|S (u)|ψ = 2ψodd (x) 1 2. [15 pts]The coherent state |α is defined by |α = e−|α| the harmonic oscillator energy eigenstates. ∞ √n α n=0 n! |n 2 /2 , where the states {|n } are First, show that for α = 0, the coherent state |α=0 is exactly equal to the harmonic oscillator ground-state, |n=0 . Then show that any other coherent state can be created by acting on the ground-state, |0 , with the ‘displacement operator’ D(α), i.e. show that |α = D(α)|0 , where D(α) := eαA † −α∗ A (1) You may need the Zassenhaus formula eB +C = eB eC e−[B,C ]/2 , which is valid only when [B, [B, C ]] = [C, [B, C ]] = 0. What is D(α2 )|α1 ? Answer: Part i) For α = 0, we have |α = e−0 α=0 ∞ ∞ 0n δn, √ |n = √ 0 |n = |n n! n! n=0 n=0 n=0 Part ii) Let B = αA† and C = −α∗ A. Then [B, C ] = −|α|2 [A† , A] = |α|2 , which commutes with everything. We can therefore use the Zassenhaus formula, which gives eαA Now we have ∗A e−α |0 = ∞ n=0 † −α∗ A = e−|α| 2 /2 (−α∗ )n n A |0 = n! † ∗A eαA e−α ∞ n=0 (−α∗ )n δn,0 |0 = |0 n! So we end up with D(α)|0 = e−|α| 2 /2 † eαA |0 = e−|α| 2 /2 ∞ n=0 αn † n 2 (A ) |0 = e−|α| /2 n! ∞ αn √ |n = |α n! n=0 Part iii) Using the Zassenhaus formula, we have D(α1 )D(α2 ) = eα1 A † −α∗ A+α A† −α∗ A 2 1 2 † ∗ † ∗ e[α1 A −α1 A ,α2 A −α2 A]/2 now α1 A† − α∗ A, α2 A† − α∗ A = −α1 α∗ [A† , A] − α∗ α2 [A, A† ] = α1 α∗ − α2 α∗ 1 2 2 1 2 1 and eα1 A † −α∗ A+α A† −α∗ A 2 1 2 This gives D(α1 )D(α2 ) = e− = D(α1 + α2 ) (α∗ α2 −α∗ α1 ) 1 2 2 D(α1 + α2 ) 3. [15 pts] Consider a system described by the Hamiltonian H = κ(A + A† ). Use your results from the previous problem to determine |ψ (t) for a system initially in the ground-state, |ψ (0) = |0 . We know that |ψ (t) = e−iHt/ |ψ (0) , so that with α(t) = −iκt, we have −α∗ (t) |ψ (t) = e(−iκA † −iκA)t |0 = −(iκt) = −iκt, so we have |ψ (t) = eαA † −α∗ A 2 |0 = |α(t) 4. [10pts each] Cohen Tannoudji, pp341-350: problems 3.6, 3.7, 3.11 3.6 For these problems, the primary task is to set up the integral which gives the desired probability: a. ∞ N2 ∞ dxdydz e−|x|/a−|y|/b−|z |/c = 1 −∞ ∞ dxe−|x|/a = 2 dxe−x/a = 2a 0 −∞ So that from symmetry we get N 2 8abc = 1 √ which gives N = 1/ 8abc. b. ∞ a P= = = = c. Based on the previous result dx 0 dy −∞ ∞ dz −∞ e−|x|/a−|y|/b−|z |/c 8abc a 1 dx e−|x|/a 2a 0 11 du e−u 20 e−1 2e and symmetry, we have P= (2) (e − 1)2 e2 d. The requested quantity is P = | px = 0, py = 0, pz = /c|ψ |2 dpx dpy dpz 0, 0, /c|ψ = = = = = = dxdydz 0, 0, /c|xyz xyz |ψ 1 e−|x|/2a−|y|/2b−|z |/2c √ dxdydz e−iz/c (2π )3/2 8abc ∞ ∞ ∞ 1 √ dx e−|x|/2a dye−|y|/2b dz e−|z |/2c−iz/c (2π )3/2 8abc −∞ −∞ −∞ √ ∞ ∞ ∞ 8 √ dx e−x/2a dye−y/2b dz e−z/2c cos(z/c) (2π )3/2 abc 0 0 0 √ 8 8abc √ (2π )3/2 abc 5 √ 8 8abc (3) 5(2π )3/2 So we have P= 64abc 512abc dpx dpy dpz = dpx dpy dpz 3 25(2π ) 25(π )3 3 3.7 a. ∞ x2 dx P= x1 b. where ψ (px , y, z ) = ∞ c. dpx p1 ∞ dx where ψ (x, y, pz ) = x1 d. where ψ (px , py , pz ) = 0 p6 dpy dpx p5 p3 p1 1 (2π )3/2 dz |ψ (px , y, z )|2 dpz |ψ (x, y, pz )|2 ψ (x, y, z ) p4 p2 P= ∞ dy 0 ∞ −ipz z/ −∞ dz e √1 2π −∞ dz |ψ (x, y, z )|2 ψ (x, y, z ). x2 P= ∞ dy −∞ ∞ −ipx x/ −∞ dxe √1 2π −∞ −∞ p2 P= ∞ dy dpz |ψ (px , py , pz )|2 ∞ ∞ ∞ −i(px x+py y +pz z )/ −∞ dx −∞ dy −∞ dz e ψ (x, y, z ) If we extend the py and pz limits to infinity we get p2 P= dpx p1 p2 dpx = ∞ −∞ ∞ ∞ dpy −∞ ∞ dpy −∞ −∞ p1 dpz ψ |px , py , pz px , py , pz |ψ dpz ψ | |px px | ⊗ |py , pz py , pz | |ψ Now the identity operator can be written as I = Ix ⊗ Iy ⊗ Iz = Ix ⊗ = Ix ⊗ ∞ −∞ ∞ ∞ dpy −∞ ∞ dy −∞ −∞ dpz |py , pz py , pz | dz |y, z y , z | This shows that ∞ dpy ∞ −∞ −∞ dpz |py , pz py , pz | = ∞ −∞ dy ∞ −∞ dz |y, z y , z | which clearly makes sense. Substituting this into the expression for P gives p2 P= dpx p1 ∞ −∞ which agrees with the answer to b. 4 dy ∞ −∞ dz ψ |px , y, z px , y, z |ψ e. Method: Treat as probability problem. Standard probability theory tells us that if u = f (x, y, z ) then the probability density p(u) is given by d3 r ρ(u|r )ρ(r ) ρ(u) = where ρ(u|r ) is the probability density over u for fixed r, and QM tells us that ρ(r ) = |ψ (r )|2 . Now we clearly must have ρ(u|r ) = aδ(u − f (r)), where the normalization constant is determined by requiring that ∞ 1= −∞ du ρ(u|r ) = a ∞ −∞ du δ(u − f (r)) = a so that a = 1. This gives d3 r |ψ (r )|2 δ(u − f (r)) ρ(u) = and then u2 u2 P= du du ρ(u) = u1 u1 5 d3 r |ψ (r )|2 δ(u − f (r)) 3.11 a. β P = dx1 α b. ∞ P = dx1 −∞ c. β P= ∞ dx1 α −∞ dx2 |ψ (x1 , x2 )|2 + ∞ dx2 |ψ (x1 , x2 )|2 dx2 |ψ (x1 , x2 )|2 β dx1 −∞ α dx2 |ψ (x1 , x2 )|2 − β β dx1 α α dx2 |ψ (x1 , x2 )|2 d. α β dx1 P= −∞ α ∞ + β dx1 α β ∞ β dx2 |ψ (x1 , x2 )|2 + dx1 β α dx2 |ψ (x1 , x2 )|2 + β α dx1 α −∞ dx2 |ψ (x1 , x2 )|2 dx2 |ψ (x1 , x2 )|2 or equivalently β P= dx1 α ∞ −∞ dx2 |ψ (x1 , x2 )|2 + β ∞ α −∞ e. p′′ P= where ψ (p1 , x2 ) = f. √1 2π β dp1 p′ α ∞ −ip1 x1 / −∞ dx1 e ∞ 1 where ψ (p1 , p2 ) = 2π −∞ dx1 g. From the results of e., we find p′ p′′′ p′′ dp1 p′ α dp2 |ψ (p1 , p2 )|2 ∞ −i(p1 x1 +p2 x2 )/ −∞ dx2 e P= dx1 dx2 |ψ (p1 , x2 )|2 p′′′′ dp1 ∞ −∞ ψ (x1 , x2 ) dx2 |ψ (p1 , x2 )|2 from the results of f., we find p′′ P= this shows that ∞ −∞ dp1 p′ ∞ −∞ dx2 |ψ (p1 , x2 )|2 = dp2 |ψ (p1 , p2 )|2 ∞ −∞ dp2 |ψ (p1 , p2 )|2 which follows because they are both equal to ψ | |p1 p1 | ⊗ I2 |ψ 6 β β ψ (x1 , x2 ) p′′ P= dx2 |ψ (x1 , x2 )|2 −2 dx1 α dx2 |ψ (x1 , x2 )|2 h. d P= dx −d d dx = −d ∞ −∞ ∞ −∞ x = X1 − X2 = ¯ dx1 ∞ −∞ dx2 δ(x − x1 + x2 )|ψ (x1 , x2 )|2 dx1 |ψ (x1 , x1 − x)|2 ∞ −∞ 7 dx1 ∞ −∞ dx2 (x1 − x2 )|ψ (x1 , x2 )|2 ...
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This note was uploaded on 12/21/2011 for the course PHY 851 taught by Professor Moore,m during the Fall '08 term at Michigan State University.

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