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851HW9_09Solutions

# 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 . Answer: u→∞ u→−∞ x|Π|x′ = x| − x′ = δ(x + x′ ) Π= Π2 = dxdx′ |x δ(x + x′ ) x′ | = dx|x −x| dx|x x| = 1 dxdx′ |x −x|x′ −x′ | = dxdx′ |x δ(−x − x′ ) −x′ | = So Π3 = Π2 · Π = Π, which generalizes to Πn = Now we have S (u) = = = = x|S (u)|ψ = 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 2 3 1; n = even Π; n = odd = 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) x|S (0)|ψ = ψeven (x) + ψodd (x) = ψ (x) u→∞ So that lim x|S (u)|ψ = 2ψeven (x) lim x|S (u)|ψ = 2ψodd (x) 1 u→−∞ 2. [15 pts]The coherent state |α is deﬁned by |α = e−|α| the harmonic oscillator energy eigenstates. 2 /2 ∞ √n α n=0 n! |n , 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 α=0 = e−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 e−α So we end up with D(α)|0 = e−|α| 2 /2 ∗A † −α∗ A = e−|α| 2 /2 eαA e−α ∞ † ∗A |0 = ∞ n=0 (−α∗ )n n A |0 = n! ∞ n=0 n=0 (−α∗ )n δn,0 |0 = |0 n! αn √ |n = |α n! n=0 ∞ eαA |0 = e−|α| † 2 /2 αn † n 2 (A ) |0 = e−|α| /2 n! Part iii) Using the Zassenhaus formula, we have D(α1 )D(α2 ) = eα1 A 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 This gives D(α1 )D(α2 ) = e− † −α∗ A+α A† −α∗ A 2 1 2 † −α∗ A+α A† −α∗ A 2 1 2 † ∗ † ∗ e[α1 A −α1 A ,α2 A −α2 A]/2 = D(α1 + α2 ) D(α1 + α2 ) (α∗ α2 −α∗ α1 ) 1 2 2 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 = −(iκt) = −iκt, so we have |ψ (t) = eαA † −α∗ A |0 |0 = |α(t) 2 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 ∞ 0 dxe−|x|/a = 2 dxe−x/a = 2a So that from symmetry we get √ which gives N = 1/ 8abc. b. −∞ N 2 8abc = 1 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 (2) P= d. The requested quantity is (e − 1)2 e2 P = | px = 0, py = 0, pz = /c|ψ |2 dpx dpy dpz 0, 0, /c|ψ = = = = = = So we have P= 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−(1/2+i)z/c (2π )3/2 abc 0 0 0 √ 8 8abc √ (2π )3/2 abc 1 + 2i √ 8 8abc (3) (1 + 2i)(2π )3/2 64abc 512abc dpx dpy dpz = dpx dpy dpz 3 5(2π ) 5(π )3 3 3.7 a. P= b. P= where ψ (px , y, z ) = c. √1 2π x2 dx x1 p2 ∞ −∞ dy ∞ −∞ dz |ψ (x, y, z )|2 dz |ψ (px , y, z )|2 dpx p1 ∞ −∞ dy ∞ −∞ ∞ −ipx x/ −∞ dxe x2 ψ (x, y, z ). ∞ P= where ψ (x, y, pz ) = d. √1 2π dx x1 0 dy 0 ∞ dpz |ψ (x, y, pz )|2 ∞ −ipz z/ −∞ dz e p2 ψ (x, y, z ) p6 p4 P= where ψ (px , py , pz ) = 1 (2π )3/2 dpx p1 p3 dpy p5 dpz |ψ (px , py , pz )|2 ψ (x, y, z ) ∞ ∞ ∞ −i(px x+py y +pz z )/ −∞ dx −∞ dy −∞ dz e If we extend the py and pz limits to inﬁnity we get p2 P= = dpx p1 p2 ∞ −∞ ∞ −∞ dpy dpy ∞ −∞ ∞ −∞ dpz ψ |px , py , pz px , py , pz |ψ dpz ψ | |px px | ⊗ |py , pz py , pz | |ψ dpx p1 Now the identity operator can be written as I = Ix ⊗ Iy ⊗ Iz = Ix ⊗ = Ix ⊗ This shows that ∞ −∞ ∞ ∞ −∞ ∞ −∞ dpy dy −∞ ∞ −∞ dpz |py , pz py , pz | dz |y, z y , z | 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 ∞ −∞ dy ∞ −∞ dz ψ |px , y, z px , y, z |ψ which agrees with the answer to b. 4 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 ρ(u) = d3 r ρ(u|r )ρ(r ) 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= so that a = 1. This gives ρ(u) = and then P= u2 ∞ −∞ du ρ(u|r ) = a ∞ −∞ du δ(u − f (r)) = a d3 r |ψ (r )|2 δ(u − f (r)) u2 du ρ(u) = u1 u1 du d3 r |ψ (r )|2 δ(u − f (r)) 5 3.11 a. P = dx1 b. P = dx1 c. P= d. β α β β α ∞ dx2 |ψ (x1 , x2 )|2 dx2 |ψ (x1 , x2 )|2 ∞ −∞ β −∞ dx1 α ∞ −∞ dx2 |ψ (x1 , x2 )|2 + β ∞ β dx1 α dx2 |ψ (x1 , x2 )|2 α β P= dx1 α + β ∞ −∞ dx2 |ψ (x1 , x2 )|2 + β dx1 α dx2 |ψ (x1 , x2 )|2 + dx1 −∞ α dx2 |ψ (x1 , x2 )|2 dx1 α dx2 |ψ (x1 , x2 )|2 or equivalently β P= e. dx1 α ∞ −∞ dx2 |ψ (x1 , x2 )|2 + ∞ −∞ p′′ β dx1 α β dx2 |ψ (x1 , x2 )|2 − β β dx1 α α dx2 |ψ (x1 , x2 )|2 P= where ψ (p1 , x2 ) = f. √1 2π p′ dp1 α dx2 |ψ (p1 , x2 )|2 ∞ −ip1 x1 / −∞ dx1 e p′′ ψ (x1 , x2 ) p′′′′ P= 1 where ψ (p1 , p2 ) = 2π −∞ dx1 g. From the results of e., we ﬁnd ∞ p′ dp1 p′′′ dp2 |ψ (p1 , p2 )|2 ψ (x1 , x2 ) ∞ −i(p1 x1 +p2 x2 )/ −∞ dx2 e p′′ ∞ −∞ P= from the results of f., we ﬁnd p′ dp1 dx2 |ψ (p1 , x2 )|2 p′′ P= this shows that ∞ −∞ p′ dp1 ∞ −∞ dp2 |ψ (p1 , p2 )|2 ∞ −∞ dx2 |ψ (p1 , x2 )|2 = dp2 |ψ (p1 , p2 )|2 which follows because they are both equal to ψ | |p1 p2 | ⊗ I2 |ψ 6 h. d P= = dx −d d ∞ −∞ ∞ −∞ dx1 ∞ −∞ dx2 δ(x − x1 + x2 )|ψ (x1 , x2 )|2 dx −d dx1 |ψ (x1 , x1 − x)|2 ∞ −∞ x = X1 − X2 = ¯ dx1 ∞ −∞ dx2 (x1 − x2 )|ψ (x1 , x2 )|2 7 ...
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