Solutions to Homework Problem 4 - Quantum Mechanics | PHYS 325, Assignments of Quantum Mechanics

Download Solutions to Homework Problem 4 - Quantum Mechanics | PHYS 325 and more Assignments Quantum Mechanics in PDF only on Docsity! 1 Physics 325 Solution to Homework Problems #4 Winter 2004 1. Your textbook Prob. 6.2 (a) We still have a harmonic oscillator, so the energies are En = (n+1/2)h̄ω′, where ω′ = √ (1 + )k/m = ω √ 1 + . √ 1 +  = (1 + )1/2 ≈ 1 +  2 −  2 8 => En = (n+ 1/2)h̄ω [ 1 +  2 −  2 8 ] (b) H = p2/2m+(1/2)k(1+ )x2 = p2/2m+(1/2)kx2 +(/2)kx2 = H(0) +H ′. Therefore, H ′ = (/2)kx2. E(1)n = 〈ψ(0)n |H ′|ψ(0)n 〉 = k 2 〈ψ(0)n |x2|ψ(0)n 〉 You are not supposed to do the integral for E(1)n . I see two ways around this. The first is to use the Virial Theorem (problem 3.53 in yout text) which shows that for the harmonic oscillator potential, 〈KE(0)〉 = 〈V (0)〉 = (1/2)〈E(0)〉. For state ψ(0)n , 〈ψ(0)n |V (0)|ψ(0)n 〉 = 〈ψ(0)n | 1 2 kx2|ψ(0)n 〉 = 1 2 E(0)n = 1 2 (n+ 1/2)h̄ω Therefore 〈ψ(0)n |x2|ψ(0)n 〉 = 1 k (n+ 1/2)h̄ω => E(1)n =  2 (n+ 1/2)h̄ω which agrees with the first order (linear term in ) result from part (a). The second approach that avoids integration is to write x2 ∝ (a+ − a−)2 as we did in class. Only the terms a+a− − a−a+ connect state ψ(0)n with itself. Including the proportionality factor and problem 2.12, you get the same result as above. 2. Your textbook Prob. 6.4 a. E(2)n = ∑ m 6=n ∣∣〈ψ(0)m |H ′|ψ(0)n 〉 ∣∣2 E (0) n − E(0)m Using the notation for Problem 6.1 from the Selected Solutions #4, we see that: 〈m|H ′|n〉 = 2 a ∫ a 0 sin(mπx/a) sin(nπx/a) αδ(x − a/2) dx = 2α a sin(mπ/2) sin(nπ/2) => 〈m|H ′|n〉 = 0 for m or n even and ∣∣〈m|H ′|n〉 ∣∣2 = 4α 2 a2 for m and n odd Using E(0)n − E(0)m = (n2 −m2)π2h̄2/(2ma2), it follows that for n odd: E(2)n = 4α2 a2 · 2ma 2 π2h̄2 ∑ m 6=n,m odd 1 n2 −m2 = 8mα2 π2h̄2 ∑ m 6=n,m odd 1 n2 −m2 It’s OK to stop here, but if you want to sum the series, note that: 1 n2 −m2 = 1 (n+m)(n−m) = 1 2n ( 1 n+m + 1 n−m ) 2 and ∑ m 6=n,m odd ( 1 n+m + 1 n−m ) = ∑ m 6=n,m odd 1 n+m + ∑ k 6=n,k odd 1 n− k For every m in the first sum above, there is a term k = 2n +m in the second sum that cancels the m term in the first sum. This eliminates the first sum and removes all k′s in the second sum except those with k < 2n+ 1 and k = 2n+ n = 3n (because this term was missing in the first sum). Therefore: ∑ m 6=n,m odd ( 1 n+m + 1 n−m ) = 1 n− 3n + 2n−1∑ k=1,k 6=n,k odd 1 n− k = − 1 2n because the sum over k above is zero (for every k in the sum, there is a term with k′ = 2n− k that cancels the term with k). Therefore ∑ m 6=n,m odd 1 n2 −m2 = − 1 4n2 => E(2)n = −2mα2 π2h̄2n2 b. E (2) 0 = ∞∑ m=1 ∣∣〈ψ(0)m |H ′|ψ(0)0 〉 ∣∣2 E (0) 0 − E (0) m where H ′ =  2 kx2 and E(0)n = (n+ 1/2)h̄ω => E(2)0 = 2k2 4 ∞∑ m=1 ∣∣〈ψ(0)m |x2|ψ(0)0 〉 ∣∣2 −mh̄ω Here, it is easiest to write x in terms of the energy raising and lowering operators: x = a+ − a− iω √ 2m => x2 = a2+ + a+a− + a−a+ + a2− −2mω2 where a+ψ(0)n = i √ h̄ω(n+ 1) ψ(0)n+1 Because a−ψ (0) 0 = 0, the only term in x 2 that contributes to E(2)0 is the a 2 + term, which raises ψ (0) 0 to ψ (0) 2 . The orthogonality of the ψ (0) n ’s then leaves only the term with m = 2: E (2) 0 = 2k2 4 ∣∣〈ψ(0)2 |x2|ψ (0) 0 〉 ∣∣2 −2h̄ω = 2k2 −8h̄ω · ∣∣〈ψ(0)2 | a2+ −2mω2 |ψ(0)0 〉 ∣∣2 = −2k2 32m2h̄ω5 ∣∣i √ h̄ω i √ 2h̄ω ∣∣2 = − 2k2h̄ 16m2ω2 = −2h̄ω 16 which agrees with the second order component of the exact solution (after substituting k = mω2). 3. Your textbook Prob. 6.6 I’ll leave off the superscripts (0) to save typing. (a) 〈ψ+|ψ−〉 = 〈α+ψa + β+ψb|α−ψa + β−ψb〉 = 〈α+ψa|α−ψa〉 + 〈β+ψb|β−ψb〉 = α∗+α− + β ∗ +β− because ψa and ψb are orthogonal and normalized. From Equation 6.21, β± = α±(E (1) ± −Waa)/Wab => α∗+α− + β ∗ +β− = α ∗ +α− + α ∗ +α− (E(1)+ −W ∗aa)(E (1) − −Waa) W ∗abWab