Q4P (a) Find the second-order correc... [FREE SOLUTION]

91影视

Introduction to Quantum Mechanics

David J. Griffiths

Physics

2nd Edition 路 469 pages

ISBN: 9780131118928

Chapter 6: Q4P (page 256)

(a) Find the second-order correction to the energies $(E_{n}^{2})$ for the potential in Problem 6.1. Comment: You can sum the series explicitly, obtaining -for odd n.
(b) Calculate the second-order correction to the ground state energy $(E_{0}^{2})$ for the potential in Problem 6.2. Check that your result is consistent with the exact solution.

Short Answer

Expert verified

$(a) E_{n}^{2} = {\begin{cases} 0, if n is even \\ - 2 m {(a / 魔苍)}^{2}, if n is odd . \end{cases} (b) E_{n}^{2} = \frac{\overset{藱}{慰^{2}} 魔蝇}{32} (- 4 n - 2) = - \overset{藱}{慰^{2}} \frac{1 魔蝇}{8} (n + \frac{1}{2}) .$

Step by step solution

(a)Finding the second order correction to the energies(En2)

$<蠄_{m}^{0} |H| 蠄_{n}^{0}> = \frac{2}{a} 伪 {鈭�}_{0}^{a} \sin (\frac{m 蟺}{a} x) 未 (x - \frac{a}{2}) \sin (\frac{n 蟺}{a} x) d x = \frac{2 伪}{a} \sin (\frac{m 蟺}{2}) \sin (\frac{n 蟺}{2})$

which is zero unless both m and n are odd-in which case it is $卤 2 伪 / a$ So Eq. 6.15 says

$E_{n}^{2} = \underset{m 鈮� n}{鈭�} 蠄_{\frac{{|<蠄_{m}^{0} |H^{'}| n^{0}>|}^{2}}{E_{n}^{0} - E_{m}^{0}}} E_{n}^{2} = \underset{m 鈮� n, o dd}{鈭�} {(\frac{2 伪}{a})}^{2} \frac{1}{(E_{n}^{0} - E_{m}^{0})}$ (6.15).

But Eq. 2.27 says

$E_{n} = \frac{魔^{2} k_{n}^{2}}{2 m} = \frac{n^{2} 蟺^{2} 魔^{2}}{2 m a^{2}} E_{n}^{0} = \frac{蟺^{2} 魔^{2}}{2 m a^{2}} n^{2}, S O$ (2.27).

$E_{n}^{2} = \{\begin{cases} 0, if n is even \\ 2 m {(\frac{2 伪}{蟺魔})}^{2} \underset{m 鈮� n, o dd}{鈭�} \frac{1}{(n^{2} - m^{2})'} if n is odd \end{cases}$

To sum the series, note that Thus, $\frac{1}{(n^{2} - m^{2})} = \frac{1}{2 n} (\frac{1}{m + n} - \frac{1}{m - n}) . Thus$

$for n = 1 : \underset{}{鈭� =} \frac{1}{2} \underset{3, 5, 7 . . .}{鈭�} (\frac{1}{m + 1} - \frac{1}{m - 1}) = \frac{1}{2} (\frac{1}{4} + \frac{1}{6} + \frac{1}{8} + . . . - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} - \frac{1}{8} . . .) = \frac{1}{2} (- \frac{1}{2}) = - \frac{1}{4} for n = 3 : \underset{}{鈭� =} \frac{1}{6} \underset{1, 5, 7 . . .}{鈭�} (\frac{1}{m + 3} - \frac{1}{m - 3}), = \frac{1}{6} (\frac{1}{4} + \frac{1}{8} + \frac{1}{10} + . . . + \frac{1}{2} - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} - \frac{1}{8} - \frac{1}{10} . . .) = \frac{1}{6} (- \frac{1}{6}) = - \frac{1}{36}$

In general, there is perfect cancellation except for the term 1/2n in the first sum, so the total is

\frac{1}{2 n} (- \frac{1}{2 n}) = - \frac{1}{{(2 n)}^{2}} .

$T h e r e f o r e : E_{n}^{2} = \{\begin{cases} 0, if n is even \\ - 2 m {(伪 / 蟺魔 n)}^{2}, if n is odd . \end{cases}$

(b) Calculating the second order correction to the ground state energy(E02)

$H^{'} = \frac{1}{2} \overset{藱}{慰} k x^{2}; <蠄_{m}^{0} |H^{'}| 蠄_{n}^{0}> = \frac{1}{2} \overset{藱}{慰} k <m |x^{2}| n> using Eqs . 2.66 and 2.69 : {\hat{a}}_{+} 蠄_{n} = \sqrt{n + 1} 蠄_{n + 1'} {\hat{a}}_{-} 蠄_{n} = \sqrt{n} 蠄_{n - 1^{鈥�}} (2.66) . x = \sqrt{\frac{魔}{2 m蝇}} ({\hat{a}}_{+} + {\hat{a}}_{-}); \hat{p} = i \sqrt{\frac{魔m蝇}{2} ({\hat{a}}_{+} - {\hat{a}}_{-})} (2.69) .$

$<m |x^{2}| n> = \frac{魔}{2 m 蝇} <m |(a_{+}^{2} + a_{+} a_{-} + a_{-} a_{+} + a_{-}^{2})| n> = \frac{魔}{2 m 蝇} [\sqrt{(n + 1) (n + 2)} <m \ n + 2> + n <m \ n> + (n + 1) <m \ n> + \sqrt{n (n - 1)} <m \ n - 2>] S O, for m 鈮� n, <蠄_{m}^{0} |H^{'}| 蠄_{n}^{0}> = (\frac{1}{2} k \overset{藱}{慰}) (\frac{魔}{2 m蝇}) [\sqrt{(n + 1) (n + 2)} 未_{m, n + 2} + \sqrt{n (n + 1)} 未_{m, n - 2}] E_{n}^{0} = {(\frac{\overset{藱}{慰魔蝇}}{4})}^{2} \underset{m 鈮� n}{鈭�} \frac{{[\sqrt{(n + 1) (n + 2)} 未_{m, n + 2} + \sqrt{n (n + 1)} 未_{m, n + 2}]}^{2}}{(n + \frac{1}{2}) \overset{藱}{魔蝇} - (m + \frac{1}{2}) \overset{藱}{魔蝇}}$

$= \frac{\overset{藱}{慰^{2} 魔蝇}}{16} \underset{m 鈮� n}{鈭�} \frac{[(n + 1) (n + 1) 未_{m, n, + 2} + n (n - 1) 未_{m, n, - 2}]}{(n - m)} = \frac{\overset{藱}{慰^{2} 魔蝇}}{32} (- n^{2} - 3 n - 2 + n^{2} - n) = \frac{\overset{藱}{慰^{2} 魔蝇}}{32} (- 4 n - 2) = - \overset{藱}{慰^{2}} \frac{1}{8} \overset{藱}{魔蝇} (n + \frac{1}{2})$