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

$(a) E_{n}^{2} = {\begin{cases} 0, if n is even \\ - 2 m {(a / ħn)}^{2}, if n is odd . \end{cases} (b) E_{n}^{2} = \frac{\dot{ο^{2}} ħ ω}{32} (- 4 n - 2) = - \dot{ο^{2}} \frac{1 ħ ω}{8} (n + \frac{1}{2}) .$

Step by step solution

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

$<ψ_{m}^{0} |H| ψ_{n}^{0}> = \frac{2}{a} α \int_{0}^{a} \sin (\frac{m π}{a} x) δ (x - \frac{a}{2}) \sin (\frac{n π}{a} x) d x = \frac{2 α}{a} \sin (\frac{m π}{2}) \sin (\frac{n π}{2})$

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

$E_{n}^{2} = \sum_{m \neq n} ψ_{\frac{{|<ψ_{m}^{0} |H^{'}| n^{0}>|}^{2}}{E_{n}^{0} - E_{m}^{0}}} E_{n}^{2} = \sum_{m \neq n, o dd} {(\frac{2 α}{a})}^{2} \frac{1}{(E_{n}^{0} - E_{m}^{0})}$ (6.15).

But Eq. 2.27 says

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

$E_{n}^{2} = \{\begin{cases} 0, if n is even \\ 2 m {(\frac{2 α}{π ħ})}^{2} \sum_{m \neq n, o dd} \frac{1}{(n^{2} - m^{2})'} if n is odd \end{cases}$

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

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

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}} .

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

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

$H^{'} = \frac{1}{2} \dot{ο} k x^{2}; <ψ_{m}^{0} |H^{'}| ψ_{n}^{0}> = \frac{1}{2} \dot{ο} k <m |x^{2}| n> using Eqs . 2.66 and 2.69 : {\hat{a}}_{+} ψ_{n} = \sqrt{n + 1} ψ_{n + 1'} {\hat{a}}_{-} ψ_{n} = \sqrt{n} ψ_{n - 1^{‐}} (2.66) . x = \sqrt{\frac{ħ}{2 mω}} ({\hat{a}}_{+} + {\hat{a}}_{-}); \hat{p} = i \sqrt{\frac{ħmω}{2} ({\hat{a}}_{+} - {\hat{a}}_{-})} (2.69) .$

$<m |x^{2}| n> = \frac{ħ}{2 m ω} <m |(a_{+}^{2} + a_{+} a_{-} + a_{-} a_{+} + a_{-}^{2})| n> = \frac{ħ}{2 m ω} [\sqrt{(n + 1) (n + 2)} <m \ n + 2> + n <m \ n> + (n + 1) <m \ n> + \sqrt{n (n - 1)} <m \ n - 2>] S O, for m \neq n, <ψ_{m}^{0} |H^{'}| ψ_{n}^{0}> = (\frac{1}{2} k \dot{ο}) (\frac{ħ}{2 mω}) [\sqrt{(n + 1) (n + 2)} δ_{m, n + 2} + \sqrt{n (n + 1)} δ_{m, n - 2}] E_{n}^{0} = {(\frac{\dot{ο ħ ω}}{4})}^{2} \sum_{m \neq n} \frac{{[\sqrt{(n + 1) (n + 2)} δ_{m, n + 2} + \sqrt{n (n + 1)} δ_{m, n + 2}]}^{2}}{(n + \frac{1}{2}) \dot{ħ ω} - (m + \frac{1}{2}) \dot{ħ ω}}$

$= \frac{\dot{ο^{2} ħ ω}}{16} \sum_{m \neq n} \frac{[(n + 1) (n + 1) δ_{m, n, + 2} + n (n - 1) δ_{m, n, - 2}]}{(n - m)} = \frac{\dot{ο^{2} ħ ω}}{32} (- n^{2} - 3 n - 2 + n^{2} - n) = \frac{\dot{ο^{2} ħ ω}}{32} (- 4 n - 2) = - \dot{ο^{2}} \frac{1}{8} \dot{ħ ω} (n + \frac{1}{2})$

(which agrees with the $ε^{2}$ term in the exact solution Problem 6.2(a)).

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Short Answer

Step by step solution

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

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

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