2nd-Order Correction

Rommel J. Jagus

Find the 2^nd-order correction to the energies $(E_n^{2})$ for the potential $H=\alpha \delta (x-\frac{a}{2})$

Solution:

$<\Psi_{m}^{0} | H | \Psi_{n}^{0}> = \frac{2} {a} \alpha \int_0^a sin(\frac{m x \pi }{a} ) \delta (x-\frac{a}{2}) sin(\frac{n x \pi }{a})$

$<\Psi_{m}^{0} | H | \Psi_{n}^{0}> =\frac{2} {a} \alpha [a sin(\frac{m \pi }{a} ) sin(\frac{n \pi }{a})]$

This is zero unless both m and n are odd in which case it is $\pm\frac{2\alpha}{a}$

$E_{n}^{2}=\sum_{m \ne n, odd} (\frac{2\alpha}{a})^2$

But

$E_{n}^{o}=\frac{\pi^2 \hbar^2 n^2}{2ma^2}$

So,

$E_{n}^{2}=2m(\frac{2\alpha}{\pi m})^2\sum_{m \ne n, odd} \frac{1}{(n^2-m^2)}$

Since $\frac{1}{n^2-m^2}=\frac{1}{2n}(\frac{1}{m+n}-\frac{1}{m-n})$

For n=1

$\sum= \frac{1}{2} \sum_{3,5,7} (\frac{1}{m+1}-\frac{1}{m-1})$

$\sum=\frac{1}{2}(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+$ … $-\frac{1}{2}-\frac{1}{4}-\frac{1}{6} - \frac{1}{8})$

$\sum= \frac{1}{2}(\frac{-1}{2}) = - \frac{1}{4}$

For n=3

$\sum= \frac{1}{6} \sum_{1,5,7} (\frac{1}{m+3}-\frac{1}{m-3})$

$\sum=\frac{1}{6}(\frac{1}{4}+\frac{1}{8}+\frac{1}{10}+$ … $-\frac{1}{2}-\frac{1}{4}-\frac{1}{8} - \frac{1}{10})$

$\sum= \frac{1}{6}(\frac{-1}{6}) = - \frac{1}{36}$

Therefore

$E_{n}^{2}=0$ if n is even

$E_{n}^{2}=-2m (\frac{\alpha}{\pi\hbar^2\n})^2$ if n is odd

Source: Problem 6.4 A in “Introduction to Quantum Mechanics” by David J. Griffiths

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