2nd-Order Correction | Quantum Science Philippines

## 2nd-Order Correction

Rommel J. Jagus

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

Solution:

[eq] <\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})[/eq]

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

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

[eq] E_{n}^{2}=\sum_{m \ne n, odd} (\frac{2\alpha}{a})^2[/eq]

But

[eq] E_{n}^{o}=\frac{\pi^2 \hbar^2 n^2}{2ma^2}[/eq]

So,

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

Since [eq]\frac{1}{n^2-m^2}=\frac{1}{2n}(\frac{1}{m+n}-\frac{1}{m-n})[/eq]

For n=1

[eq] \sum= \frac{1}{2} \sum_{3,5,7} (\frac{1}{m+1}-\frac{1}{m-1}) [/eq]

[eq]\sum=\frac{1}{2}(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+[/eq]…[eq] -\frac{1}{2}-\frac{1}{4}-\frac{1}{6} – \frac{1}{8})[/eq]

[eq]\sum= \frac{1}{2}(\frac{-1}{2}) = – \frac{1}{4}[/eq]

For n=3

[eq] \sum= \frac{1}{6} \sum_{1,5,7} (\frac{1}{m+3}-\frac{1}{m-3}) [/eq]

[eq]\sum=\frac{1}{6}(\frac{1}{4}+\frac{1}{8}+\frac{1}{10}+[/eq]…[eq]-\frac{1}{2}-\frac{1}{4}-\frac{1}{8} – \frac{1}{10})[/eq]

[eq]\sum= \frac{1}{6}(\frac{-1}{6}) = – \frac{1}{36}[/eq]

Therefore

[eq] E_{n}^{2}=0 [/eq] if n is even

[eq] E_{n}^{2}=-2m (\frac{\alpha}{\pi\hbar^2\n})^2[/eq] if n is odd

Source: Problem 6.4 A in â€œIntroduction to Quantum Mechanics” by David J. Griffiths