Normalization constant

Using Brillouin-Wigner perturbation theory we will proof that $Z=\langle N|N\rangle ^{-1}={\frac {\partial E_{n}}{\partial \epsilon _{n}}}$

In this theory, the exact state and exact energy can be written as follows:

$|N\rangle =|n\rangle +\sum _{k=1}^{+\infty }\lambda ^{k}\left(\sum _{m_{1}}'...\sum _{m_{k}}'|m_{1}\rangle {\frac {1}{E_{n}-\epsilon _{m_{1}}}}\langle m_{1}|H'|m_{2}\rangle .{\frac {1}{E_{n}-\epsilon _{m_{2}}}}\langle m_{2}|H'|m_{3}\rangle ...{\frac {1}{E_{n}-\epsilon _{m_{k-1}}}}\langle m_{k-1}|H'|m_{k}\rangle .{\frac {1}{E_{n}-\epsilon _{m_{k}}}}\langle m_{k}|H'|n\rangle \right)\;({\mathbf {1}})$

$E_{n}=\epsilon _{n}+\lambda \langle n|H'|n\rangle +$

$+\sum _{k=1}^{+\infty }\lambda ^{k+1}\left(\sum _{m_{1}}'...\sum _{m_{k}}'\langle n|H'|m_{1}\rangle {\frac {1}{E_{n}-\epsilon _{m_{1}}}}\langle m_{1}|H'|m_{2}\rangle .{\frac {1}{E_{n}-\epsilon _{m_{2}}}}\langle m_{2}|H'|m_{3}\rangle ...{\frac {1}{E_{n}-\epsilon _{m_{k-1}}}}\langle m_{k-1}|H'|m_{k}\rangle .{\frac {1}{E_{n}-\epsilon _{m_{k}}}}\langle m_{k}|H'|n\rangle \right)$

Taking the derivative of $E_{n}$ with respect ${\mathbf {\epsilon }}_{n}$ to, using the chain rule ,we get:

${\frac {\partial E_{n}}{\partial \epsilon _{n}}}=1+0+$

$+{\frac {\partial E_{n}}{\partial \epsilon _{n}}}\sum _{k=1}^{+\infty }\lambda ^{k+1}{\frac {\partial }{\partial E_{n}}}\left(\sum _{m_{1}}'...\sum _{m_{k}}'\langle n|H'|m_{1}\rangle {\frac {1}{E_{n}-\epsilon _{m_{1}}}}\langle m_{1}|H'|m_{2}\rangle .{\frac {1}{E_{n}-\epsilon _{m_{2}}}}\langle m_{2}|H'|m_{3}\rangle ...{\frac {1}{E_{n}-\epsilon _{m_{k-1}}}}\langle m_{k-1}|H'|m_{k}\rangle .{\frac {1}{E_{n}-\epsilon _{m_{k}}}}\langle m_{k}|H'|n\rangle \right)$

$=1-{\frac {\partial E_{n}}{\partial \epsilon _{n}}}\sum _{k=1}^{+\infty }\lambda ^{k+1}\sum _{m_{1}}'...\sum _{m_{k}}'\langle n|H'|m_{1}\rangle {\frac {1}{E_{n}-\epsilon _{m_{1}}}}\langle m_{1}|H'|m_{2}\rangle .{\frac {1}{E_{n}-\epsilon _{m_{2}}}}\langle m_{2}|H'|m_{3}\rangle ...$

$...{\frac {1}{E_{n}-\epsilon _{m_{p-1}}}}\langle m_{p-1}|H'|m_{p}\rangle .{\frac {1}{(E_{n}-\epsilon _{m_{p}})^{2}}}\langle m_{p}|H'|m_{p+1}\rangle ...{\frac {1}{E_{n}-\epsilon _{m_{k-1}}}}\langle m_{k-1}|H'|m_{k}\rangle .{\frac {1}{E_{n}-\epsilon _{m_{k}}}}\langle m_{k}|H'|n\rangle$

From this we can solve for ${\frac {\partial E_{n}}{\partial \epsilon _{n}}}$

$({\frac {\partial E_{n}}{\partial \epsilon _{n}}})^{-1}=\sum _{k=1}^{+\infty }\lambda ^{k+1}\sum _{m_{1}}'...\sum _{m_{k}}'\langle n|H'|m_{1}\rangle {\frac {1}{E_{n}-\epsilon _{m_{1}}}}\langle m_{1}|H'|m_{2}\rangle .{\frac {1}{E_{n}-\epsilon _{m_{2}}}}\langle m_{2}|H'|m_{3}\rangle ...$

$...{\frac {1}{E_{n}-\epsilon _{m_{p-1}}}}\langle m_{p-1}|H'|m_{p}\rangle .{\frac {1}{(E_{n}-\epsilon _{m_{p}})^{2}}}\langle m_{p}|H'|m_{p+1}\rangle ...{\frac {1}{E_{n}-\epsilon _{m_{k-1}}}}\langle m_{k-1}|H'|m_{k}\rangle .{\frac {1}{E_{n}-\epsilon _{m_{k}}}}\langle m_{k}|H'|n\rangle \qquad \qquad \qquad \qquad ({\mathbf {2}})$

Now let's evaluate $Z=\langle N|N\rangle ^{-1}$ from $({\mathbf {1}})$