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 $({\mathbf {1}})$

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

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

$=1+\sum _{p=1}^{+\infty }\sum _{q=1}^{+\infty }\lambda ^{p+q}\sum _{m_{1}}'...\sum _{m_{p}}'\sum _{m'_{1}}'...\sum _{m'_{q}}'\langle n|H'|m_{p}\rangle {\frac {1}{E_{n}-\epsilon _{m_{p}}}}\langle m_{p}|H'|m_{p-1}\rangle {\frac {1}{E_{n}-\epsilon _{m_{p-1}}}}...\langle m_{2}|H'|m_{1}\rangle {\frac {1}{E_{n}-\epsilon _{m_{1}}}}\langle m_{1}|m'_{1}\rangle {\mathbf {.}}$

${\mathbf {.}}{\frac {1}{E_{n}-\epsilon _{m'_{1}}}}\langle m'_{1}|H'|m'_{2}\rangle ...{\frac {1}{E_{n}-\epsilon _{m'_{q-1}}}}\langle m'_{q-1}|H'|m'_{q}\rangle {\frac {1}{E_{n}-\epsilon _{m'_{q}}}}\langle m'_{q}|H'|n\rangle$

We have $\langle m_{1}|m'_{1}\rangle =\delta _{m_{1}m'_{1}}$, therefore the summing over $m'_{1}$ is equivalent to setting $m'_{1}=m_{1}$. We get:

$Z^{-1}=1+\sum _{p=1}^{+\infty }\sum _{q=1}^{+\infty }\lambda ^{p+q}\sum _{m_{1}}'...\sum _{m_{p}}'\sum _{m'_{2}}'...\sum _{m'_{q}}'\langle n|H'|m_{p}\rangle {\frac {1}{E_{n}-\epsilon _{m_{p}}}}\langle m_{p}|H'|m_{p-1}\rangle {\frac {1}{E_{n}-\epsilon _{m_{p-1}}}}...\langle m_{2}|H'|m_{1}\rangle {\frac {1}{(E_{n}-\epsilon _{m_{1}})^{2}}}{\mathbf {.}}$

${\mathbf {.}}\langle m_{1}|H'|m'_{2}\rangle ...{\frac {1}{E_{n}-\epsilon _{m'_{q-1}}}}\langle m'_{q-1}|H'|m'_{q}\rangle {\frac {1}{E_{n}-\epsilon _{m'_{q}}}}\langle m'_{q}|H'|n\rangle$

Let's define $k=p+q-1$ and exchange the indexes as follows:

$m_{p}\rightarrow m_{1};\;m_{p-1}\rightarrow m_{2};\;...;m_{1}\rightarrow m_{p}$

$m'_{2}\rightarrow m_{p+1};\;m'_{3}\rightarrow m_{p+2};\;...;m'_{q}\rightarrow m_{p+q-1}=m_{k}$

Doing so we can see that $Z^{-1}$ exactly equals to $({\frac {\partial E_{n}}{\partial \epsilon _{n}}})^{-1}$ given in (2). Therefore:

$Z=\langle N|N\rangle ^{-1}={\frac {\partial E_{n}}{\partial \epsilon _{n}}}$