Normalization constant: Difference between revisions
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In this theory, the exact state and exact energy can be written as follows: | In this theory, the exact state and exact energy can be written as follows: | ||
<math>|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 )</math> | <math>|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 ) \; (\bold 1)</math> | ||
<math>E_{n}=\epsilon _{n}+ \lambda \langle n|H'|n \rangle + </math> | <math>E_{n}=\epsilon _{n}+ \lambda \langle n|H'|n \rangle + </math> | ||
<math> + \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 ) </math> | <math> + \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 ) </math> | ||
where <math>\sum '</math>does not allow the running indexes equal to n. | |||
Taking the derivative of <math>E_{n}</math> with respect <math>\bold \epsilon _{n}</math> to, using the chain rule ,we get: | |||
<math>\frac {\partial E_{n}}{\partial \epsilon _{n}}=1+0+ </math> | |||
<math> + \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 ) </math> | |||
<math>=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 ...</math> | |||
<math>...\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 </math> | |||
From this we can solve for <math>\frac {\partial E_{n}}{\partial \epsilon _{n}}</math> | |||
<math> (\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 ...</math> | |||
<math>...\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 (\bold 2)</math> | |||
Now let's evaluate <math>Z=\langle N|N \rangle ^{-1}</math> from <math>(\bold 1)</math> | |||
<math>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 ) \bold .</math> | |||
<math> \bold . \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 )=</math> | |||
<math>=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 \bold .</math> | |||
<math> \bold . \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 </math> | |||
We have <math>\langle m_{1}|m'_{1} \rangle= \delta _{m_{1}m'_{1}}</math>, therefore the summing over <math>m'_{1}</math> is equivalent to setting <math>m'_{1}=m_{1}</math>. We get: | |||
<math> 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} \bold .</math> | |||
<math> \bold . \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 </math> | |||
Let's define <math>k=p+q-1</math> and exchange the indexes as follows: | |||
<math>m_{p} \rightarrow m_{1}; \; m_{p-1} \rightarrow m_{2}; \; ... ; m_{1} \rightarrow m_{p} </math> | |||
<math>m'_{2} \rightarrow m_{p+1}; \; m'_{3} \rightarrow m_{p+2}; \; ... ; m'_{q} \rightarrow m_{p+q-1}=m_{k} </math> | |||
Doing so we can see that <math>Z^{-1}</math> exactly equals to <math>(\frac {\partial E_{n}}{\partial \epsilon_{n}})^{-1}</math> given in (2). Therefore: | |||
<math>Z=\langle N|N \rangle ^{-1} = \frac {\partial E_{n}}{\partial \epsilon_{n}}</math> | |||
Latest revision as of 16:54, 18 April 2009
Using Brillouin-Wigner perturbation theory we will proof that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |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 ) \; (\bold 1)}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{n}=\epsilon _{n}+ \lambda \langle n|H'|n \rangle + }
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle + \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 ) }
where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum '} does not allow the running indexes equal to n.
Taking the derivative of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_{n}} with respect Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold \epsilon _{n}} to, using the chain rule ,we get:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac {\partial E_{n}}{\partial \epsilon _{n}}=1+0+ }
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle + \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 ) }
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =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 ...}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ...\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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac {\partial E_{n}}{\partial \epsilon _{n}}}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\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 ...}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ...\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 (\bold 2)}
Now let's evaluate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z=\langle N|N \rangle ^{-1}} from Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\bold 1)}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 ) \bold .}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold . \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 )=}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =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 \bold .}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold . \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle m_{1}|m'_{1} \rangle= \delta _{m_{1}m'_{1}}} , therefore the summing over Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m'_{1}} is equivalent to setting Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m'_{1}=m_{1}} . We get:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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} \bold .}
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold . \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k=p+q-1} and exchange the indexes as follows:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_{p} \rightarrow m_{1}; \; m_{p-1} \rightarrow m_{2}; \; ... ; m_{1} \rightarrow m_{p} }
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z^{-1}} exactly equals to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\frac {\partial E_{n}}{\partial \epsilon_{n}})^{-1}} given in (2). Therefore:
Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z=\langle N|N \rangle ^{-1} = \frac {\partial E_{n}}{\partial \epsilon_{n}}}