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First, we start off with the question of why <math> < f_{p}|f_{p'}> =\delta (p-p^{'})</math> And what exactly does this mean?
To begin, we know that we can operate on an eigenfunction as follows: <math>\widehat{p}|f_{p}>=p|f_{p}></math> where we have simply multiplied “p” onto each part of the vector
So for any “Q” we have <math>\widehat{Q}|q_{n}>=q_{n}|q_{n}></math>
and we know that <math><q_{n}|q_{n}>=1 </math>
so for two vectors <math> q_{n} , q_{m} </math> we will have <math><q_{m}|q_{n}>=\delta _{m,n}</math>
So to solve the case for <math> f_{p}</math>, we know <math> f_{p}=Ae^{i\frac{p}{\hbar}x}</math> where the usual “k” has been replaced with <math>\frac{p}{\hbar}</math>
Then <math>A^{2} \int_{-\infty}^{\infty}f_{p}^{*}\: f_{p^{'}}\; dx </math>
<math>= A^{2} \int_{-\infty }^{\infty}e^{-i\frac{p}{\hbar}x}\: \; e^{i\frac{p}{\hbar}x}\; dx </math>
<math> \int_{-\infty}^{\infty}e^{-i\frac{p-p^{'}}{\hbar}x}\; dx</math>
<math>=A^{2} 2\pi \int_{-\infty}^{\infty}\delta \left ( \frac{p-p^{'}}{\hbar} \right )dp </math>
and then we use a u-substitution <math>p=\hbar k </math>so <math>dp=\hbar dk</math> so we find
<math>=2\pi \hbar A^{2}\int \delta (k-k^{'})dk</math>
so we find <math>A=\frac{1}{\sqrt{2\pi \hbar}}</math>
so <math> f_{p}=\frac{1}{\sqrt{2\pi \hbar}}e^{i\frac{p}{\hbar}x} </math>
then applying this
<math> <f_{p}|\Psi >=\frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}e^{i\frac{p}{\hbar}x}\; \Psi (x,t) \; dx </math>
The above expression is the all important equation representing momentum space, <math>\Phi(p,t)</math>
*This section posted in segments, so it's not complete yet*
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