Responsible party: poit0009, Hydra

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Please try to include the following

• main points understood, and expand them - what is your understanding of what the points were.
  • expand these points by including many of the details the class discussed.
• main points which are not clear. - describe what you have understood and what the remain questions surrounding the point(s).
  • Other classmates can step in and clarify the points, and expand them.
• How the main points fit with the big picture of QM. Or what is not clear about how today's points fit in in a big picture.
• wonderful tricks which were used in the lecture.
  

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>=A^{2} \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>

The equation for <math>\Psi(p,t)</math> is a result of a Fourier transform, ⇒<math> \sum C_{n}\Psi_{n}</math> Here, the sum of all coefficients are equal to 1 <math>\sum C_{n}=1 </math> which holds true when the wavefunction is normalized.

*This section posted in segments, so it's not complete yet*

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