===== Oct 28 (Wed) 3.4 generalized probability, delta-func normalization =====
** 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.\\
\\

==== Main points discussed ====
  * Normalization of non-normalizable wave function: why <math> < f_{p}|f_{p'}> =\delta (p-p^{'})</math> makes sense?
  * generalized probability interpretation.
  * momentum-space based wave function, energy-space based wave function, real-space based wave function (what we have been dealing with)

===Normalization of non-normalizable wave function===
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.

>In the past, for some operator “Q,” its eigenvalue and eigenvector, we can express this relation as <math>\widehat{Q}|q_{n}>=q_{n}|q_{n}></math>
>
>and we tried to normalize the eigenvector so that <math><q_{n}|q_{n}>=1 </math>
>
>or more in general, for two eigenvectors <math> q_{n}</math> and <math>q_{m} </math> we will have <math><q_{m}|q_{n}>=\delta _{m,n}</math>.

For our current case: <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 can we normalize this wave function by setting |A| properly so that <math> \int_{-\infty}^{\infty}f_{p}^{*}\: f_{p^{'}}\; dx = 1</math>?

Let's try.

<math>1 = |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>

This is, as you may remember, one way to express the delta function (times 2pi), where <math>\frac{p-p^{'}}{\hbar}</math> should be the argument of the delta function.

<math>=|A|^{2}  2\pi \delta \left ( \frac{p-p^{'}}{\hbar} \right )
</math> 

Since this form of delta function is not very familiar to us, we need to use substitution to bring it to a more familiar form, where the argument should look like (k-k') form.  Also delta function mean something only when it goes inside an integral, and that integral should be over //p// originally, but when we do the substitution to //k//, dp has to be converted to dk in a proper way.

Following this plan, we use a substitution <math>p=\hbar k </math>so <math>dp=\hbar dk</math> so we find

<math>=2\pi\hbar |A|^{2} \delta (k-k^{'})</math> where the implied integral variable has changed from //p// to //k//.

This suggests that there is no way to accomplish <math> \int_{-\infty}^{\infty}f_{p}^{*}\: f_{p^{'}}\; dx = 1</math> no matter what value we choose for //A//.

If we give up this idea, and normalize <math>f_p</math> to the delta function, then <math>A=\frac{1}{2\pi\hbar}</math>, but at this point, this is an arbitrary decision.  We could have chosen so that <math> \int_{-\infty}^{\infty}f_{p}^{*}\: f_{p^{'}}\; dx = 10000\delta(k-k')</math> equally justifiably as <math> \int_{-\infty}^{\infty}f_{p}^{*}\: f_{p^{'}}\; dx = \delta(k-k')</math>.

In any case, for now, 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>

>Let's remember that we expanded wave functions in terms of stationary-state wave functions (linear combination), <math>|\psi>=\sum C_n \psi_n(x)</math>, where C_n's can be calculated by <math>C_n=<\psi_n|\psi></math>.
>
>Furthermore, we learned that |C_n|^2 can be INTERPRETED to be the probability that the particle would be observed to be in the state with energy E_n.  To make sure this interpretation makes sense, we also checked that 
><math>\sum C_{n}=1 </math> and
><math>\sum |C_{n}|^{2}E_{n}=<E>_{\Psi}</math>.
>To prove that this is possible, we start with <math> \hat{H}\Psi=E\Psi</math>  where <math>\hat{H}=\left( -\frac{\hbar^{2}}{2m}dx+V \right )</math>  is the Hamiltonian.
>
>With the definition of expectation value, <math> <E>=\int \Psi^{*}(x,t) \: \hat{H} \; \Psi(x,t)\; dx</math>
>
><math>=\int \sum C_{n}\Psi_{n}^{*}\; \hat{H}\; \sum C_{m}\Psi_{m}\; dx</math>
>
>=<math> E_{n}\sum  C_{n}^{*}C_{m}\int \Psi_{n}^{*}\Psi_{m}\; dx</math>
>
>Here <math>\Psi_{m}^{*}\Psi_{n}=\delta_{m,n}</math>
>
>So it is proven that <math>\sum |C_{n}|^{2}E_{n}=<E></math>


Instead of <math>\psi_n</math>'s, can we use <math>f_p</math>'s to do the same (except the summation will be integral)?

Namely, <math>\Psi(x,t)=\int \Phi(p,t)f_p dp</math> where <math>\Phi(p,t)</math> plays the role of C_n's.  It should be calculable by <math>\Phi(p,t)=<f_p|Psi(x,t)</math>.  Furthermore, <math>|Phi(p,t)|^2</math> should be interpreted as probability (density).  For the last part, we need to be able to show that 
<math>\int |\Phi(p,t)|^2 dp = 1</math> and
<math>\int p |\Phi(p,t)|^2 dp = <p></math>.

It turns out that if you choose so that <math> \int_{-\infty}^{\infty}f_{p}^{*}\: f_{p^{'}}\; dx = 1</math> everything works out.  (Students are strongly encouraged to show this claim is true.  Some of them may appear in the next quiz!  (//Yuichi//)


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






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