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Main class wiki page: home

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.
  

• 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)

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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