Oct 28 (Wed) 3.4 generalized probability, delta-func normalization [Physics Intranet]

Oct 28 (Wed) 3.4 generalized probability, delta-func normalization

Responsible party: poit0009, Hydra

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previous lecture note: Oct 26 (Mon) Section 3.5 to ?
next lecture note: Oct 30 (Fri) Uncertainty Principle

Main class wiki page: Physics 4101.001 QM wiki page

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 $< f_{p}|f_{p'}> =\delta (p-p^{'})$ 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 $< f_{p}|f_{p'}> =\delta (p-p^{'})$ And what exactly does this mean?

To begin, we know that we can operate on an eigenfunction as follows: $\widehat{p}|f_{p}>=p|f_{p}>$ 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 $\widehat{Q}|q_{n}>=q_{n}|q_{n}>$

and we tried to normalize the eigenvector so that $<q_{n}|q_{n}>=1$

or more in general, for two eigenvectors $q_{n}$ and $q_{m}$ we will have $<q_{m}|q_{n}>=\delta _{m,n}$ .

For our current case: $f_{p}$ , we know $f_{p}=Ae^{i\frac{p}{\hbar}x}$ where the usual “k” has been replaced with $\frac{p}{\hbar}$

Then can we normalize this wave function by setting |A| properly so that $\int_{-\infty}^{\infty}f_{p}^{*}\: f_{p^{'}}\; dx = 1$ ?

Let's try.

$1 = |A|^{2} \int_{-\infty}^{\infty}e^{-i\frac{p}{\hbar}x}\: e^{i\frac{p'}{\hbar}x}\; dx$

$=|A|^{2} \int_{-\infty}^{\infty}e^{-i\frac{p-p^{'}}{\hbar}x}\; dx$

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

$=|A|^{2} 2\pi \delta \left ( \frac{p-p^{'}}{\hbar} \right )$

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 $p=\hbar k$ so $dp=\hbar dk$ so we find

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

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

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

In any case, for now, we find $A=\frac{1}{\sqrt{2\pi \hbar}}$

so $f_{p}=\frac{1}{\sqrt{2\pi \hbar}}e^{i\frac{p}{\hbar}x}$

then applying this

$<f_{p}|\Psi >=\frac{1}{\sqrt{2\pi \hbar}}\int_{-\infty}^{\infty}e^{i\frac{p}{\hbar}x}\; \Psi (x,t) \; dx$

The above expression is the all important equation representing momentum space, $\Phi(p,t)$

Let's remember that we expanded wave functions in terms of stationary-state wave functions (linear combination), $|\psi>=\sum C_n \psi_n(x)$ , where C_n's can be calculated by $C_n=<\psi_n|\psi>$ .

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
$\sum C_{n}=1$ and
$\sum |C_{n}|^{2}E_{n}=<E>_{\Psi}$ .
To prove that this is possible, we start with $\hat{H}\Psi=E\Psi$ where $\hat{H}=\left( -\frac{\hbar^{2}}{2m}dx+V \right )$ is the Hamiltonian.

With the definition of expectation value, $<E>=\int \Psi^{*}(x,t) \: \hat{H} \; \Psi(x,t)\; dx$

$=\int \sum C_{n}\Psi_{n}^{*}\; \hat{H}\; \sum C_{m}\Psi_{m}\; dx$

= $E_{n}\sum C_{n}^{*}C_{m}\int \Psi_{n}^{*}\Psi_{m}\; dx$

Here $\Psi_{m}^{*}\Psi_{n}=\delta_{m,n}$

So it is proven that $\sum |C_{n}|^{2}E_{n}=<E>$

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

It turns out that if you choose so that $\int_{-\infty}^{\infty}f_{p}^{*}\: f_{p^{'}}\; dx = 1$ 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*

To go back to the lecture note list, click Lecture Notes
previous lecture note: Oct 26 (Mon) Section 3.5 to ?
next lecture note: Oct 30 (Fri) Uncertainty Principle