# Sakurai 1.32

## Homework Statement

Evaluate the expectation value of p and p² using the momentum-space wave function

## Homework Equations

Momentum-space wave function:

$$\sqrt{\frac{d}{\hbar\sqrt{\pi}}}e^{\frac{-\left(p'-\hbar k\right)^2d^2}{2\hbar^2}}$$

## The Attempt at a Solution

I can get $$\langle p \rangle$$, so that's not a problem.

$$\langle p^2 \rangle = \int_{-\infty}^{\infty}\sqrt{\frac{d}{\hbar\sqrt{\pi}}}e^{\frac{-\left(p'-\hbar k\right)^2d^2}{2\hbar^2}}p'^2\sqrt{\frac{d}{\hbar\sqrt{\pi}}}e^{\frac{-\left(p'-\hbar k\right)^2d^2}{2\hbar^2}}dp'$$

$$= \frac{d}{\hbar\sqrt{\pi}}\int_{-\infty}^{\infty}e^{\frac{-\left(p'-\hbar k\right)^2d^2}{2\hbar^2}}p'^2e^{\frac{-\left(p'-\hbar k\right)^2d^2}{2\hbar^2}}dp'$$

$$= \frac{d}{\hbar\sqrt{\pi}}\int_{-\infty}^{\infty}e^{\frac{-\left(p'-\hbar k\right)^2d^2}{\hbar^2}}p'^2dp'$$

This has the form:

$$\int_{-\infty}^{\infty}e^{-a\left(x-b\right)^2}xdx = b\sqrt{\frac{\pi}{a}}$$

So I'll integrate it by parts:

$$u=p'$$
$$du=dp'$$
$$dv=e^{\frac{-\left(p'-\hbar k\right)^2d^2}{\hbar^2}}p' dp'$$
$$v=\hbar k\frac{\hbar}{d}\sqrt{\pi}$$

$$p'\hbar k\frac{\hbar}{d}\sqrt{\pi}|_{-\infty}^{\infty}-\int_{-\infty}^{\infty}\hbar k\frac{\hbar}{d}\sqrt{\pi}dp'$$

$$=p'\hbar k\frac{\hbar}{d}\sqrt{\pi}|_{-\infty}^{\infty}-p'\hbar k\frac{\hbar}{d}\sqrt{\pi}|_{-\infty}^{\infty}=0$$

But it should be:

$$\langle p^2 \rangle = \frac{\hbar^2}{2d^2}+\left(\hbar k\right)^2$$

What am I doing wrong?

gabbagabbahey
Homework Helper
Gold Member
$$= \frac{d}{\hbar\sqrt{\pi}}\int_{-\infty}^{\infty}e^{\frac{-\left(p'-\hbar k\right)^2d^2}{\hbar^2}}p'^2dp'$$

This has the form:

$$\int_{-\infty}^{\infty}e^{-a\left(x-b\right)^2}xdx = b\sqrt{\frac{\pi}{a}}$$

No it doesn't. It has the form

$$\int_{-\infty}^{\infty}e^{-a\left(x-b\right)^2}x^2dx$$

No it doesn't. It has the form

$$\int_{-\infty}^{\infty}e^{-a\left(x-b\right)^2}x^2dx$$

What I meant was I could get it to that form by integrating by parts. However, doing so resulted in zero.

The trick for this problem is this:

$$x^2=x^2-2bx+b^2+2bx-b^2=\left(x-b\right)^2+2bx-b^2$$

Then

$$c\int_{-\infty}^{\infty}x^2e^{-a(x-b)^2}dx=c\int_{-\infty}^{\infty}\left(\left(x-b\right)^2+2bx-b^2\right)e^{-a(x-b)^2}dx$$

Expand that and there will be integrals that can be looked up in tables. Do the algebra, and get the correct answer.

Hmm, I'm not sure about that substitution there (for instance, why is it only applied to the leading $$x^2$$ term and not the $$e^{-a(x-b)^2}$$ term?

Anyways, the reason I'm responding is there is an integral of this form in Griffiths QM, problem 1.3c, which is to find $$<x^2>$$ of the function:

$$\rho(x) = Ae^{-\lambda(x-a)^2}$$

Which is obviously the same integral. The way you are proposing is a lot of work-- I think the quick way is to differentiate both sides of the known result of the integral without the $$x^2$$... I believe this is called Leibniz' rule? (not sure, correct me if I'm wrong).

Anyways, it looks like this, and I thought it was pretty cool.

You know (can calculate):

$$\int^\infty_{-\infty} e^{-\lambda x^2} dx = (\dfrac{\pi}{\lambda})^\dfrac{1}{2}$$

Now if you differentiate both sides wrt lambda:

$$=> \dfrac{\partial}{\partial\lambda} \int^\infty_{-\infty} e^{-\lambda x^2} dx = \dfrac{\partial}{\partial\lambda} (\dfrac{\pi}{\lambda})^\dfrac{1}{2}$$

$$=> \int^\infty_{-\infty} x^2 e^{-\lambda x^2} dx = \dfrac{\sqrt{\pi}}{2\lambda^{ \small\dfrac{3}{2} }}$$

And, if you can look past my horrible latex exponent skills, the answer just pops right out!

I think its pretty cool.

Edit to add: Oh, I see what you did there with the x^2 term. Hmm, that integral seems more complicated than the one you started with to me!

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Edit to add: Oh, I see what you did there with the x^2 term. Hmm, that integral seems more complicated than the one you started with to me!

I could not find this integral in an integral table:

$$c\int_{-\infty}^{\infty}x^2e^{-a(x-b)^2}dx$$

However, if I break it up, like this:

$$c\int_{-\infty}^{\infty}\left(\left(x-b\right)^2+2bx-b^2\right)e^{-a(x-b)^2}dx$$

Each of those can be done (that is, "looked up") . And I already worked it all out and got the correct answer. I'll post it here when I get a chance so the next poor unsuspecting sap trying to figure this out may find this in a search and will have a glimmer of hope that life isn't just about doom and gloom.

gabbagabbahey
Homework Helper
Gold Member
If you are curious as to how to actually perform the integrations yourself (as opposed to looking them up in tables), just make the substitution $\overline{x}=\sqrt{a}(x-b)$, use integration by parts, and the following trick:

$$\int_{-\infty}^{\infty}e^{-\overline{x}^2}d\overline{x}=\left(\int_{-\infty}^{\infty}e^{-\overline{x}^2}d\overline{x}\int_{-\infty}^{\infty}e^{-\overline{y}^2}d\overline{y}\right)^{1/2}=\left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(\overline{x}^2+\overline{y}^2)}d\overline{x}d\overline{y}\right)^{1/2}=\left(\int_{0}^{\infty}\int_{0}^{2\pi}e^{-\overline{r}^2}\overline{r}d\overline{r}d\overline{\theta}\right)^{1/2}=\sqrt{2\pi}$$