Sakurai 1.32

  • #1
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Homework Statement



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

Homework Equations



Momentum-space wave function:

[tex]\sqrt{\frac{d}{\hbar\sqrt{\pi}}}e^{\frac{-\left(p'-\hbar k\right)^2d^2}{2\hbar^2}}[/tex]

The Attempt at a Solution



I can get [tex]\langle p \rangle [/tex], so that's not a problem.

[tex]\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'[/tex]

[tex]= \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'[/tex]

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

This has the form:

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

So I'll integrate it by parts:

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

[tex]p'\hbar k\frac{\hbar}{d}\sqrt{\pi}|_{-\infty}^{\infty}-\int_{-\infty}^{\infty}\hbar k\frac{\hbar}{d}\sqrt{\pi}dp'[/tex]

[tex]=p'\hbar k\frac{\hbar}{d}\sqrt{\pi}|_{-\infty}^{\infty}-p'\hbar k\frac{\hbar}{d}\sqrt{\pi}|_{-\infty}^{\infty}=0[/tex]

But it should be:

[tex]\langle p^2 \rangle = \frac{\hbar^2}{2d^2}+\left(\hbar k\right)^2[/tex]

What am I doing wrong?
 

Answers and Replies

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

This has the form:

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

No it doesn't. It has the form

[tex]\int_{-\infty}^{\infty}e^{-a\left(x-b\right)^2}x^2dx[/tex]
 
  • #3
338
0
No it doesn't. It has the form

[tex]\int_{-\infty}^{\infty}e^{-a\left(x-b\right)^2}x^2dx[/tex]

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:

[tex]x^2=x^2-2bx+b^2+2bx-b^2=\left(x-b\right)^2+2bx-b^2[/tex]

Then

[tex]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[/tex]

Expand that and there will be integrals that can be looked up in tables. Do the algebra, and get the correct answer.
 
  • #4
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Hmm, I'm not sure about that substitution there (for instance, why is it only applied to the leading [tex]x^2[/tex] term and not the [tex]e^{-a(x-b)^2}[/tex] 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 [tex]<x^2>[/tex] of the function:

[tex]\rho(x) = Ae^{-\lambda(x-a)^2}[/tex]

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 [tex]x^2[/tex]... 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):

[tex]\int^\infty_{-\infty} e^{-\lambda x^2} dx = (\dfrac{\pi}{\lambda})^\dfrac{1}{2}[/tex]

Now if you differentiate both sides wrt lambda:

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

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

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!
 
Last edited:
  • #5
338
0
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:

[tex]
c\int_{-\infty}^{\infty}x^2e^{-a(x-b)^2}dx
[/tex]

However, if I break it up, like this:

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

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.
 
  • #6
gabbagabbahey
Homework Helper
Gold Member
5,002
7
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 [itex]\overline{x}=\sqrt{a}(x-b)[/itex], use integration by parts, and the following trick:

[tex]\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}[/tex]
 

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