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Sakurai 1.32

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data

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

    2. Relevant 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]

    3. 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?
     
  2. jcsd
  3. Oct 28, 2009 #2

    gabbagabbahey

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    No it doesn't. It has the form

    [tex]\int_{-\infty}^{\infty}e^{-a\left(x-b\right)^2}x^2dx[/tex]
     
  4. Oct 28, 2009 #3
    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.
     
  5. Oct 28, 2009 #4
    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: Oct 28, 2009
  6. Oct 29, 2009 #5
    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.
     
  7. Oct 29, 2009 #6

    gabbagabbahey

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    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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