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Expectation Values and Operators

  1. Mar 13, 2010 #1
    I've never seen an expectation value taken and would greatly appreciate seeing a step by step of how it is done. Feel free to use any wavefunction, this is the one I've been trying to do:

    In the case of [tex]\Psi=c[/tex]1[tex]\Psi[/tex]1[tex] + c[/tex]2[tex]\Psi[/tex]2[tex] + ... + c[/tex]n[tex]\Psi[/tex]n

    And the operator A(hat) => A(hat)[tex]\Psi[/tex]1 = a1[tex]\Psi[/tex]1; A(hat)[tex]\Psi[/tex]2 = a2[tex]\Psi[/tex]2; A(hat)[tex]\Psi[/tex]n = an[tex]\Psi[/tex]n

    Calculate: [tex]\left\langle\Psi\left|A(hat)\right|\right\Psi\rangle[/tex]

    img286.png
     
  2. jcsd
  3. Mar 13, 2010 #2

    jtbell

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    First step: substitute your formula for [itex]\Psi[/itex] into the integral, for [itex]\Psi^*[/itex] as well as for [itex]\Psi[/itex]. Don't multiply out the "products" yet. Use parentheses to keep things together.
     
  4. Mar 13, 2010 #3
    [tex]\int[/tex] c*n[tex]\Psi[/tex]n[tex] \left| A(hat) \right| c[/tex]n[tex] \Psi[/tex]n

    I'm having trouble putting subscript into the LaTex format, is there a way to do this without breaking pup the tex formatting?
     
  5. Mar 13, 2010 #4

    Fredrik

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    c^*_n gives you [itex]c^*_n[/itex]
     
  6. Mar 13, 2010 #5

    jtbell

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    You didn't substitute the entire [itex]\Psi^*[/itex] and [itex]\Psi[/itex]:

    [tex]\Psi = c_1 \Psi_1 + c_2 \Psi_2 + \ldots + c_n \Psi_n[/tex]

    [tex]\Psi^* = c_1^* \Psi_1^* + c_2^* \Psi_2^* + \ldots + c_n^* \Psi_n^*[/tex]

    Like I said, use parentheses as necessary.

    By the way, you can see the LaTeX code for an equation by clicking on it. (I use a Mac, so I don't know whether it's left-click or right-click under Windows.) Here's your original integral to use as a model:

    [tex]\langle A \rangle = \int^{+\infty}_{-\infty} {\Psi^* \hat A \Psi d\tau}[/tex]
     
    Last edited: Mar 13, 2010
  7. Mar 14, 2010 #6
    What do you mean by substitute the 'entire' [tex]\Psi[/tex] and [tex]\Psi[/tex]*, simply that I didn't denote the first wavefunction with asterisks fully? Or do you mean literally put in c21, c2, c3, etc.?

    Anyways here was an attempt at this.

    [tex]\int(c^*_n\Psi^*_n\hat A)(\hat Ac_n\Psi_n)[/tex]

    [tex]c^*_n c_n\int(a^*_n\Psi^*_n)(a_n\Psi_n)[/tex]

    [tex]c^*_n c_n a^*_n a_n\int(\Psi^*_n)(\Psi_n)[/tex]

    The constants don't pull through like that but how do I simplify this equation? I'm so lost on this.
     
  8. Mar 14, 2010 #7

    jtbell

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    I mean "replace [itex]\Psi[/itex] with [itex]c_1 \Psi_1 + c_2 \Psi_2 + \ldots + c_n \Psi_n[/itex]" and similarly for [itex]\Psi^*[/itex]. Like this:

    [tex]\langle A \rangle = \int^{+\infty}_{-\infty} {\Psi^* \hat A \Psi d\tau}[/tex]

    [tex]\langle A \rangle = \int^{+\infty}_{-\infty}
    {(c_1^* \Psi_1^* + c_2^* \Psi_2^* + \ldots + c_n^* \Psi_n^*) \hat A (c_1 \Psi_1 + c_2 \Psi_2 + \ldots + c_n \Psi_n) d\tau}[/tex]

    Now move the operator [itex]\hat A[/itex] inside the parentheses on the right, and apply it to each term in the sum.
     
  9. Mar 14, 2010 #8

    Fredrik

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    You should use two different letters for the indices since there are two different sums. And I don't know where that extra A came from in your last post.

    [tex]\langle A\rangle_\psi=\langle \psi,A\psi\rangle=\langle\sum_n c_n\psi_n,A\sum_m c_m \psi_m\rangle=\sum_n\sum_m c_n^*c_m\langle\psi_n,A\psi_m\rangle[/tex]

    [tex]=\sum_n\sum_m c_n^*c_m a_m\langle\psi_n,\psi_m\rangle=\sum_n|c_n|^2 a_n[/tex]

    D'oh, I'm too slow. Jtbell beat me too it, but now at least you get to see it in a different notation.
     
  10. Mar 14, 2010 #9
    Thank you both for the contrasting notations. Is there anything I can do to return the help?

    [tex]=\sum_n\sum_m c_n^*c_m a_m\langle\psi_n,\psi_m\rangle=\sum_n|c_n|^2 a_n[/tex]


    Very elegant. The absolute value of the complex conjugates is a clever touch. I've not used complex number enough to recognize to do that.

    The an is still representing the eigenvalues of the A operator, am, right? You simply changed the notation to show that it is the same as if the operator were applied to the [tex]\psi_n[/tex] ?
     
    Last edited: Mar 14, 2010
  11. Mar 14, 2010 #10

    Fredrik

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    Yes, I skipped a step or two. It's [tex]\langle\psi_n,A\psi_m\rangle=\langle\psi_n,a_m\psi_m\rangle=A_m\langle\psi_n,\psi_m\rangle=A_m\delta_{nm}[/tex].
     
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