Expectation Values and Operators

  • Thread starter Nezva
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  • #1
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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
 

Answers and Replies

  • #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.
 
  • #3
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[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?
 
  • #4
Fredrik
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c^*_n gives you [itex]c^*_n[/itex]
 
  • #5
jtbell
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[tex]\int[/tex] c*n[tex]\Psi[/tex]n[tex] \left| A(hat) \right| c[/tex]n[tex] \Psi[/tex]n
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]
 
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  • #6
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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.
 
  • #7
jtbell
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What do you mean by substitute the 'entire' [tex]\Psi[/tex] and [tex]\Psi[/tex]*
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.
 
  • #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.
 
  • #9
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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] ?
 
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  • #10
Fredrik
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The an is still representing the eigenvalues of the A operator, am, right?
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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