Expectation Values and Operators

[Nezva]

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 \Psi=c₁\Psi₁+ c₂\Psi₂+ ... + c_n\Psi_n

And the operator A(hat) => A(hat)\Psi₁ = a₁\Psi₁; A(hat)\Psi₂ = a₂\Psi₂; A(hat)\Psi_n = a_n\Psi_n

Calculate: \left\langle\Psi\left|A(hat)\right|\right\Psi\rangle

 

[jtbell]

First step: substitute your formula for \Psi into the integral, for \Psi^* as well as for \Psi. Don't multiply out the "products" yet. Use parentheses to keep things together.

 

[Nezva]

\int c*_n\Psi_n\left| A(hat) \right| c_n\Psi_n

I'm having trouble putting subscript into the LaTex format, is there a way to do this without breaking pup the tex formatting?

 

[Fredrik]

c^*_n gives you c^*_n

 

[jtbell]

\int c*_n\Psi_n\left| A(hat) \right| c_n\Psi_n

You didn't substitute the entire \Psi^* and \Psi:

\Psi = c_1 \Psi_1 + c_2 \Psi_2 + \ldots + c_n \Psi_n

\Psi^* = c_1^* \Psi_1^* + c_2^* \Psi_2^* + \ldots + c_n^* \Psi_n^*

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:

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

 

[Nezva]

What do you mean by substitute the 'entire' \Psi and \Psi*, 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.

\int(c^*_n\Psi^*_n\hat A)(\hat Ac_n\Psi_n)

c^*_n c_n\int(a^*_n\Psi^*_n)(a_n\Psi_n)

c^*_n c_n a^*_n a_n\int(\Psi^*_n)(\Psi_n)

The constants don't pull through like that but how do I simplify this equation? I'm so lost on this.

 

[jtbell]

What do you mean by substitute the 'entire' \Psi and \Psi*

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

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

\langle A \rangle = \int^{+\infty}_{-\infty}<br /> {(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}

Now move the operator \hat A inside the parentheses on the right, and apply it to each term in the sum.

 

[Fredrik]

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.

\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

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

D'oh, I'm too slow. Jtbell beat me too it, but now at least you get to see it in a different notation.

 

[Nezva]

Thank you both for the contrasting notations. Is there anything I can do to return the help?

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


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 a_n is still representing the eigenvalues of the A operator, a_m, right? You simply changed the notation to show that it is the same as if the operator were applied to the \psi_n ?

 

[Fredrik]

The a_n is still representing the eigenvalues of the A operator, a_m, right?

Yes, I skipped a step or two. It's \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}.