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 $$\Psi=c$$₁$$\Psi$$₁$$+ c$$₂$$\Psi$$₂$$+ ... + c$$_n$$\Psi$$_n

And the operator A(hat) => A(hat)$$\Psi$$₁ = a₁$$\Psi$$₁; A(hat)$$\Psi$$₂ = a₂$$\Psi$$₂; 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:

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

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

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 $$\langle\psi_n,A\psi_m\rangle=\langle\psi_n,a_m\psi_m\rangle=A_m\langle\psi_n,\psi_m\rangle=A_m\delta_{nm}$$.