# For 2 electrons in a simple harmonic oscillator (SHO) potential, is $\langle x^2\rangle$ the same as $\langle(x_2 - x_1)^2\rangle$?

We're dealing with 2 electrons in a simple harmonic oscillator (SHO) potential. We're given the creation and annihilation operators as well as the position operator and I have to find the expectation value $\langle(x_2 - x_1)^2\rangle$ where $x_2$ and $x_1$ correspond to the positions of the two electrons. My problem is that I don't know how to input those two different positions into the position operator, which by the way is defined as:

$$\hat{x}_i=\sqrt{\frac{\hbar}{2m\omega}}(\hat{a}_i+\hat{a}_i^\dagger)$$

So there you have it, I am stuck because I'm unclear on notation.

EDIT I am still stuck, I don't know what the $x_1$ or $x_2$ operators would look like. The definition above of the position operator, as I read it, is telling me that I should make the following substitution to get the $x_1$ and $x_2$ operators:

$$\hat{x}_1=\sqrt{\frac{\hbar}{2m\omega}}(\hat{a}_1+\hat{a}_1^\dagger)$$ $$\hat{x}_2=\sqrt{\frac{\hbar}{2m\omega}}(\hat{a}_2+\hat{a}_2^\dagger)$$

But this doesn't make a lot of sense to me, because now thelowering operator would be this:

$$\hat{a}_i\vert n \rangle_i=\sqrt{n} \vert n-1\rangle \rightarrow \hat{a}_1\vert n \rangle_1=\sqrt{n} \vert n-1\rangle$$

$$\hat{a}_i\vert n \rangle_i=\sqrt{n} \vert n-1\rangle \rightarrow \hat{a}_2\vert n \rangle_2=\sqrt{n} \vert n-1\rangle$$

And the raising operator would be this:

$$\hat{a}_i^\dagger\vert n \rangle_i=\sqrt{n+1} \vert n+1\rangle \rightarrow \hat{a}_1\vert n \rangle_1=\sqrt{n+1} \vert n+1\rangle$$

$$\hat{a}_i^\dagger\vert n \rangle_i=\sqrt{n+1} \vert n+1\rangle \rightarrow \hat{a}_2\vert n \rangle_2=\sqrt{n+1} \vert n+1\rangle$$

which to me looks like they do exactly the same. Now someone much more knowledgeable than me said that this would be the wavefunction on which we operate:

$$\vert \Psi\rangle = \frac{1}{\sqrt{2}}\lbrace\vert \Psi_0(x_1)\rangle\vert\psi_1(x_2)\rangle + \vert\Psi_0(x_2)\rangle\vert\psi_1(x_1)\rangle\rbrace$$ where $\vert\Psi_0(x_1)\rangle$ means the ground state of the first particle and $\vert\Psi_1(x_1)\rangle$ is the first excited state of the first electron, you get the idea.

So this is fine by me, I still don't know how I am to solve the problem because I don't know what $x_1$ or $x_2$ would look like nor what the difference between those operators is. Am I making sense here? I am being thrown off by subscripts, I can't see any difference between those two operators, and so I can't do the problem.

It is not a full solution to the problem, but my two cents: In Heisenberg picture the dynamics is encoded in the observables. The evolution equation for an observable $A=A(t)$ is

$$\frac{dA(t)}{dt}=\frac{i}{\hbar}[H,A(t)]-\frac{\partial A(t)}{\partial t}$$

As you are interested in expectation values:

$$\langle\frac{dA(t)}{dt}\rangle=\langle\frac{i}{\hbar}[H,A(t)]\rangle-\langle\frac{\partial A(t)}{\partial t}\rangle$$

Being more specific $A(t)=(x_2-x_1)^2$. It is not an explicit function of the, so the partial derivative vanish. Then you have

$$\langle \frac{d(x_2-x_1)^2}{dt}\rangle=\langle \frac{i}{\hbar}[H,(x_2-x_1)^2] \rangle$$

Now write both $H$ and $(x_1-x_2)^2$ in terms of $a$ and $a^{\dagger}$ and you will have your differential equation.

What's the problem with this answer? The solution will depend on a initial condition about $(x_1-x_2)^2$

If the electrons aren't interacting, I think you can safely expand $\langle(x_2 - x_1)^2\rangle$ as

\begin{align} \langle(x_2 - x_1)^2\rangle &= \langle x_2^2 + x_1^2 - 2x_2 x_1\rangle\\ &= \langle x_2^2\rangle + \langle x_1^2\rangle -2\langle x_2\rangle\langle x_1\rangle. \end{align}

Then you can use the raising and lowering operator representation and let it act on the two body wavefunction. That should give you the right answer.

• I am still stuck, I placed more details on the initial post. I just need to know what the operator $x_1$ would look like as opposed to $x_2$ and I think I'd be set to finally operate on the wavefunction. – user17338 Feb 28 '13 at 23:08