# Determine the mean of the position <x> in a state

1. Aug 11, 2010

### Ylle

1. The problem statement, all variables and given/known data
Hello everyone...

I'm kinda stuck with a problem I'm trying to do.
The problem states:

Express the operator $\hat{x}$ by the ladder operators $a_{+}$ and $a_{-}$, and determine the mean of the position $\left\langle x \right\rangle$ in the state $\left| \psi \right\rangle$.

2. Relevant equations
$$\left| \psi \right\rangle = \frac{1}{\sqrt{2}}(\left| 3 \right\rangle + \left| 2 \right\rangle)$$

Hamiltonian for a one dimensional harmonic oscillator:

$$\hat{H} = \frac{\hat{p}^{2}}{2m}+\frac{1}{2}mw^{2}\hat{x}^{2},$$
where $w$ is the oscillators frequency, and $x$ and $p$ are the operators for position and momentum. The normalized energy eigenfunctions for $H$ is denoted $\left| n \right\rangle$, where $n = 0,1,2,....$ so that:

$$\hat{H}\left| n \right\rangle = (n + \frac{1}{2})\hbarw\left| \psi \right\rangle$$

3. The attempt at a solution
The first is easy, since:

$$\hat{x} = \sqrt{\frac{\hbar}{2mw}}(a_{+} + a_{-}).$$

My problem is finding the mean of the position.

I tried to do it like this:

$$\left\langle x \right\rangle = \sqrt{\frac{\hbar}{2mw}}\int \psi^{*}_{n}(a_{+} + a_{-})\psi_{n} dx$$

And that didn't go well. It got very confusing, so I was not sure if I was on the right track or not. So here I am.

I know the answer should be:

$$\left\langle x \right\rangle = \sqrt{\frac{3}{2}}\sqrt{\frac{\hbar}{mw}},$$
but again, I'm kinda lost atm.

So I was hoping any of you could give me a clue. Something in my heads tells me it's pretty simple, but I really can't figure it out right now, so :)

Regards

2. Aug 11, 2010

### diazona

Hint: what are
$$a_+\lvert n\rangle$$
and
$$a_-\lvert n\rangle$$
equal to?

3. Aug 11, 2010

### Ylle

$$a_+ \lvert n\rangle = \sqrt{n+1} \lvert n+1\rangle$$
and
$$a_- \lvert n\rangle = \sqrt{n} \lvert n-1\rangle$$

I think I did something like that, but as I said. I may have messed it up. But I was right with the thought I had ? I just need to redo the math perhaps ?

4. Aug 12, 2010

### vela

Staff Emeritus
The idea is to do the problem without resorting to doing any integrals. Just use the orthogonality of the eigenstates. For example,

$$\langle 1|a_+|0 \rangle = \langle 1|\sqrt{1}|1\rangle = 1$$

5. Aug 12, 2010

### Ylle

Hmmm... I think I may have it now. But what if the ladder operators are used on state |3> fx. ? At least the a+ operator. Then it would raise it to state 4, but that state isn't represented here, so will that just equal 0, or... ?

6. Aug 12, 2010

### vela

Staff Emeritus
Yes, that term will drop out.

7. Aug 12, 2010

### Ylle

Thank you very much. I thought it was what I did to start with, but I just messed it up.
But I got it now :)

Thank you.