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

[Ylle]

Homework Statement

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.

Homework 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

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

 

[diazona]

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

 

[Ylle]

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

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 ?

 

[vela]

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

 

[Ylle]

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

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... ?

 

[vela]

Yes, that term will drop out.

 

[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.