# Harmonic Oscillator

1. Dec 18, 2013

### White_M

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

I need to show that for an eigen state of 1D harmonic oscillator the expectation values of the position X is Zero.

2. Relevant equations

Using

a+=$\frac{1}{\sqrt{2mhw}}$($\hat{Px}$+iwm$\hat{x}$)
a-=$\frac{1}{\sqrt{2mhw}}$($\hat{Px}$-iwm$\hat{x}$)

3. The attempt at a solution

<x>=<n|x|n>=$\sqrt{\frac{h}{2mw}}$<n|(a-+a+)|n=??

2. Dec 18, 2013

### Staff: Mentor

That doens't look like the standard form, see http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator#Ladder_operator_method

You won't get $\hat{x}$ from $a^+ + a^-$ with those.

What is your question exactly? Can you figure out what
$$\left( a^- + a^+ \right) \left|n\right\rangle$$
results in?

3. Dec 19, 2013

### White_M

I used the book "Fundamentals of Quantum Mechanics for Solid State Electronics Optics" - C.Tang
Here is the link to the book :http://en.bookfi.org/book/1308543 [Broken] (page 79 (pdf)).

Using the a+ and a- in the link you gave I get:
<n|x|n>=$\sqrt{\frac{h}{2mw}}$<n|a-+a+|n>

Now using:
a+|n>=$\sqrt{n+1}$|n+1>
<n|a-=a+|n>=$\sqrt{n+1}$|n+1>

I get:
<n|x|n>=2*$\sqrt{\frac{h}{2mw}}${<n|$\sqrt{n+1}$|n+1>}

Now can I say that since all {n} vectors are orthogonal the expression<n|$\sqrt{n+1}$|n+1>=$\sqrt{n+1}$<n|n+1>=0?

Thanks

Last edited by a moderator: May 6, 2017
4. Dec 19, 2013

### vela

Staff Emeritus
Yes, you can.