Solve 1D Harmonic Oscillator: Expectation Value of X is Zero

[White_M]

Homework Statement

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

Homework Equations

Using

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

The Attempt at a Solution

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

 

[DrClaude]

Using

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

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.

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

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

 

[White_M]

That doens't look like the standard form.

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 (page 79 (pdf)).

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

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

 

[vela]

Yes, you can.