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.