# Quantum Mechanics: Harmonic Oscillator Variance.

## Homework Statement

The problem wants me to calculate (Δx)^2 and (Δp)^2 to find the uncertainty principle. Delta x is the variance and the problem gives the formula as..

$$Δx= <n|x^{2}|n>-<n|x|n>^{2}$$

## Homework Equations

$$x=\sqrt{\frac{\hbar}{2m \omega}}(A^{-}+A^{+})$$
Where A+ and A- are the raising and lowering operator respectively.

## The Attempt at a Solution

So I plugged in this expression for x into the above expression for the variance.
I have the answer in front of me, the math for a similar argument (when we have two different eigenvalues, k and n) is..

http://imageshack.us/a/img266/5854/qmproblem3.png [Broken]

Meaning, that it vanishes unless n=k+1 or k-1

My main confusion is this..

First we have..
$$<k|A^{-}+A^{+}|n>$$
So I can distribute and get...
$$<k|A^{-}|n>+<k|A^{+}|n>$$
But since A raises or lowers the eigenvalue...
$$<k|A^{-}|n>=(n-1)<k|n>$$
Likewise..
$$<k|A^{+}|n>=(n+1)<k|n>$$

Why does the image I linked to change the eigenvector? I thought it only changed the eigenvalue. With the answer I got n would have to be equal to k for it not to vanish.

If I can understand this concept I can do the problem without any problem.

When I was trying to find $<n|x^{2}|n>$
I came to.. (Using a commutation relation)
$$\frac{m \omega \hbar}{2}<n|(A^+)^2+(A^-)^2+ 2A^{+}A^{-}-1|n>$$
Which I then distributed as I did above, and it boils down to the same math.
This is supposed to be
$$\frac{m \omega \hbar}{2}(2n+1)$$

Can anyone help?

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