# 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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## Answers and Replies

Nevermind! I realized that it does indeed change the eigenvector as well as the eigenvalue. Of course this must be true because it changes the state of the system, and you cannot have an eigenvector of one state with an eigenvalue of another. With this in mind, I have gotten the problem.