- #1
Xyius
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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..
[tex]Δx= <n|x^{2}|n>-<n|x|n>^{2}[/tex]
Homework Equations
[tex]x=\sqrt{\frac{\hbar}{2m \omega}}(A^{-}+A^{+})[/tex]
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
Meaning, that it vanishes unless n=k+1 or k-1
My main confusion is this..
First we have..
[tex]<k|A^{-}+A^{+}|n>[/tex]
So I can distribute and get...
[tex]<k|A^{-}|n>+<k|A^{+}|n>[/tex]
But since A raises or lowers the eigenvalue...
[tex]<k|A^{-}|n>=(n-1)<k|n>[/tex]
Likewise..
[tex]<k|A^{+}|n>=(n+1)<k|n>[/tex]
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 [itex]<n|x^{2}|n>[/itex]
I came to.. (Using a commutation relation)
[tex]\frac{m \omega \hbar}{2}<n|(A^+)^2+(A^-)^2+ 2A^{+}A^{-}-1|n>[/tex]
Which I then distributed as I did above, and it boils down to the same math.
This is supposed to be
[tex]\frac{m \omega \hbar}{2}(2n+1)[/tex]
Can anyone help?
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