- #1
Denver Dang
- 143
- 1
Homework Statement
I'm having some trouble calculating the 2nd order energy shift in a problem.
I am given the pertubation:
\hat{H}'=\alpha \hat{p},
where $\alpha$ is a constant, and \hat{p} is given by:
p=i\sqrt{\frac{\hbar m\omega }{2}}\left( {{a}_{+}}-{{a}_{-}} \right),
where {a}_{+} and {a}_{-} are the usual ladder operators.
Homework Equations
Now, according to my book, the 2nd order energy shift is given by:
E_{n}^{2}=\sum\limits_{m\ne n}{\frac{{{\left| \left\langle \psi _{m}^{0} \right|H'\left| \psi _{n}^{0} \right\rangle \right|}^{2}}}{E_{n}^{0}-E_{m}^{0}}}
The Attempt at a Solution
Now, what I have tried to do is to calculate the term inside the power of 2. And so far I have done this:
\begin{align}<br /> & E_{n}^{1}=\alpha i\sqrt{\frac{\hbar m\omega }{2}}\int{\psi _{m}^{*}\left( {{{\hat{a}}}_{+}}-{{{\hat{a}}}_{-}} \right)}\,{{\psi }_{n}}\,dx=\alpha i\sqrt{\frac{\hbar m\omega }{2}}\left( \int{\psi _{m}^{*}\,{{{\hat{a}}}_{+}}{{\psi }_{n}}\,dx-\int{\psi _{m}^{*}\,{{{\hat{a}}}_{-}}{{\psi }_{n}}\,dx}} \right) \\ <br /> & =\alpha i\sqrt{\frac{\hbar m\omega }{2}}\left( \sqrt{n+1}\int{\psi _{m}^{*}\,{{\psi }_{n+1}}\,dx-\sqrt{n}\int{\psi _{m}^{*}\,{{\psi }_{n-1}}\,dx}} \right) <br /> \end{align}<br />
As you can see, I end up with the two integrals. But I don't know what to do next ? 'Cause if m > n, and only by 1, then the first integral will be 1, and the other will be zero. And if n > m, only by 1, then the second integral will be 1, and the first will be zero. Otherwise both will be zero.
And it seems wrong to have to make two expressions for the energy shift for n > m and m > n.
So am I on the right track, or doing it totally wrong ?
Thanks in advance.Regards
