Hi,(adsbygoogle = window.adsbygoogle || []).push({});

I'm new to this subject, so bear with me. We consider the harmonic oscillator with a pertubation:

[tex]\hat{H}' = \alpha\hat{p}.[/tex]

(What kind of a perturbation is that anyway, it's not a disturbance in the potential, what does it correspond to physically.)

Now I have to calculate the first and second order energy corrections. I express p by the ladder operators:

[tex]p=i\sqrt{\tfrac{\hslash m\omega}2}(a_+-a_-).[/tex]

I find that the first order correction is 0, and that the second order correction is [tex]E_n^2=m\alpha^2/2[/tex], that is, the energy shift is INDEPENDENT of n (i.e. it is the same for all excited states of the oscillator), but this can't be true, can it?

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# Perturbation theory

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