Harmonic Osclillator Purturbation Matrix Elements

[teroenza]

Homework Statement

I am trying to follow Sakurai's use of perturbation theory on a harmonic oscillator,

Homework Equations

Perturbation:
v=\epsilon x^2 , \epsilon << 1

Matrix elements:
V_{km}=<k|v|m>

The Attempt at a Solution

The book says that all other matrix elements besides V_{00}, V_{20}, and of the form V_{k0}=<k|v|0> vanish. I don't understand why. I see that the perturbation and the ground state have even parity, and that the SHO eigenstates alternate between even and odd parity with quantum number n. That should kill off the odd n states, but why should the even ones vanish too for k above 2?

 

[mfb]

They might be negligible.
I don't understand your perturbation - the harmonic oscillator has x^2, and your perturbation has the same shape?

 

[teroenza]

Yes, sorry if that was not clear. The unperturbed potential is V_{0}=1/2 m \omega x^2, and the perturbation is the same thing multiplied by \epsilon.

 

[mfb]

Okay. Then it's easy to predict how the energy levels will change. I would expect expressions like V_k2 and so on to be non-zero as well, but they could be too small to be relevant (and I did not calculate it).

 

[teroenza]

Ok. Sakurai says they "vanish", which might mean that they are negligible. The motivation behind this is my trying to solve for the second order energy correction to a perturbation proportional to x^4. I thought I could make most of the terms in the sum go away if I was able to follow whatever procedure Sakurai did in the book for the x^2 perturbation.

 

[Orodruin]

Of the $V_{k0}$ elements, the only non-zero ones are for k=0 and k=2. Hint: Rewrite x in terms of creation and annihilation operators.

 

[teroenza]

Got it. Thank you.For any state higher than 2, they all go to zero because the creation operator to the second power can only get you to the |2> state.

 

[Orodruin]

Spot on. In general, any state $|n\rangle$ will have a non-zero matrix element with itself, $|n+2\rangle$, and $|n-2\rangle$.