Time independent perturbation - potential with inner function

[ffia]

Homework Statement

I need to find the corrected ground state for perturbed harmonic oscillator (1D) with perturbation of the form V(x) = λ sin(κx), κ>0.

My problem is I have no idea how to handle a potential that has its operator as an inner function.

Homework Equations

The perturbed Hamiltonian of the system:
H = H[itex]_{0}[/itex] + λV = [itex]\frac{1}{2}[/itex]mω[itex]^{2}[/itex]x[itex]^{2}[/itex] + λsin(κx).

At least the first correction to the ground state is defined
[itex]\sum\frac{<m| λV |0>}{E_{m}-E_{n}}[/itex] where the sum is taken over m, m goes from 1 to infinity? (m and 0 are states, sorry I couldn't get the latex notation working there)

The Attempt at a Solution

I tried to use the ladder operators.
In terms of the ladder operators
x = [itex]\sqrt{\frac{h}{2mω}}{}(a+a^\dagger)[/itex]
But the problem is I don't know how to operate the states when the operators are inside sines argument. I could use the taylor series of sin, but it would leave me with (a+a^\dagger)[itex]^{2n+1}[/itex] and that doesn't seem good either.

Thank you in advance, I'm really stuck here.

 

[mark726]

Thank you for your post. Handling a potential with an operator as an inner function can be tricky, but there are some techniques you can use to find the corrected ground state in this case.

Firstly, you are correct in using the ladder operators to simplify the Hamiltonian. As you mentioned, you can express the position operator x in terms of the ladder operators a and a†. This allows you to rewrite the potential V(x) as a function of the ladder operators as well. In this case, V(x) = λsin(κx) becomes V(a,a†) = λsin(κ\sqrt{\frac{h}{2mω}}(a+a^\dagger)).

Next, you can use the fact that the ladder operators a and a† commute with each other, but not with the potential V(a,a†). This means that you can use the Baker-Campbell-Hausdorff formula to expand the potential in terms of the ladder operators, which will then allow you to operate on the states.

Finally, you can use the eigenvalue equation for the harmonic oscillator, H|n> = E_n|n>, to simplify the expression for the corrected ground state. This will involve using the relation [H,a] = ℏωa and [H,a†] = -ℏωa†.

I hope this helps guide you in finding the corrected ground state. Good luck with your calculations!