1D SHO in external electric field

[big-ted]

Homework Statement

I have a charged particle in a 1D harmonic oscillator, on which an externally applied electric field acts, such that the Hamiltonian can be written:

\frac{p^2}{2m}+\frac{kx^2}{2}-qEx

The problem asks to first find the first (trivial) and second order corrections to the energy levels via perturbation theory, which I have done. Now the problem asks to verify that the second order correction agrees with the exact solution of the Schrodinger equation. The hint is to use a change of variable

x'=x-(\frac{qE}{m \omega ^2})

The Attempt at a Solution

Using the suggested change of variable, it is straight-forward to show that the Hamiltonian can be written as that of the conventional, unperturbed oscillator, only in x' rather than x. My question is, where do I go now? What I essentially want to do is calculate the inner product

\left\langle\Psi(x)\right|H^{new}\left|\Psi(x)\right\rangle

But I want to do this for a general case of all \Psi(x), and I have an operator, H^{new} that acts on x', and not on x. My only guess is that I can rewrite the known solutions of the 1D SHO, (The Hermite Polynomials) in terms of x' and integrate directly, but this gets very messy very quickly, even for the ground state! It strikes me I'm missing some obvious trick.

Can anyone suggest an approach here?


Aside: This is a (reworded) question out of a very common quantum text. I've refrained from specifying which in the interest of avoiding this thread popping up in an internet search for the question number. Hopefully my rewording makes sense!

Thanks in advance!

 

[vela]

The problem asks to first find the first (trivial) and second order corrections to the energy levels via perturbation theory, which I have done. Now the problem asks to verify that the second order correction agrees with the exact solution of the Schrodinger equation. The hint is to use a change of variable

x'=x-(\frac{qE}{m \omega ^2})

The Attempt at a Solution

Using the suggested change of variable, it is straight-forward to show that the Hamiltonian can be written as that of the conventional, unperturbed oscillator, only in x' rather than x.

You should have an extra term so that H_{new} = H + \mbox{constant}.

My question is, where do I go now? What I essentially want to do is calculate the inner product

\left\langle\Psi(x)\right|H^{new}\left|\Psi(x)\right\rangle

I'm not sure why you'd want to do that. You want to find the new eigenstates and their energy levels, not the expectation value of the Hamiltonian.

But I want to do this for a general case of all \Psi(x), and I have an operator, H^{new} that acts on x', and not on x. My only guess is that I can rewrite the known solutions of the 1D SHO, (The Hermite Polynomials) in terms of x' and integrate directly, but this gets very messy very quickly, even for the ground state! It strikes me I'm missing some obvious trick.

Can anyone suggest an approach here?

You said that the new Hamiltonian is essentially that of the harmonic oscillator, right? You already know what the eigenstates are and their energies.

 

[diazona]

I remember that problem It's been a while...

Think about this: the energy eigenstates of the harmonic oscillator form a complete set. That means that any function can be expressed as a linear combination of them. Of course, this applies to both the old Hamiltonian and the new Hamiltonian; in other words, the eigenstates of the new Hamiltonian, expressed in terms of x', are just as valid of a basis as the eigenstates of the old Hamiltonian, expressed in terms of x.

What I'm trying to say is that, when you calculate the energy levels, you find the expectation value of H (the old H) in one of its own eigenstates. So when you try to calculate the new energy levels, you should be finding the expectation value of the new H in one of its eigenstates, not an eigenstate of the old H. Roughly speaking, you should calculate

\langle\psi'_n(x')\vert H^\text{new}\vert \psi'_n(x')\rangle

instead of

\langle\psi_n(x)\vert H^\text{new}\vert \psi_n(x)\rangle

like you were trying to do. (I don't know if this notation is making things clear at all)

Anyway, you should have found that the new Hamiltonian, with the electric field term, expressed in terms of x', takes the same form as the original Hamiltonian (E=0), except for an additive constant. That additive constant doesn't affect the experimental consequences, but it does change the energy levels. As for the rest of the new H, since it takes the same form as the original H, its eigenstates also take the same form, just in terms of x' instead of x.

Hope that helps a bit...

 

[big-ted]

Argh! Vela had it right. I'd forgotten the constant when I completed the squares! If you guys hadn't already replied I'd go ahead and delete the thread. I guess instead my humiliation is engraved in html for all to see!Thanks for your help!