# Constant force perturbation of the quantum SHO

1. Jan 22, 2008

### nathan12343

[SOLVED] Constant force perturbation of the quantum SHO

1. The problem statement, all variables and given/known data
We're supposed to consider the Hamiltonian for the simple harmonic oscillator:

$$\hat{H}_{0} = \hat{p}^{2}/2m + m\omega^2\hat{x}^2/2$$

With a perturbation, so that $$\hat{H} = \hat{H}_0 + \hat{H}'$$, where $$\hat{H}' = F\hat{x}$$

I've already solved for the first and second order energy corrections, as well as the first order correction to the wave function. The last part of the question is to solve for the exact energies using the change of variables $$x' \equiv x + F/m\omega^2$$

2. Relevant equations

See above

3. The attempt at a solution

When I substitute for x in the perturbed Hamiltonian, I get

$$\hat{H} = \hat{p}^2/2m + m\omega^2\hat{x'}^2/2 - F^2/2m\omega^2$$

Which is of the same form as the unperturbed Hamiltonian, except for a constant, which would shift the energies down by the constant compared to the unperturbed case. However, when I solved for the second order correction to the energies, I got that there wasn't a constant shift, but instead the spacing of the energy levels increased by $$F^2(n+1/2)/m\omega^2$$ Since the energy correction looks a lot like the extra constant that was introduced to the Hamiltonian, I'm inclined to think that my reasoning is incorrect somewhere, and that I should get that the exact energies are equal to the 0th plus 2nd order energies, but that's just a guess.

Does anyone know where I'm going wrong?

2. Jan 22, 2008

### Rainbow Child

I think you mean

$$\hat{H}'= -F\hat{x}$$ wrong sign

in order to get $$\hat{H} = \hat{p}^2/2m + m\omega^2\hat{x'}^2/2 - F^2/2m\omega^2$$

The 2nd order corrections is exactly

$$E_n^2=-\frac{F^2}{2\,m\,\omega^2}$$

EDIT: $$\hat{H}'$$ was correct.

Last edited: Jan 23, 2008
3. Jan 23, 2008

### nathan12343

No, it's $$H' = F\hat{x}$$, you have to plug in for x, which means solving for x in the equation for x' given above, there's some cancellation, and $$\hat{H}$$ comes out as I've given it above.

One part of the problem had us calculate $$\left\langle m\right|H'\left|n\right\rangle$$, which came out to:
$$F\sqrt{\bar{h}/2m\omega}(\sqrt{n+1}\delta_{m,n+1} + \sqrt{n}\delta_{m,n-1})$$

Where I've denoted the nth stationary state of the unperturbed SHO by the n ket

So, the second order correction is

$$\sum\left|\left\langle m\right|H'\left|n\right\rangle \right|^2/(E_{n}^{0} - E_{m}^{0})$$

Where the sum is to be taken over all m $$\neq$$ n.

The kronecker deltas in $$\left\langle m \left|\hat{H'}\right|n\right\rangle$$ kill off all the terms in the sum except for those which $$m = n \pm 1$$

And, if you do the algebra, The second order energy correction comes out to:

$$F^2(n + 1/2)/m\omega^2$$

I'm not seeing how you got your answer, Rainbow Child, can you explain?

P.S. Anyone know how to say hbar in LaTex? That doesn't look right

Last edited: Jan 23, 2008
4. Jan 23, 2008

### Rainbow Child

I found the same result

$$F\sqrt{\hbar/2m\omega}(\sqrt{n+1}\delta_{m,n+1} + \sqrt{n}\delta_{m,n-1})$$

thus

$$m=n+1, \quad F^2\frac{\hbar}{2m\omega}\frac{n+1}{E_n^0-E_{n+1}^0}=F^2\frac{\hbar}{2m\omega}\frac{n+1}{(-\hbar \omega)}=-F^2\frac{n+1}{2m\omega^2}$$
$$m=n-1, \quad F^2\frac{\hbar}{2m\omega}\frac{n}{E_n^0-E_{n-1}^0}=F^2\frac{\hbar}{2m\omega}\frac{n}{(\hbar \omega)}=F^2\frac{n}{2m\omega^2}$$

and

$$E_n^2=-\frac{F^2}{2m\omega^2}$$

P.S.1 My mistake for the sign!

P.S.2 In LaTeX the solutions is always the obvious \hbar=$$\hbar$$.

5. Jan 23, 2008

### nathan12343

Wow...you're totally, right, thanks a lot!!