# First Order Perturbation Theory - QM

1. Sep 27, 2016

### BOAS

1. The problem statement, all variables and given/known data
The ground state energy of the 1D harmonic oscillator with angular frequency $\omega$ is $E_0 = \frac{\hbar \omega}{2}$. The angular frequency is perturbed by a small amount $\delta \omega$. Use first order perturbation theory to estimate the ground state energy of the perturbed system.

2. Relevant equations

3. The attempt at a solution

I have calculated the energy shift, but I don't think my units make sense.

My perturbation: $\Delta V(x) = V(x) - V_0(x)$, where $V_0(x)$ is my unperturbed potential. $V(x) = \frac{m (\omega + \delta \omega)^2 x^2}{2} = \frac{mx^2}{2} (\omega^2 + \delta \omega^2 + 2 \omega \delta \omega)$.

$\Delta V(x) = \frac{m x^2}{2} (\delta \omega^2 + 2 \omega \delta \omega)$

From first order perturbation theory, $\Delta E_n^{(1)} = \langle \Delta V \rangle ^{(0)}_n$ i.e the first order correction to the energy is the expectation value of the perturbation in the unperturbed state.

Using the ground state wave function for the harmonic oscillator $\psi_0 = (\frac{m \omega}{\pi \hbar})^{1/4} e^{-\frac{m \omega x^2}{\hbar}}$ I calculate the expectation value of the perturbation.

$\langle \Delta V \rangle = \int^{\infty}_{-\infty} \psi* \Delta V \psi$

I'm skipping the details of this integral, but I find that it is equal to $\langle \Delta V \rangle = \frac{m}{2} (\delta \omega^2 + 2 \omega \delta \omega)$.

This doesn't seem right to me as it does not have units of energy. Have I made a mistake in my reasoning, or does the error likely lie in how I have evaluated that integral? If it's the latter, I can try to find the error myself.