2.6: Example

Consider a simple harmonic oscillator in its ground state, to which we apply a perturbation \(\hat{V} = \lambda x^2\). We know the unperturbed wavefunction \(|n_0 \rangle = [m \omega_0/ \pi \hbar]^{\frac{1}{4}} \text{ exp}\{−m\omega_0x^2/2\hbar\}\), so we can evaluate the first order shift in energy according to the perturbation theory:

\[\Delta E_0 = \langle n_0| \lambda x^2 |n_0 \rangle = \lambda \sqrt{m\omega_0/\pi \hbar} \int x^2 \text{ exp}\{−m\omega_0x^2 /\hbar \}dx = \frac{\lambda}{2}\frac{\hbar}{m\omega_0} \nonumber\]

In this case we know the exact shift, since the perturbation is simply an additional harmonic potential, giving a total \(k = m\omega^2_0 + 2\lambda\) and an exact ground state energy of \(\frac{1}{2} \hbar \sqrt{\omega^2_0 + 2 \lambda / m} \). It is easy to verify that to first order in \(\lambda\) these expressions are identical.

To determine the amount of mixing of states, we need to evaluate matrix elements like \(\langle n_0|\lambda x^2 |n_i \rangle\). We won’t evaluate these here, but we will note that for odd \(i\) the integral is zero - the symmetric perturbation only mixes in symmetric excited states.