4.2: Example

Consider the 2D harmonic oscillator $V_0 = \frac{1}{2}m \omega^2 (x^2 + y^2 )$. If we measure the energy and find it to be $2\hbar \omega$, then the state could be $|n_x = 1$, $n_y = 0 \rangle$ or $|n_x = 0$, $n_y = 1\rangle$ or any linear combination. To fully define any state we require any two quantum numbers: $n_x$, $n_y$ and $E = (n_x + n_y + 1)\hbar \omega$.

Suppose we measure the energy and find $3\hbar \omega$: there is a partial collapse of the wavefunction and there are three degenerate possibilities. Suppose we then apply a perturbation $\Delta V = \lambda x^2$ (see 2.6). This breaks the symmetry and collapses the wavefunction onto either $|1, 1\rangle |2, 0\rangle$ or $|0, 2\rangle$. The perturbation matrix (see 3.1) $\langle n_x, n_y|\Delta V |n_x, n_y\rangle$ is diagonal provided we choose the basis with $x$ along the direction of the perturbation, and it has eigenvalues $(n_x + \frac{1}{2} ) \lambda \hbar /m \omega$. If we then measure the energy and find $E = 3\hbar \omega + \lambda \hbar/ 2m \omega$ then we know that the state is $|0, 2 \rangle$: a complete collapse onto a single wavefunction.

Aside: Consider mixing with the non-degenerate states. By symmetry $\langle 1, 0| \lambda x^2 |2, 0 \rangle = 0$: the perturbation does not mix $n_x = 0$ and $n_x = 1$ states, nor does it affect $n_y$ (see 2.3). Thus applying the perturbation may induce a transition from $|0, 2 \rangle$ to $|2, 2 \rangle, |4$, $2 \rangle$ etc. but not to $n_x$ = odd or $n_y \neq 2$. This gives rise to selection rules