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