3.6: Wavefunction Collapse onto degenerate levels

Refer back to the postulates of quantum mechanics: We know that acting with an operator $\hat{A}$ on an eigenstate $|\alpha_n \rangle$ of that operator gives us an eigenvalue $A_n$, which corresponds to a measurable quantity.

There is no guarantee that $|\alpha_n \rangle$ is the only eigenstate of $\hat{A}$ which has this eigenvalue (e.g. energy levels in hydrogen). Different states with the same eigenvalue are referred to as degenerate.

Assume we find two orthogonal, degenerate eigenstates of $\hat{A}$: $|\alpha_1\rangle$ and $|\alpha 2\rangle$. i.e. $\hat{A}|\alpha_1\rangle = A_1|\alpha_1\rangle$ and $\hat{A}|\alpha_2\rangle = A_1|\alpha_2\rangle$. We also see that

$$\hat{A} (\cos \theta |\alpha_1\rangle + \sin \theta |\alpha_2\rangle ) = A_1 (\cos \theta |\alpha_1\rangle + \sin \theta |\alpha_2\rangle ) \nonumber$$

for any $\theta$. We use $\cos \theta$ for the expansion instead of the normal $c_i$ to emphasise the similarity between eigenstates and vectors. It also allows for easy normalisation since $\cos^2 \theta + \sin^2 \theta = 1$.

Thus any linear combination of degenerate eigenstates produces another eigenstate. There is still only twofold degeneracy, because there are only two orthogonal states, $(\sin \theta |\alpha_1 \rangle − \cos \theta |\alpha_2 \rangle)$ being the other one. The complete set of orthonormal eigenstates for $\hat{A}$ is thus not a unique quantity, since we can choose any $\theta$ to generate a pair of degenerate eigenstates.

A consequence of this is that when a measurement is made of $\hat{A}$ which finds $A_1$, there is not a complete collapse of the wavefunction.

Consider measuring observable A in a system in a general state $|\Phi \rangle$. By expanding $|\Phi \rangle$ in the eigenstates of $\hat{A}$: $|\Phi \rangle = \sum_i c_i |\alpha_i \rangle$ we find the probability that the measurement will yield result $A_1$ is

$$|\langle \alpha_1|\Phi \rangle |^2 + |\langle \alpha_2|\Phi \rangle |^2 \equiv |c_1|^2 + |c_2|^2 \nonumber$$

The measurement has determined that we are either in state $\alpha_1$ or $\alpha_2$, but not which. Thus there is a partial collapse of the wavefunction onto a linear combination of them:

$$(\cos \theta |\alpha_1 \rangle + \sin \theta |\alpha_2 \rangle ); \quad \cos \theta = \frac{c_1}{\sqrt{|c_1|^2 + |c_2|^2}} \nonumber$$

which is itself an eigenvector of $\hat{A}$.

Thus, in the case of degenerate final states, the final wavefunction after the measurement does depend on the initial wavefunction. The generalisation of this to the case of many degenerate states is straightforward.