Wavefunction – What Happens to the Wave Function of a Particle Immediately After Measuring Its Energy If Degenerate?

hamiltonianquantum mechanicswavefunction

For this question, I will be adhering to the Copenhagen interpretation (since that's what I've learned in university so far). For the sake of brevity/clarity, also, assume the Hamiltonian here has finite eigenvalues and eigenvectors, say $n$ of them.

If the Hamiltonian operator of a particular physical system, $\hat{H}$, has eigenvalues $\lambda_1,…,\lambda_n$ with corresponding eigenstates $\Psi_1,…,\Psi_n$, and we are considering a particle in said physical system described by the state $\Psi$, then we can expand $\Psi$ as a superposition of the energy eigenstates, i.e. $\Psi=\sum_{k=1}^nc_k\Psi_k$ for some complex constants $c_k$.

Upon measurement of the energy of the particle in state $\Psi$, the only possible values we can measure are the eigenvalues $\lambda_k$, and we measure them with probability $|c_k|^2$ (assuming $\Psi$ is normalized). In this sense, if we measure the eigenvalue $\lambda_1$ (for example), then the state $\Psi$ is said to have collapsed to the state $\Psi_1$.

What happens if there is degeneracy among the eigenstates and eigenvalues? That is, say there are two eigenstates $\Psi_1$ and $\Psi_2$ that each have the same eigenvalue, say $\lambda_1$. If we measure an energy of $\lambda_1$, how do we know if $\Psi$ has been collapsed to the state $\Psi_1$ or $\Psi_2$?

Best Answer

Almost. You do know its state. It's $(c_1\Psi_1+c_2\Psi_2)$ (apart from a normalization constant). Remember that your choice of basis vectors to represent the degenerate subspace is arbitrary. There's nothing in the physics distinguishing the two you happened to choose. That superposition state is just as good as any of the other states in that degenerate subspace. This is what it means to project onto the degenerate subspace and renormalize.

Now here's a messier variant on the problem: Suppose the spectrum of the Hamiltonian is continuous, and the energy measurement has a finite precision (with a known error distribution). Now what's the state after the measurement?

There actually is a good answer to this, but it's unlikely to be taught in most courses. You basically end up with a quantum mechanical version of Bayes' theorem. The basic version you're taught in a course is just the limit of an ideal measurement, such that the state of the measurement device is guaranteed to 100% correlate with the thing being measured.