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 $\Psi_2$?

  • $\begingroup$ Hint: does it have to be the case that the state is either $\Psi_1$ or $\Psi_2$? $\endgroup$ Nov 23, 2015 at 0:31
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    $\begingroup$ You project the wave function onto the degenerate subspace. Basically, all of the coefficients in the expansion become zero except for those for the degenerate states, and these coefficients remain the same (up to renormalization). $\endgroup$
    – march
    Nov 23, 2015 at 0:33
  • $\begingroup$ Hmm, so the resulting state will again be a superposition of $\Psi_1$ and $\Psi_2$ with coefficients $c_1$ and $c_2$? Does this not contradict the postulate that measurement collapses the wave function to a definite state? $\endgroup$
    – Ducky
    Nov 23, 2015 at 1:19
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    $\begingroup$ @Ducky, you seem to be missing the crucial point: $(\alpha \Psi_1 + \beta \Psi_2)$ is a state of definite energy. $\endgroup$ Nov 23, 2015 at 2:07
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    $\begingroup$ Of course! Even though we may not know its state, its energy will be the same no matter the state, by degeneracy. Thanks for the answers! $\endgroup$
    – Ducky
    Nov 23, 2015 at 2:13

1 Answer 1


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.

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    $\begingroup$ Can you please give some reference for the quantum mechanical Bayes' theorem you mentioned? $\endgroup$
    – Sidd
    Aug 1, 2016 at 23:01
  • $\begingroup$ Yes but in this case, if I understand correctly, the additional uncertainty from the quantum Bayes theorem is subjective uncertainty. The state in fact is some fixed linear combination of eigenvectors associated with some particular value of the energy measurement operator.... Right? $\endgroup$
    – Econ
    Mar 24, 2022 at 17:06

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