Most introductory QM texts say that measurement collapses the quantum state into one of the eigenstates of the observable's operator (and with a probability amplitude given by the corresponding coefficient, if represented in the basis of eigenvectors). However, this isn't a complete description for the case of degenerate eigenstates or continuous observables.

For example, if a particle has the state $|\psi⟩=\alpha|1⟩ + \beta|2⟩ +\gamma|3⟩$, expressed in terms of orthonormal eigenvectors where $|1⟩$ and $|2⟩$ have the same (degenerate) eigenvalue, then the measurement will collapse it into either the state $|3⟩$, with probability $|⟨3|\psi⟩|^2$, or into the superposition $(1-\gamma^*\gamma)^{-\frac 1 2}(\alpha|1⟩ + \beta|2⟩)$. It will not collapse into $|1⟩$ or $|2⟩$.

Similarly for continuous variables: for quantum optics (with individual photons) to reproduce the results of classical wave optics (where an interference pattern is the Fourier transform of the aperture function) then the effect of a measurement on the wavefunction must sometimes be equivalent to multiplication by an aperture function, rather than collapse into a delta-function eigenstate. (It is not only that there is classical uncertainty related to measurement precision.)

Is there a more general axiom or postulate that defines how the quantum state collapses when the eigenstates are degenerate and/or not discrete? Or can the general behaviour be derived entirely from the discrete non-degenerate Born rule?

Edit: I’m not asking what happens after collapse (there are other questions that ask that), I’m asking whether there is a general axiom that unifies the different cases (including continuous and degenerate) and what it’s relationship is to the Born rule?


2 Answers 2


Given that the system is in a pure state represented by the unit vector $\Psi$, the more general formulation of the Born rule answers the following two questions:

  1. What is the probability that we measure an observable $A$ to take its value in some set $E\subseteq \mathbb R$?
  2. If we measure $A$ to takes its value in $E$, what will be the state of the system after the measurement is performed?

The standard answer to both questions can be framed in terms of the projection-valued measure $\mu_A$ corresponding to the self-adjoint operator $A$. In essence, $\mu_A$ is a function which eats a (Borel-measurable) subset $E\subseteq\mathbb R$ and spits out a projection operator $\mu_A(E)$. From there, the answers to the questions are

  1. $\mathrm{Prob}_\Psi(A,E) := \Vert \mu_A(E) \Psi\Vert^2$
  2. After the measurement, $\Psi \mapsto \mu_A(E)\Psi\big/\Vert \mu_A(E)\Psi\Vert$

The physical intution is that $\mu_A(E)$ is the operator which projects a state vector into the eigenspace of $A$ which is consistent with a measurement outcome in $E$.

  • In the simplest case, $E=\{\lambda\}$ is a singleton set containing a single non-degenerate eigenvalue of $A$. If the corresponding eigenvector is $|\phi\rangle$, then the projection operator is $\mu_A\big(\{\lambda\}\big) = |\phi\rangle\langle\phi|$.

  • If the eigenspace corresponding to $\lambda$ is more than 1D, then we would have $$\mu_A\big(\{\lambda\}\big) = \sum_i |\phi_i\rangle\langle\phi_i|$$ where the vectors $|\phi_i\rangle$ span the eigenspace of $A$ with eigenvalue $\lambda$.

  • Assuming that the spectrum of $A$ is discrete, we can more generally write $$\mu_A(E) = \sum_{\lambda \in E} \mu_A\big(\{\lambda\}\big)$$

  • If the spectrum of $A$ is continuous, the problem becomes a bit more subtle. Essentially the same rule applies, but the previous expression would take the form $$\mu_A(E) = \int_{E} \sum_i|\phi_i(\lambda)\rangle\langle\phi_i(\lambda)| \mathrm d\lambda$$ where $|\phi_i(\lambda)\rangle$ is the $i^{th}$ generalized (non-normalizable) eigenvector of $A$ with eigenvalue $\lambda$, and the integration is performed over all $\lambda\in E$. If the spectrum of $A$ is non-degenerate, then the summation can be dropped.


The general rule is that the state vector is orthogonally projected onto the subspace of eigenvectors of the measured eigenvalue.

$\alpha|1⟩ + \beta|2⟩$ is an eigenvalue of the measurement operator, and is no more or less a superposition than $|1⟩$ and $|2⟩$ are. It doesn't make sense to say that a state vector is a superposition full stop, only that it's a superposition of certain other vectors, and it just means that it's a nontrivial linear combination of those vectors.

Multiplication by an aperture function (that is everywhere 0 or 1) is another special case of projection onto a subspace. If $δ(x)$ is an eigenvector with a fixed eigenvalue for all $x\in S$, then any wave function that vanishes outside $S$ is an eigenvector with the same eigenvalue. You aren't obliged to work in a basis of delta functions and declare other functions to be mere superpositions.


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