# Is there an axiom for degenerate (or continuous) quantum collapse? [duplicate]

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?

• Perhaps you want to look up positive operator valued measures, which are as far as I know the most general form of quantum measurements. Jan 28, 2023 at 6:37
• There are many questions here on Physics SE already concerning these issues... Jan 28, 2023 at 10:29
• @TobiasFünke not sure what you mean by "these issues", but I searched to no avail for another question asking this. Jan 28, 2023 at 12:58
• Does this answer your question? What happens to the wave function of a particle immediately after measuring its energy if degenerate? Jan 28, 2023 at 23:52
• Does this answer your question? Collapse of state vector for degenerate eigenvalues
– hft
Jan 29, 2023 at 1:43

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.