Do quantum measurements destroy the symmetrization requirement? The symmetrization requirement requires that the combined wave function of a system of particles be either symmetric under an exchange of two particles (bosons) or antisymmetric (fermions). For instance, if you have two spin 1/2 fermions in the singlet spin configuration, then (assuming the two particles are in the same orbitals), the (total) wave function is antisymmetric (as required). But if you actually go and measure their spins, then the wave function collapses into a state with one particle having spin up and the other spin down. Now the (spin space) wave function is no longer antisymmetric, and hence the total (spatial + spin) wave function isn't either. In general, the symmetrization requirement means that two particles must be entangled in some way (meaning you can't write the total wave function as a simple product of single-particle functions (so long as you're not talking about a bunch of bosons in the same state)), but this entanglement is usually destroyed under measurement. So what's going on here?


1 Answer 1


Indeed this is a subtle point. The key point is that a measurement doesn't tell you which electron has spin up, just that one of them does. A second important fact is that only the total state needs to be anti-symmetric.

Let's take an example with two electrons. We will define the state $|x \uparrow, y \uparrow \rangle$ to mean electron $1$ is at position $x$ with spin up along some axis, and electron $2$ is at position $y$ with spin up along the same axis. Let's say they are in an anti-symmetrized state like so \begin{equation} |\Psi\rangle = \frac{1}{2}\left( |x\uparrow, y \uparrow\rangle + |x \downarrow, y\downarrow\rangle - |y \uparrow, x \uparrow \rangle - |y \downarrow, x \downarrow \rangle \right) \end{equation} In this state, neither electron is in a definite spin state.

Now we do a measurement, and find an electron at position $x$ is in the spin up state. You might think this means the state after measurement becomes \begin{equation} |\Psi\rangle = |x \uparrow, y\uparrow \rangle \ \ {\rm (NOT\ CORRECT)} \end{equation} In fact this assumes that we have measured the first electron to be at position $x$ with spin up, but since electrons are indistinguishable there is no way to be sure whether we have measured the first or second electron. Therefore the correct state after measurement is \begin{equation} |\Psi\rangle = \frac{1}{\sqrt{2}} \left(|x \uparrow, y\uparrow \rangle - |y \uparrow, x\uparrow \rangle \right) \end{equation} which is anti-symmetric, and has been projected to the subspace where one of the electrons is at position $x$ with spin up (consistent with our measurement). This logic generalizes fairly straightforwardly to systems with $N$ identical fermions.

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    $\begingroup$ Thanks for this amazing answer. So I guess then that there is a general principle that no matter what measurement you perform on a collection of particles, they are still gonna end up in some (anti)symmetric state, so long as they were in one before the measurement? $\endgroup$
    – User3141
    Aug 2, 2021 at 19:33
  • $\begingroup$ @FelisSuper For identical fermions, every state in the Hilbert space is totally anti-symmetric. (And for identical bosons every state is symmetric). Both unitary time evolution and measurements take you from one state in the Hilbert space to another, so (for identical fermions) the state is always totally antisymmetric. $\endgroup$
    – Andrew
    Aug 2, 2021 at 19:37
  • $\begingroup$ @FelisSuper As an analogy (although it's not too far off from being exactly right), you can have a vector space comprised of general $2\times 2$ matrices. But anti-symmetric $2\times 2$ matrices form a subspace of this general space, and so anti-symmetric $2\times 2$ matrices also form a vector space. With the caveat that you should really make this statement including a map from "vector space" to "Hilbert space" and have more dimensions, you can roughly say that fermions live in this anti-symmetric subspace. (This is also true for bosons if you replace "antisymmetric" with "symmetric") $\endgroup$
    – Andrew
    Aug 2, 2021 at 19:40
  • $\begingroup$ What if the initial state was: |x up,y down>-|y down,x up> $\endgroup$
    – richard
    Aug 29, 2023 at 9:06

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