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Suppose an initial quantum state $\psi = a_1\phi_1 + a_2\phi_2 + ... + a_n\phi_n$, where $\phi_i$ is the eigenfunction with eigenvalue $\lambda_i$ of some measurement operator. Post-measurement, we will find the system in state $\phi_i$ with probability $|a_i|^2$.

What happens to the phase post-measurement? The principle that immediate subsequent measurements should always return the same value would be satisfied no matter the resulting phase. We might find the system in any state $b\phi_i$, so long as $|b|^2=1$. I am sure the postulates of quantum mechanics specify something about this, but I haven't managed to find any text that addresses it. What should $b$ be?

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3 Answers 3

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In quantum mechanics, states are represented by rays in Hilbert space, or more accurately, the space of states is projective Hilbert space - for example, for a finite dimensional system, the space is $H_n / \sim \ \cong \mathbb{C}P^{n-1}$, where for $u, v \in H_n$, $u \sim v$ if $u = \alpha w$ for some non-zero complex number $\alpha$.

Now usually we prefer to work with the plain Hilbert space rather than the projective one, choosing to impose the quotient whenever useful - simply because we have many more useful tools at our disposal while working with Hilbert spaces.

However, you must always remember that the actual space of states is the projective Hilbert space, which means that the statement "We might find the system in any state $b\phi_i$ as long as $|b|^2 = 1$" is meaningless, because there aren't separate states $b\phi_i$ - neither is it that all these states are the "same" - the real reason is that there is only one state $\phi_i$ in projective Hilbert space.

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  • $\begingroup$ There is an important conceptual distinction that must be made. There are two different entities that are sometimes called "states", which seems to have caused confusion in the whole discussion. One entity, the one you are talking about in your answer, is an element of the projective Hilbert space (let's call that the physical state for now). The other entity, the one the question was really about I think, is the mathematical object that enters in the standard Schrodinger equation. That object is an element of the regular Hilbert space, and most people, and the OP, refer to it as the state... $\endgroup$
    – ReasonMeThis
    Commented Dec 29, 2020 at 6:09
  • $\begingroup$ … Let's call that object the standard state for now, until we sort out the terminology, and to distinguish it from the physical state. The crucial difference between the two is revealed starkly by the fact that it is specifically the standard state, NOT the physical state, that we are talking about when we say that the energy eigenstate evolves by a multiplicative phase factor. It is the standard state that we most commonly work with, most often call simply "state", most often use in linear combinations and inner products etc. And for that object the phase is certainly important. $\endgroup$
    – ReasonMeThis
    Commented Dec 29, 2020 at 6:22
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Wavefunction collapse is just a fiction that we employ because it would be a hassle to describe measurements realistically as entanglement of the observer with the thing being observed, with decoherence.

Phase in quantum mechanics isn't an observable. You can only determine the phase of something relative to something else. The phase $b_1$ of the state after you've measured the system to be in state 1 doesn't have any meaning by itself. You would need to compare it with some other phase, such as the phase $b_2$ of the system that's entangled with a person who measured it to be in state 2. If you could do this, then it would be meaningful to say, for example, that $\operatorname{arg}(b_2/b_1)$ has some value. To do this, you would have to do something like measuring interference between the person in state 1 and the person in state 2. But the whole reason that collapse is a good approximation is that decoherence makes it impossible for us to detect this kind of interference, so that person 1 might as well stop keeping track of the existence of the other possibility.

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Post-measurement, we will find the system in state $\phi_i$ with probability $|a_i|^2$.

Almost, the correct final state is $$a_i\phi_i,$$ it's just the result of applying the projection operator. If we wish, we can then normalize it to $$\frac{a_i}{|a_i|}\phi_i,$$ but we should only do it if we know we won't be comparing or superposing it with other states. When we normalize it, we divide it by a real number, which does not remove the phase. The overall phase is not important only if we don't plan to compare/superpose the state with other states.

One way to see that the final state is $a_i\phi_i$, or if we wish its normalized cousin with the phase intact, is to imagine first that all but the $i$th coefficients $a_j$ are 0 and consider the overall post-measurement state of system+apparatus. By continuity, immediately post-measurement the overall state is exactly the same as immediately pre-measurement (we are talking about instantaneous collapses in this question). Therefore we should assign the post-measurement state of the system to also be what it was immediately pre-measurement, $a_i\phi_i$. Anything else would be a bizarre ad hoc unnecessary step.

For the general case, with non-zero other coefficients, the same should be true by linearity, because collapsing the state just means keeping only one of the resulting branches.

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  • $\begingroup$ But do we end up in state $\phi$ or $i\phi$? $\endgroup$
    – Retracted
    Commented Dec 28, 2020 at 4:06
  • $\begingroup$ $\phi_i$, as you said, not $i\phi_i$. The phase does matter... well, sort of... more precisely, we need to keep track of the phases, in order to keep track of relative phases. No reason to deliberatelly drop them when we are sure we don't need them anymore, though @NiharKarve is correct that technically you can (not must) drop them. $\endgroup$
    – ReasonMeThis
    Commented Dec 28, 2020 at 4:26
  • $\begingroup$ @retracted darn, I realized I made a silly mistake in my answer. Not about the phase being unimportant, it's definitely important to keep track of it (for example, we don't say that it's incorrect or meaningless to say that an energy eigenstate evolves by a multiplicative factor $e^ {i\omega t}$!, which is what the view expressed in the other answer seems to imply)... $\endgroup$
    – ReasonMeThis
    Commented Dec 29, 2020 at 0:48
  • $\begingroup$ …My mistake was that for some reason my brain decided that the coefficients $a_i$ are real, which is of course not necessarily true. So the correct final state is $a_i\phi_i$, it's just the result of applying the projection operator. If we wish, we can normalize it, although we certainly don't have to, and we can only do it if we know we won't be comparing it with other states. When we normalize it, we divide it by a real number, which does not remove the phase. The overall phase is not important only if we don't compare/superpose the state with other states, $\endgroup$
    – ReasonMeThis
    Commented Dec 29, 2020 at 0:58
  • $\begingroup$ @ReasonMeThis If we're being pedantic, an energy eigenket evolves by a multiplicative phase factor, but the state to which that ket corresponds does not evolve at all (or evolves trivially, if you prefer). The distinction is typically not operationally important, but it does have the benefit of being true. $\endgroup$
    – J. Murray
    Commented Dec 29, 2020 at 2:16

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