What happens to phase after wavefunction collapse? 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?
 A: 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.
A: 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.
A: 
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.
