Collapse of quantum state upon measurement - why not some other unit scalar multiple I have learnt that upon measurement, the state of the system changes to the eigenstate corresponding to the observed eigenvalue. Specifically, we obtain the eigenstate that is the normalized projection of the original state onto the eigenspace. However, if you multiply a state by a scalar of absolute value 1, it will still represent the same state but I thought the scalars can affect future calculations (e.g. when interference is brought into the picture).
So my question is: why not a unit scalar multiple of the projected state instead when we performed a measurement?
 A: There is a fascinating discussion in the final sections of Feynman's lectures on physics (available online for free) that actually shows that in some cases the complex phase can play a significant physical role, in superconductors. However, this does not invalidate the general wisdom that a complex phase does not matter, however, as you will see if you decide to read it: In Feynman's explanation of superconductors the wave function takes on a more classical meaning, producing weird results :) This is not even a completely correct description of the phenomenon (this would require quantum field theory), but it was very enlightening to me and I just thought it was cool and should be mentioned. 
Now, more to the point: Generally, the reason why we say that a complex phase $e^{i\theta}$ cannot be physical is because, when we consider a general expectation value of an operator $\hat{O}$ in the state $|\psi\rangle$, we have 
$$\langle \psi|\hat{O}|\psi\rangle=\langle \psi|e^{-i\theta}\hat{O}e^{i\theta}|\psi\rangle$$
So that any overall phase cancels immediately. It is important here to always keep in mind what is observable and what is not. On a side note, it is important to realize that relative phases  such as the one occurring in the state $|\phi\rangle =\frac{1}{\sqrt{2}}(|0\rangle +e^{i\theta}|1\rangle$ do produce important physical effects (see, for instance, this wikipedia article). It can be quickly understood why this should be the case by remembering how interference works.
