Lunct said:
When measured/viewed particles in superposition collapse and take the form of one state. So how do we know they are in superposition in the first place.
Taken at face value (which may not be quite what you meant, as I'll explain shortly),
@PeroK gives the correct answer, which I quote again because this fact can never be repeated too much:
PeroK said:
This question is based on a fundamental misunderstanding of superposition. Every state is a single state and, at the same time, a superposition of other states. Essentially a state is a vector and a vector, as you know, is a single vector and linear combination of other vectors.
However, here is a modification of your question which I think you may have intended (and if not, I'm pretty sure it will teach you something useful anyhow!).
Suppose we have an observable ##A## which has two eigenstates ##|a\rangle## and ##|b\rangle## corresponding to eigenvalues ##a## and ##b##, respectively. We now make a series of ##A##-measurements on a beam of particles and get a series of readings like this:$$aababaabbbbaababaabbaaaabbbaba$$ Your teacher says that the particles in the beam have all been prepared in a superposition state of the form ##\alpha|a\rangle + \beta|b\rangle## and that upon measurement each particle's state 'collapses' randomly to either ##|a\rangle## or ##|b\rangle## with probabilities ##\alpha^{2}## and ##\beta^{2}##, respectively, resulting in the observed series of ##a## and ##b## values.
But You Ask: How do we know that the particles were originally in the assumed superposition state? Couldn't we just assume that the particles in the beam are each in either the state ##|a\rangle## or ##|b\rangle##, resulting in a mixture of these states in proportions ##\alpha^{2}## and ##\beta^{2}##, rather than a superposition? After all, we never get to 'directly observe' the superposition state ##\alpha|a\rangle + \beta|b\rangle## because the measurement process keeps projecting the damn thing onto a
single eigenstate!
The Answer: You can't distinguish a
superposition of eigenstates of ##A## from a
mixture of those eigenstates using the observable ##A## itself, because it will give just the same statistics in each case! The way to distinguish the
superposition ##\alpha|a\rangle + \beta|b\rangle## from the corresponding
mixture (which, by the way, we represent as ##\alpha^{2}|a\rangle\langle a| + \beta^{2}|b\rangle\langle b|##) is to
measure instead a different observable ##B## that (to use a technical term) does not
commute with ##A##.
The
difference between the two statistical patterns of ##B##-measurements on the
superposition ##\alpha|a\rangle + \beta|b\rangle## versus the
mixture ##\alpha^{2}|a\rangle\langle a| + \beta^{2}|b\rangle\langle b|## is precisely what we mean by the term
interference effect, motivated by our good old friend the
double-slit experiment. (In that experiment ##A## corresponds to the pair of detectors placed at the slits, while ##B## corresponds to the screen placed some distance away.)
Lunct said:
And do we know why they collapse into one state?
Forgetting any quibbles about how you've phrased the question, the answer is, no we don't.
