quantum mechanics - Two-Person Problem

The following is a thought experiment I've been stuck on. (No, this isn't homework. I made it up.) Here it is:
Two-Person Problem
Let's say we have a superposed particle going through a tube. There are two people on each side of the tube, as well as two detectors, one 1 yard away from the start of the tube, and the other 2 yards away from the start of the tube, each on different sides. Also they are two people on each side. Now, when the particle reaches the first detector, it is collapsed to person A, but still superposed to person B. Let's say that the result was $|0 \rangle$. My question is, is it possible to be collapsed and superposed at the same time? Even worse, when the particle reaches the second detector, is it possible it could have a result of $|1 \rangle$? And no, in the experiment they are not allowed to communicate with each other.

What's happening here!

Answer

Look up the "Wigner's Friend" thought experiment: your scenario is the one at the point when the friend knows the state of the cat but before Wigner comes back and asks his friend whether his friend's cat is dead.

Generally such thought experiments involving "conscious observers" are not tackled much anymore as a conscious observer is a hugely complicated, uncharacterised system. Instead we replace Wigner and his friend by quantum observables: simple operators that take the quantum state as an input, return a real valued measurement and somehow (the answer to this somehow is the quantum measurement problem ) straight after the "observable's" application, the quantum system is in the eigenstate of the observable's operator that corresponds to the value measured. The quantum measurement problem need not worry us here. We simply replace the two people by observables and assume that an eigenstate of an observable prevails straight afterward its application, whether or not the quantum state is "collapsed" there. (As an aside, I sense a hope in some writings that the quantum measurement problem is indeed resolvable and may be so even in my lifetime - look up Einselection for example).

So now, observable $\hat{W}_1$ acts on a pure quantum state $\psi$ (a pure "superposition", as you call it) and forces it into one of its eigenstates. Now, from the standpoint of the second person (who, as you say, knows the first observable $\hat{W}_1$ has been applied, but does know the eigenstate prevailing after application of $\hat{W}_1$), the quantum state is now in a mixed quantum state. The question of consciousness does not enter: the quantum state has been acted on by $\hat{W}_1$, it's simply that the second person does not know the outcome.

So now, the accepted way to think of the state from the second person's standpoint is as a classical statistical mixture of pure quantum states. Suppose the second observer wants to impart observable $\hat{W}_2$ some time after $\hat{W}_1$ is applied. In principle (this is not done in practice, rather the density Matrix Formalism is used), to foretell the probability distributions of outcomes, the second observer must calculate the pure quantum state's evolution (by, say the Schrödinger equation) for each possible outcome of the observable $\hat{W}_1$ (i.e. for each possible eigenstate $\left|\left.\psi_{1,j}\right>\right.$ output by the observable $\hat{W}_1$), work out the the probability distibutions arising from the application of $\hat{W}_2$ on each of these evolved eigenstates and combine these distributions following the rules of classical probability. As I said, there is a much simpler way of doing all this through the density matrix formalism: the density matrix is:

$$\rho = \sum_j p_{1,j} \,\left|\left.\psi_{1,j}\right>\right. \left<\left.\psi_{1,j}\right|\right.$$

where $\left|\left.\psi_{1,j}\right>\right.$ are the eigenstates of the first observable $\hat{W}_1$ and $p_{1,j}$ their probabilities given the pure quantum state that was input to the first observable. Instead of evolving the pure state by the Schrödinger equation, the density matrix evolves following the Liouville-von Neumann equation:

$$i \hbar \frac{\partial \rho}{\partial t} = [\hat{H},\rho]$$

where $\hat{H}$ is the quantum system's Hamiltonian, and when we get to the time where the second person does their experiment, i.e. imparts observable $\hat{W}_2$, we calculate the $n^{th}$ moment of the statistical distribution of the measurement outcomes as:

$$m_n={\rm tr}\left(\rho\,\hat{W}_2^n\right)$$

whence we can derive the full probability density function for the measurement outcomes from $\hat{W}_2$