Tripartite Bell experiment, calculating probabilites Imagine a tripartite photon Bell scenario with, let's say, a prepared state such as W-state or GHZ state:
$$|GHZ>=\frac{1}{\sqrt 2}(|000>+|111>)$$
$$|W>=\frac{1}{\sqrt 3}(|001>+|010>+|100>)$$
Here |0> means that the particle is polarized horizontally and |1> means it is polarized vertically. 
Now imagine the particles are emitted from a source and particle 1 (the first one in each ket) goes to observer A, particle 2 to observer B and particle 3 to observer C.
Each observer has a device which can turn the plane of polarization by the angles $\alpha$ (for A), $\beta$ (for B) and $\gamma$ (for C). They are set in such a way that if the angle is zero only horizontally polarized light is transmitted. Behind each device a measurement on the polarization will be done. 
Now if I have a fixed setting with angles $\alpha_1$, $\beta_1$ and $\gamma_1$ and the W-state, how do I calculate the probability that observer A measures horizontally polarized photons, B measures horizontally polarized and C aswell. What is the probability that A measures vertical, B measures horizontal and C measures vertical (and so on)?
I don't know which basis I should choose to describe to polarizers and the states... I hope somebody can help me, thanks.
 A: http://www.users.csbsju.edu/~frioux/polarize/POLAR-sup.pdf
Passage of a photon in state
$|\psi\rangle = a_0 |0\rangle + a_1 |1\rangle$
through a filter at angle $\alpha$ wrt $|1\rangle$ leaves the photon in state
$|\psi_f \rangle = 1/A(a_0 \sin \alpha |0\rangle + a_1 \cos \alpha |1\rangle).$
Where $A = \sqrt(a_0^2 \sin^2 \alpha + a_1^2 \cos^2 \alpha) $.
We thus observe horizontal polarization with probability
$P_0 = a_0^2 \sin^2 \alpha / A^2$.
Having observed horizontal polarization on the first particle, the state is now projected onto the particle 1 = $|0\rangle$ axis of the Hilbert space. In the GHZ case, for example, it is now certain that the full state is $|000\rangle$. Thus, particle 2 is mapped to
$|\psi_2\rangle  = \sin \beta |0\rangle + \cos \beta |0\rangle$
so the probability of observing horizontal polarization is $\sin^2 \beta$. Particle 3 is in the same state with $\gamma$ replacing $\beta$.
Note that the probabilities of either observation depend on the outcome of the particle 1 measurement. You'll need to consider all cases.
A: You need to invert an 8 by 8 matrix, but fortunately it's got all kinds of symmetries.  Or equally fortunately, there exists Mathematica.
I write $E$ and $F$ instead of $|0\rangle$ and $|1\rangle$.  For each angle $\xi$, put $X_\xi=\cos(\xi)E+\sin(\xi)F$, $Y_\xi=-\sin(\xi)E+\cos(\xi)F$.  
A basis of eigenstates for the measurement you're going to make is the set of eight vectors of the form $A_\alpha\otimes B_\beta\otimes C_\gamma$, where each of $A,B,C$ can be either $X$ or $Y$.  Call this basis ${\cal V}(\alpha,\beta,\gamma)$.
Let $M$ be the matrix that converts the basis ${\cal V}(0,0,0)$ to the basis $\cal{V}(\alpha,\beta,\gamma)$, so that, for example, $M_{11}=\cos(\alpha)\cos(\beta)\cos(\gamma)$.  
Let $N$ be the inverse of $M$.  The rows of $N$ will tell you how to write your given state as a linear combination of the eigenstates. 
For example, the coefficients for the state you've called GHZ are $(1/\sqrt{2})(N_{1*}+N_{8*})$ (assuming you've chosen the natural ordering for your basis). Thus the probability of finding three horizontals is $$(1/2)(N_{11}+N_{81})^2=(1/2)\big(\cos(\alpha)\cos(\beta)\cos(\gamma)+\sin(\alpha)\sin(\beta)\sin(\gamma)\big)^2$$  The probability of finding  (horiz,horiz,vert) is $$(1/2)(N_{12}+N_{82})^2=(1/2)\big(\sin(\alpha)\sin(\beta)\cos(\gamma)-\cos(\alpha)\cos(\beta)\sin(\gamma)\big)^2$$ etc.
