# 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.

• The eight possible settings for the set of three polarizers give you eight possible measurement outcomes. Take the eight states in which those outcomes are certain, write your initial state as a linear combination of those eight states, and take the squared length of the coefficient on the state whose probability you want to measure. Commented Sep 27, 2016 at 16:27

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.

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.