# Measuring the polariztion of a pair of entangled photons

1. Jan 5, 2013

### Cylten

Hi,

I'm quite new to quantum mechanics, learning about it in my free time in a life-long learning fashion :) I've been trying to find a solution to a problem for some time, and the results I included below appear to be OK, but I have doubts about the method I used, so any help or guidance would be much appreciated.

I imagine the following experiment: suppose there is a source that emits two entangled photons along the z axis, A and B, both linearly polarized, but in an orthogonal direction. Photon A encounters a linear polarizer aligned with the x and y axes and a detector behind it. Photon B encounters a linear polarizer rotated by $\phi$ related to the first one and another detector behind it. There is some apparatus, C, connected to both detectors. I think after the experiment, the apparatus should be in a superposition of detecting one, the other, both, or neither photons:
$$|C\rangle = x\left|C_{A\&B}\right> + y\left|C_A\right>+ z\left|C_B\right>+ w\left|C_0\right>.$$
I'm trying to find the coefficients (amplitudes) x, y, z and w, and I know the order in which the photons interact with the polarizers and detectors should not matter.

I've tried various ways to do this, and the most promising was the following:

In isolation, the polarization of photon A can be described in the 2-dimensional Hilbert space $H_A$. Let the observable associated with the x,y-aligned polarizer be P with the following eigenvectors:
$$P = \begin{pmatrix}1&0\\ 0&-1\end{pmatrix} ~~~~~|+p\rangle = \begin{pmatrix}1\\0\end{pmatrix} ~~~~~|-p\rangle = \begin{pmatrix}0\\1\end{pmatrix}.$$
The polarization of photon B can be described in a similar space $H_B$. Let the observable associated with the $\phi$-rotated polarizer be $P_\phi$ with the following eigenvectors (using the equivalents of $|+p\rangle$ and $|-p\rangle$ as the basis):
$$P_\phi = \begin{pmatrix}\cos2\phi& \sin2\phi\\ \sin2\phi& -\cos2\phi\end{pmatrix} ~~~~~|+\phi\rangle = \begin{pmatrix}\cos\phi\\ \sin\phi\end{pmatrix} ~~~~~|-\phi\rangle = \begin{pmatrix}-\sin\phi\\ \cos\phi\end{pmatrix}.$$

The (linear) polarization of the entangled photons, if A is polarized in the $a$ direction, can be described in $H_A\otimes H_B$ as
$$|\Psi\rangle = \cos a |+p\rangle\otimes|-p\rangle + \sin a |-p\rangle\otimes|+p\rangle,$$
and, using the $|\pm p\rangle \otimes|\pm p\rangle$ basis, this can be expressed as
$$|\Psi\rangle = \begin{pmatrix}0\\ \cos a\\ \sin a\\ 0\end{pmatrix}.$$
Now in order to find the outcome of the experiment, I considered $P\otimes P_\phi$ as an observable in $H_A\otimes H_B$ (tensor product of Hermitians is Hermitian). Its eigenvectors are the tensor products of the eigenvectors of $P$ and $P_\phi$, which are:
\begin{aligned} |+p\rangle\otimes|+\phi\rangle =& \begin{pmatrix}\cos\phi\\ \sin\phi\\ 0\\ 0\end{pmatrix} \\ |+p\rangle\otimes|-\phi\rangle =& \begin{pmatrix}-\sin\phi\\ \cos\phi\\ 0\\ 0\end{pmatrix} \\ |-p\rangle\otimes|+\phi\rangle =& \begin{pmatrix}0\\ 0\\ \cos\phi\\ \sin\phi\end{pmatrix} \\ |-p\rangle\otimes|-\phi\rangle =& \begin{pmatrix}0\\ 0\\ -\sin\phi\\ \cos\phi\end{pmatrix} \end{aligned}

And from these, the amplitudes appear the be:
\begin{aligned} \langle+p+\phi|\Psi\rangle = \sin\phi\cos a &= x \\ \langle+p-\phi|\Psi\rangle = \cos\phi\cos a &= y \\ \langle-p+\phi|\Psi\rangle = \cos\phi\sin a &= z \\ \langle-p-\phi|\Psi\rangle = -\sin\phi\sin a &= w, \end{aligned}
as we expect both photons to go through the polarizers at $\phi=\pi/2$ and $a=0$.

This result also seems to be in accordance with the reasoning around Bell's theorem,
which, as I understand it, suggests that the average correlation between the two detectors, defined as
$$\text{Cor} = \frac{ \left(\begin{matrix}\text{number of experiments}\\ \text{showing correlation}\end{matrix}\right) - \left(\begin{matrix}\text{number of experiments}\\ \text{with no correlation}\end{matrix}\right) } {\text{number of experiments}}.$$
as a function of $\phi$ is not linear, but sinusoid. Based on the above amplitudes, this is
\begin{aligned} \text{Cor} = \sin^2\phi\cos^2a + \sin^2\phi\sin^2a - \cos^2\phi\cos^2a - \cos^2\phi\sin^2a &\\ = \sin^2\phi - \cos^2\phi = -\cos(2\phi)&, \end{aligned}
which seems to be OK as we expect complete correlation at $\phi=\pi/2$ (when the polarizers are aligned orthogonally) and again at $\phi=3\pi/2$.

Does this mean that the above values of x,y,z and w are correct? What I find most confusing is that the eigenvectors of $P\otimes P_\phi$ are (I think) not states the system can actually be in, as, for example, $|\Psi\rangle = |+p\rangle\otimes |+\phi\rangle$ would mean that A is polarized in the $|+p\rangle$ direction and B is polarized in the $|+\phi\rangle$ direction, which seems to contradict the entanglement. Can this be fixed or does this mean that the whole argument is invalid?

2. Jan 5, 2013

### jk22

once you measured the system it is no more entangled and become a separable state namely one of the eigenstates with the corresponding probabilities