Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Measuring the polariztion of a pair of entangled photons

  1. Jan 5, 2013 #1
    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?

    Many thanks in advance!
     
  2. jcsd
  3. Jan 5, 2013 #2
    once you measured the system it is no more entangled and become a separable state namely one of the eigenstates with the corresponding probabilities
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Measuring the polariztion of a pair of entangled photons
Loading...