- #1
Cylten
- 1
- 0
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!
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!
