Understanding Entanglement and Born's Rule in Photon Polarization Experiments

[mbond]

Let be an entangled pair of photons 1 and 2, with the same polarization. The wave function is
$|12>=\cos\psi|HH>+\sin\psi|VV>$ with $\psi$ the angle of polarization. The first $H$ ($V$) in $|HH>$
($|VV>$) is photon 1, and the second one is photon 2.

Alice observes photon 1 with a polarizer oriented $\theta_A$. The corresponding projector is
$P_A=_1|\theta_A><\theta_A|_1$ with $|\theta_A>=\cos\theta_A|H>+\sin\theta_A|V>$. The index $_1$ is to tell the projector applies only to photon 1 (not sure about the notation...). According to Born's rule, the probability Alice detects photon 1 is:

$\mathcal{P}_A=<21|P_A|12>$

$\hphantom{\mathcal{P}_A}=(\cos\psi<H|<H|\theta_A>+\sin\psi<V|<V|\theta_A>)$
$(\cos\psi<\theta_A|H>|H>+\sin\psi<\theta_A|V>|V>)$

$\hphantom{\mathcal{P}_A}=\cos^2\psi\cos^2\theta_A+\sin^2\psi\sin^2\theta_A$

After Alice has observed photon 1 (detecting it or not), the wave function $|12>$ no longer exists (it has collapsed). The probability that Bob subsequently detects photon 2 with a polarizer oriented $\theta_B$ is then $\mathcal{P}_A$ times a $\cos^2$ from the Malus law:

$\mathcal{P}_B=(\cos^2\psi\cos^2\theta_A+\sin^2\psi\sin^2\theta_A)\cos^2(\theta_A-\theta_B)$

Is that correct? I would be grateful if someone pointed me the errors...

 

[DrClaude]

Looks ok.

However, the bra corresponding to $\ket{12}$ should be $\bra{12}$ (one does not change the order of particles).

 

[mbond]

Let be 2 non-entangled photons $|\phi_i>=\cos\phi_i|H>+\sin\psi_i|V>$ with i=1 or 2. The wave function of the 2 photons is then
$|\psi_1\psi_2>=\cos\psi_1\cos\phi_2|HH>+\cos\psi_1\sin\psi_2|HV>+\sin\psi_1\cos\psi_2|VH>+\sin\psi_1\sin\psi_2|VV>$

Alice makes the wave function interact with a polarizer $<\theta_A|=\cos\theta_A<H|+\sin\theta_A<V|$ and is left with the wave function
$|\psi_A>=<\theta_A|\psi_1\psi_2>=cos(\theta_A-\psi_1)|\psi_2>$
The probability she detects a photon is given by Born's rule:
$<\psi_1\psi_2|\theta_A><\theta_A|\psi_1\psi_2>=\cos^2(\theta_A-\psi_1)<\psi_2|\psi_2>=\cos^2(\theta_A-\psi_1)$
A problem is that the whole wave function looks collapsed while it is only $|\psi_1>$.

If Bob measures $|\psi_A>$ with a polarizer $\theta_B$, he gets a probability
$<\psi_A| \theta_B><\theta_B|\psi_A>=\cos^2(\theta_A-\psi_1)\cos^2(\theta_B-\psi_2)$
which is the probability that both Alice and Bob detect a photon.

When the photons are entangled, Alice is left with
$|\psi_A>=\cos\psi\cos\theta_A|H>+\sin\psi\sin\theta_A|V>$
If Bob measures $|\psi_A>$ he has the probability:
$<\psi_A|\theta_B><\theta_B|\psi_A>=(\cos\psi\cos\theta_A\cos\theta_B+\sin\psi\sin\theta_A\sin\theta_B)^2$
but there is no Malus law there!?

Either Born's rule doesn't work with 2 photons, or, more likely, I don't apply it correctly...

 

[vanhees71]

If the photons are entangled, the measurement outcomes are not independent anymore, because entanglement describes strong correlations, i.e., the joint probability for both Bob and Alice detecting a photon doesn't factorize anymore as in your case with unentangled photons, which describes a state, where the photons are uncorrelated.

To see whether you calculation is correct, I'd need to know the specific entangled state the photons are prepared in. Also your notation is a bit vague!

 

[DrClaude]

The Born rule gives you the probability of the outcome of a measurement, but not the state after the measurement. That is given by the eigenstate of the observable corresponding to the measurement result.

 

[vanhees71]

I think, in principle the calculations are correct, but one should use the correct notation. If I understand it right, what's considered is the joint probability to find the photons at a certain linear-polarization state at A's and B's detector. The correct formula for that probability is
$$P(\Theta_a,\Theta_b)=\langle \Psi|\hat{P}_{A}(\Theta_A) \otimes \hat{P}_{B}(\Theta_B)|\Psi \rangle,$$
for any two-photon state
$$|\Psi \rangle = \sum_{h_1,h_2} \Psi(h_1,h_2) |h_1 \rangle_A \otimes |h_2 \rangle,$$
where $h_j \in \{V,H\}$ run over the polarization basis used to write down these vectors. The $\hat{P}$ are the corresponding single-photon-polarization projection operators.

 

[mbond]

Thank you for the answers.
I note $|HH>\equiv|H>\otimes|H>$, the same with V. With this notation, the wave function of the entangled pair is $|\psi>=\cos\psi|HH>+\sin\psi|VV>$. The projectors of Alice and Bob are, respectively $\hat{P}_A=|\theta_A><\theta_A|$ and $\hat{P}_B=|\theta_B><\theta_B|$ with $|\theta_A>=\cos\theta_A|H>+\sin\theta_A|V>$, the same with B.

I would like to calculate the probability that Alice and Bob observe the same result (the entangled photons go through their respective polarizer, or are absorbed). (When $\theta_A-\theta_B=\pm120°$ I expect that Alice and Bob get the same outcome 25% of the time, independently of $\psi$).

 

[vanhees71]

Then you just have to do the algebra and calulate
$$P(h_A,h_B)=\langle \psi|\hat{P}_A \otimes \hat{P}_B|\psi \rangle.$$
I guess the result will not only depend on the difference, $\Theta_A-\Theta_B$, because that's the case only for the singlet state, $|\psi_{\text{sing}}=(|HV \rangle-|VH \rangle)/\sqrt{2}$.

 

[DrChinese]

I would like to calculate the probability that Alice and Bob observe the same result (the entangled photons go through their respective polarizer, or are absorbed). (When $\theta_A-\theta_B=\pm120°$ I expect that Alice and Bob get the same outcome 25% of the time, independently of $\psi$).

For theta=+/-120 degrees on Type I photon entangled pairs per your example: Yes. they get the same output 25% of the time when a polarizing beam splitter (PBS) is used. When a *polarizer* is used (i.e. half absorbed), the stats are different:

No result: 12.5% (of course you actually can't actually count these because they were both absorbed)
Same result (Alice and Bob record hits): 12.5% (actual would be about 14.3%)
Alice only: 37.5% (actual would be about 42.9%)
Bob only: 37.5% (actual would be about 42.9%)

 

[mbond]

"Then you just have to do the algebra and calculate $P=<\psi|\hat{P}_A\otimes\hat{P}_B|\psi>$"

Let me do it:
$P=(\cos\psi<HH|+\sin\psi<VV|)|\theta_A><\theta_A|\otimes|\theta_B><\theta_B|(\cos\psi|HH>+\sin\psi|VV>)$
$\hphantom{P}=(\cos\psi<H|\theta_A><H|+\sin\psi<V| \theta_A><V|)|\theta_B><\theta_B|$
$(\cos\psi<\theta_A|H>|H>+\sin\psi<\theta_A|V>|V>)$
$\hphantom{P}=(\cos\psi\cos\theta_A<H|+\sin\psi\sin\theta_A<V|)|\theta_B><\theta_B|(\cos\psi\cos\theta_A>|H>+\sin\psi\sin\theta_A|V>)$
$\hphantom{P}=(\cos\psi\cos\theta_A\cos\theta_B+\sin\psi\sin\theta_A\sin\theta_B)^2$
But there is no Malus law $\cos^2(\theta_A-\theta_B)$ there!? (except for $\psi=\pi/4$).

EDIT: I think my error is to use a $\psi$ angle; there is no axis there, so no angle, the probabilities for the entangled photons to be HH or VV are equal so there is a $1/\sqrt{2}$ instead.

Thank you very much for the help.

 

[vanhees71]

Why do you expect Malus's Law to hold here? The entangled state implies strong correlations between the outcomes of the measurements done at the two photons.

 

[mbond]

The experiments to test the Bell inequalities such as the Aspect's one are explained using the XIX century Malus law of classical optics. I think, with entangled photons, using Born's rule is much better, even if it happens to give the same result.

 

[vanhees71]

But Malus's Law applies to the simple measurement of a linearly polarized photon with a polarization filter in a direction of an angle $\alpha$ relative to the polarization direction of the photon to be $\cos^2 \alpha$. That doesn't imply that the joint measurements must obey this law.

Usually they discuss as the initial state the case where $\psi=\pi/4$, i.e., $\cos \psi=\sin \psi=\sqrt{2}/2$. Then your formula gives obviously $P=1/2 \cos^2(\theta_A-\theta_B)$. This is easy to understand in this case: If you put a polarization filter in $\theta_A$ direction in the way of A's photon it's projected to the state
$\cos \theta_A |H \rangle + \sin \theta_A |V \rangle$ and so is also the photon at B due to the prepared entengled state. When you put a polarization filter in $\theta_B$ direction to B's photon, then you get, provided A's photon went through the probability $|\cos \theta_A \cos \theta_B +\sin \theta_A \sin \theta_B|^2=\cos^2 (\theta_A-\theta_B)$. But now this case occurs with a probability of $1/2$, which explains your result.

 

[mbond]

$P_{AB}\equiv P(\theta_A,\theta_B)=\frac{1}{2}\cos^2(\theta_A-\theta_B)$, right from Born's rule, is the probability that both Alice and Bob observe a photon. The probability that Alice's photon is absorbed and not Bob's is $P_{\bar{A}B}=P(\theta_A+90°,\theta_B)$. One can check that $P_{AB}+P_{\bar{A}B}+P_{A\bar{B}}+P_{\bar{A}\bar{B}}=1$. The probability Alice and Bob observe the same thing is $P_{AB}+P_{\bar{A}\bar{B}}=\cos^2(\theta_A-\theta_B)$.