Problem with the polarization of entangled photons

[Christian Thom]

Consider this thought experiment : we use a source of identically vertically polarized photons, such as produced by a type 0 SPDC. One beam go to Alice and the second to Bob.
1. Whatever measurement Bob makes on its beam, if Alice use a vertically polarized detector, all photons are detected, due to the nature of the source.
2. If Bob use a detector with a polarization @ 45°, about half of the photons are detected.
3. Now Alice places its detector @ 45 ° too. The twin photons of those who are detected at Bob's are also detected at Alice's, since they are entangled, but they would have been also detected with a vertically polarized detector, as seen in 1.

So here is the problem : what is the polarization of these photons since they pass at 100 % in two differently oriented polarizers ? Please point out where is my mistake.

 

[DrClaude]

Consider this thought experiment : we use a source of identically vertically polarized photons, such as produced by a type 0 SPDC.

Pairs of identically polarized photons are not in an entangled state.

 

[vanhees71]

but in type-0 or type-I SPDC you have
$$|\Psi \rangle=\frac{1}{\sqrt{2}}(|H_i H_s \rangle+ \exp(\mathrm{i} \phi) |V_i V_s \rangle)$$
which are entangled states.

 

[Hill]

vertically polarized photons

If they are vertically polarized, then $|\Psi \rangle=|V_i V_s \rangle$, which is not entangled.

 

[vanhees71]

Ok, in type-0 SPDC this you get if the pump photon is in the state $|V_p \rangle$, and this doesn't produce entangled two-photon state. If you use a $45^{\circ}$-polarized pump photon you get the entangled state given #3.

 

[DrChinese]

Consider this thought experiment : we use a source of identically vertically polarized photons, such as produced by a type 0 SPDC. One beam go to Alice and the second to Bob.
1. Whatever measurement Bob makes on its beam, if Alice use a vertically polarized detector, all photons are detected, due to the nature of the source.
2. If Bob use a detector with a polarization @ 45°, about half of the photons are detected.
3. Now Alice places its detector @ 45 ° too. The twin photons of those who are detected at Bob's are also detected at Alice's, since they are entangled, but they would have been also detected with a vertically polarized detector, as seen in 1.

So here is the problem : what is the polarization of these photons since they pass at 100 % in two differently oriented polarizers ? Please point out where is my mistake.

Just to add to the correct answers above by @DrClaude, @vanhees71 , @Hill :

It is possible to generate vertically polarized photon pairs using SPDC. They are entangled in some bases, but they won’t be polarization entangled.

Measuring both of them at 45 degrees- your #3 - will show 0 correlation. They show 100% correlation only when measured as V.

 

[Christian Thom]

Thank you all for your answers that clarifie the situation. I guess the impossibility to have this kind of entanglement is not limited to spdc, but generalizes to other methods of producing entangled pairs.

 

[vanhees71]

It's all about the polarization state of the pump photon relative to the axis of the birefringent crystal used for SPDC. Let's denote the ordinary axis of the crystal as the $x$-axis (and the polarization state with $|H \rangle$) and the extraordinary axis as the $y$-axis (and the polarization state with $|V \rangle$). In type-0 SPDC a H-polarized pump photon splits in two photons ("idler and signal photon") in the state $|H_i H_s \rangle$ and a V-polarized one in $\exp(\mathrm{i} \varphi) |V_i V_s \rangle$. Here $\varphi$ is the phase difference between the one or the other case.

For an arbitrarily polarized pump photon with state $\alpha |H \rangle + \beta V \rangle$ you get $\alpha |H_i H_s \rangle + \beta \exp(\mathrm{i} \varphi) |V_i V_s \rangle$. Of course $|\alpha|^2+|\beta|^2=1$, and thus you get an entangled state if neither $|\alpha|^2=1$ nor $|\beta|^2=1$.