I Spontaneous parametric down-conversion entanglement using BBO

[Paul159]

Hello,

I have a question about the creation of the Bell's entanglement state $1/\sqrt{2} (|HH> + |VV>)$using type I BBO crystals (https://en.wikipedia.org/wiki/Spontaneous_parametric_down-conversion).

Two crystals are put orthogonal to each other and each of them emits a photon pair ($|HH>$ or $|VV>$). Then at the intersection of the two emitted cones we have the Bell's state. But I don't understand why.
Indeed, if I do measurement of the photons with two polarizers, one at 90° and the other 0°, I don't understand why I will have 0 correlation. For me I could detect for example the signal photon of the $|HH>$ state and the idler photon of the $|VV>$.

I hope my "question" is clear.

 

[PeterDonis]

Two crystals are put orthogonal to each other and each of them emits a photon pair ($|HH>$ or $|VV>$). Then at the intersection of the two emitted cones we have the Bell's state.

Where are you getting this from? Not from the article you linked to; that just talks about producing one photon pair using one crystal.

 

[Paul159]

https://arxiv.org/ftp/arxiv/papers/1001/1001.4182.pdf (look at the end of page 4).

 

[PeterDonis]

https://arxiv.org/ftp/arxiv/papers/1001/1001.4182.pdf (look at the end of page 4).

Thanks for the reference. It looks to me like the key properties of this setup that allows it to produce the Bell-type state are:

(1) The two crystals act on the two orthogonal polarization components ($H$ and $V$);

(2) The down-conversion processes in each crystal are coherent, so an input photon polarized at 45 degrees (halfway between $H$ and $V$) will create quantum uncertainty about which crystal is doing the downconversion; this is what creates the Bell-type entangled state.

 

[Paul159]

Yes of course, I understand now. Thanks !