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I've seen a few articles (ex: 1, 2) that use either two BBO crystals consecutively or a BBO and a KDP crystal to create entangled photon pairs from Spontaneous Parametric Down-Conversion (SPDC). And from this answer, I understand that it's used as a trick because type I conversion doesn't produce entangled photons.

I just don't get what they're achieving by using 2 crystals. Can someone spell this out for me? From my understanding, by using two crystals, you have more chances to create PDC, but I'm not sure how by using two crystals in orthogonal directions that you're producing entangled pairs, contrary to just one crystal.

Thanks for the help all!

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  • $\begingroup$ Although it might not be quite at the right level, take a look at my explanation from a previous question: physics.stackexchange.com/questions/172159/… . $\endgroup$
    – Rococo
    Commented Sep 9, 2016 at 21:55
  • $\begingroup$ I read your answer. I still don't know why you need two crystals. Why doesn't 1 suffice? Is this dependent on the cut of the crystals? If you put the beam at 45o from the n_e or n_o axis of the crystal, then could you achieve the same effect? I really, just don't get how 2 crystals helps you get polarization entanglement. $\endgroup$
    – Sean
    Commented Sep 12, 2016 at 14:36

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There are some good answers to related questions. So I probably don't need to go into too much detail. Spontaneous parametric down-conversion will usually produce a pair of entangled photons. The questions is, in terms of which degree of freedom are they entangled? Type I phase matching produces the two photons with the same polarization. As a result they are not entangled in terms of polarization. However, they are entangled in terms of there spatial degrees of freedom. That means that if one photon propagates in a particular direction, its correlated photon would propagate in a direction that would ensure momentum conservation.

To get polarization entanglement, as used in many EPR experiments, one needs type II phase matching. Alternatively, one can use some scheme using two BBO's, which is what you are refering to. The two BBO crystals are placed one behind the other and they are oriented differently so that the down-converted photons produced by one have an different polarization from those produce by the other. The combination now gives you a state that is entangled in polarization.

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  • $\begingroup$ Are you saying the photons produce from the first crystal are entangled with the second? Because that doesn't make any sense to me if they were created at totally different instances in time. Just to clarify, You have two down converted photons from the first crystal, in type 2, they're orthogonally polarized. So we have H&V created at T1. Now you're saying to create entanglement we can create a second pair at T2, that are also H&V and suddenly they're entangled? $\endgroup$
    – Sean
    Commented Sep 12, 2016 at 14:38
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    $\begingroup$ The way I understand it is that the pump photon can either be down-converted in the first crystal or in the second crystal. That produces a superposition which gives rise to entanglement. However, since this is based on Type I phase matching you'll have HH from the first and VV from the second, giving the state (HH+VV)/$\sqrt{2}$. $\endgroup$ Commented Sep 13, 2016 at 4:12
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I will assume familiarity with bra-ket notation, which makes this much easier to explain.

Schematically, what a single type-1 SPDC crystal does is it takes photons that are polarized along one axis and splits them into two photons of lower energy and the opposite polarization. Photons that are polarized along the perpendicular axis just pass through the crystal undisturbed. So, if the downconversion axis of the crystal is labelled 'V' and the other axis is 'H' (for vertical and horizontally polarized photons), this corresponds to the following rules: $$ |V_p\rangle\rightarrow |H_1 H_2 \rangle$$ $$|H_p\rangle\rightarrow |H_p\rangle $$

Here 'p' labels the 'pump' photons, which are just the laser photons, and 1,2 label the two photons of lower energy that a laser photon splits into.

(By the way, just for completion I should mention two complications: this splitting process actually only happens very rarely for a polarization aligned with the downconversion axis, and in general you can also have crystals where one V changes into two Hs. Neither of these are relevant for our purposes.)

This means, using the linearity of quantum mechanics, that a superposition state with both horizonal and vertical components will evolve into the following state:

$$\alpha|H_p\rangle+\beta|V_p\rangle \rightarrow \alpha|H_p\rangle+\beta|H_1 H_2\rangle$$

So the photon ends up in a superposition of being downconverted and just passing through. This is somewhat interesting, but it does not have any polarization entanglement. In particular, if you select for only photons that are downconverted, which is what happens in experiments in practice (you use filters to get rid of all the photons at the pump frequency), you end up with the clearly separable state $|H_1 H_2\rangle$.

However, if you add another SPDC crystal that is right after the first one and oriented at 90 degrees relative to it, it will do downconversion according to the rules $$ |V_p\rangle\rightarrow |V_p \rangle$$ $$|H_p\rangle\rightarrow |V_1 V_2\rangle $$

since the downconversion axis is now along H. And this means that the initial pump photon in a general state will go through the following transformations:

$$\alpha|H_p\rangle+\beta|V_p\rangle \rightarrow \alpha|H_p\rangle+\beta|H_1 H_2\rangle \rightarrow \alpha|V_1 V_2\rangle+\beta|H_1 H_2\rangle$$

where we are ignoring the very tiny probability of any nonlinear process happening to the $|H_1 H_2\rangle$ photons as they pass through the second crystal. This is an entangled state. It is maximally entangled when $|\alpha|=|\beta|$, which conveniently just corresponds to an initial polarization that is equally between H and V, which is to say diagonal. So, without considering the various experimental complications that can limit this in practice, all you need to do is shine diagonal linearly polarized light through the two BBO crystals and out come the maximally entangled downconverted photons.

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  • $\begingroup$ If we keep in mind that the "splitting process actually only happens very rarely", shouldn't we rather have something like $$\alpha | H_p > + \beta |V_p> \to \dots \to \alpha\left[ (1-\lambda) |H_p> + \lambda e^{i \phi} | H_1 H_2 > \right] + \beta\left[ (1-\lambda) |V_p> + \lambda e^{i \phi} | V_1 V_2 > \right] $$ for some small $\lambda > 0$ (and some phase $\phi$ for completeness). Hence, the entangled state we are after is only achieved after we filter out the pump photon part. My point is that this filter is then actually essential! Am I correct? $\endgroup$
    – iolo
    Commented Jul 7, 2022 at 14:05
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    $\begingroup$ Hi iolo, that is right. The state in this answer (which, rereading after six years, seems a little sloppy to me) has some implicit post-selection on the photons that are downconverted. Experimentally, this happens very naturally by putting some filter after the BBO crystals that filters out the pump wavelength. $\endgroup$
    – Rococo
    Commented Jul 11, 2022 at 14:55

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