I'm reading various introductions to SPDC, and they all seem to take the state of the entangled photons to be the Bell state that is asymmetric under exchange: $$\Big( \left| H\right>_1\left| V\right>_2-\left| V\right>_1\left| H\right>_2 \Big)\frac{1}{\sqrt2}$$ but I haven't run across anyone explaining why. I wouldn't expect the asymmetric case for photons (bosons).
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$\begingroup$ Please define your notation. There seems to be four polarizations in each term, while there is only supposed to be two photons. $\endgroup$– flippiefanusCommented Sep 1, 2023 at 2:56
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$\begingroup$ @flippiefanus Thanks. A case of typing without thinking. Corrected. $\endgroup$– garypCommented Sep 1, 2023 at 14:19
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$\begingroup$ The asymmetric Bell state you mention is common to Type II PDC crystals. Another technique is used that produces the symmetric state HH+VV. It is called Type I PDC, and uses 2 crystals that are oriented 90 degrees apart. This excellent reference explains it: arxiv.org/abs/quant-ph/0205171 $\endgroup$– DrChineseCommented Sep 1, 2023 at 14:27
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$\begingroup$ see Bartlett, S. D., et al. "Unitary transformations for testing Bell inequalities." Physical Review A 63.4 (2001): 042310 available here $\endgroup$– ZeroTheHeroCommented Sep 4, 2023 at 22:54
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3 Answers
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It makes sense that the state should have a minus sign, because being a polarization basis, one would expect the state to be invariant with respect to rotation.
This state is the singlet under rotation. One can test it by imposing the transformation of the polarization states under a rotation: $|H\rangle_n \rightarrow |H\rangle_n\cos(\phi)-|V\rangle_n\sin(\phi)$ and $|V\rangle_n \rightarrow |H\rangle_n\sin(\phi)+|V\rangle_n\cos(\phi)$, where $n$ is either 1 or 2. The result after the transformation is again the same as before.
The other three Bell states transform as a triplet under rotation.
answered Sep 2, 2023 at 4:06
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The 4 Bell States are: psi+/- & phi+/-: HV-VH, HV+VH, HH+VV, HH-VV. The particular Bell state that comes from PDC is a function of the particular SPDC crystal setup - and there are a number of these. Generally, modern Type I setups generate HH+VV while Type II generate HV-VH.
Phi+ is HH+VV: Eq. 1 from https://arxiv.org/abs/quant-ph/0205171
Psi- is HV-VH: Eq. 1 from https://arxiv.org/abs/quant-ph/0201134
There is nothing special about asymmetric vs symmetric, and in fact it possible to switch from one to the other using a half wave plate. See eq. 1 & 9 https://arxiv.org/abs/quant-ph/0103168 I have never seen the term singlet used for photon entanglement, only electrons, but I guess singlet might apply. Regardless, all of these Bell states maintain entanglement upon rotation. So in that sense I disagree with @flippiefanus to the extent he implies otherwise.
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The (anti)symmetrization requirement of identical particles only applies to meaningless labels, not to measurable properties. Here $H$ and $V$ label polarizations and $1$ and $2$ label positions, so the wave function needn't be symmetric with respect to interchange of them.
But the index of each particle in the tensor product is a meaningless label, so really that wave function should be
$$\frac12 \Big( \left| H\right>_1\left| V\right>_2 + \left| V\right>_2\left| H\right>_1 - \left| V\right>_1\left| H\right>_2 - \left| H\right>_2\left| V\right>_1 \Big)$$
People often omit the extra terms in cases where particle statistics doesn't affect the result, such as Bell-type experiments and quantum computing.
