Spontaneous Parametric Down Conversion (SPDC): why the asymmetric Bell state?

Question

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).

Answer 1

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.

Answer 2

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.

Answer 3

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.

Answer 4

Like @flippiefanus mentioned, this singlet state is invariant under rotation as it carries an overall angular momentum of zero. One of the triplet states $HH-VV$ also has this property for rotations about z once more since $J_z=0$ for this state. This property makes the singlets robust against depolarization and it comes under what's known as decoherence-free subspace which is why they are first among equals of the Bell states.

That being said, the condition on the symmetric nature of the photons is on their overall wavefunction and what you're seeing is just the polarization part. In fact, the spatial wavefunction of the pairs generated for triplets are symmetric and for the singlet is antisymmetric to maintain overall symmetric nature of the photons.

This can readily be tested by a Hong-Ou-Mandel interference where two photons at the output will be bunched for the triplet case and anti-bunched for the singlet case. You may refer to my answer on the interference of polarised light for more details. In fact, this trick was used to implement an all-linear-optical deterministic Bell-state analyzer by Kwiat and Weinfurter.