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I'm thinking about entangled photon pairs generation, specifically about Type-I SPDC where you use a pair of non-linear crystals such as BBO with their optical axes crossed and then the pump beam is sent in a linearly polarized state that is $45^\circ$ tilted from either axis. Namely, if the input state is $|+\rangle = \frac{1}{\sqrt{2}}(|H\rangle+|V\rangle)$, then after the pair of crystals we get an output state $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|HH\rangle+|VV\rangle)$, which is clearly entangled and one of the Bell states. If the input state is aligned with either of the axis, say $|H\rangle$ or $|V\rangle$, then the output state is just either $|VV\rangle$ or $|HH\rangle$ and certainly not entangled. If for example you control the polarization state of the pump beam with a half waveplate, then by rotating it you can switch from an entangled to a non-entangled output (I have tested this experimentally myself).

I'm trying to think, however, what would be the output state after the SPDC if the input pump is circularly polarized, say $|R\rangle = \frac{1}{\sqrt{2}}(|H\rangle+i|V\rangle)$. Intuitively, I think this should work because $|+\rangle$ and $|R\rangle$ are both mutually unbiased with respect to the basis states defined by the pair of non-linear crystals. The output state should thus still be entangled, and I'd predict the output state to be $|\psi\rangle = \frac{1}{\sqrt{2}}(i|HH\rangle+|VV\rangle)$, which is still one of the Bell states (although not one of the canonical ones). However, when I asked a colleague he argued that if a $|\pm\rangle$ pump creates entangled pairs but $|H\rangle$ and $|V\rangle$ doesn't, then in the circularly polarized pump case, where the polarization is oscillating between these states, then the efficiency with which pump photons will be converted to entangled pairs would be extremely small. I haven't seen anyone try this out yet experimentally and I will eventually, but in the mean time I'm curious about what is the expected prediction.

QuantumExplorer
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4 Answers4

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I think the actual nonlinear crystals available in the market would not be suitable to realize SPDC with circularly polarized light.

The SPDC process are governed by a interaction that could be written as follows [1]

$$ V = \iiint \chi_{yzy}^{(2)} \hat a_3 \hat a_1^\dagger \hat a_2^\dagger {\color{red} e^{i(k_3-k_1-k_2)x}} d^3\mathbf r +\quad \text{H.c}. $$

The phase of the red term above is called wave vector mismatch and it is related to the intensity of the produced beams [2]. Basically, when $\Delta k = k_3-k_1-k_2\neq 0$, the intensity of the produced beams reduces a lot.

The perfect matching condition $\Delta k=0$ is achieved using birefringent crystals, setting the input signal polarization as the extraordinary polarization (i.e., parallel to the plane formed by the optical axis of the crystal and the direction of propagation), and in type I SPDC, the two produces beams have the same polarization. In order to achieve the phase matching, it is needed to realize the angle tuning, in other to find the precise angle between the direction of propagation and optical axis, in order to achieve the phase match.

It seems not possible to realize the angle tuning for circular polarized signal, since the polarization vector is varying during not the propagation inside the crystal. Sometimes it will be aligned to the extraordinary rays, sometimes it will not, and the phase mismatch will reduce the intensity of the produced beams significantly.

But you could insist on it. Maybe you know that some crystals satisfy the quasi-phase matching condition, i.e., they are constructed in such a way that the red term above could be ignored. It could be that the crystal is constructed as a periodically poled material, such that it includes a term in the phase-matching condition $\Delta k = k_3-k_2-k_1+k_m$ due to spatial variation of the nonlinear coupling coefficient [3].

For a linear polarized light, it makes sense that a oscillating nonlinear coupling coefficient works. When the amplitude of the signal is decreasing due to phase mismatch, the coefficient change its sign, and the amplitude of the signal now increases monotonically. This effect is achieved by varying the optical axis orientation by $\pi$ periodically.

But this explanation seems to not make sense for a circular polarized light. If the polarization vector vary in space, the mere inversion of the optical axis periodically would not be enough to compensate the phase mismatch. Maybe a special crystal, with rotating optical axis, would be suitable for a circular polarized SPDC, but I can't find anything like that in literature.

It seems too much labor for too little results. If your source produces only circular polarized light, you could add a linear polarizer before SPDC, do the SPDC with linear polarized light, and then you could use wave plates and phase shifters after SPDC to produce any relative phase you want. The state that you want to produce is locally equivalent to a Bell state, so it can be achieved with local operations after SPDC.

[1] Introduction to Quantum Optics. Mankei Tsang. Chapter 12. Page 100.

[2] * Nonlinear Optics. Robert Boyd. Third Edition*. Chapter 2.3. pg 79-84.

[3] * Nonlinear Optics. Robert Boyd. Third Edition*. Chapter 2.4. pg 74-88.

Ruffolo
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Quote from the source you mentioned:

Normally the output state from type I PDC is not entangled: to get the required phase matching in the nonlinear material, the pump polarization must be fixed.

This reads to me that the type-1 SPDC is only with a certain polarization direction of the pump beam able to emit photon pairs. If you use a circularly polarized pump beam, the efficiency of the photon pair emission will drop rapidly.

I'm trying to think, however, what would be the output state after the SPDC if the input pump is circularly polarized

The crux of type 1 is that the randomness at the output is not given due to the non-orthogonal alignment of the photon pair. This means that the measurable correlation is 1 and the process is unsuitable for quantum cryptography.

HolgerFiedler
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I don't think I agree with your colleague. The reason why a pump in either $|H\rangle$ or $|V\rangle$ cannot produce polarization entanglement is because it wouldn't be able to produce both linear states of polarization in the down-converted light in such a crossed Type-I phase-matching configuration. Consider a general state of polarization for the pump, given by $|H\rangle\alpha+|V\rangle\beta$ where $\alpha$ and $\beta$ are complex factors with the same magnitude. Then the down-converted state should have a state given by $|V\rangle|V\rangle\alpha +|H\rangle|H\rangle\beta\exp(i\varphi)$ where $\varphi$ is an extra phase that I include to model the possible phase shift due to the fact that one crystal in the crossed configuration is behind the other one. The result is still entangled, despite the unknown relative phase.

flippiefanus
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Your colleague is wrong. The Hamiltonian for the type-I process with two crossed crystals is

$$ \hat{H}_{\mathrm{spdc}} = \alpha_V (\hat{a}^{\dagger}_H)^2 + e^{i\varphi}\alpha_H (\hat{a}^{\dagger}_V)^2 $$

where the $\alpha$ are coherent state amplitudes. There is a relative phase between the two processes that depends both on the differential propagation phase $\varphi$ (between the pump and single photons), as well as the phase between the two pump polarizations (encoded in the complex amplitudes). To get entanglement, this phase needs to be kept constant, but there is nothing special about a real superposition of pump polarizations. The phase-matching condition in the crystal defines a preferred basis, and doesn't "see" the phase.

Furthermore, there is no need for you to conduct the experiment as it has already been done, see Figure 2 of Kwiat et al. Physical Review A 60(2) (1999).

fulis
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