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.
