If I had a laser beam with 50 % left circular polarised and 50 % right circular polarised photons, what is the overall polarisation of the beam?
I can guess either unpolarised or linear polarisation.
(Not homework: genuine question out of curiosity).
The answer depends on whether the photons are entangled or not.
Nonentangled Photons
This is the case that in classical optics is described as depolarized light or, if the ratios are other than 50:50, partially polarized light.
In this case, the light is a classical mixture of two pure quantum states and each photon has a von Neumann entropy is exactly one bit per photon. In classical optics words, the light is said to be perfectly depolarized; a 50:50 mixture maximizes the entropy, the degree of polarization, as discussed in my answer here is nought and the density matrix is $\frac{1}{2}\mathrm{id}$. Other probabilities in the mixture give us partial polarization with a nonzero degree of polarization. A probability mix of $p:1-p$ give us a lower von Neumann entropy and the degree of polarization is then $\frac{1}{2}-\min(p,\,1-p)$.
Entangled Case
Suppose that the photons are entangled in pairs. Now we have bipartite or two particle states. In this case one cannot talk in terms of the usual Jones or Mueller calculus nor in terms of everyday polarization because the quantum state space is four dimensional. Recall that all these polarization tools apply to a two dimensional Hilbert space (Jones matrices act on $2\times 1$ column vectors, for example). The new Hilbert space is spanned by the vectors $|L\,L\rangle$, $|L\,R\rangle$, $|R\,R\rangle$ and $|R\,L\rangle$. The augmented Jones calculus now acts on $4\times 1$ vectors, the augmented Jones matrices are members of the group ${\rm SU}(4)$, density matrices are $4\times 4$ Hermitian matrices and the augmented Stokes vectors are sixteen dimensional (being the dimension of the space of $4\times4$ Hermitian density matrices).
Now it could be that all the entangled pairs are in precisely the same quantum state, say $|L\,R\rangle$ or $|R\,L\rangle$. In this case, the quantum state is pure; you might even say perfectly polarized but this latter term is not advisable as it would confuse most people.
But we could equally well have classical mixtures of pure pairwise entangled states; in this case you can work out a generalized 16-dimensional Stokes vector and a corresponding degree of pureness and/ or von Neumann entropy. In your 50:50 case, only the states $|L\,R\rangle$ and $|R\,L\rangle$ are present, so you could now use a $2\times 2$ density matrix and a 4-dimensional Stokes/ Mueller calculus to describe the situation, but the basis states that this calculus would describe are quite different from those usually described in this way.
Ok I think I have something.
1) if single photons are observed, then we’ll see 50% L and 50% R.
2) if field sensitive measurements are performed, like the direction of the current in a conductive plate, the a linear polarisation will be observed. The angle will depend on the phase difference between the two fields.