Which experiment can differentiate a statistical mixture from a superposition state?

Question

I am in trouble with polarization and entanglement.

Let's consider three cases :

Case 1) : Statistical mixture of $|H\rangle$ and $|V\rangle$ polarized photons

Case 2) : Photons in a superposition state $1/\sqrt{2}(|H\rangle+|V\rangle)$

Case 3) : Photons which are entangled with twin ones in $1/\sqrt{2}(|H,H\rangle+|V,V\rangle)$ state

Which experiments can be conducted to differentiate the case

1. statistical mixture from the case

2. superposition state ?

Which experiments can be conducted to differentiate the case

1. statistical mixture from the case

2. entangled photons in superposition state ?

Using a $45^\circ$ polarizer I think you can differentiate case 1/case 2 but not case 1/case 3

I don't know how to differentiate case 1/case 3 except maybe using quantic tomography and Wigner function. Is it true ? Is there any other simpler way ?

Thanks a lot for your answer and sorry for this maybe dummy question...

Answer 1

If a state in a finite dimensional space is pure, it will be an eigenstate of some hermitian operator. Thus measuring this operator on your test state will result in this outcome 100% of the time.

You correctly concluded that in your Case2 the operation is polarization at $45^\circ$. In your case 3 you have a composite state so it lives in the space of states with polarization $L=1$ and $L=0$. It looks like your state (because it is symmetric under exchange of the first and second particle) is probably in the $L=1$ subspace only, and I would think that $\vert \psi\rangle\langle \psi\vert$ can be expressed of quadrupole moments, and would be an eigenstate of some linear combination of these quadrupole moments. How to measure such moments for polarization I do not know.

Note that in a finite dimensional space any state $\vert\psi\rangle$ is pure, whether it is a single-particle or a composite state. Actually doing the measurement is something else but there is literature on this:

• Park, J.L. and Band, W., 1971. A general theory of empirical state determination in quantum physics: Part I & Part II. Foundations of Physics, 1(3), pp.211-226.

Band and Park have a series of paper on this general topic, most of which are precursor to the more general topic of quantum tomography for state reconstruction. For instance, in the case of a spin-$1/2$ system, the density matrix can be completely reconstructed by measuring $\sigma_x,\sigma_y$ and $\sigma_z$, and then it's a matter of just testing if this density matrix describes a pure or a mixed state.

Answer 2

As you've pointed out, a polarizer (or, more usefully, a polarizing beam splitter) at 45° orientation will separate cases 1 and 2.

Case 3 (entangled photons) cannot be distinguished from case 1 (statistical mixture) using observables from only the first photon. This is because the reduced density matrix for case 3 reads \begin{align} \rho_A & = \mathrm{Tr}_B\big[|\Psi⟩⟨\Psi|\big] \\ & = \frac12 \mathrm{Tr}_B\bigg[{\rm (|HH⟩+|VV⟩)(⟨HH|+⟨VV|)}\bigg] \\ & = \frac12 \mathrm{Tr}_B\bigg[{\rm |HH⟩⟨HH|+|HH⟩⟨VV|+|VV⟩⟨HH|+|VV⟩⟨VV|}\bigg] \\ & = \frac12 \bigg[{\rm |H⟩⟨H|+|V⟩⟨V|}\bigg] , \end{align} i.e., the maximally-mixed state that describes a statistical mixture. Since the reduced density matrix $\rho_A$ fully determines the outcome of all local experiments, no such experiment can distinguish between the two cases.

That said, case 3 can be distinguished easily from case 1 by putting an H/V polarizing beam splitter on both of the twin systems, and correlating the two outputs.

(Of course, that doesn't guarantee you that the system is in an entangled state, as that protocol cannot distinguish case 3 from a statistical mixture $\rho = \frac12 \big[{\rm |HH⟩⟨HH|+|VV⟩⟨VV|}\big]$ with classical correlations; to benchmark the entanglement you would need to show a Bell-inequality violation, or a full quantum state tomography if you're feeling fancy.)