Difference between Mixed and Pure states

Question

Suppose that there is a system of two photons 1 and 2, each of which is in a mixed state $1/2|R\rangle\langle R| + 1/2 |L \rangle\langle L|$, where $|R \rangle$ and $\langle L|$ are two orthonormal pure polarization states. (So the composite state would be a product of two mixed states.)

How is this case different from the case where each of 1 and 2 is in a pure state $1/\sqrt{2}(|R\rangle+|L\rangle)$? Can you distinguish these two cases experimentally? Because it looks like the measurement outcomes should be the same in both cases: either both are in $|R\rangle$, both are in $|L\rangle$, or one is in $|R\rangle$ and the other in $|L\rangle$. Please educate me.

Answer 1

There is no difference for measurement outcomes in the $\{|R\rangle, |L \rangle\}$ basis. However, you will see a difference if you look at a rotated basis e.g. consider the axes:

$$|+\rangle = \frac{1}{\sqrt{2}} \left(|R\rangle + |L\rangle\right) \\ |-\rangle = \frac{1}{\sqrt{2}} \left(|R\rangle - |L\rangle \right) $$

A measurement along these axes yields $|-\rangle$ 50% of the time for the mixed state, but never yields $|-\rangle$ for the pure state.

The difference between mixed and pure states in general has to do with whether correlations are due to entanglement. Mixed states are classical combinations (e.g. no Bell inequalities, no 'spooky action at a distance', no interference), whereas correlations in a pure state are due to entanglement.

You can see interference in this example, a measurement along the $|-\rangle$ axis has probability 0 because the contribution to the state from the $|R\rangle$ and |$L\rangle$ states cancel along that axis. But you don't get cancelling probabilities for the mixed state, only additive.

Answer 2

The thing is whether those two photons are independent or not.

A single photon can be $|L\rangle$ or $|R\rangle$, which are pure states, or any linear combination of them (normalized).

However, whtn you have two photons as a compound system, you can have these 4 pure states: $$|LL\rangle, |LR\rangle, |RL\rangle, |RR\rangle$$

This notation is obviously $|LR\rangle = |L\rangle_1 \otimes |R\rangle_2 $

And your final state can be any combination of those 4 primitive states.

If you are working with the two photons together, you should write any state as a lienar combination of your basis.

So if you want to say that each photon is in $\frac{1}{\sqrt2} \left(|L\rangle + |R\rangle \right)$, it's okay, but you should write it as a linear combination of the 4 basis vectors, because it will be much easier to work with them.