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The Wikipedia page for "Density Matrix" (https://en.wikipedia.org/wiki/Density_matrix) takes each of a pair of entangled photons as an example of a mixed state:

A radioactive decay can emit two photons traveling in opposite directions, in the quantum state $|R,L\rangle+|L,R\rangle/\sqrt2$. The two photons together are in a pure state, but if you only look at one of the photons and ignore the other, the photon behaves just like unpolarized light.

I find this a bit puzzling for the following reason. A mixed state is characterized by statistical probabilities, which is due to our ignorance, essentially, as opposed to genuine quantum indeterminacy. But the nature of the uncertainty associated with the polarization of each of the EPR photons is, to my understanding, not statistical; it is genuine quantum uncertainty. So, how should I understand the above-quoted passage exactly?

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  • $\begingroup$ "A mixed state is characterized by statistical probabilities, which is due to our ignorance, essentially, as opposed to genuine quantum indeterminacy." -- Nope. $\endgroup$ Jun 23, 2020 at 21:30

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Short answer: as you said the nature of the uncertainty associated with the polarization of each of the photons is not classical since they are prepared in a pure state. The uncertainty after observing each photon alone is purely quantum.

Long-and-maybe-boring answer:

The most general representation of a quantum system is written in terms of the density operator

\begin{equation} \varrho = \sum_i p_i |\psi_i\rangle \langle \psi_i| \, , \end{equation}

which is built in such a way that it naturally encompasses both quantum and classical probabilities. Essentially, the connection between mixed states and entanglement is made by the notion of the reduced state or reduced density matrix. When a composite system is in a product state $|\psi \rangle_A \otimes |\psi \rangle_B$ , it makes sense to say the state of $A$ is simply $|\psi \rangle_A$. However, if $A$ and $B$ are entangled, then what is exactly the “state” of $A$? To see how it works, consider first a bipartite state of $AB$ of the form

\begin{equation} |\psi \rangle_{AB}= \sum_i c_i |i\rangle_A |i\rangle_B \end{equation}

Now let $\mathcal{O}_A$ be an operator which acts only on system $A$. That is, an operator which has the form $\mathcal{O}_A = \mathcal{O}_A \otimes I_B$. The expectation value of $\mathcal{O}_A$ in the state will be

\begin{equation} \langle \mathcal{O}_A \rangle = \langle \psi | \mathcal{O}_A |\psi\rangle = \sum_i |c_i|^2\langle i | \mathcal{O}_A |i\rangle \end{equation}

See that there is no way to attribute a state $|\psi \rangle_A$ for system $A$ such that the above result can be expressed as $\langle i | \mathcal{O}_A |i\rangle$. In other words, there is no pure state we can associate with $A$. Instead, if we wish to associate a quantum state to $A$, it will have to be a mixed state, described by a density matrix of the form

\begin{equation} \varrho_A = \sum_i |c_i|^2 |i \rangle \langle i| \end{equation}

with the expectation value $\langle A \rangle = \textrm{tr}(A\varrho_A)$. The first equation has exactly the same form as last equation with the classical probabilities $p_i$ replaced by quantum coefficients $|c_i|^2$. But there is absolutely nothing classical here. We started with a pure state and we are talking about a purely quantum effect. Notwithstanding, we observe that in general the state of $A$ will be mixed. This result has extremely important consequences and tells us that when $AB$ is entangled, the reduced state of A and B will be mixed.

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A mixed state is very different than an entangled state, but in this case the entanglement can only be detected by measuring both photons.

If you only look at a single photon, the state is effectively a classical mixture, because the entanglement cannot be detected in any way.

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