TITLE: What is the difference between pure Bell states and Bell mixed states?
QUESTION [1 upvotes]: I consider the Bell pair $(|00\rangle + |11\rangle)/\sqrt 2$ in the tensor product $H_1 \otimes H_2$
It is a pure state in this vector space and its density matrix is
\begin{pmatrix}
\frac{1}{2} & 0 & 0 & \frac{1}{2} \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
\frac{1}{2} & 0 & 0 & \frac{1}{2}
\end{pmatrix}
Once the pairs are prepared in this state the experimentalist can measure observables on this set of pairs. If Alice and Bob share them, they also can measure local observables.
I consider a different case: the experimentalist produces pairs described by
\begin{pmatrix}
\frac{1}{2} & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & \frac{1}{2}
\end{pmatrix}
in the same basis. It is now a mixed state.
Will this change something? (averages, fringe visibility, correlations...)

REPLY [2 votes]: In both cases, if Alice and Bob measure their qubits in the Z basis, then they will see perfect correlations between their results: they will either both measure the qubits to be in $|0\rangle$, or they will both measure the qubits in $|1\rangle$.
However, if they measure in other bases, then their results will depend on whether the state is a true Bell state or a mixed state. For example, Alice and Bob could each apply a Hadamard gate to their qubit, and then measure. The unitary for applying the Hadamard to both qubits is
$$
U = \frac{1}{2}\pmatrix{1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1}
$$
If we transform our starting density matrix according to this unitary, we see two different possible outcomes:
$$
U \begin{pmatrix} 1/2 & 0 & 0 & 1/2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1/2 & 0 & 0 & 1/2\end{pmatrix}U^\dagger = \begin{pmatrix} 1/2 & 0 & 0 & 1/2 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1/2 & 0 & 0 & 1/2\end{pmatrix}
$$
In this case, if we started with a true Bell state, then after our basis rotation Alice and Bob will still find that their qubits are perfectly correlated.
In contrast, if we start with a mixed state, we find:
$$
U \begin{pmatrix} 1/2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1/2\end{pmatrix}U^\dagger = \begin{pmatrix} 1/4 & 0 & 0 & 1/4 \\ 0 & 1/4 & 1/4 & 0 \\ 0 & 1/4 & 1/4 & 0 \\ 1/4 & 0 & 0 & 1/4\end{pmatrix}
$$
Here, after the basis rotation, Alice and Bob find no more correlations between their measurements: they each randomly measure $|0\rangle$ or $|1\rangle$.
This concept is precisely what forms the basis for experimental verification of entangled states! In particular, experimentalists prepare an entangled state, and then measure in rotated bases and show that correlations are still preserved in these rotated bases.

Edit: In light of the comment, can we interpret this as interference? Yes!
Let's evaluate the resulting state independently for if we start in $|00\rangle$ or if we start in $|11\rangle$. Recall that the Hadamard gate has the following action on a single qubit: $H |0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$, and $H|1\rangle = \frac{1}{\sqrt{2}} (|0\rangle - |1\rangle)$.
If we apply a Hadamard to each qubit starting in $|00\rangle$, we find:
$$
|00\rangle \to \frac{1}{2}(|0\rangle + |1\rangle)(|0\rangle + |1\rangle) = \frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)
$$
If we apply a Hadamard to each qubit starting in $|11\rangle$, we find:
$$
|11\rangle \to \frac{1}{2}(|0\rangle - |1\rangle)(|0\rangle - |1\rangle) = \frac{1}{2}(|00\rangle - |01\rangle - |10\rangle + |11\rangle)
$$
In both cases, we have a 1/4 probability for each possible outcome. If we started with a statistical mixture of these two possible input states, we would still end up randomly with any of the four possible outcomes, with probability 1/4 each.
However, starting with a coherent superposition, the resulting state is the sum of these two output wavefunctions. Note that the $|10\rangle$ and $|01\rangle$ outputs have two different signs -- when adding these wavefunctions together, we get perfect destructive interference such that we would never observe these outcomes!
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