# What is the difference between pure Bell states and Bell mixed states?

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...)

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!

• There is a off diagonal term in the pure density matrix which disappears by tracing out. Is it a kind or interference term at the global level? Commented Oct 30, 2020 at 8:58
• It absolutely can be understood as a type of destructive interference! I edited my answer to highlight this interference, hopefully that helps! Commented Oct 30, 2020 at 22:13

I would not say “something changes”. Basically any kind of joint or local measurements is performed by means of the same experimental setups. What rather changes is the mathematical description of your measurement, i.e. the formulation of post-measurement states (for both joint measurements or local measurements, in the latter case you resort to reduced density matrices) and probabilities for measurement outcomes.

• As we have two different states (one is pure and the other mixed) i think that they describe different things (or else they belong to an équivalent class of operators). Have we the same outputs for averages , correlations , fringe visibility and so on? i think that it is the case for local measurements. but something must differ at the level of the pairs. Commented Oct 29, 2020 at 16:43
• Experimentally one never manages to initialize a perfectly pure state. All of your lab systems you describe with a mixed state, as there is always some imperfect probability for a successful preparation. And of course the two states are different and describe different bipartite systems, but that does not affect how you perform your joint or local measurements. The (anti)correlations of measurement outcomes may then of course be different Commented Oct 29, 2020 at 16:47
• In this Bell case could you give me an example of correlation outcomes which would differ? we always have pairs of 00 or pairs of 11 in the two cases. what else can differ? Commented Oct 29, 2020 at 17:07
• With the naked eye, and without explicitly performing any calculation, I can only tell the probability for joint measurement outcomes would be the same given both density matrices, the pure and mixed one, with the same probability for the allowed tuples "00" and "11". In order to describe the properties of measurements on one subsystem, given that the other is left unobserved, the latter has to be traced out to find a reduced density matrix. So you would compute the latter for the two density matrices. Commented Oct 29, 2020 at 18:11
• More generally, the off-diagonal elements of the density matrix tell how your system oscillates between eigenstates in time. In fact, given the Heisenberg equation of motion for a density matrix $\frac{d\rho}{dt}=\frac{1}{i\hbar}\left[H,\rho\right]$, the probability for measurement outcomes of states are constant over time as long as $H$ and $\rho$ commute, which is the case for density matrices with vanishing off-diagonal elements Commented Oct 29, 2020 at 18:16

the answer to my question was already given by Phoenix87 in Distinguishing density operators with the same diagonal elements

One of the matrices is a projector. a projector on the Bell state i wrote, with 1 as eigen value for it. 00> is an eigenvector of the mixed density matrix but not of the pure projector. so they describe different measurements on the states