Reduced States Are Mixed

Two qubits can be perfectly entangled and yet each one, examined in isolation, looks completely random. This is one of the most striking features of entanglement, and it has an exact mathematical expression: the partial trace.

From pure to mixed

The Bell state $|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}\big(|00\rangle + |11\rangle\big)$ is a pure state — it is a single well-defined vector in the four-dimensional two-qubit Hilbert space. Its density matrix is

\rho = |\Phi^+\rangle\langle\Phi^+| = \frac{1}{2}\begin{pmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 \end{pmatrix},

where rows and columns are ordered $|00\rangle, |01\rangle, |10\rangle, |11\rangle$ . This matrix has rank 1 (only one non-zero eigenvalue), which is the signature of a pure state.

Now suppose you hand qubit 1 to Alice and keep qubit 0. What state do you actually hold? To answer that you trace out Alice's qubit — you sum over her basis states:

\rho_0 = \text{Tr}_1(\rho) = \langle 0|_1\,\rho\,|0\rangle_1 + \langle 1|_1\,\rho\,|1\rangle_1.

Carrying out the sum gives

\rho_0 = \frac{1}{2}\,|0\rangle\langle 0| + \frac{1}{2}\,|1\rangle\langle 1| = \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \frac{I}{2}.

This is the maximally mixed state: both diagonal entries are $\tfrac{1}{2}$ and all off-diagonal entries vanish. By the Born rule, measuring in the computational basis yields $|0\rangle$ with probability $\tfrac{1}{2}$ and $|1\rangle$ with probability $\tfrac{1}{2}$ — a perfectly flat distribution, identical to a fair coin toss.

Purity

A useful scalar summary is the purity $\text{Tr}(\rho^2)$ . For a pure single-qubit state $|\psi\rangle\langle\psi|$ , purity equals 1. For $\rho_0 = I/2$ ,

\text{Tr}\!\left(\rho_0^2\right) = \text{Tr}\!\left(\tfrac{1}{4}I\right) = \tfrac{1}{2},

the minimum possible value for a single qubit, confirming that this is as mixed as a qubit can be.

What measurement reveals

The partial trace prediction can be tested directly: prepare the Bell state and measure qubit 0 without touching qubit 1. The measurement outcomes should split 50/50 between 0 and 1, matching $\rho_0 = I/2$ exactly.

Try it

Build the Bell state and measure qubit 0 only. The grader checks the marginal distribution of the measured qubit — it should be flat (50/50).

Run your code to see the quantum state.

After running, look at the probability bars: P(0) = 0.5 and P(1) = 0.5. That flat distribution is the direct experimental signature of the maximally mixed reduced state $I/2$ .