Two qubits, A and B, are entangled and share a single joint state \rho_{AB}. Now B is carried off to another galaxy and you keep only A. Here is the question that runs this whole page: what is the state of the qubit in your hand? It must have some description — every measurement you make on A alone has to come out somehow — but a plain ket \alpha|0\rangle + \beta|1\rangle can't be it, because A is tangled up with a qubit you no longer hold.

The answer is a \rho_A called the reduced density matrix, and it is built by an operation called the partial trace that mathematically "discards" B:

It is the honest local description of A: it reproduces the statistics of every measurement you could perform on your qubit, using only your qubit.

What the partial trace does

Recall that an ordinary density matrix lives on the full two-qubit space, spanned by |00\rangle, |01\rangle, |10\rangle, |11\rangle. The full trace \operatorname{Tr}(\rho) sums the diagonal and collapses everything to a single number (always 1). The partial trace over B is the same idea done to only B's degrees of freedom: it traces out B and leaves a matrix that still acts on A.

The whole operation is fixed by what it does to a basis operator. Writing a two-system operator as a tensor product |a\rangle\langle a'| \otimes |b\rangle\langle b'| (the A part times the B part), the rule is

Read it as: leave the A part alone (|a\rangle\langle a'| passes straight through), and trace the B part into the number \langle b'|b\rangle. Because the computational basis is orthonormal, \langle b'|b\rangle is 1 when b'=b and 0 otherwise — so only the terms where B's bra and ket match survive. Extend it to a general \rho_{AB} by linearity, term by term.

Worked example: one half of a Bell pair

Take the maximally entangled Bell state |\Phi^+\rangle = \tfrac{1}{\sqrt2}\big(|00\rangle + |11\rangle\big), whose density matrix is \rho_{AB} = |\Phi^+\rangle\langle\Phi^+|. Multiplying out,

Now trace out B, one term at a time, using the rule above (each ket |xy\rangle = |x\rangle_A \otimes |y\rangle_B):

• \operatorname{Tr}_B(|00\rangle\langle 00|) = \langle 0|0\rangle\,|0\rangle\langle 0| = |0\rangle\langle 0|,
• \operatorname{Tr}_B(|00\rangle\langle 11|) = \langle 1|0\rangle\,|0\rangle\langle 1| = 0 — the off-diagonal term dies,
• \operatorname{Tr}_B(|11\rangle\langle 00|) = \langle 0|1\rangle\,|1\rangle\langle 0| = 0 — likewise,
• \operatorname{Tr}_B(|11\rangle\langle 11|) = \langle 1|1\rangle\,|1\rangle\langle 1| = |1\rangle\langle 1|.

Only the two diagonal terms live. Adding them with the \tfrac12 out front,

This is the maximally mixed qubit — a perfect coin flip. Locally your half of a Bell pair is indistinguishable from a qubit that is truly random: measure it in any basis and you get 0 or 1 with probability \tfrac12 each, with no bias to exploit. And yet the global state |\Phi^+\rangle is pure and perfectly definite. The purity went somewhere — into the correlations between A and B, which is exactly what the partial trace threw away.

Worked example: a product state stays pure

Contrast an unentangled pair. Prepare |\psi\rangle = |0\rangle \otimes |{+}\rangle, so \rho_{AB} = |0\rangle\langle 0| \otimes |{+}\rangle\langle{+}| factors cleanly into an A part and a B part. The partial trace acts only on the B factor:

since \operatorname{Tr}(|{+}\rangle\langle{+}|) = 1. The reduced state is just |0\rangle\langle 0| — the pure state |0\rangle. That makes sense: nothing about A ever depended on B, so losing B costs you nothing. Whatever B is doing, A is cleanly, definitely |0\rangle.

Mixed reduced state = the fingerprint of entanglement

These two examples are not a coincidence — they are the general law. For a pure joint state |\psi\rangle_{AB}:

A clean, computable test for "mixed" is the purity \operatorname{Tr}(\rho_A^2): it equals 1 exactly when \rho_A is pure, and drops below 1 when it is mixed. So the operational signature of entanglement is \operatorname{Tr}(\rho_A^2) < 1. Check it on our two cases:

• Bell half: \operatorname{Tr}\big((\tfrac12 I)^2\big) = \operatorname{Tr}(\tfrac14 I) = \tfrac14 \cdot 2 = \tfrac12 < 1 — mixed, hence entangled. ✓
• Product half: \operatorname{Tr}\big((|0\rangle\langle 0|)^2\big) = \operatorname{Tr}(|0\rangle\langle 0|) = 1 — pure, hence not entangled. ✓

The bars below plot the two eigenvalues of \rho_A in each case. A pure state puts all its weight on one eigenvalue (a spike at 1); a mixed state spreads the weight out. The flatter the spread, the more entangled the original pair — and nothing is flatter than \tfrac12, \tfrac12.