Reduced Density Matrices

When a system is composed of two parts $A$ and $B$, we often care only about $A$. The reduced density matrix $\rho_A$ is the state that reproduces every measurement statistic on $A$ alone — and it is, in general, mixed even when the whole $AB$ system is pure.

The setup

Let the joint state be $\rho_{AB}$, living on the tensor-product space $\mathcal{H}_A \otimes \mathcal{H}_B$. We want a description of $A$ that gives the correct answer for any observable acting only on $A$, i.e. an observable of the form $M_A \otimes I_B$. We demand

There is exactly one operator $\rho_A$ satisfying this. It is obtained by the partial trace over $B$:

Computing the partial trace by entries

Label the joint basis $|a\rangle_A \otimes |b\rangle_B$. The partial trace over $B$ sums over the diagonal in the $B$ index while keeping the $A$ indices free:

Concretely, you "contract" the two $B$ slots: hold the $A$ indices $a, a'$ fixed and add up the matrix elements where the $B$ index is the same on both sides. The result is a smaller matrix acting only on $A$, and it automatically satisfies $\operatorname{Tr}(\rho_A) = 1$ and all the density-operator properties.

The surprising part

If $\rho_{AB} = |\Phi^+\rangle\langle\Phi^+|$ with $|\Phi^+\rangle = \tfrac{1}{\sqrt2}(|00\rangle + |11\rangle)$ — a pure, maximally entangled state — tracing out $B$ gives

the maximally mixed qubit. A globally pure state can have a maximally mixed part. This is not a flaw in the bookkeeping: it is the defining feature of entanglement, which the next two lessons explore.

Try it

Take the Bell state $|\Phi^+\rangle$, form $\rho_{AB} = |\Phi^+\rangle\langle\Phi^+|$, trace out qubit $B$, and return $(\rho_A)_{00}$. You should find $1/2$ — the reduced state is maximally mixed.

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