The Partial Trace

The partial trace is the operation that produces a subsystem's state from a joint state. It is the quantum generalization of marginalizing a joint probability distribution.

Definition on product operators

For an operator that factorizes, $M = A \otimes B$ , the partial trace over the second system is defined by

\operatorname{Tr}_B(A\otimes B) = A\,\operatorname{Tr}(B).

It traces the $B$ factor down to a number and leaves the $A$ factor alone. Because every operator on $\mathcal{H}_A\otimes\mathcal{H}_B$ is a linear combination of such product operators, this single rule plus linearity defines $\operatorname{Tr}_B$ on everything.

The index formula

In components, with the joint basis $|a,b\rangle$ ,

\big(\operatorname{Tr}_B\rho\big)_{a a'} = \sum_b \rho_{(a,b),(a',b')}\Big|_{b'=b} = \sum_b \langle a,b|\rho|a',b\rangle.

You sum over the diagonal of the traced-out system while keeping the surviving system's indices open. Tracing out $A$ instead is symmetric: $\big(\operatorname{Tr}_A\rho\big)_{b b'} = \sum_a \langle a,b|\rho|a,b'\rangle$ .

Key properties

Trace-preserving: $\operatorname{Tr}\big(\operatorname{Tr}_B\rho\big) = \operatorname{Tr}(\rho)$ , so $\rho_A$ stays a valid density operator.
Recovers factors of product states: if $\rho_{AB} = \rho_A\otimes\rho_B$ then $\operatorname{Tr}_B(\rho_{AB}) = \rho_A\operatorname{Tr}(\rho_B) = \rho_A$ , since $\rho_B$ has trace 1.
Completely positive: it maps valid states to valid states even when applied to part of a larger system — a property essential to the theory of open systems and noise channels.

Marginalization analogy

If $\rho$ were a classical joint distribution $p(a,b)$ written as a diagonal matrix, the partial trace over $B$ would compute $\sum_b p(a,b) = p(a)$ — the ordinary marginal. The partial trace is exactly this operation lifted to operators, where it also correctly handles the off-diagonal coherences that have no classical counterpart.

Try it

Build the product state $\rho_{AB} = \rho_A\otimes\rho_B$ with $\rho_A = \operatorname{diag}(0.8, 0.2)$ and $\rho_B = \operatorname{diag}(0.5, 0.5)$ via the Kronecker product, then trace out $B$ and return the $(0,0)$ entry. It should return $0.8$ , recovering $\rho_A$ exactly.

Run your code to see the quantum state.