Partial trace - Knowlpedia

Let $H_A$ and $H_B$ be finite-dimensional complex Hilbert spaces, and let $X$ be an operator on the tensor product $H_A\otimes H_B$ . The partial trace over $B$ is the unique linear map

\operatorname{Tr}_B:\ \mathcal{L}(H_A\otimes H_B)\to \mathcal{L}(H_A)

characterized by the identity

\operatorname{Tr}\!\big((M_A\otimes I_B)\,X\big)=\operatorname{Tr}\!\big(M_A\,\operatorname{Tr}_B(X)\big) \quad\text{for all operators } M_A\in \mathcal{L}(H_A),

where $\operatorname{Tr}$ is the usual trace (see trace-operator).

Equivalent characterizations

Basis formula

If $\{\,|j\rangle\,\}$ is any orthonormal basis of $H_B$ , then

\operatorname{Tr}_B(X)=\sum_j (I_A\otimes \langle j|)\,X\,(I_A\otimes |j\rangle).

This expression is independent of the chosen orthonormal basis.

Basic properties

For operators $A$ on $H_A$ and $B$ on $H_B$ ,

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

More generally:

$\operatorname{Tr}_B$ is linear.
$\operatorname{Tr}(\operatorname{Tr}_B(X))=\operatorname{Tr}(X)$ (trace-preserving).
$\operatorname{Tr}_B$ is positive and completely positive.

Reduced states

If $\rho_{AB}$ is a density-operator on $H_A\otimes H_B$ , then

\rho_A := \operatorname{Tr}_B(\rho_{AB})

is a density operator on $H_A$ , called the reduced state (or marginal) of subsystem $A$ . Even when $\rho_{AB}$ is a pure-state-quantum, the reduced state $\rho_A$ can be mixed-state-quantum.

Example: maximally entangled state

If $\dim H_A=\dim H_B=d$ and

|\Phi\rangle=\frac{1}{\sqrt d}\sum_{i=1}^d |i\rangle\otimes |i\rangle, \quad \rho_{AB}=|\Phi\rangle\langle\Phi|,

then $\operatorname{Tr}_B(\rho_{AB})=I_A/d$ is maximally mixed.