Partial trace (${\mathrm{Tr}}_B$)

The partial trace over $B$ is the linear operation characterized byApplied to a density operator, it produces a reduced density matrix.

The vectors need not form orthonormal bases. With the inner product linear in its second entry, expand the rank-one operator $|\Psi ⟩⟨\Psi |$ and take the trace on the second factor. This gives the displayed reduced positive operator. For every observable $A$ on the first factor,Positivity also follows directly from $⟨v,S_{1}v⟩=\Vert {\sum }_j⟨v,{\xi }_j⟩{\eta }_j{\Vert }^2$. Thus the physical density operator is $S_1/\Vert \Psi {\Vert }^2$ when the nonzero source vector is not normalized. The Gram matrix of the second-factor vectors supplies the cross coefficients; replacing those coefficients by Kronecker deltas without checking orthonormality changes the subsystem state. This formula is useful when an efficient product expansion uses nonorthogonal vectors, even though the partial trace itself is basis-independent.

The partial trace of a positive operator is positive by this sum of nonnegative expectations in an orthonormal basis of the traced factor. Its trace equals the original trace, using the product orthonormal basis. Basis independence follows from $\mathrm{Tr}[({\mathrm{Tr}}_B\rho )A]=\mathrm{Tr}[\rho (A\otimes I)]$, which uniquely determines the reduced operator. Thus reductions of density matrices remain density matrices. For a positive trace-class operator the same argument uses convergent nonnegative basis sums.

For an operator on a tensor-product Hilbert space, averaging unitary conjugations on subsystem $B$ with normalized Haar measure gives ${\mathcal{E}}_A(O)=({\mathrm{Tr}}_{B}O/d_B)\otimes I_B$. It fixes precisely the operators acting trivially on $B$ and is contractive in operator norm. Writing $O-UO{U}^{†}=[O,U]U^{†}$ converts localization error into a commutator bound.