6.8 Partial trace, yet again

The partial trace is the only map \rho^{\mathcal{AB}}\mapsto\rho^{\mathcal{A}} such that95 \operatorname{tr}[X\rho^{\mathcal{A}}] = \operatorname{tr}[(X\otimes\mathbf{1})\rho^{\mathcal{AB}}] holds for any observable X acting on \mathcal{A}. This condition concerns the consistency of statistical predictions. Any observable X on \mathcal{A} can be viewed as an observable X\otimes\mathbf{1} on the composite system \mathcal{AB}, where \mathbf{1} is the identity operator acting on \mathcal{B}. When constructing \rho^{\mathcal{A}} we had better make sure that for any observable X the average value of X in the state \rho^\mathcal{A} is the same as the average value of X\otimes\mathbf{1} in the state \rho^{\mathcal{AB}}. This is indeed the case for the partial trace.

For example, let us go back to the state in Equation (9.7.1) and assume that Alice measures some observable X on her part of the system. Technically, such an observable can be expressed as X\otimes \mathbf{1}, where \mathbf{1} is the identity operator acting on the subsystem \mathcal{B}. The expected value of this observable in the state |\psi_{\mathcal{AB}}\rangle is \operatorname{tr}(X\otimes\mathbf{1})|\psi_{\mathcal{AB}}\rangle\langle\psi_{\mathcal{AB}}|, i.e. \begin{aligned} \operatorname{tr}[(X\otimes \mathbf{1}) \rho^{\mathcal{AB}}] &= \operatorname{tr}\left[ (X\otimes\mathbf{1}) \left( \sum_{ij} |\widetilde\psi_i\rangle\langle\widetilde\psi_j| \otimes |b_i\rangle\langle b_j| \right) \right] \\&= \sum_{ij} \left[ \operatorname{tr}\left(X |\widetilde\psi_i\rangle\langle\widetilde\psi_j|\right) \right] \underbrace{\left[\operatorname{tr}\left(|b_i\rangle\langle b_j|\right)\right]}_{\delta_{ij}} \\&= \sum_i \operatorname{tr}\big[X |\widetilde\psi_i\rangle\langle\widetilde\psi_i|\big] \\&= \operatorname{tr}\left[ X \underbrace{\sum_i p_i|\psi_i\rangle\langle\psi_i|}_{\rho^{\mathcal{A}} = \operatorname{tr}_{\mathcal{B}}\rho^{\mathcal{AB}}} \right] \\&= \operatorname{tr}[X\rho^{\mathcal{A}}] \end{aligned} as required.

!!!TO-DO!!! The uniqueness of the partial trace, for now see Nielsen & Chuang Box 2.6.


  1. One can repeat the same argument for \rho^{\mathcal{AB}}\mapsto\rho^{\mathcal{B}}: the partial trace is the unique map \rho^{\mathcal{AB}}\mapsto\rho^{\mathcal{B}} such that \rho^{\mathcal{B}} satisfies \operatorname{tr}[Y\rho^{\mathcal{B}}] = \operatorname{tr}[(1\otimes Y)\rho^{\mathcal{AB}}] for any observable Y on \mathcal{B}.↩︎