6.6 Partial trace, revisited

If you are given a matrix you calculate the trace by summing its diagonal entries. How about the partial trace? Suppose someone writes down for you a density matrix of two qubits in the standard basis, \{|00\rangle, |01\rangle, |10\rangle, |11\rangle\}, and asks you to find the reduced density matrices of the individual qubits. The tensor product structure of this (4\times 4) matrix means that it is has a block form: \rho^{\mathcal{AB}} = \left[ \begin{array}{c|c} P & Q \\\hline R & S \end{array} \right] where P,Q,R,S are (2\times 2) sized sub-matrices. The two partial traces can then be evaluated as92 \begin{aligned} \rho^\mathcal{A} &= \operatorname{tr}_{B}\rho^{\mathcal{AB}} = \left[ \begin{array}{c|c} \operatorname{tr}P & \operatorname{tr}Q \\\hline \operatorname{tr}R & \operatorname{tr}S \end{array} \right] \\\rho^\mathcal{B} &= \operatorname{tr}_{A}\rho^{\mathcal{AB}} = P+S. \end{aligned} The same holds for a general \rho^{\mathcal{AB}} on any \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}} with corresponding block form ((m\times m) blocks of (n\times n)-sized sub-matrices, where m and n are the dimensions of \mathcal{H}_{\mathcal{A}} and \mathcal{H}_{\mathcal{B}}, respectively).

  1. Take any of the Bell states, write its (4\times 4)-density matrix explicitly, and then trace over each qubit. In each case you should get the maximally mixed state.↩︎