6.5 Subsystems of entangled systems

We have already trumpeted that one of the most important features of the density operator formalism is its ability to describe the quantum state of a subsystem of a composite system. Let me now show you how it works.

Given a quantum state of the composite system \mathcal{AB}, described by some density operator \rho^{\mathcal{AB}}, we obtain reduced density operators \rho^{\mathcal{A}} and \rho^{\mathcal{B}} of subsystems \mathcal{A} and \mathcal{B}, respectively, by the partial trace: \begin{aligned} \rho^{\mathcal{AB}} &\longmapsto \underbrace{\rho^\mathcal{A}=\operatorname{tr}_{\mathcal{B}}\rho^{\mathcal{AB}}}_{\mathrm{partial\,trace\,over}\,\mathcal{B}}\qquad \\\rho^{\mathcal{AB}} &\longmapsto \underbrace{\rho^\mathcal{B}=\operatorname{tr}_{\mathcal{A}}\rho^{\mathcal{AB}}}_{\mathrm{partial\,trace\,over}\,\mathcal{A}} \end{aligned} We define the partial trace over \mathcal{B}, or \mathcal{A}, first on a tensor product of two operators A\otimes B as \begin{aligned} \operatorname{tr}_{\mathcal{B}} (A\otimes B) &= A(\operatorname{tr}B) \\\operatorname{tr}_{\mathcal{A}} (A\otimes B) &= (\operatorname{tr}A) B, \end{aligned} and then extend to any operator on \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}} by linearity.

Here is a simple example. Suppose a composite system \mathcal{AB} is in a pure entangled state, which we can always write as |\psi_{\mathcal{AB}}\rangle = \sum_{i} c_{i} |a_i\rangle\otimes|b_i\rangle, where |a_i\rangle and |b_j\rangle are two orthonormal bases (e.g. the Schmidt bases), and where \sum_i |c_i|^2 = 1 (due to the normalisation). The corresponding density operator of the composite system is the projector \rho^{\mathcal{AB}}= |\psi_{\mathcal{AB}}\rangle\langle\psi_{\mathcal{AB}}|, which we can write as \rho^{\mathcal{AB}} = |\psi_{\mathcal{AB}}\rangle\langle\psi_{\mathcal{AB}}| = \sum_{ij} c_i c^\star_j |a_i\rangle\langle a_j| \otimes |b_i\rangle\langle b_j| Let us compute the reduced density operator \rho^{\mathcal{A}} by taking the partial trace over \mathcal{B}: \begin{aligned} \rho^\mathcal{A} &= \operatorname{tr}_{\mathcal{B}}\rho^{\mathcal{AB}} \\&= \operatorname{tr}_{\mathcal{B}} |\psi_{\mathcal{AB}}\rangle\langle\psi_{\mathcal{AB}}| \\&= \operatorname{tr}_{\mathcal{B}} \sum_{ij} c_i c^\star_j |a_i\rangle\langle a_j| \otimes |b_i\rangle\langle b_j| \\&= \sum_{ij} c_i c^\star_j |a_i\rangle\langle a_j|(\operatorname{tr}|b_i\rangle\langle b_j|) \\&= \sum_{ij} c_i c^\star_j |a_i\rangle\langle a_j| \underbrace{\langle b_i|b_j\rangle}_{\delta_{ij}} \\& = \sum_{i} |c_i|^2 |a_i\rangle\langle a_i| \end{aligned} where we have used the fact that \operatorname{tr}|b_i\rangle\langle b_j| = \langle b_j|b_i\rangle=\delta_{ij}. In the |a_i\rangle basis, the reduced density matrix \rho^{\mathcal{A}} is diagonal, with entries p_i=|c_i|^2. We can also take the partial trace over \mathcal{A} and obtain \rho^\mathcal{B} = \sum_{i} |c_i|^2 |b_i\rangle\langle b_i|. In particular, for the maximally entangled states in the (d\times d)-dimensional Hilbert spaces \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}}, |\psi_{\mathcal{AB}}\rangle = \frac{1}{\sqrt{d}} \sum_{i}^d |a_i\rangle|b_i\rangle, and the reduced density operators, \rho^\mathcal{A} and \rho^\mathcal{B}, are the maximally mixed states: \rho^\mathcal{A}=\rho^\mathcal{B}=\frac{1}{d}\mathbf{1}. It follows that the quantum states of individual qubits in any of the Bell states are maximally mixed: their density matrix is \frac12\mathbf{1}. A state such as \frac{1}{\sqrt 2} \left( |00\rangle + |11\rangle \right) guarantees perfect correlations when each qubit is measured in the standard basis: the two equally likely outcomes are (0 and 0) or (1 and 1), but any single qubit outcome, be it 0 or 1 or anything else, is completely random.