8.5 Subsystems of entangled systems

Earlier, we claimed 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 us now show you how this 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 the subsystems \mathcal{A} and \mathcal{B} (respectively) by the partial trace: \begin{aligned} \rho_{\mathcal{AB}} &\longmapsto \rho_{\mathcal{A}} = \underbrace{\operatorname{tr}_{\mathcal{B}}\rho_{\mathcal{AB}}}_{\mathrm{partial\,trace\,over}\,\mathcal{B}}\qquad \\\rho_{\mathcal{AB}} &\longmapsto \rho_{\mathcal{B}} = \underbrace{\operatorname{tr}_{\mathcal{A}}\rho_{\mathcal{AB}}}_{\mathrm{partial\,trace\,over}\,\mathcal{A}} \end{aligned} We will revisit the notion of partial trace quite a few times, but for now we simply 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 |\psi_{\mathcal{AB}}\rangle. We can always write this 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, from Exercise 5.14.13), 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_{i,j} 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_{i,j} c_i c^\star_j |a_i\rangle\langle a_j| \otimes |b_i\rangle\langle b_j| \\&= \sum_{i,j} c_i c^\star_j |a_i\rangle\langle a_j|(\operatorname{tr}|b_i\rangle\langle b_j|) \\&= \sum_{i,j} 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} So, in the |a_i\rangle basis, the reduced density matrix \rho_{\mathcal{A}} is diagonal, with entries p_i=|c_i|^2. Similarly, if we take the partial trace over \mathcal{A}, then we get \rho_{\mathcal{B}}=\sum_{i} |c_i|^2 |b_i\rangle\langle b_i|.

In particular, if \dim\mathcal{H}_{\mathcal{A}}=\dim\mathcal{H}_{\mathcal{B}}=d, then the maximally mixed state |\psi_{\mathcal{AB}}\rangle = \frac{1}{\sqrt{d}} \sum_{i}^d |a_i\rangle|b_i\rangle, in the (d\times d)-dimensional Hilbert space \mathcal{H}_{\mathcal{A}}\otimes\mathcal{H}_{\mathcal{B}} is such that the reduced density operators \rho_{\mathcal{A}} and \rho_{\mathcal{B}} are also the maximally mixed states of their respective subsystems: \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 \frac{1}{2}\mathbf{1}.

A bipartite 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 outcomes are “0 and 0” or “1 and 1” (which are equally likely), and we will never observe e.g. “0 and 1”, but the outcome of either single-qubit subsystem is completely random.