8.5 Subsystems of entangled systems

Partial trace

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.