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 5.9.11), 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}
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.