The Reduced Density Matrix

From a global pure state to a local mixed one

When system $S$ and environment $E$ become entangled, the joint state $\rho_{SE}$ still contains everything. But an experimenter who only acts on $S$ can never see $E$ . The object that reproduces all statistics of system-only measurements is the reduced density matrix

\rho_S = \mathrm{Tr}_E(\rho_{SE}),

obtained by the partial trace over the environment. It is defined so that for every system observable $M$ ,

\langle M \rangle = \mathrm{Tr}_S(\rho_S\, M) = \mathrm{Tr}_{SE}\bigl(\rho_{SE}\,(M\otimes \mathbb{1}_E)\bigr).

Demanding this equality for all $M$ fixes $\rho_S$ uniquely — the partial trace is the only map that gets every local expectation value right.

Computing the partial trace

Pick an orthonormal basis $\{|e\rangle\}$ for the environment. The partial trace sandwiches $\rho_{SE}$ between environment basis states and sums:

\rho_S = \sum_e \langle e|_E\, \rho_{SE}\, |e\rangle_E .

For a pure global state $|\Psi\rangle = \sum_{s,e} \psi_{se}\,|s\rangle_S |e\rangle_E$ , this gives the compact result

(\rho_S)_{ab} = \sum_e \psi_{ae}\,\psi_{be}^{*},

i.e. $\rho_S = \psi\,\psi^\dagger$ where $\psi$ is the matrix of amplitudes with system index as row and environment index as column. (This is exactly the Schmidt picture: the eigenvalues of $\rho_S$ are the squared Schmidt coefficients.)

The maximally entangled case

Take the Bell state $|\Phi^+\rangle = \tfrac{1}{\sqrt2}(|00\rangle + |11\rangle)$ with the first qubit the system. The amplitude matrix is $\psi = \tfrac{1}{\sqrt2}\,\mathbb{1}_2$ , so

\rho_S = \psi\,\psi^\dagger = \tfrac12 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \tfrac12\,\mathbb{1}.

The system, though part of a perfectly definite pure global state, is locally in the maximally mixed state. All information about it has migrated into the correlations with $E$ .

Properties you can rely on

The reduced density matrix is always a legitimate state: it is Hermitian, positive semidefinite, and has unit trace, $\mathrm{Tr}(\rho_S) = 1$ . Its eigenvalues are probabilities. If the global state is a product $|\psi_S\rangle\otimes|\phi_E\rangle$ , then $\rho_S = |\psi_S\rangle\langle\psi_S|$ stays pure — no entanglement, no mixing. The more entangled $S$ and $E$ are, the more mixed $\rho_S$ becomes, until at maximal entanglement it is proportional to the identity.

Try it

This is a numerical exercise — your code should return a number. Build the $2\times2$ reduced density matrix of the system qubit from $|\Phi^+\rangle$ by partial trace, then return its purity $\mathrm{Tr}(\rho_S^2)$ .

Run your code to see the quantum state.