Mixed States from Entanglement

Entanglement and mixedness are two sides of the same coin. Here we make the link precise: a subsystem of an entangled pure state is always mixed, and the amount of mixedness measures the amount of entanglement.

The Schmidt decomposition

Any pure bipartite state can be written in a special diagonal form, the Schmidt decomposition:

with Schmidt coefficients $\lambda_k \ge 0$ summing to $1$, and orthonormal sets $\{|k\rangle_A\}$, $\{|k\rangle_B\}$. The number of nonzero $\lambda_k$ is the Schmidt rank.

Reduced states from the Schmidt form

Tracing out either subsystem of the Schmidt form gives a diagonal density matrix whose entries are the Schmidt coefficients:

Both reduced states share the same spectrum $\{\lambda_k\}$. Two immediate consequences:

• Product (unentangled) state $\iff$ pure reduced state. A single nonzero $\lambda_k = 1$ means $|\psi\rangle = |k\rangle_A|k\rangle_B$ factorizes, and $\rho_A$ is the pure state $|k\rangle_A\langle k|_A$.
• Entangled state $\iff$ mixed reduced state. Two or more nonzero $\lambda_k$ means the state cannot factorize, and $\rho_A$ is mixed.

Mixedness measures entanglement

The purity of the reduced state quantifies the entanglement: $\operatorname{Tr}(\rho_A^2) = \sum_k \lambda_k^2$, which is $1$ for a product state and drops as entanglement grows. The entanglement entropy $S(\rho_A) = -\sum_k \lambda_k \log_2 \lambda_k$ packages the same information, reaching its maximum for a maximally entangled state like $|\Phi^+\rangle$, whose reduced state is $I/2$.

Try it

For the partially entangled state $|\psi\rangle = \cos t\,|00\rangle + \sin t\,|11\rangle$ at $t = \pi/6$, the Schmidt coefficients are $\cos^2 t$ and $\sin^2 t$, so $\rho_A = \operatorname{diag}(\cos^2 t, \sin^2 t)$. Compute the purity $\operatorname{Tr}(\rho_A^2)$. A value below $1$ confirms the subsystem is mixed — the fingerprint of entanglement.

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