How much entanglement does a state hold? For a pure state of two parts the answer is the entropy of either part, $E=S(\hat{\rho}_{A})=S(\hat{\rho}_{B})$: zero for a product, one ebit for a Bell pair, never more than $\log_{2}$ of the smaller dimension. For two qubits there is a shortcut, the concurrence $C=2|c_{00}c_{11}-c_{01}c_{10}|$, which runs from 0 for products to 1 for Bell states and fixes the entropy. Operations on each part with messages between them cannot increase entanglement, so it behaves as a resource: teleportation spends it, and many weakly entangled pairs can be distilled into fewer strong ones. For mixed states the question is harder, and for two qubits a test called the partial transpose settles it.

Money can be moved between accounts and spent, but not printed by bookkeeping. Entanglement is a currency of that kind: local operations and phone calls can spend it or reshape it, never create it.

For $\cos\theta\ket{00}+\sin\theta\ket{11}$ against $\theta$: the entanglement entropy in ebits and the concurrence $|\sin2\theta|$. Both are 0 for the products at the ends and 1 for the Bell state at , and each fixes the other.

For a pure state of two parts, and for two qubits $\ket{\psi}=\sum c_{ij}\ket{ij}$:

• $E=0$ exactly for products; each Bell state has $E=1$ ebit; $E\le\log_{2}\min(d_{A},d_{B})$.

Tracing out $B$ makes the reduced state the matrix $c$ times its adjoint. Its determinant is therefore the squared modulus of $\det c$, a quarter of $C^{2}$; its trace is one. Two numbers with sum one and product $$ are the roots of a quadratic, which gives the weights.

Trace $B$ out of $\ket{\psi}\bra{\psi}$.

The determinant of a product is the product of the determinants.

Trace and determinant of $\hat{\rho}_{A}$.

The roots of $\lambda^{2}-\lambda+\tfrac14C^{2}=0$.

The entanglement entropy depends on $C$ alone.

The entanglement of a pure state of two parts is the entropy of either part. A Bell pair leaves each qubit completely mixed: one ebit.

A product state leaves each part pure, of entropy zero, so its entanglement is zero.

Operations on each part alone, even with messages between the two sides, cannot increase entanglement on average. It can be spent or reshaped, not created that way.

For two qubits the concurrence is twice the modulus of $c_{00}c_{11}-c_{01}c_{10}$: zero for products, one for Bell states.

Great work! You now know how to:

• measure the entanglement of a pure state by the entropy of a part
• compute the concurrence of two qubits and the weights it fixes
• say why entanglement behaves as a resource