Quantifying entanglement

Last section’s CNOT built the Bell state $\frac{\ket{00}+\ket{11}}{\sqrt2}$, and we proved it cannot be written as a product. But “cannot be written as a product” is only a qualitative verdict. Twist the parameters a little:

$$\ket{\psi_1}=\frac{\ket{00}+\ket{11}}{\sqrt2},\qquad \ket{\psi_2}=0.99\ket{00}+\sqrt{1-0.99^2}\,\ket{11}\tag{9.3.1}$$

Neither can be written as a product; both are entangled. Yet $\ket{\psi_2}$ is almost $\ket{00}$ — intuition says its entanglement is “much fainter”. Intuition needs to become a number: the Bell experiments of section 9.4 and the teleportation of section 9.5 both consume entanglement, and how much is consumed, and whether there is enough, must be computable.

First, lay the foundation: separable versus entangled

The general two-qubit pure state is

$$\ket{\psi}=\alpha\ket{00}+\beta\ket{01}+\gamma\ket{10}+\delta\ket{11}\tag{9.3.2}$$

If it can be written as $\ket{a}\otimes\ket{b}$ it is called separable (each qubit minding its own business); otherwise it is entangled. The criterion fits on one line: a product state’s coefficient matrix $\begin{pmatrix}\alpha&\beta\\\gamma&\delta\end{pmatrix}$ has rank 1, i.e.

$$\alpha\delta-\beta\gamma=0 \quad\Longleftrightarrow\quad \text{separable}\tag{9.3.3}$$

A non-zero determinant means entanglement — and it even hints that the “amount of entanglement” should have something to do with the size of $|\alpha\delta-\beta\gamma|$. We will come back to that at the end.

Where the old tools fall short: the wavefunction has no “local readout”

Chapter 3’s state vector describes the whole system. But in an entangled state, a single qubit simply has no state vector of its own — there is no $\ket{a}$ that correctly predicts every experiment performed on particle A alone. This is exactly why section 3.10 introduced the density operator: everything local lives in the reduced density matrix,

$$\rho_A=\mathrm{Tr}_B\,\ket{\psi}\bra{\psi}\tag{9.3.4}$$

(Tracing out B = averaging away all of B’s possible outcomes, with their weights.) The key observation:

The Bell state's reduced density matrix: globally pure, locally maximally mixedbasic~6 minexpand

Goal: compute $\rho_A$ for $\ket{\Phi^+}=\frac{\ket{00}+\ket{11}}{\sqrt2}$.

Step 1: write the full density operator.

$$\ket{\Phi^+}\bra{\Phi^+} =\frac12\Big(\ket{00}\bra{00}+\ket{00}\bra{11}+\ket{11}\bra{00}+\ket{11}\bra{11}\Big)\tag{9.3.5}$$

Step 2: trace out B. The rule: $\mathrm{Tr}_B$ keeps only terms whose B-side “left and right labels match” — the partial trace of $\ket{i}_A\ket{j}_B\bra{k}_A\bra{l}_B$ is $\delta_{jl}\ket{i}_A\bra{k}_A$. Term by term:

• $\ket{00}\bra{00}$: B side is $\ket{0}\bra{0}$, trace 1, leaving $\ket{0}\bra{0}_A$;
• $\ket{00}\bra{11}$: B side is $\ket{0}\bra{1}$, trace 0 — the whole term vanishes;
• $\ket{11}\bra{00}$: vanishes for the same reason;
• $\ket{11}\bra{11}$: leaves $\ket{1}\bra{1}_A$.

$$\rho_A=\frac12\ket{0}\bra{0}+\frac12\ket{1}\bra{1}=\frac{I}{2}\tag{9.3.6}$$

Step 3: read off the physics. $\rho_A=I/2$ is the maximally mixed state — the centre of last section’s Bloch sphere. Looking at particle A alone, every axis measures fifty-fifty; statistically it is indistinguishable from “a coin that has been tossed but not yet looked at”. The cross terms (the coherences) were killed in the partial trace by the orthogonality on B’s side: the global coherence has moved into the correlations, and locally only noise remains.

Compare a separable state $\ket{a}\otimes\ket{b}$: the partial trace gives $\rho_A=\ket{a}\bra{a}$ directly, still pure.

Quantifying “mixedness”: Schmidt decomposition and entanglement entropy

Any bipartite pure state can be brought, via the singular value decomposition, into Schmidt form:

$$\ket{\psi}=\sum_k \sqrt{\lambda_k}\,\ket{u_k}_A\ket{v_k}_B, \qquad \lambda_k\ge0,\ \sum_k\lambda_k=1\tag{9.3.7}$$

where $\ket{u_k}$ and $\ket{v_k}$ are orthonormal bases on the A and B sides respectively. For two qubits, $k$ runs to at most 2. The payoff is immediate:

$$\rho_A=\sum_k\lambda_k\ket{u_k}\bra{u_k},\qquad \rho_B=\sum_k\lambda_k\ket{v_k}\bra{v_k}\tag{9.3.8}$$

The two reduced density matrices have the same eigenvalues — the Schmidt coefficients $\lambda_k$. A single non-zero $\lambda$ ⇔ product state; two non-zero $\lambda$‘s ⇔ entangled. Weigh the mixedness with chapter 3’s von Neumann entropy:

$$E(\psi)\equiv S(\rho_A)=-\mathrm{Tr}\,\rho_A\log_2\rho_A =-\sum_k\lambda_k\log_2\lambda_k\tag{9.3.9}$$

This is the entanglement entropy (base 2, measured in bits). Three benchmark points:

State	Schmidt coefficients	$E$
Product state $\ket{a}\otimes\ket{b}$	$1,\ 0$	$0$
$\cos\chi\ket{00}+\sin\chi\ket{11}$	$\cos^2\chi,\ \sin^2\chi$	$0<E<1$
Bell state	$\tfrac12,\ \tfrac12$	$1$ (maximal)

A state with entanglement entropy 1 is called maximally entangled; “spending 1 ebit” means using up one such pair — the unit in which the protocols of sections 9.5 and 9.6 keep all their accounts.

The shortcut formula: concurrence

Entropy requires a partial trace and then a diagonalisation; for two qubits there is a shortcut. The criterion determinant $\alpha\delta-\beta\gamma$ from the opening delivers the concurrence:

$$C=2\,|\alpha\delta-\beta\gamma|,\qquad 0\le C\le 1\tag{9.3.12}$$

$C=0$ is separable, $C=1$ maximally entangled. Its relation to the Schmidt coefficients is $C=2\sqrt{\lambda_1\lambda_2}$, and it corresponds monotonically to the entropy:

$$E=h\!\left(\frac{1+\sqrt{1-C^2}}{2}\right),\qquad h(p)\equiv-p\log_2 p-(1-p)\log_2(1-p)\tag{9.3.13}$$

Check it: $\ket{\Phi^+}$ has $\alpha=\delta=\tfrac{1}{\sqrt2}$, $\beta=\gamma=0$, so $C=2\cdot\tfrac12=1$; the state $\cos\chi\ket{00}+\sin\chi\ket{11}$ gives $C=\sin 2\chi$ — as $\chi$ turns from 0 to 45°, the entanglement climbs smoothly from 0 to 1.

To “see” these correlations, visit Lab module 08: entanglement and Bell non-locality — the spectacle of two particles whose individual readouts are perfectly random yet strictly correlated when combined is waiting there.

∎Key formulas

• Separability criterion (pure states)

  $$\alpha\delta-\beta\gamma=0\ \Leftrightarrow\ \ket{\psi}=\ket{a}\otimes\ket{b}$$

  Coefficient matrix has rank 1; non-zero means entangled

• Schmidt decomposition

  $$\ket{\psi}=\sum_k\sqrt{\lambda_k}\ket{u_k}_A\ket{v_k}_B$$

  Both reduced ρ share the same spectrum {λ_k}

• Entanglement entropy

  $$E=S(\rho_A)=-\sum_k\lambda_k\log_2\lambda_k$$

  Product states 0, Bell states 1 (unit: ebit)

• Concurrence

  $$C=2|\alpha\delta-\beta\gamma|=2\sqrt{\lambda_1\lambda_2}$$

  Two-qubit pure-state shortcut, monotonically tied to E

?Self-check4 questions

1. 1.

   What is the entanglement entropy of the state (∣00⟩ + ∣01⟩ + ∣10⟩ + ∣11⟩)/2?

   • 0 — it is the product state ∣+⟩⊗∣+⟩
   • 1 — it is maximally entangled
   • 1/2
   • 2

   Submit
2. 2.

   Measuring only particle A of the Bell state ∣Φ⁺⟩ (along any axis), what do you observe?

   • Completely random ±1, 50% each, regardless of the measurement axis
   • A definite result along z, random along x
   • The outcome depends on whether particle B has been measured, so it can be used for faster-than-light communication
   • Always +1

   Submit
3. 3.

   What is the concurrence C of the state cos30°∣00⟩ + sin30°∣11⟩? (Give three decimal places.)

   1% relative tolerance

   Submit
4. 4.

   What is the entanglement entropy E of the state √0.9∣00⟩ + √0.1∣11⟩, in ebits? (Give two decimal places.)

   ebit3% relative tolerance

   Submit

What comes next

We can now weigh entanglement, but a more fundamental account is still owed. Einstein’s old objection is not answered by “the entropy equals 1”: perhaps the two particles secretly agreed on all their answers before separating — the so-called correlations being nothing but two copies of the same crib sheet? This “local hidden variable” hypothesis sounds unassailable, yet it can be hauled before the tribunal of experiment. Next section: the Bell inequality, and how quantum mechanics received its loophole-free final verdict in 2015.