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Topic 3 of 12

Quantum Entanglement

Entanglement is the defining feature that separates quantum mechanics from classical physics. Two entangled particles share a joint quantum state that cannot be described as a product of individual states — measuring one instantly determines the other's state, regardless of distance. This resource drives quantum teleportation, cryptography, and computing.

Key Concepts

Entangled State

A multi-qubit state

|\psi\rangle

is entangled if it cannot be written as

|\psi_A\rangle \otimes |\psi_B\rangle

. Example:

|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)

— no single-qubit states

|\psi_A\rangle, |\psi_B\rangle

multiply to give this.

Bell States

The four maximally entangled two-qubit states:

|\Phi^\pm\rangle = \frac{1}{\sqrt{2}}(|00\rangle \pm |11\rangle)

and

|\Psi^\pm\rangle = \frac{1}{\sqrt{2}}(|01\rangle \pm |10\rangle)

. They form an orthonormal basis for the two-qubit Hilbert space and are the key resource in teleportation and superdense coding.

Schmidt Decomposition

Any pure bipartite state

|\psi\rangle_{AB}

can be written as

\sum_i \lambda_i |i_A\rangle|i_B\rangle

where

\lambda_i \geq 0

and

\sum_i \lambda_i^2 = 1

. The Schmidt rank (number of nonzero

\lambda_i

) equals 1 for product states and

>1

for entangled states.

No-Cloning Theorem

It is impossible to create a perfect copy of an unknown quantum state. If a machine could clone

|\psi\rangle

it would violate the linearity of quantum mechanics. This theorem is foundational for quantum cryptography — an eavesdropper cannot copy qubits without detection.

CHSH Inequality

A Bell inequality testing local hidden variable theories:

|E(a,b) - E(a,b') + E(a',b) + E(a',b')| \leq 2

classically. Quantum mechanics allows up to

2\sqrt{2} \approx 2.828

(Tsirelson bound). Experimental violations (Aspect 1982, Hensen 2015) confirm quantum non-locality.

Quantum Non-locality

Entangled particles exhibit correlations that cannot be explained by any local realistic hidden-variable theory (Bell's theorem). This does not allow faster-than-light communication, since Alice's measurement outcomes appear random — only the correlations are non-classical.

Key Equations

Four Bell States

|\Phi^\pm\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle \pm |11\rangle), \quad |\Psi^\pm\rangle = \tfrac{1}{\sqrt{2}}(|01\rangle \pm |10\rangle)

Maximally entangled two-qubit basis states.

Schmidt Decomposition

|\psi\rangle_{AB} = \sum_{i=1}^{r} \lambda_i |i_A\rangle|i_B\rangle, \quad \lambda_i > 0,\; \sum_i \lambda_i^2 = 1

Canonical form of a bipartite pure state; Schmidt rank r > 1 iff entangled.

CHSH Tsirelson Bound

|E(a,b) - E(a,b') + E(a',b) + E(a',b')| \leq 2\sqrt{2} \approx 2.828

Maximum quantum violation of the CHSH Bell inequality.

Entanglement from CNOT+H

\text{CNOT}\cdot(H\otimes I)|00\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle+|11\rangle) = |\Phi^+\rangle

Standard circuit to generate the |Φ+⟩ Bell pair from |00⟩.

Worked Example

Proving a State is Entangled

Problem

Show that $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ cannot be written as $|a\rangle \otimes |b\rangle$ .

Solution

Expanding the product:

= \alpha\gamma|00\rangle + \alpha\delta|01\rangle + \beta\gamma|10\rangle + \beta\delta|11\rangle

Matching coefficients with $|\Phi^+\rangle$ : we need $\alpha\delta = 0$ and $\beta\gamma = 0$ , but $\alpha\gamma = \beta\delta = 1/\sqrt{2} \neq 0$ .

If $\alpha\delta = 0$ then $\alpha=0$ or $\delta=0$ . If $\alpha=0$ then $\alpha\gamma=0 \neq 1/\sqrt{2}$ . Contradiction.

Answer |Φ+⟩ is entangled — no product decomposition exists.

Practice

Exercises

5 problems

1 of 5

For $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ , what is $P(\text{first qubit} = |0\rangle)$ ?

Your answer

2 of 5

For $|\Phi^+\rangle$ , given first qubit measured as $|0\rangle$ , what is $P(\text{second qubit} = |0\rangle)$ ?

Your answer

3 of 5

What is the Schmidt rank of $|\Phi^+\rangle$ ?

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4 of 5

The Tsirelson bound on CHSH = $2\sqrt{2}$ . Give this value to 3 decimal places.

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5 of 5

How many distinct Bell states (maximally entangled two-qubit states) are there?

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Key Takeaways

A state is entangled if it cannot be factored as $|\psi_A\rangle \otimes |\psi_B\rangle$ ; the Schmidt rank exceeds 1.
The four Bell states are the maximally entangled two-qubit states and form a complete orthonormal basis.
The no-cloning theorem prevents perfect copying of unknown quantum states, securing quantum cryptography.
Bell inequality violations (CHSH up to $2\sqrt{2}$ ) prove quantum correlations are non-classical.
Entanglement is a resource — it enables teleportation, superdense coding, and quantum cryptographic security.