2  Entanglement

Entanglement is the most profoundly non-classical phenomenon in quantum mechanics. Einstein called it “spooky action at a distance.” It’s the foundation of quantum computing’s power—and the core primitive in QPL.

2.1 Two-Qubit Systems

Before diving into entanglement, we need to understand how to describe multiple qubits mathematically.

2.1.1 Tensor Product

When we have two independent quantum systems, we combine them using the tensor product (denoted \(\otimes\)).

Example: System A in state \(|0\rangle_A\) and system B in state \(|1\rangle_B\) combine to:

\[ |0\rangle_A \otimes |1\rangle_B = |01\rangle \]

Shorthand notation: We often write \(|01\rangle\) instead of \(|0\rangle_A \otimes |1\rangle_B\).

2.1.2 Computational Basis for Two Qubits

The 4 basis states for a 2-qubit system:

\[ |00\rangle, \quad |01\rangle, \quad |10\rangle, \quad |11\rangle \]

As vectors in \(\mathbb{C}^4\):

\[ |00\rangle = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, \quad |01\rangle = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, \quad |10\rangle = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \quad |11\rangle = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \]

2.1.3 General Two-Qubit State

The most general 2-qubit state is:

\[ |\psi\rangle = \alpha_{00}|00\rangle + \alpha_{01}|01\rangle + \alpha_{10}|10\rangle + \alpha_{11}|11\rangle \]

with normalization: \(|\alpha_{00}|^2 + |\alpha_{01}|^2 + |\alpha_{10}|^2 + |\alpha_{11}|^2 = 1\).

2.2 Separable vs Entangled States

Not all 2-qubit states are created equal. There’s a fundamental distinction:

2.2.1 Separable States

A state is separable (or product state) if it can be written as:

\[ |\psi\rangle = |\phi\rangle_A \otimes |\chi\rangle_B \]

where \(|\phi\rangle_A\) is a single-qubit state for system A and \(|\chi\rangle_B\) is a single-qubit state for system B.

Example:

\[ |\psi\rangle = |0\rangle_A \otimes |+\rangle_B = |0\rangle_A \otimes \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)_B = \frac{1}{\sqrt{2}}(|00\rangle + |01\rangle) \]

This state is separable - the two qubits are independent.

2.2.2 Entangled States

A state is entangled if it cannot be written as a tensor product of single-qubit states.

Definition: Entanglement

A 2-qubit state \(|\psi\rangle\) is entangled if and only if it cannot be expressed as:

\[ |\psi\rangle = |\phi\rangle_A \otimes |\chi\rangle_B \]

for any single-qubit states \(|\phi\rangle_A\) and \(|\chi\rangle_B\).

2.3 Bell States: Maximally Entangled Pairs

The four Bell states are the most important entangled states in quantum computing:

\[ \begin{align*} |\Phi^+\rangle &= \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle) \\[1em] |\Phi^-\rangle &= \frac{1}{\sqrt{2}}(|00\rangle - |11\rangle) \\[1em] |\Psi^+\rangle &= \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle) \\[1em] |\Psi^-\rangle &= \frac{1}{\sqrt{2}}(|01\rangle - |10\rangle) \end{align*} \]

2.3.1 Why Can’t Bell States Be Factored?

Let’s prove that \(|\Phi^+\rangle\) cannot be written as a product state.

Proof: \(|\Phi^+\rangle\) Is Entangled

Claim: \(|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\) cannot be written as \(|\phi\rangle_A \otimes |\chi\rangle_B\).

Proof by contradiction:

Assume \(|\Phi^+\rangle = |\phi\rangle_A \otimes |\chi\rangle_B\) where:

\[ |\phi\rangle_A = \alpha|0\rangle + \beta|1\rangle, \quad |\chi\rangle_B = \gamma|0\rangle + \delta|1\rangle \]

Then:

\[ |\phi\rangle_A \otimes |\chi\rangle_B = (\alpha|0\rangle + \beta|1\rangle) \otimes (\gamma|0\rangle + \delta|1\rangle) \]

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

Comparing with \(|\Phi^+\rangle = \frac{1}{\sqrt{2}}|00\rangle + 0|01\rangle + 0|10\rangle + \frac{1}{\sqrt{2}}|11\rangle\):

\[ \alpha\gamma = \frac{1}{\sqrt{2}}, \quad \alpha\delta = 0, \quad \beta\gamma = 0, \quad \beta\delta = \frac{1}{\sqrt{2}} \]

From \(\alpha\delta = 0\): either \(\alpha = 0\) or \(\delta = 0\).

• If \(\alpha = 0\): then \(\alpha\gamma = 0 \neq \frac{1}{\sqrt{2}}\) ⚡ contradiction
• If \(\delta = 0\): then \(\beta\delta = 0 \neq \frac{1}{\sqrt{2}}\) ⚡ contradiction

Therefore, \(|\Phi^+\rangle\) cannot be factored. \(\square\)

2.4 Physical Meaning of Entanglement

When two qubits are entangled in the state \(|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\):

1. Before measurement: Neither qubit has a definite state. They exist in a correlated superposition.

2. Measuring qubit A:

   • 50% chance of measuring 0 → qubit B instantly becomes \(|0\rangle\)
   • 50% chance of measuring 1 → qubit B instantly becomes \(|1\rangle\)

3. Perfect correlation: The measurement outcomes are perfectly correlated, even if the qubits are separated by light-years.

This is what troubled Einstein - the “spooky action at a distance.” However, no information travels faster than light (you can’t control which outcome you get).

2.5 Schmidt Decomposition

Every bipartite pure state can be expressed in a special canonical form.

Schmidt Decomposition Theorem

Any pure state \(|\psi\rangle\) of a bipartite system A-B can be written as:

\[ |\psi\rangle = \sum_{i=1}^{r} \lambda_i |u_i\rangle_A \otimes |v_i\rangle_B \]

where: - \(\{|u_i\rangle_A\}\) is an orthonormal basis for system A - \(\{|v_i\rangle_B\}\) is an orthonormal basis for system B - \(\lambda_i > 0\) are real numbers (Schmidt coefficients) with \(\sum_i \lambda_i^2 = 1\) - \(r\) is the Schmidt rank (number of non-zero terms)

This decomposition is unique (up to reordering and degeneracies).

2.5.1 Schmidt Rank and Entanglement

The Schmidt rank tells us about entanglement:

• \(r = 1\): State is separable (product state)
• \(r > 1\): State is entangled
• \(r = \min(\dim A, \dim B)\): State is maximally entangled

2.5.2 Example: Schmidt Decomposition of \(|\Phi^+\rangle\)

The Bell state \(|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\) is already in Schmidt form:

\[ |\Phi^+\rangle = \frac{1}{\sqrt{2}}|0\rangle_A \otimes |0\rangle_B + \frac{1}{\sqrt{2}}|1\rangle_A \otimes |1\rangle_B \]

Schmidt coefficients: \(\lambda_1 = \lambda_2 = \frac{1}{\sqrt{2}}\)

Schmidt rank: \(r = 2\) (maximal for 2 qubits)

2.5.3 Example: Separable State

Consider \(|\psi\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |01\rangle) = |0\rangle_A \otimes \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)_B\).

Schmidt decomposition:

\[ |\psi\rangle = 1 \cdot |0\rangle_A \otimes |+\rangle_B \]

Schmidt rank: \(r = 1\) (separable)

2.6 Entanglement Entropy

How do we quantify how much entanglement a state has?

2.6.1 Reduced Density Matrix

For a bipartite state \(|\psi\rangle_{AB}\), the reduced density matrix for system A is:

\[ \rho_A = \text{Tr}_B(|\psi\rangle\langle\psi|) \]

where \(\text{Tr}_B\) means “trace out system B” (sum over all states of B).

For a state in Schmidt form \(|\psi\rangle = \sum_i \lambda_i |u_i\rangle_A \otimes |v_i\rangle_B\):

\[ \rho_A = \sum_i \lambda_i^2 |u_i\rangle\langle u_i| \]

The eigenvalues of \(\rho_A\) are the Schmidt coefficients squared: \(\{\lambda_1^2, \lambda_2^2, \ldots\}\).

2.6.2 Von Neumann Entropy

The entanglement entropy is the von Neumann entropy of the reduced density matrix:

\[ S(\rho_A) = -\text{Tr}(\rho_A \log_2 \rho_A) = -\sum_i \lambda_i^2 \log_2(\lambda_i^2) \]

Properties of Entanglement Entropy

• \(S = 0\): State is separable (no entanglement)
• \(S > 0\): State is entangled
• \(S = \log_2 d\): State is maximally entangled (for \(d\)-dimensional subsystems)
• For 2 qubits: \(S_{\max} = \log_2 2 = 1\) bit

2.6.3 Example: Entropy of \(|\Phi^+\rangle\)

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

Schmidt coefficients: \(\lambda_1 = \lambda_2 = \frac{1}{\sqrt{2}}\)

Eigenvalues of \(\rho_A\): \(\frac{1}{2}, \frac{1}{2}\)

Entropy:

\[ S = -\frac{1}{2}\log_2\left(\frac{1}{2}\right) - \frac{1}{2}\log_2\left(\frac{1}{2}\right) = -\frac{1}{2}(-1) - \frac{1}{2}(-1) = 1 \]

Result: \(S = 1\) bit - maximally entangled! ✅

Proof: Bell States Have Maximal Entropy

Claim: All four Bell states have entanglement entropy \(S = 1\) bit.

Proof:

All Bell states have Schmidt decomposition with coefficients \(\lambda_1 = \lambda_2 = \frac{1}{\sqrt{2}}\).

For example: - \(|\Phi^+\rangle = \frac{1}{\sqrt{2}}|0\rangle|0\rangle + \frac{1}{\sqrt{2}}|1\rangle|1\rangle\) - \(|\Psi^+\rangle = \frac{1}{\sqrt{2}}|0\rangle|1\rangle + \frac{1}{\sqrt{2}}|1\rangle|0\rangle\)

For all: \(\lambda_1^2 = \lambda_2^2 = \frac{1}{2}\)

Therefore:

\[ S = -\frac{1}{2}\log_2\frac{1}{2} - \frac{1}{2}\log_2\frac{1}{2} = 1 \text{ bit} \quad \square \]

2.6.4 Example: Partially Entangled State

Consider:

\[ |\psi\rangle = \frac{\sqrt{3}}{2}|00\rangle + \frac{1}{2}|11\rangle \]

Schmidt coefficients: \(\lambda_1 = \frac{\sqrt{3}}{2}, \lambda_2 = \frac{1}{2}\)

Entropy:

\[ S = -\frac{3}{4}\log_2\frac{3}{4} - \frac{1}{4}\log_2\frac{1}{4} \approx 0.811 \text{ bits} \]

This is partially entangled (\(0 < S < 1\)).

2.7 Bell’s Theorem and Non-Locality

Entanglement leads to correlations that cannot be explained by any local hidden variable theory.

2.7.1 EPR Paradox

In 1935, Einstein, Podolsky, and Rosen (EPR) argued that quantum mechanics must be incomplete. They believed:

1. Elements of reality exist for unmeasured properties
2. Locality: Measuring one particle can’t instantly affect a distant particle

These assumptions lead to local hidden variable theories.

2.7.2 Bell’s Inequality

In 1964, John Bell proved that local hidden variable theories predict:

\[ |E(a,b) - E(a,c)| \le 1 + E(b,c) \]

where \(E(x,y)\) is the correlation between measurements in directions \(x\) and \(y\).

Quantum mechanics violates this inequality!

For Bell states measured in appropriate bases:

\[ |E(a,b) - E(a,c)| = 2\sqrt{2} \approx 2.83 > 1 + E(b,c) \]

This has been experimentally verified countless times. Nature is non-local - entanglement is real.

Implications for Quantum Computing

Bell’s theorem proves that quantum correlations are fundamentally different from classical correlations. This is the source of quantum computing’s power:

• Quantum algorithms exploit entanglement for exponential speedups
• Classical computers cannot efficiently simulate entangled systems
• QPL makes entanglement a first-class primitive to harness this power

2.8 Measurement of Entangled States

When we measure one qubit of an entangled pair, the entire state collapses.

2.8.1 Example: Measuring \(|\Phi^+\rangle\)

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

Measure qubit A in Z-basis:

• Outcome 0 (probability \(\frac{1}{2}\)):
  • State collapses to \(|00\rangle\)
  • Qubit B is now in state \(|0\rangle\)
• Outcome 1 (probability \(\frac{1}{2}\)):
  • State collapses to \(|11\rangle\)
  • Qubit B is now in state \(|1\rangle\)

Key observation: Measuring qubit A instantaneously determines the state of qubit B, regardless of distance!

2.8.2 Partial Measurement Formula

For a general state \(|\psi\rangle = \sum_{ij} c_{ij}|i\rangle_A|j\rangle_B\):

Measuring qubit A and getting outcome \(i\):

\[ |\psi\rangle \xrightarrow{\text{measure A}=i} \frac{1}{\sqrt{P(i)}} \sum_j c_{ij}|i\rangle_A|j\rangle_B = |i\rangle_A \otimes \left(\frac{1}{\sqrt{P(i)}} \sum_j c_{ij}|j\rangle_B\right) \]

where \(P(i) = \sum_j |c_{ij}|^2\) is the probability of outcome \(i\).

2.9 Summary

In this chapter, you learned:

✅ Tensor product combines multiple quantum systems ✅ Separable states can be factored: \(|\phi\rangle_A \otimes |\chi\rangle_B\) ✅ Entangled states cannot be factored ✅ Bell states are maximally entangled 2-qubit states ✅ Schmidt decomposition is the canonical form for bipartite states ✅ Entanglement entropy quantifies entanglement: \(S = -\sum_i \lambda_i^2 \log_2 \lambda_i^2\) ✅ Bell’s theorem proves quantum non-locality ✅ Measuring one qubit collapses the entangled partner

2.10 Exercises

2.10.1 Exercise 2.1: Identifying Entanglement

Which of the following states are entangled? Prove your answer.

(a) \(|\psi\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |01\rangle)\)

(b) \(|\psi\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |10\rangle)\)

(c) \(|\psi\rangle = \frac{1}{2}(|00\rangle + |01\rangle + |10\rangle + |11\rangle)\)

2.10.2 Exercise 2.2: Schmidt Decomposition

Find the Schmidt decomposition for:

\[ |\psi\rangle = \frac{1}{2}|00\rangle + \frac{1}{2}|01\rangle + \frac{1}{2}|10\rangle + \frac{1}{2}|11\rangle \]

Calculate the Schmidt rank and entanglement entropy.

2.10.3 Exercise 2.3: Entanglement Entropy

Calculate the entanglement entropy for:

\[ |\psi\rangle = \frac{2}{\sqrt{5}}|00\rangle + \frac{1}{\sqrt{5}}|11\rangle \]

Is this state maximally entangled?

2.10.4 Exercise 2.4: Bell State Transformations

Prove that all four Bell states can be transformed into each other using single-qubit gates (Pauli gates):

\[ \begin{align*} X|0\rangle = |1\rangle, \quad X|1\rangle = |0\rangle \\ Z|0\rangle = |0\rangle, \quad Z|1\rangle = -|1\rangle \end{align*} \]

Hint: Start with \(|\Phi^+\rangle\) and apply \(X\) or \(Z\) gates to derive the others.

2.10.5 Exercise 2.5: Partial Measurement

Given the state \(|\Psi^+\rangle = \frac{1}{\sqrt{2}}(|01\rangle + |10\rangle)\):

(a) If qubit A is measured in the Z-basis and outcome 0 is obtained, what is the post-measurement state?

(b) What is the probability of measuring qubit B as 1 after the measurement in part (a)?

2.10.6 Exercise 2.6: Creating Bell States

In a quantum circuit, a Bell state \(|\Phi^+\rangle\) is created from \(|00\rangle\) using:

1. Hadamard gate on qubit A: \(H|0\rangle_A = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)_A\)
2. CNOT gate with A as control and B as target

Verify that this produces \(|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)\).

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Next: Chapter 3: n-Qubit Systems - Scaling to many qubits