Part IX, Chapter 3 | Page 1 of 3

Quantum Entanglement Fundamentals

Product states, Bell states, Schmidt decomposition, and entanglement entropy

Product States vs. Entangled States

Consider a composite quantum system consisting of two subsystems A and B, with Hilbert spaces$\mathcal{H}_A$ and $\mathcal{H}_B$. The joint Hilbert space is the tensor product $\mathcal{H}_{AB} = \mathcal{H}_A \otimes \mathcal{H}_B$. A state of this composite system is called a product state (or separable state) if it can be written as:

$$|\psi\rangle_{AB} = |\phi\rangle_A \otimes |\chi\rangle_B$$

In a product state, each subsystem has a definite individual quantum state. Measuring A tells us nothing about B beyond what we already know.

A state is entangled if it cannot be written as a product state. For example, consider two qubits (two-level systems with basis $\{|0\rangle, |1\rangle\}$):

$$|\psi\rangle = \frac{1}{\sqrt{2}}\left(|0\rangle_A|1\rangle_B + |1\rangle_A|0\rangle_B\right)$$

This state cannot be factored into $|\phi\rangle_A \otimes |\chi\rangle_B$ for any choice of $|\phi\rangle$ and $|\chi\rangle$. The two qubits are entangled.

For mixed states, the definition extends: a density matrix $\hat{\rho}_{AB}$ is separable if it can be written as a convex combination of product states:

$$\hat{\rho}_{AB} = \sum_i p_i\; \hat{\rho}_A^{(i)} \otimes \hat{\rho}_B^{(i)}, \qquad p_i \geq 0,\;\sum_i p_i = 1$$

Determining whether a given mixed state is separable is, in general, computationally hard (NP-hard for large systems).

Bell States: Maximally Entangled Pairs

The four Bell states form an orthonormal basis for the two-qubit Hilbert space consisting entirely of maximally entangled states:

$$|\Phi^+\rangle = \frac{1}{\sqrt{2}}\left(|00\rangle + |11\rangle\right), \qquad |\Phi^-\rangle = \frac{1}{\sqrt{2}}\left(|00\rangle - |11\rangle\right)$$

$$|\Psi^+\rangle = \frac{1}{\sqrt{2}}\left(|01\rangle + |10\rangle\right), \qquad |\Psi^-\rangle = \frac{1}{\sqrt{2}}\left(|01\rangle - |10\rangle\right)$$

Key properties of the Bell states:

Orthonormality: $\langle\Phi^+|\Phi^-\rangle = \langle\Phi^+|\Psi^\pm\rangle = 0$, etc. They form the Bell basis.
Maximal entanglement: The reduced density matrix of either qubit is $\hat{\rho}_A = \hat{\rho}_B = \frac{1}{2}\mathbb{I}$ (maximally mixed).
Perfect correlations: For $|\Phi^+\rangle$, measuring qubit A in the computational basis always gives the same result as qubit B.
Interconvertibility: Any Bell state can be obtained from any other by local unitary operations on one qubit (Pauli gates).

The singlet state $|\Psi^-\rangle$ is special: it is the unique state invariant under identical rotations applied to both qubits ($U \otimes U|\Psi^-\rangle = |\Psi^-\rangle$ for any $U \in SU(2)$). This makes it the natural state for testing Bell inequalities.

Schmidt Decomposition

The Schmidt decomposition theorem provides a canonical form for any pure bipartite state and is the fundamental tool for analyzing entanglement.

Theorem (Schmidt Decomposition): For any pure state$|\psi\rangle_{AB} \in \mathcal{H}_A \otimes \mathcal{H}_B$, there exist orthonormal bases$\{|i\rangle_A\}$ for $\mathcal{H}_A$ and $\{|i\rangle_B\}$ for$\mathcal{H}_B$ such that:

$$\boxed{|\psi\rangle_{AB} = \sum_{i=1}^{r} \sqrt{\lambda_i}\; |i\rangle_A \otimes |i\rangle_B}$$

where $\lambda_i > 0$, $\sum_i \lambda_i = 1$, and $r \leq \min(\dim\mathcal{H}_A, \dim\mathcal{H}_B)$.

Proof Sketch

Expand $|\psi\rangle_{AB}$ in arbitrary orthonormal bases: $|\psi\rangle = \sum_{jk} c_{jk}|j\rangle_A|k\rangle_B$. The coefficient matrix $C = (c_{jk})$ admits a singular value decomposition$C = UDV^\dagger$ where $D = \text{diag}(\sqrt{\lambda_1}, \sqrt{\lambda_2}, \ldots)$. Defining $|i\rangle_A = \sum_j U_{ji}|j\rangle_A$ and $|i\rangle_B = \sum_k V_{ki}^*|k\rangle_B$yields the Schmidt form. The $\lambda_i$ are the singular values squared, uniquely determined up to ordering.

Schmidt Number as Entanglement Measure

The Schmidt number (or Schmidt rank) $r$ — the number of non-zero Schmidt coefficients — provides a basic classification of entanglement:

$r = 1$: The state is a product state (not entangled)
$r > 1$: The state is entangled
$r = d = \min(\dim\mathcal{H}_A, \dim\mathcal{H}_B)$ with all $\lambda_i = 1/d$: Maximally entangled

Density Matrix of a Subsystem

Given a pure state $|\psi\rangle_{AB}$ of the composite system, the state of subsystem A alone is obtained by tracing out subsystem B:

$$\hat{\rho}_A = \text{Tr}_B\!\left(|\psi\rangle_{AB}\langle\psi|\right) = \sum_k \langle k|_B\left(|\psi\rangle_{AB}\langle\psi|\right)|k\rangle_B$$

Using the Schmidt decomposition, this becomes diagonal in the Schmidt basis:

$$\hat{\rho}_A = \sum_{i=1}^{r} \lambda_i\; |i\rangle_A\langle i|, \qquad \hat{\rho}_B = \sum_{i=1}^{r} \lambda_i\; |i\rangle_B\langle i|$$

Crucially, $\hat{\rho}_A$ and $\hat{\rho}_B$ share the same eigenvalues $\{\lambda_i\}$. If the state is entangled ($r > 1$), both reduced states are mixed even though the global state is pure. This is the hallmark of entanglement.

Entanglement Entropy

The entanglement entropy quantifies the degree of entanglement in a pure bipartite state using the von Neumann entropy of the reduced density matrix:

$$\boxed{S(\hat{\rho}_A) = -\text{Tr}\!\left(\hat{\rho}_A \ln\hat{\rho}_A\right) = -\sum_{i=1}^{r} \lambda_i \ln\lambda_i}$$

Fundamental properties:

Symmetry: $S(\hat{\rho}_A) = S(\hat{\rho}_B)$ for any pure bipartite state (since $\hat{\rho}_A$ and $\hat{\rho}_B$ share eigenvalues)
Zero iff separable: $S = 0$ if and only if $|\psi\rangle_{AB}$ is a product state
Maximum: $S = \ln d$ for a maximally entangled state in $d \times d$ dimensions
Concavity: Entanglement entropy is a concave function of $\hat{\rho}_A$
Monotonicity under LOCC: Local operations and classical communication cannot increase entanglement entropy (on average)

Example: Bell State Entanglement

For the Bell state $|\Phi^+\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$, the Schmidt coefficients are $\lambda_1 = \lambda_2 = 1/2$, so:

$$\hat{\rho}_A = \frac{1}{2}|0\rangle\langle 0| + \frac{1}{2}|1\rangle\langle 1| = \frac{\mathbb{I}}{2}$$

$$S(\hat{\rho}_A) = -\frac{1}{2}\ln\frac{1}{2} - \frac{1}{2}\ln\frac{1}{2} = \ln 2$$

This is the maximum possible entanglement for two qubits. The reduced state is maximally mixed — measuring qubit A alone gives completely random results.

Renyi Entropies

A family of entanglement measures generalizing the von Neumann entropy:

$$S_\alpha(\hat{\rho}_A) = \frac{1}{1-\alpha}\ln\text{Tr}\!\left(\hat{\rho}_A^\alpha\right) = \frac{1}{1-\alpha}\ln\!\left(\sum_i \lambda_i^\alpha\right)$$

The von Neumann entropy is recovered as $S_1 = \lim_{\alpha\to 1}S_\alpha$. The Renyi-2 entropy$S_2 = -\ln\text{Tr}(\hat{\rho}_A^2)$ is often easier to measure experimentally.

Purity and the Entanglement Connection

The purity of the reduced density matrix provides another way to detect entanglement. For a subsystem with density matrix $\hat{\rho}_A$:

$$\gamma(\hat{\rho}_A) = \text{Tr}\!\left(\hat{\rho}_A^2\right) = \sum_i \lambda_i^2$$

Range: $1/d \leq \gamma \leq 1$ where $d = \dim\mathcal{H}_A$. If $\gamma = 1$, the subsystem is in a pure state (no entanglement). If $\gamma = 1/d$, the subsystem is maximally mixed (maximum entanglement).

For the global pure state $|\psi\rangle_{AB}$, there is a direct relationship:

$$\text{Tr}\!\left(\hat{\rho}_A^2\right) = \text{Tr}\!\left(\hat{\rho}_B^2\right) = \sum_i \lambda_i^2$$

Both subsystems have the same purity. The linear entropy$S_L = 1 - \text{Tr}(\hat{\rho}_A^2) = 1 - \sum_i\lambda_i^2$ provides a simpler (though less fine-grained) entanglement measure than the von Neumann entropy.

Detecting Entanglement: Witnesses and the PPT Criterion

In practice, determining whether a given state is entangled can be challenging. Two important tools:

PPT (Positive Partial Transpose) Criterion:

$$\hat{\rho}_{AB}^{T_B} \geq 0 \quad \Longleftarrow \quad \hat{\rho}_{AB} \text{ is separable}$$

If the partial transpose has any negative eigenvalue, the state is definitely entangled. For $2\times 2$ and $2\times 3$ systems, PPT is both necessary and sufficient (Horodecki theorem).

Entanglement Witness:

$$\hat{W}: \quad \text{Tr}(\hat{W}\hat{\rho}_{\text{sep}}) \geq 0 \;\;\forall\text{ separable states}, \qquad \text{Tr}(\hat{W}\hat{\rho}_{\text{ent}}) < 0 \;\;\text{for some entangled state}$$

An entanglement witness is an observable that detects entanglement experimentally. For every entangled state, there exists a witness that detects it (by the Hahn-Banach theorem applied to convex sets).

Historical Note: Schrodinger coined the term "Verschrankung" (entanglement) in 1935, calling it "the characteristic trait of quantum mechanics, the one that enforces its entire departure from classical lines of thought." Einstein, together with Podolsky and Rosen, used entanglement that same year to argue that quantum mechanics must be incomplete — a debate that would not be experimentally resolved for another 50 years.

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