Mathematical and Physical Foundations of Quantum Information

The Schmidt decomposition stands as one of the most powerful analytical tools for characterizing entanglement in bipartite quantum systems, bridging the abstract structure of Hilbert spaces with concrete, measurable properties of composite quantum states. For any pure state $|\psi\rangle_{AB}$ living in the tensor product Hilbert space $\mathcal{H}_A \otimes \mathcal{H}_B$ with $\dim(\mathcal{H}_A) = d_A$ and $\dim(\mathcal{H}_B) = d_B$, there always exist orthonormal bases $\{|i\rangle_A\}_{i=1}^{d_A}$ for subsystem $A$ and $\{|j\rangle_B\}_{j=1}^{d_B}$ for subsystem $B$ such that the state can be rewritten in the diagonal Schmidt form: $$|\psi\rangle_{AB} = \sum_{k=1}^{\min(d_A, d_B)} \sqrt{\lambda_k} |k\rangle_A \otimes |k\rangle_B$$ where the Schmidt coefficients $\{\sqrt{\lambda_k}\}$ are non-negative real numbers satisfying $\sum_k \lambda_k = 1$, and the number of non-zero Schmidt coefficients is called the Schmidt rank of the state. The Schmidt rank is a fundamental invariant of pure-state entanglement: a state has Schmidt rank 1 if and only if it is a product state (separable, no entanglement), while states with Schmidt rank greater than 1 are entangled. For the Bell state $|\Phi^+\rangle = (|00\rangle + |11\rangle)/\sqrt{2}$, the Schmidt decomposition is immediate: $\lambda_1 = \lambda_2 = 1/2$, so the Schmidt rank is 2, confirming it is maximally entangled for two qubits.

To compute the Schmidt decomposition for an arbitrary pure state, one can perform a singular value decomposition (SVD) on the coefficient matrix of the state. For a two-qubit state written as $|\psi\rangle = \sum_{i,j=0}^1 c_{ij} |ij\rangle$, the $2\times2$ complex matrix $C$ with entries $c_{ij}$ has SVD $C = U \text{diag}(\sqrt{\lambda_1}, \sqrt{\lambda_2}) V^\dagger$, where $U$ and $V$ are unitary matrices and $\lambda_k$ are the squared singular values. The Schmidt basis for subsystem $A$ is given by the columns of $U$, and the basis for subsystem $B$ by the columns of $V$. For example, consider the two-qubit state $|\psi\rangle = 0.6|00\rangle + 0.8|11\rangle$: the coefficient matrix is $C = \begin{pmatrix} 0.6 & 0 \\ 0 & 0.8 \end{pmatrix}$, which is already diagonal, so the Schmidt coefficients are $0.6$ and $0.8$, with Schmidt rank 2. If we instead take $|\psi\rangle = (|00\rangle + |01\rangle + |10\rangle + |11\rangle)/2$, the coefficient matrix is $C = \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}$, which has singular values $1$ and $0$, so the Schmidt rank is 1, confirming this is a product state: $|\psi\rangle = |+\rangle \otimes |+\rangle$ where $|+\rangle = (|0\rangle + |1\rangle)/\sqrt{2}$.

```python import numpy as np

def schmidt_decomposition(state_vector, dim_A=2, dim_B=2): """ Compute Schmidt decomposition of a bipartite pure state. Args: state_vector: 1D numpy array of length dim_A * dim_B dim_A: dimension of subsystem A dim_B: dimension of subsystem B Returns: schmidt_coeffs: list of non-negative Schmidt coefficients basis_A: orthonormal basis for subsystem A (columns of U) basis_B: orthonormal basis for subsystem B (columns of V) """

Reshape state vector into coefficient matrix

C = state_vector.reshape(dim_A, dim_B)

Perform singular value decomposition

U, S, Vh = np.linalg.svd(C)

Schmidt coefficients are singular values, normalized to sum to 1

schmidt_coeffs = S / np.linalg.norm(S) return schmidt_coeffs, U, Vh.T

Example 1: Bell state |Φ+> = (|00> + |11>)/√2

bell_state = np.array([1/np.sqrt(2), 0, 0, 1/np.sqrt(2)]) coeffs, basis_A, basis_B = schmidt_decomposition(bell_state) print("Bell state Schmidt coefficients:", coeffs) # Output: [0.7071 0.7071], Schmidt rank 2

Example 2: Product state |+>⊗|+>

plus_state = np.array([0.5, 0.5, 0.5, 0.5]) coeffs2, _, _ = schmidt_decomposition(plus_state) print("Product state Schmidt coefficients:", coeffs2) # Output: [1. 0.], Schmidt rank 1

Example 3: Non-maximally entangled state 0.6|00> + 0.8|11>

entangled_state = np.array([0.6, 0, 0, 0.8]) coeffs3, _, _ = schmidt_decomposition(entangled_state) print("Non-maximal entangled state Schmidt coefficients:", coeffs3) # Output: [0.6 0.8], Schmidt rank 2 ```

For mixed states, the Schmidt decomposition does not apply directly, but we can use the concept of partial trace to characterize entanglement between subsystems. The partial trace over subsystem $B$, denoted $\text{Tr}_B(\rho_{AB})$, is an operation that removes the degrees of freedom of subsystem $B$ from the composite density matrix, yielding the reduced density matrix of subsystem $A$. Mathematically, for an orthonormal basis $\{|j\rangle_B\}$ of subsystem $B$, the partial trace is defined as: $$\text{Tr}_B(\rho_{AB}) = \sum_j (\mathbb{I}_A \otimes \langle j|) \rho_{AB} (\mathbb{I}_A \otimes |j\rangle)$$ If the composite state $\rho_{AB}$ is separable (i.e., $\rho_{AB} = \sum_k p_k \rho_A^k \otimes \rho_B^k$), then the reduced density matrix $\rho_A = \text{Tr}_B(\rho_{AB})$ is a valid quantum state of subsystem $A$ independent of measurements performed on $B$. For entangled states, however, the reduced density matrix is generally mixed, even if the composite state is pure. For the Bell state $|\Phi^+\rangle$, the composite density matrix is $\rho_{AB} = |\Phi^+\rangle\langle\Phi^+|$, and the reduced density matrix of subsystem $A$ is: $$\rho_A = \text{Tr}_B(\rho_{AB}) = \sum_{j=0}^1 \langle j| \rho_{AB} |j\rangle = \frac{1}{2}|0\rangle\langle0| + \frac{1}{2}|1\rangle\langle1| = \frac{\mathbb{I}}{2}$$ This is the maximally mixed state, with no pure quantum coherence: any measurement on subsystem $A$ will yield outcomes with equal probability $1/2$, regardless of the basis chosen. This is a hallmark of entanglement: the subsystems of an entangled pure state are always in mixed states, with no local coherence.

The degree of mixing of the reduced density matrix can be quantified using the von Neumann entropy, which generalizes the classical Shannon entropy to quantum states. For a density matrix $\rho$ with eigenvalues $\{\lambda_i\}$, the von Neumann entropy is defined as: $$S(\rho) = -\text{Tr}(\rho \log_2 \rho) = -\sum_i \lambda_i \log_2 \lambda_i$$ For a pure state, $\rho$ has a single eigenvalue 1 and all others 0, so $S(\rho) = 0$. For the maximally mixed state of a $d$-dimensional system, all eigenvalues are $1/d$, so $S(\rho) = \log_2 d$. For bipartite pure states, the von Neumann entropy of the reduced density matrix of either subsystem is exactly the entanglement entropy of the state, a measure of the amount of entanglement between the two subsystems. For the Bell state, $S(\rho_A) = S(\rho_B) = \log_2 2 = 1$ ebit, the maximum possible entanglement for two qubits. For the non-maximally entangled state $0.6|00\rangle + 0.8|11\rangle$, the reduced density matrix is $\rho_A = 0.36|0\rangle\langle0| + 0.64|1\rangle\langle1|$, so the entanglement entropy is $S(\rho_A) = -0.36\log_2 0.36 - 0.64\log_2 0.64 \approx 0.97$ ebit, slightly less than the maximum.

> 📖 *Sidebar (Print Edition: Boxed Highlight; Digital Edition: Hover Note)* > Entanglement vs. Classical Correlation > A common misconception is that all correlated states are entangled. Consider the classical correlated state $\rho = 0.5|00\rangle\langle00| + 0.5|11\rangle\langle11|$, which has the same measurement statistics as the Bell state when measuring both qubits in the computational basis: both yield (0,0) or (1,1) with equal probability. However, this state is separable (it is a classical mixture of product states), and its reduced density matrix for subsystem $A$ is also $\mathbb{I}/2$, identical to the Bell state. The key difference is revealed by measurements in non-computational bases: for the Bell state, measuring both qubits in the $X$ basis yields perfectly correlated outcomes, while for the classical correlated state, the outcomes are uncorrelated in the $X$ basis. Entanglement is a stronger form of correlat