Tensor Products and Composite Systems

When two quantum systems A and B are combined, their state space is the tensor product $\mathcal{H}_A \otimes \mathcal{H}_B$. If $\dim(\mathcal{H}_A) = m$ and $\dim(\mathcal{H}_B) = n$, then $\dim(\mathcal{H}_A \otimes \mathcal{H}_B) = mn$. The tensor product is the natural way to build the Hilbert space of a composite system.

A basis for the tensor product is obtained by taking all pairwise tensor products of basis kets. If $\{|i\rangle_A\}$ and $\{|j\rangle_B\}$ are bases, then $\{|i\rangle_A \otimes |j\rangle_B\}$ is a basis for $\mathcal{H}_A \otimes \mathcal{H}_B$. We often omit the tensor symbol and write $|i\rangle_A |j\rangle_B$ or $|ij\rangle$.

The tensor product is bilinear: it distributes over addition and scalar multiplication in each factor. However, not every state in the tensor product space can be written as a single product of states from the factors. Those that cannot are entangled.

Tensor products appear throughout quantum mechanics: combining the spin and position of a particle, combining multiple qubits, and combining a system with its environment. Mastering tensor-product arithmetic is essential for quantum computing and many-body physics.

A product state has the form $|\Psi\rangle = |\psi\rangle_A \otimes |\phi\rangle_B$. Measurements on system A depend only on $|\psi\rangle_A$, and measurements on system B depend only on $|\phi\rangle_B$. There are no correlations between the two systems beyond those that follow from independent local states.

If $|\psi\rangle_A = \sum_i a_i |i\rangle_A$ and $|\phi\rangle_B = \sum_j b_j |j\rangle_B$, then the product state is $|\Psi\rangle = \sum_{i,j} a_i b_j |ij\rangle$. The coefficient of $|ij\rangle$ factorizes as the product $a_i b_j$. This factorization is the defining feature of a product state.

Product states are easy to manipulate. The inner product of two product states factorizes: $(\langle\psi|_A \otimes \langle\phi|_B)(|\chi\rangle_A \otimes |\xi\rangle_B) = \langle\psi|\chi\rangle_A \langle\phi|\xi\rangle_B$. Operator actions also factorize when the operator is itself a tensor product.

The set of product states spans the tensor product space, but most states are not product states. This abundance of non-product states is what gives composite quantum systems their enormously larger state space compared to classical composite systems.

A state of a composite system is entangled if it cannot be written as a product state. Entangled states exhibit correlations that have no classical explanation. The most famous examples are the Bell states of two qubits.

The Bell state $|\Phi^+\rangle = (|00\rangle + |11\rangle)/\sqrt{2}$ is entangled. There are no single-qubit amplitudes $a_0, a_1, b_0, b_1$ such that $(a_0|0\rangle + a_1|1\rangle)(b_0|0\rangle + b_1|1\rangle)$ equals $|\Phi^+\rangle$, because the factorization would require $a_0 b_1 = 0$ and $a_1 b_0 = 0$ simultaneously with $a_0 b_0 = a_1 b_1 = 1/\sqrt{2}$, which is impossible.

Entanglement is a resource for quantum computation and communication. It enables quantum teleportation, superdense coding, and the violation of Bell inequalities. Quantifying entanglement is a central topic in quantum information theory.

A quick test for pure bipartite states is the Schmidt decomposition: any bipartite pure state can be written $\sum_k s_k |u_k\rangle_A |v_k\rangle_B$ with nonnegative Schmidt coefficients $s_k$. The state is entangled if more than one Schmidt coefficient is nonzero.

If $A$ acts on $\mathcal{H}_A$ and $B$ acts on $\mathcal{H}_B$, their tensor product $A \otimes B$ acts on $\mathcal{H}_A \otimes \mathcal{H}_B$ by $(A \otimes B)(|\psi\rangle_A \otimes |\phi\rangle_B) = (A|\psi\rangle_A) \otimes (B|\phi\rangle_B)$. The action is extended by linearity to all states in the tensor product space.

The matrix representation of $A \otimes B$ is the Kronecker product of the matrices of $A$ and $B$. If $A$ is m x m and $B$ is n x n, then $A \otimes B$ is mn x mn. Its entries satisfy $(A \otimes B)_{(i,j),(k,l)} = A_{ik} B_{jl}$.

A local operator on subsystem A only is written $A \otimes I_B$, and a local operator on B is $I_A \otimes B$. A general operator on the composite system may be a sum of tensor products, such as $H = H_A \otimes I_B + I_A \otimes H_B + H_{\text{int}}$, where $H_{\text{int}}$ represents interactions.

Tensor-product operators inherit properties from their factors. For example, $(A \otimes B)^\dagger = A^\dagger \otimes B^\dagger$, and if $A$ and $B$ are unitary then so is $A \otimes B$. These rules are used constantly when analyzing multi-qubit circuits.

For n qubits, the computational basis consists of the $2^n$ states $|x_1 x_2 \dots x_n\rangle$, where each $x_i$ is 0 or 1. A general n-qubit pure state is a normalized superposition $\sum_{x\in\{0,1\}^n} c_x |x\rangle$ with $\sum_x |c_x|^2 = 1$.

Dirac notation compresses these expressions dramatically. The two-qubit state $(|00\rangle + |01\rangle + |10\rangle + |11\rangle)/2$ is the uniform superposition over all four basis states. It can be written as $|+\rangle^{\otimes 2}$, the tensor product of two $|+\rangle$ states.

Operators on n qubits are $2^n$ x $2^n$ matrices. Common gates include single-qubit gates acting on one factor, such as $X \otimes I$, and two-qubit entangling gates such as CNOT. The CNOT gate flips the second qubit conditioned on the first.

The exponential growth of dimension with the number of qubits is the source of both the power and the difficulty of quantum computing. A classical description of an arbitrary n-qubit state requires $2^n$ complex amplitudes.

The four Bell states form an orthonormal basis for the two-qubit Hilbert space and are maximally entangled. They are defined as $|\Phi^\pm\rangle = (|00\rangle \pm |11\rangle)/\sqrt{2}$ and $|\Psi^\pm\rangle = (|01\rangle \pm |10\rangle)/\sqrt{2}$.

In each Bell state, measuring one qubit in the computational basis immediately determines the outcome of the other qubit. For $|\Phi^+\rangle$, both measurements yield the same bit; for $|\Psi^-\rangle$, they yield opposite bits. These perfect correlations persist no matter how far apart the qubits are.

The Bell states are eigenstates of the operators $\sigma_x \otimes \sigma_x$ and $\sigma_z \otimes \sigma_z$. They violate Bell inequalities, proving that no local hidden-variable theory can reproduce all quantum predictions.

Bell states are created from the computational basis by applying a Hadamard gate to the first qubit followed by a CNOT. This simple circuit is the prototype for generating entanglement in quantum information processing.

Start with two qubits in $|00\rangle$. Apply a Hadamard gate to the first qubit and then a CNOT controlled by the first qubit. Show that the resulting state is $|\Phi^+\rangle = (|00\rangle + |11\rangle)/\sqrt{2}$.

Step 1 -- Hadamard on first qubit. $H|0\rangle = (|0\rangle + |1\rangle)/\sqrt{2}$. Since the second qubit is unchanged, the two-qubit state becomes $(|00\rangle + |10\rangle)/\sqrt{2}$.

Step 2 -- Apply CNOT. The CNOT gate flips the second qubit when the first is $|1\rangle$. Thus $|00\rangle \mapsto |00\rangle$ and $|10\rangle \mapsto |11\rangle$. The state becomes $(|00\rangle + |11\rangle)/\sqrt{2}$.

Step 3 -- Identify. This is exactly the Bell state $|\Phi^+\rangle$. The circuit has transformed a product state into an entangled state using only one single-qubit gate and one two-qubit gate.