Tensor-Product Spaces

A single qubit lives in $\mathbb{C}^2$ . But quantum computing and most of physics involve many subsystems at once. The tensor product is the rule that combines the Hilbert spaces of individual parts into the Hilbert space of the whole — and it is where entanglement first appears.

Combining two systems

If system $A$ has Hilbert space $\mathcal{H}_A$ and system $B$ has $\mathcal{H}_B$ , the composite system $AB$ has the tensor-product space

\mathcal{H}_{AB} = \mathcal{H}_A \otimes \mathcal{H}_B.

Dimensions multiply: $\dim(\mathcal{H}_A \otimes \mathcal{H}_B) = \dim\mathcal{H}_A \cdot \dim\mathcal{H}_B$ . Two qubits therefore live in $\mathbb{C}^2 \otimes \mathbb{C}^2 = \mathbb{C}^4$ , and $n$ qubits in $\mathbb{C}^{2^n}$ . This exponential growth in dimension is the resource quantum computers exploit.

The product basis

If $\{|i\rangle_A\}$ is a basis for $\mathcal{H}_A$ and $\{|j\rangle_B\}$ for $\mathcal{H}_B$ , then the set of products $\{|i\rangle_A \otimes |j\rangle_B\}$ is a basis for the composite space. For two qubits the four basis states are

|0\rangle\otimes|0\rangle,\;\; |0\rangle\otimes|1\rangle,\;\; |1\rangle\otimes|0\rangle,\;\; |1\rangle\otimes|1\rangle,

written compactly as $|00\rangle, |01\rangle, |10\rangle, |11\rangle$ . The leftmost label is qubit 0.

Tensor product of vectors (the Kronecker product)

On column vectors the tensor product is the Kronecker product: each entry of the first vector scales the entire second vector. For two qubits,

\begin{pmatrix} a_0 \\ a_1 \end{pmatrix} \otimes \begin{pmatrix} b_0 \\ b_1 \end{pmatrix} = \begin{pmatrix} a_0 b_0 \\ a_0 b_1 \\ a_1 b_0 \\ a_1 b_1 \end{pmatrix}.

The four entries are the amplitudes of $|00\rangle, |01\rangle, |10\rangle, |11\rangle$ in that order.

Inner products factorize

The inner product on the composite space respects the factor structure:

\big(\langle a|\otimes\langle b|\big)\big(|c\rangle\otimes|d\rangle\big) = \langle a|c\rangle\,\langle b|d\rangle.

So orthonormality of the factor bases gives orthonormality of the product basis, and norms multiply for product states.

Product states versus entangled states

A composite state is a product state if it factorizes as $|\psi\rangle_A \otimes |\phi\rangle_B$ . Not every state does. The general two-qubit state

|\Psi\rangle = c_{00}|00\rangle + c_{01}|01\rangle + c_{10}|10\rangle + c_{11}|11\rangle

factorizes if and only if $c_{00}c_{11} = c_{01}c_{10}$ . When this condition fails, the state is entangled — it cannot be described as "qubit $A$ is in some state and qubit $B$ is in some state." The Bell state $\tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$ has $c_{00}c_{11} = 1/2 \ne 0 = c_{01}c_{10}$ , so it is entangled. Entanglement is impossible without the tensor-product structure.

Worked example

The state $|0\rangle \otimes |+\rangle$ leaves qubit 0 in $|0\rangle$ and puts qubit 1 in $|+\rangle = \tfrac{1}{\sqrt2}(|0\rangle + |1\rangle)$ . Taking the Kronecker product,

\begin{pmatrix} 1 \\ 0 \end{pmatrix} \otimes \frac{1}{\sqrt2}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \frac{1}{\sqrt2}\begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \end{pmatrix} = \frac{1}{\sqrt2}\big(|00\rangle + |01\rangle\big).

This is a product state, prepared by applying $H$ to qubit 1 alone.

Try it

Prepare the product state $|0\rangle \otimes |+\rangle = \tfrac{1}{\sqrt2}(|00\rangle + |01\rangle)$ on two qubits and return the circuit. The grader checks the full statevector.

Run your code to see the quantum state.