Product vs Entangled: A Test

Not every two-qubit state that looks complicated is entangled. The key question is whether the state can be factored into independent parts for each qubit — what physicists call a product state.

The separability criterion

A general two-qubit pure state can be written

where $|\alpha|^2 + |\beta|^2 + |\gamma|^2 + |\delta|^2 = 1$. The state is a product state — meaning qubit 0 and qubit 1 are completely independent — if and only if it can be factored as $(a|0\rangle + b|1\rangle)\otimes(c|0\rangle + d|1\rangle)$. Expanding that product gives $ac|00\rangle + ad|01\rangle + bc|10\rangle + bd|11\rangle$, so we need $\alpha = ac,\ \beta = ad,\ \gamma = bc,\ \delta = bd$. Multiplying the corner pair gives $\alpha\delta = abcd$; multiplying the other pair gives $\beta\gamma = abcd$. Therefore:

Equivalently, the $2\times 2$ coefficient matrix $\begin{pmatrix}\alpha & \beta \\ \gamma & \delta\end{pmatrix}$ has determinant zero. If the determinant is non-zero, the state is entangled.

Two worked examples

Example 1 — product state. Consider

Here $\alpha = \beta = \gamma = \delta = \tfrac{1}{2}$. Then $\alpha\delta = \tfrac{1}{4}$ and $\beta\gamma = \tfrac{1}{4}$. Because they match, this is a product state — indeed $|{+}{+}\rangle = |+\rangle \otimes |+\rangle$.

Example 2 — entangled state. The Bell state

has $\alpha = \delta = \tfrac{1}{\sqrt{2}}$ and $\beta = \gamma = 0$. Then $\alpha\delta = \tfrac{1}{2}$ but $\beta\gamma = 0$. The determinant is $\tfrac{1}{2} \neq 0$, so the state is entangled: there is no way to write it as $|\psi\rangle_0 \otimes |\phi\rangle_1$.

Try it

Apply the determinant test to the Bell state $|\Phi^+\rangle = \tfrac{1}{\sqrt{2}}(|00\rangle + |11\rangle)$. Return true if the state is entangled, or false if it is a product state.

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