Exam Prep — Entangled or Not

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Method

The determinant test

Given any 2-qubit state, write it in the computational basis:

$$\ket{\psi} = \alpha\ket{00} + \beta\ket{01} + \gamma\ket{10} + \delta\ket{11}$$

Arrange the coefficients in a 2×2 matrix and compute the determinant:

$$M = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix} \qquad \det(M) = \alpha\delta - \beta\gamma$$

$\alpha\delta = \beta\gamma$ → product state (separable)
$\alpha\delta \neq \beta\gamma$ → entangled

That's it. One multiplication on each side, one comparison. Total exam time: under 30 seconds.

Worked Examples

Example 1 — Real coefficients

Is $\ket{\psi} = \frac{1}{2}\ket{00} + \frac{1}{2}\ket{01} + \frac{1}{2}\ket{10} + \frac{1}{2}\ket{11}$ entangled?

Step 1 — Read off coefficients $$\alpha = \tfrac{1}{2}, \quad \beta = \tfrac{1}{2}, \quad \gamma = \tfrac{1}{2}, \quad \delta = \tfrac{1}{2}$$ Step 2 — Compute products $$\alpha\delta = \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{4} \qquad \beta\gamma = \tfrac{1}{2} \cdot \tfrac{1}{2} = \tfrac{1}{4}$$ Step 3 — Compare $$\alpha\delta = \beta\gamma = \tfrac{1}{4} \implies \textbf{product state}$$

Sanity check: $\ket{\psi} = \frac{1}{\sqrt{2}}(\ket{0}+\ket{1}) \otimes \frac{1}{\sqrt{2}}(\ket{0}+\ket{1}) = \ket{+}\otimes\ket{+}$. Correct.

Example 2 — Complex coefficients

Is $\ket{\psi} = \frac{1}{2}\ket{00} + \frac{i}{2}\ket{01} + \frac{i}{2}\ket{10} - \frac{1}{2}\ket{11}$ entangled?

Step 1 — Read off coefficients $$\alpha = \tfrac{1}{2}, \quad \beta = \tfrac{i}{2}, \quad \gamma = \tfrac{i}{2}, \quad \delta = -\tfrac{1}{2}$$ Step 2 — Compute products $$\alpha\delta = \tfrac{1}{2} \cdot \bigl(-\tfrac{1}{2}\bigr) = -\tfrac{1}{4}$$ $$\beta\gamma = \tfrac{i}{2} \cdot \tfrac{i}{2} = \tfrac{i^2}{4} = -\tfrac{1}{4}$$ Step 3 — Compare $$\alpha\delta = \beta\gamma = -\tfrac{1}{4} \implies \textbf{product state}$$

Factored form: $\frac{1}{\sqrt{2}}(\ket{0}+i\ket{1}) \otimes \frac{1}{\sqrt{2}}(\ket{0}+i\ket{1}) = \ket{{+i}}\otimes\ket{{+i}}$.

Example 3 — Missing terms (Bell state)

Is $\ket{\psi} = \frac{1}{\sqrt{2}}\ket{00} + \frac{1}{\sqrt{2}}\ket{11}$ entangled?

Step 1 — Read off coefficients (missing terms = 0) $$\alpha = \tfrac{1}{\sqrt{2}}, \quad \beta = 0, \quad \gamma = 0, \quad \delta = \tfrac{1}{\sqrt{2}}$$ Step 2 — Compute products $$\alpha\delta = \tfrac{1}{\sqrt{2}} \cdot \tfrac{1}{\sqrt{2}} = \tfrac{1}{2} \qquad \beta\gamma = 0 \cdot 0 = 0$$ Step 3 — Compare $$\alpha\delta = \tfrac{1}{2} \neq 0 = \beta\gamma \implies \textbf{entangled}$$

This is $\ket{\Phi^+}$, one of the four Bell states. Any state with only $\ket{00}$ and $\ket{11}$ terms (or only $\ket{01}$ and $\ket{10}$ terms) is entangled — because one product is nonzero and the other is zero.

Watch Out

Common pitfalls

Coefficient ordering. It's $\alpha\ket{00} + \beta\ket{01} + \gamma\ket{10} + \delta\ket{11}$. The matrix is $\bigl(\begin{smallmatrix}\alpha & \beta \\ \gamma & \delta\end{smallmatrix}\bigr)$. First qubit picks the row, second picks the column. Swapping $\beta$ and $\gamma$ flips the answer.
Missing terms = 0. If the state is $\frac{1}{\sqrt{2}}\ket{01} + \frac{1}{\sqrt{2}}\ket{10}$, then $\alpha = 0$ and $\delta = 0$. Don't skip them — write them down explicitly.
Complex multiplication. Remember $i \cdot i = -1$, and $e^{i\theta} \cdot e^{i\phi} = e^{i(\theta+\phi)}$. A sign error in complex multiplication changes the answer.
Global phase doesn't matter. Multiplying the entire state by $e^{i\theta}$ multiplies both products by the same factor. The equality $\alpha\delta = \beta\gamma$ is preserved. Don't waste time removing global phase first.

If It's a Product State — How to Factor

When $\alpha\delta = \beta\gamma$, the state factors as $(a\ket{0}+b\ket{1}) \otimes (c\ket{0}+d\ket{1})$, which expands to:

ac\ket{00} + ad\ket{01} + bc\ket{10} + bd\ket{11}

So $\alpha = ac$, $\beta = ad$, $\gamma = bc$, $\delta = bd$. To find $a, b, c, d$:

Factoring recipe

Pick $a$ and $b$ from the first column: $a = \alpha$, $b = \gamma$ (up to a shared factor). Or equivalently, $a : b = \alpha : \gamma$.
Find $c$ and $d$: If $a \neq 0$, then $c = \alpha / a$ and $d = \beta / a$. If $a = 0$, use $b$: $c = \gamma / b$, $d = \delta / b$.
Normalize each factor separately so that $|a|^2 + |b|^2 = 1$ and $|c|^2 + |d|^2 = 1$.

Factoring example

Factor $\ket{\psi} = \frac{1}{2}\ket{00} + \frac{i}{2}\ket{01} + \frac{i}{2}\ket{10} - \frac{1}{2}\ket{11}$ (from Example 2).

Read off rows Row 0 (first qubit = \(\ket{0}\)): coefficients \(\frac{1}{2}, \frac{i}{2}\)  →  proportional to \((1, i)\) Row 1 (first qubit = \(\ket{1}\)): coefficients \(\frac{i}{2}, -\frac{1}{2}\)  →  proportional to \((i, -1) = i \cdot (1, i)\) Extract factors Both rows are proportional to \((1, i)\), confirming a product state. Row ratio: \(1 : i\)  →  first qubit is \(\frac{1}{\sqrt{2}}(\ket{0} + i\ket{1})\) Column pattern: \((1, i)\)  →  second qubit is \(\frac{1}{\sqrt{2}}(\ket{0} + i\ket{1})\) Result $$\ket{\psi} = \frac{1}{\sqrt{2}}(\ket{0}+i\ket{1}) \otimes \frac{1}{\sqrt{2}}(\ket{0}+i\ket{1}) = \ket{{+i}} \otimes \ket{{+i}}$$

Practice

Problem 1

Is the following state entangled or a product state?

$$\ket{\psi} = \frac{1}{\sqrt{2}}\ket{00} + \frac{1}{\sqrt{2}}\ket{01}$$

Solution

Coefficients: $\alpha = \frac{1}{\sqrt{2}}$, $\beta = \frac{1}{\sqrt{2}}$, $\gamma = 0$, $\delta = 0$.

$$\alpha\delta = \frac{1}{\sqrt{2}} \cdot 0 = 0 \qquad \beta\gamma = \frac{1}{\sqrt{2}} \cdot 0 = 0$$

$$\alpha\delta = \beta\gamma = 0 \implies \textbf{product state}$$

Factored: $\ket{0} \otimes \frac{1}{\sqrt{2}}(\ket{0}+\ket{1}) = \ket{0}\otimes\ket{+}$.

Problem 2

Is the following state entangled or a product state?

$$\ket{\psi} = \frac{1}{\sqrt{2}}\ket{01} - \frac{1}{\sqrt{2}}\ket{10}$$

Solution

Coefficients: $\alpha = 0$, $\beta = \frac{1}{\sqrt{2}}$, $\gamma = -\frac{1}{\sqrt{2}}$, $\delta = 0$.

$$\alpha\delta = 0 \cdot 0 = 0 \qquad \beta\gamma = \frac{1}{\sqrt{2}} \cdot \bigl(-\frac{1}{\sqrt{2}}\bigr) = -\frac{1}{2}$$

$$\alpha\delta = 0 \neq -\frac{1}{2} = \beta\gamma \implies \textbf{entangled}$$

This is $\ket{\Psi^-}$, one of the four Bell states.

Problem 3

Is the following state entangled or a product state?

$$\ket{\psi} = \frac{1}{2}\ket{00} - \frac{i}{2}\ket{01} + \frac{i}{2}\ket{10} + \frac{1}{2}\ket{11}$$

Solution

Coefficients: $\alpha = \frac{1}{2}$, $\beta = -\frac{i}{2}$, $\gamma = \frac{i}{2}$, $\delta = \frac{1}{2}$.

$$\alpha\delta = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$

$$\beta\gamma = \bigl(-\frac{i}{2}\bigr)\bigl(\frac{i}{2}\bigr) = -\frac{i^2}{4} = -\frac{(-1)}{4} = \frac{1}{4}$$

$$\alpha\delta = \beta\gamma = \frac{1}{4} \implies \textbf{product state}$$

Factored: $\frac{1}{\sqrt{2}}(\ket{0}+i\ket{1}) \otimes \frac{1}{\sqrt{2}}(\ket{0}-i\ket{1}) = \ket{{+i}}\otimes\ket{{-i}}$.

Problem 4

Is the following state entangled or a product state?

$$\ket{\psi} = \frac{1}{2}\ket{00} + \frac{1}{2}\ket{01} + \frac{1}{2}\ket{10} - \frac{1}{2}\ket{11}$$

Solution

Coefficients: $\alpha = \frac{1}{2}$, $\beta = \frac{1}{2}$, $\gamma = \frac{1}{2}$, $\delta = -\frac{1}{2}$.

$$\alpha\delta = \frac{1}{2} \cdot \bigl(-\frac{1}{2}\bigr) = -\frac{1}{4}$$

$$\beta\gamma = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$

$$\alpha\delta = -\frac{1}{4} \neq \frac{1}{4} = \beta\gamma \implies \textbf{entangled}$$

Rewriting: $\frac{1}{\sqrt{2}}(\ket{0}\ket{+} + \ket{1}\ket{-})$. This is a CZ-entangled state — the second qubit's basis depends on the first.

Problem 5

Is the following state entangled or a product state?

$$\ket{\psi} = \frac{1+i}{2\sqrt{2}}\ket{00} + \frac{1+i}{2\sqrt{2}}\ket{01} + \frac{1-i}{2\sqrt{2}}\ket{10} + \frac{1-i}{2\sqrt{2}}\ket{11}$$

Solution

Coefficients: $\alpha = \frac{1+i}{2\sqrt{2}}$, $\beta = \frac{1+i}{2\sqrt{2}}$, $\gamma = \frac{1-i}{2\sqrt{2}}$, $\delta = \frac{1-i}{2\sqrt{2}}$.

$$\alpha\delta = \frac{(1+i)(1-i)}{(2\sqrt{2})^2} = \frac{1 - i^2}{8} = \frac{1-(-1)}{8} = \frac{2}{8} = \frac{1}{4}$$

$$\beta\gamma = \frac{(1+i)(1-i)}{(2\sqrt{2})^2} = \frac{2}{8} = \frac{1}{4}$$

$$\alpha\delta = \beta\gamma = \frac{1}{4} \implies \textbf{product state}$$

Factored: note that $\frac{1+i}{\sqrt{2}} = e^{i\pi/4}$ and $\frac{1-i}{\sqrt{2}} = e^{-i\pi/4}$.

$$\ket{\psi} = \frac{1}{\sqrt{2}}\bigl(e^{i\pi/4}\ket{0} + e^{-i\pi/4}\ket{1}\bigr) \otimes \frac{1}{\sqrt{2}}\bigl(\ket{0}+\ket{1}\bigr)$$

The first qubit is a T-rotated $\ket{+}$ state. The second qubit is $\ket{+}$.