Checkpoint: Statistics of a Qubit

A qubit is not simply 0 or 1 — it is a superposition whose measurement statistics can be fully predicted from the state vector. This checkpoint asks you to apply the probability, mean, and variance formulas you have built up in Module 3 directly to a qubit state.

From amplitudes to probabilities

A general single-qubit state is

$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle,$

with $\alpha, \beta \in \mathbb{C}$ and the normalisation condition $|\alpha|^2 + |\beta|^2 = 1$. When the qubit is measured in the computational basis, the Born rule gives

$P(0) = |\alpha|^2, \qquad P(1) = |\beta|^2.$

These are the only two possible outcomes; they form a valid probability distribution because they are non-negative and sum to one.

Eigenvalues and the measurement random variable

To compute a mean and variance we need numerical eigenvalues. A standard convention assigns eigenvalue $0$ to outcome $|0\rangle$ and eigenvalue $1$ to outcome $|1\rangle$, so the measurement defines a random variable $X$ with distribution

$P(X = 0) = |\alpha|^2, \qquad P(X = 1) = |\beta|^2.$

Mean. By definition,

$\langle X \rangle = 0 \cdot |\alpha|^2 + 1 \cdot |\beta|^2 = |\beta|^2.$

The mean equals the probability of measuring $|1\rangle$ — a useful shortcut for this particular eigenvalue assignment.

Mean square. Because $0^2 = 0$ and $1^2 = 1$,

$\langle X^2 \rangle = 0^2 \cdot |\alpha|^2 + 1^2 \cdot |\beta|^2 = |\beta|^2.$

Note that $\langle X^2 \rangle = \langle X \rangle$ for this binary variable with eigenvalues $0$ and $1$.

Variance. Applying $\operatorname{Var}(X) = \langle X^2 \rangle - \langle X \rangle^2$,

$\operatorname{Var}(X) = |\beta|^2 - |\beta|^4 = |\beta|^2\bigl(1 - |\beta|^2\bigr) = |\beta|^2\,|\alpha|^2.$

This is maximised at $|\alpha|^2 = |\beta|^2 = 1/2$ (the uniform superposition, e.g. $H|0\rangle$), where $\operatorname{Var}(X) = 1/4$, and it vanishes when the state is either $|0\rangle$ or $|1\rangle$ — a perfectly predictable outcome carries zero variance.

A worked example

Consider $|\psi\rangle = \tfrac{1}{\sqrt{3}}|0\rangle + \sqrt{\tfrac{2}{3}}|1\rangle$. The amplitudes give

$|\alpha|^2 = \frac{1}{3}, \qquad |\beta|^2 = \frac{2}{3}.$

Check normalisation: $\tfrac{1}{3} + \tfrac{2}{3} = 1$. Then

$\langle X \rangle = \frac{2}{3},$

$\operatorname{Var}(X) = \frac{2}{3} \cdot \frac{1}{3} = \frac{2}{9} \approx 0.222.$

The standard deviation is $\sigma = \sqrt{2/9} = \sqrt{2}/3 \approx 0.471$.

Try it

This is a numerical exercise — your code should return a number, not a circuit. The qubit is in state $|\psi\rangle = \tfrac{1}{\sqrt{3}}|0\rangle + \sqrt{\tfrac{2}{3}}|1\rangle$. Eigenvalues are $0$ for $|0\rangle$ and $1$ for $|1\rangle$. Compute the variance of a single measurement outcome.

Sign in on the full site to ask questions and join the discussion.