Distributions of Measurement Outcomes

Every time you measure a qubit you get a definite result — either $0$ or $1$ — but which one you get is fundamentally random. Quantum mechanics does not hide a secret predetermined answer; the randomness is intrinsic. What the theory provides instead is a complete probability distribution over the possible outcomes, derived from the state of the system.

The qubit as a two-outcome distribution

A qubit in a pure state is written

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

where $\alpha$ and $\beta$ are complex numbers called amplitudes. When you measure in the computational basis $\{|0\rangle, |1\rangle\}$ , there are exactly two possible outcomes. The Born rule assigns probabilities to them:

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

Because one outcome must occur, these probabilities must sum to one:

$|\alpha|^2 + |\beta|^2 = 1.$

This is the normalisation condition — it is not an extra assumption but a requirement baked into how quantum states are defined. A properly normalised state automatically gives a valid probability distribution.

Writing it as a distribution table

The measurement outcome is a random variable $M$ that takes the values $0$ and $1$ . We can write its distribution as a table:

| Outcome $m$ | Probability $P(M = m)$ | |:-----------:|:---------------------:| | $0$ | $|\alpha|^2$ | | $1$ | $|\beta|^2$ |

This two-row table is the complete probability distribution of a single qubit measurement. Every statistical quantity — expectation value, variance, entropy — can be computed from it using the same rules that apply to any discrete distribution.

For example, the expectation value of the measurement outcome (treating $0$ and $1$ as numbers) is

The mean is just the probability of the outcome $1$ , a fact worth remembering.

A worked example

Suppose $\alpha = \tfrac{1}{\sqrt{2}}$ and $\beta = \tfrac{i}{\sqrt{2}}$ . Check normalisation:

$|\alpha|^2 + |\beta|^2 = \left|\tfrac{1}{\sqrt{2}}\right|^2 + \left|\tfrac{i}{\sqrt{2}}\right|^2 = \tfrac{1}{2} + \tfrac{1}{2} = 1. \checkmark$

The distribution is $P(0) = \tfrac{1}{2}$ , $P(1) = \tfrac{1}{2}$ — a perfect 50/50 split, the same as a fair coin toss. The complex phase in $\beta$ does not affect probabilities, only interference in later operations.

Now consider $\alpha = \tfrac{\sqrt{3}}{2}$ , $\beta = \tfrac{1}{2}$ (both real, and $\tfrac{3}{4} + \tfrac{1}{4} = 1$ , so normalised). Here $P(0) = \tfrac{3}{4}$ and $P(1) = \tfrac{1}{4}$ . Outcome $0$ is three times more likely than outcome $1$ .

From one qubit to many

When several qubits are measured, the joint outcomes form a distribution over all $2^n$ bit strings $\{00\ldots0,\, 00\ldots1,\, \ldots,\, 11\ldots1\}$ . The total probability still sums to one across all $2^n$ outcomes — the same Born-rule logic applies to the multi-qubit amplitudes. This distribution over bit strings is what a quantum algorithm ultimately computes: the algorithm shapes the amplitudes so that the correct answer has high probability, and a final measurement samples from that distribution.

Try it

This is a numerical exercise — return a number. A qubit is in the state $|\psi\rangle = \tfrac{\sqrt{3}}{2}|0\rangle + \tfrac{1}{2}|1\rangle$ . What is the probability of measuring the outcome $0$ ?

Deriving it: $P(0) = |\alpha|^2 = \left(\tfrac{\sqrt{3}}{2}\right)^2 = \tfrac{3}{4} = 0.75$ .

Run your code to see the quantum state.