Reading the State Vector Output

When you run a circuit in the simulator, you get back a table with one row per basis state. Each row shows three numbers: an amplitude, a probability, and a phase. Knowing what each column means lets you read the output confidently rather than just stare at numbers.

Amplitudes

The state vector of an $n$ -qubit system is a list of $2^n$ complex numbers called amplitudes, one for every possible classical bit string. For a single qubit the two amplitudes are labeled by $|0\rangle$ and $|1\rangle$ :

$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle, \qquad \alpha,\beta \in \mathbb{C}.$

The simulator displays $\alpha$ and $\beta$ either as decimal numbers (when they happen to be real) or as rectangular or polar complex numbers. The entire set of amplitudes completely describes everything you could ever know about the qubit — there is no hidden extra information.

Probabilities

The Born rule converts amplitudes to observable probabilities. The probability of measuring basis state $|b\rangle$ is

$P(b) = |\langle b|\psi\rangle|^2 = |\alpha_b|^2,$

the squared absolute value of the corresponding amplitude. Because all outcomes are mutually exclusive and exhaustive, the probabilities must sum to one:

$\sum_b |\alpha_b|^2 = 1.$

This is why you will often see the simulator report that the amplitudes are "normalised." For example, the equal-superposition state $|+\rangle = \tfrac{1}{\sqrt{2}}|0\rangle + \tfrac{1}{\sqrt{2}}|1\rangle$ has $|\alpha_0|^2 = |\alpha_1|^2 = \tfrac{1}{2}$ , so both outcomes are equally likely.

Phase

The phase of an amplitude $\alpha = r e^{i\theta}$ is the angle $\theta$ . Phase has no direct effect on individual measurement probabilities (only $r^2$ appears in the Born rule), but it controls how amplitudes interfere with each other when gates are applied later. The simulator usually displays the phase in degrees or radians alongside the magnitude $r$ .

Putting it together

When you read a state vector table, the workflow is straightforward:

Find the basis state of interest in the left column.
Read off its amplitude in the next column.
Square the magnitude to get the probability.
Note the phase angle if you care about how this state will interfere with future gates.

A circuit that does nothing (the identity) leaves the qubit in $|0\rangle$ with amplitude $1$ , probability $1$ , and phase $0°$ . Apply a Hadamard gate and both amplitudes become $\tfrac{1}{\sqrt{2}} \approx 0.7071$ , both probabilities become $0.5$ , and both phases are $0°$ . Apply a $Z$ gate afterward and the amplitude of $|1\rangle$ picks up a phase of $180°$ (it becomes $-\tfrac{1}{\sqrt{2}}$ ) while its probability stays $0.5$ — a concrete illustration that phase matters for interference even though it is invisible in a single measurement.

Try it

The $|+\rangle$ state has amplitude $\tfrac{1}{\sqrt{2}}$ for $|0\rangle$ . Use the Born rule to compute the probability of measuring $|0\rangle$ and return it.

Run your code to see the quantum state.