Interference from Phases

Quantum amplitudes are complex numbers, and their phase — the argument of the complex number — carries real physical information. A lone phase difference between $|0\rangle$ and $|1\rangle$ is completely invisible in a computational-basis measurement: only the absolute squares of the amplitudes determine probabilities. But once we sandwich the phase gate between two Hadamards, that hidden phase becomes a measurable amplitude difference. This is interference.

Relative phase is invisible by itself

Consider the state $|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ . Applying the $T$ gate — which adds a phase of $e^{i\pi/4}$ to the $|1\rangle$ component — gives

T|+\rangle = \frac{1}{\sqrt{2}}\!\left(|0\rangle + e^{i\pi/4}|1\rangle\right).

The probabilities are still 50/50: $|1/\sqrt{2}|^2 = 1/2$ for both outcomes. The phase $e^{i\pi/4}$ is there, but a $Z$ -basis measurement cannot see it.

H–phase–H reveals the phase

To make the phase visible we apply $H$ after $T$ , which mixes the two amplitudes together. Starting from $|0\rangle$ :

|0\rangle \xrightarrow{H} \frac{|0\rangle+|1\rangle}{\sqrt{2}} \xrightarrow{T} \frac{|0\rangle + e^{i\pi/4}|1\rangle}{\sqrt{2}} \xrightarrow{H} \frac{(1+e^{i\pi/4})|0\rangle + (1-e^{i\pi/4})|1\rangle}{2}.

The probability of measuring $|0\rangle$ is now

P(0) = \frac{|1+e^{i\pi/4}|^2}{4} = \frac{2 + 2\cos(\pi/4)}{4} = \frac{1+\cos(\pi/4)}{2} = \cos^2\!\left(\frac{\pi}{8}\right) \approx 0.854,

using the identity $|1+e^{i\theta}|^2 = 2+2\cos\theta$ and the half-angle formula $\frac{1+\cos\theta}{2} = \cos^2(\theta/2)$ . Similarly $P(1) = \sin^2(\pi/8) \approx 0.146$ .

The two outcomes are no longer equally likely. The $\pi/4$ phase introduced by $T$ has been turned into a measurable amplitude imbalance by the second $H$ gate.

General rule

For any phase gate $P(\theta)$ (which multiplies $|1\rangle$ by $e^{i\theta}$ ), the $H$ – $P(\theta)$ – $H$ sandwich on $|0\rangle$ yields

P(0) = \cos^2\!\left(\frac{\theta}{2}\right), \qquad P(1) = \sin^2\!\left(\frac{\theta}{2}\right).

At $\theta = 0$ (no phase) you get $P(0)=1$ , the qubit stays in $|0\rangle$ . At $\theta = \pi$ (the $Z$ gate) you get $P(0)=0$ , the qubit flips to $|1\rangle$ — which is exactly the identity $HZH = X$ we saw in an earlier lesson.

Try it

Build the circuit $H \cdot T \cdot H$ on qubit 0. The statevector grader will check that your final state has the correct complex amplitudes. A circuit that does nothing, or that skips the second $H$ , will fail — only the full sandwich produces the asymmetric output.

Run your code to see the quantum state.

After running, look at the State vector panel: the amplitude for $|0\rangle$ should have magnitude $\cos(\pi/8) \approx 0.924$ and the amplitude for $|1\rangle$ magnitude $\sin(\pi/8) \approx 0.383$ . The Probabilities panel should show roughly 85 % and 15 %.