Quantum phase and interference

Global phase changes no prediction

Multiplying an entire state by $e^{i\gamma}$ does not change measurement probabilities. For any measurement vector $\lvert v\rangle$,

$$|\langle v|e^{i\gamma}\psi\rangle|^2=|e^{i\gamma}|^2|\langle v|\psi\rangle|^2=|\langle v|\psi\rangle|^2.$$

Relative phase is different. Multiplying just one component changes its relationship with the others.

Work the circuit through

Prepare $\lvert0\rangle$, apply H, then $P(\phi)=\operatorname{diag}(1,e^{i\phi})$, then H again:

$$\lvert0\rangle\longmapsto \frac{\lvert0\rangle+\lvert1\rangle}{\sqrt2} \longmapsto\frac{\lvert0\rangle+e^{i\phi}\lvert1\rangle}{\sqrt2} \longmapsto\frac{1+e^{i\phi}}2\lvert0\rangle+\frac{1-e^{i\phi}}2\lvert1\rangle.$$

The final probabilities are

$$P(0)=\cos^2(\phi/2),\qquad P(1)=\sin^2(\phi/2).$$

Before the second H, the Z probabilities are always 50/50. The final H recombines amplitudes and exposes the phase difference.

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Superposition & interference

The two Hadamard gates recombine amplitudes. A relative phase between them changes the result even though the intermediate Z-basis probabilities stay at 50/50.

Make a prediction first

H then H returns 0. What do you expect if a Z gate is placed between them?

Experiment controls

|0⟩ — H — P(φ) — H — M

Relative phase φ0°

H · HH · Z · H

P(φ) = diag(1, e^(iφ)). P(180°) is the Z gate.

Random seed

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Exact probability

|0⟩

100.0%

|1⟩

0.0%

Exact ideal-model predictions · run a measurement experiment to compare.

MODEL · Ideal pure-state vector, complex double precisionORDER · q0 is the leftmost bit

Take it further

Set the phase to 180°. Explain why the final probability of 1 becomes 100%.

Prediction: H then H returns 0. What do you expect if a Z gate is placed between them? Challenge: Set the phase to 180°. Explain why the final probability of 1 becomes 100%.

Three checks

• At 0°, the amplitudes for 1 cancel; the result is 0.
• At 180°, the amplitudes for 0 cancel; the result is 1.
• At 90°, the output probabilities are equal.

Challenge: explain why applying a global minus sign at the middle of the circuit cannot replace the Z gate. Z changes the relative sign; a global minus multiplies both amplitudes.

Reference: IBM Quantum — Single systems.