Interference

You will chain three gates on one qubit: H, then Z, then H again. The middle gate (Z) changes something called the relative phase, which is a property of the quantum state that does not show up directly when you measure. The final H gate converts that hidden change into a result you can see.

After the first H, the qubit is in an equal superposition, and measuring would give 50/50. The Z gate then flips the sign of the $|1⟩$ branch. If you measured right now, you would still see 50/50 because sign does not affect probability directly. But the second H recombines the two branches, and the flipped sign causes destructive interference on the $|0⟩$ branch and constructive interference on the $|1⟩$ branch. The qubit ends up deterministically in $|1⟩$. A change that was invisible has become completely visible.

• After the first H, the histogram shows a balanced 50/50 split.
• After Z, the histogram looks exactly the same, but the Bloch sphere has rotated. The state changed even though probabilities did not.
• After the second H, the qubit is deterministically in $|1⟩$. The hidden phase difference became a visible bit flip.
• Compare this to applying H twice without Z in between: H-H returns the qubit to $|0⟩$, not $|1⟩$.

The qubit ends in $|1⟩$ with certainty. Every shot returns 1. This confirms that the Z gate's phase flip, although invisible after step 2, was converted into a deterministic outcome by the final Hadamard.