Project: Guess the Hidden Gate

Every gate has a unique fingerprint: the way it transforms states. If you are handed a black box that applies one of several known gates, you can figure out which gate it is by carefully choosing your probe state and observing how the output changes. This idea — called gate tomography in its full form — underpins quantum verification, benchmarking, and several quantum algorithms.

Why the probe state matters

Consider the $Z$ gate and the identity $I$. Their matrices are

If you probe with $|0\rangle$: both give $|0\rangle$ back — completely useless for telling them apart. If you probe with $|1\rangle$: $Z|1\rangle = -|1\rangle$, which differs from $I|1\rangle = |1\rangle$ only by a global phase ($-1$). By the Born rule, global phases are unobservable, so a computational-basis measurement still cannot distinguish them.

The winning probe is $|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$:

Now the two gates produce different states, $|-\rangle$ versus $|+\rangle$ — but those states are still 50/50 in a $Z$-basis measurement. To turn the phase difference into an amplitude difference you can detect, apply a second Hadamard:

The complete circuit $H \to \text{gate?} \to H$ outputs $|1\rangle$ if the gate was $Z$ and $|0\rangle$ if it was $I$. One measurement shot is enough to decide with certainty.

Comparing X and Z

The $X$ gate swaps $|0\rangle$ and $|1\rangle$, so $X|+\rangle = |+\rangle$ — $|+\rangle$ is an eigenstate of $X$ with eigenvalue $+1$. Probing with $|+\rangle$ cannot distinguish $X$ from $I$. Instead, probe with $|0\rangle$: $X|0\rangle = |1\rangle$ versus $I|0\rangle = |0\rangle$, which is immediately measurable. The right probe always depends on the gate set you are trying to tell apart.

| Gate | Effect on $|0\rangle$ | Effect on $|+\rangle$ | |-------|----------------------|----------------------| | $I$ | $|0\rangle$ | $|+\rangle$ | | $X$ | $|1\rangle$ | $|+\rangle$ | | $Z$ | $|0\rangle$ | $|-\rangle$ | | $H$ | $|+\rangle$ | $|0\rangle$ |

No single probe state distinguishes all four simultaneously, but pairing $|0\rangle$ and $|+\rangle$ does — this is why real tomography uses multiple probe states.

Try it

Build the discrimination circuit for $Z$: prepare $|+\rangle$, apply $Z$, rotate back with $H$. The grader checks the final statevector — it should be $|1\rangle$, confirming the hidden gate is $Z$ and not $I$.

After running, the amplitude display should show $1.0$ on the $|1\rangle$ component. Swap the $Z$ for an $X$ in your circuit and notice you get $|0\rangle$ — a completely different answer, which is how you would rule $X$ out as the hidden gate.

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