T phase gate

A small worked example, computed in your browser. It is not a result from this record's papers.

T, an eighth turn of phase

One qubit, put into |+⟩ = (|0⟩ + |1⟩)/√2 first, because on |0⟩ or |1⟩ alone this gate changes nothing a measurement or another gate can ever detect.

1. Step 1 of 2 · H

   Puts the qubit into an equal superposition. Without this there is no second amplitude for the gate's phase to be measured against, and a phase on the whole state is not observable.

   Probability spreads from 1 outcome to 2. All 2 are equally likely.

2. Step 2 of 2 · T

   Multiplies the |1⟩ half of the state by a phase of π/4, and leaves |0⟩ alone. That is half of S again. T is also the gate that takes a Clifford circuit out of classical reach, which is why fault-tolerant cost is counted in T gates. The two outcomes stay equally likely — the difference is carried entirely in the phase between them, which is what the panel beside the bars shows.

   The outcome probabilities do not move at all. What this step changes is phase — 2 distinct relative phases across the state now — which no measurement can see until a later step turns it back into probability.

Readout Measuring gives 0 or 1, each half the time — exactly as it did before the gate. A single measurement of this qubit cannot detect this gate at all. Its effect becomes visible only when the phase is turned back into probability by a later interference step, which is what every algorithm that uses it does.