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Gates · Phase gate

T† gate

The Hermitian conjugate (and inverse) of the non-Clifford T gate, applying a -π/4 phase to |1⟩ and appearing throughout fault-tolerant circuits as the T-count partner of T.

Exact & formalUpdated

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

T†, T turned the other way

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.

q0Step 1 of 2 · HHStep 2 of 2 · TDGTDG

Step 1 of 2 · H

Probability of each outcome after this step

  • 050%
  • 150%
  1. 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. Multiplies the |1⟩ half of the state by a phase of −π/4, printed as 7π/4, and leaves |0⟩ alone. It is what undoes a T applied earlier, and it costs a fault-tolerant circuit exactly as much as a T does. 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.

Open in Studio
  • Steps1
  • Wires1
  • Qubits1
  • Depth1 gate
  • Parameternone (fixed -π/4 phase)
  • Papers2

References

Exact & formal
  • Unitary / matrix equivalence
  • Direct mathematics
  • Textbook / standard citation
Method
Compare the T† matrix against diag(1, e^{-iπ/4}) and multiply it by the T matrix to confirm the product is the identity.
Result
Pass · T† = diag(1, e^{-iπ/4}) exactly, and T·T† = T†·T = I.
OpenQASM 3 standard gate set ↗
Kind
curated reference
Reviewed by
Leona Quantum curation pass
License
CC BY 4.0-compatible reference metadata
phasenon-cliffordsingle qubitinversefault tolerant

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