S, T and Their Inverses

Two single-qubit gates you will meet often are the S gate and the T gate. Both act only on the $|1\rangle$ component of a state, leaving $|0\rangle$ unchanged — they are called phase gates because they shift the relative phase between the two basis amplitudes.

The S gate and its inverse

The $S$ gate (also written $P(\pi/2)$) applies a phase of $i = e^{i\pi/2}$ to $|1\rangle$:

Its action on basis states is straightforward:

On a superposition $\alpha|0\rangle + \beta|1\rangle$, the effect is to rotate $\beta$ by $90°$ in the complex plane: $S(\alpha|0\rangle + \beta|1\rangle) = \alpha|0\rangle + i\beta|1\rangle$.

The inverse $S^\dagger$ (or $S^{-1}$) undoes that rotation by applying $-i = e^{-i\pi/2}$:

You can verify that $S^\dagger S = I$ by multiplying the two matrices:

The T gate and its inverse

The $T$ gate applies half the phase of $S$ — a $45°$ rotation, $e^{i\pi/4}$:

This means $S = T^2$: two $T$ gates compose to one $S$ gate. Similarly, $T^\dagger$ applies the conjugate phase $e^{-i\pi/4}$:

Why inverses matter

Every quantum gate is unitary — its inverse is just its conjugate transpose (Hermitian adjoint), often written $U^\dagger$. Because of this, quantum operations are always reversible: you can run any circuit backward by applying each gate's adjoint in reverse order. For phase gates the inverse is simply the gate with the negated phase angle. Applying $S$ followed immediately by $S^\dagger$ is therefore the identity transformation, and the qubit is left exactly as it started.

Try it

Apply $S$ to qubit 0, then undo it with $S^\dagger$. The grader checks the whole unitary matrix of your circuit, so doing nothing will not pass — you must actually apply both gates, and they must cancel to the identity.

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