Phase and rotation gates

S, T, P and the RX/RY/RZ family: fine control of a qubit, and why T matters.

Beginner12 min read4 sections · 2 questions+45

By the end you will be able to

01Relate S, T and P(λ) to RZ.
02Pick the rotation gate that prepares a given single-qubit state.
03Explain 'universal gate set' in one sentence.

P (λ) = (10 0 e^{iλ}), S = P (\frac{π}{2}), T = P (\frac{π}{4}), R_{z} (θ) = e^{- i θ Z /2}

P and RZ differ only by a global phase, which no measurement can see. S² = Z and T² = S.

Rotations let you place a qubit anywhere on the Bloch sphere: RY(θ) sets the latitude, RZ(φ) the longitude. Together with CNOT, the set {H, T, CNOT} is universal: any n-qubit unitary can be approximated to any precision from those three alone. That is why the T gate, not the more common S, decides the cost of fault-tolerant circuits.

Hands-on

Prepare a target state

The goal is the state with probability 0.75 on |1⟩ and phase π/2 on it. The slider on RY sets the probability (cos²(θ/2) = 0.25 → θ = 2π/3); the RZ adds the phase. Check the Amplitudes view.

OpenQASM 3

OPENQASM 3.0;
include "stdgates.inc";
qubit[1] q;
ry(2*pi/3) q[0];
rz(pi/2) q[0];

Open in the IDE Open in the Playground

You want	Use
Flip the probabilities	x
An equal superposition	h (or ry(pi/2))
Any probability p on \|1⟩	ry(2·acos(√(1−p)))
A phase without changing probabilities	p(λ) / rz(θ) / s / t

Knowledge check

2 questions · one try each counts

Question 1

Which pair of gates is equal up to a global phase?

Question 2

Why is {H, S, CNOT} not universal while {H, T, CNOT} is?

You want

Use

Flip the probabilities

An equal superposition

h (or ry(pi/2))

Any probability p on |1⟩

ry(2·acos(√(1−p)))

A phase without changing probabilities

p(λ) / rz(θ) / s / t