Phase Gates (S and T)

Overview

So far, you’ve seen gates that visibly change probabilities in the computational basis (like X). You’ve also seen a gate (Z) that can leave computational-basis probabilities unchanged while still changing the state.

Phase gates continue that theme. They are essential because:

• phase is a real degree of freedom in quantum states,
• phase controls interference,
• and small phase rotations are building blocks for general single-qubit evolution.

In this lesson we focus on two standard phase gates:

• S (a $\pi/2$ phase rotation around $z$),
• T (a $\pi/4$ phase rotation around $z$).

Intuition

A general single-qubit state is

$$|\psi\rangle = \alpha|0\rangle + \beta|1\rangle.$$

A phase gate is a gate that keeps the magnitude of amplitudes but changes the relative phase between basis components.

A particularly important class is “phase on $|1\rangle$”:

• $|0\rangle$ stays the same,
• $|1\rangle$ picks up a complex factor of magnitude 1, like $e^{i\varphi}$.

That means probabilities in the computational basis are unchanged (because $|e^{i\varphi}|^2 = 1$), but the state’s phase structure changes, which affects what happens under later basis changes and measurements.

Bloch-sphere picture:

• Z was a 180° rotation about the $z$ axis.
• S and T are smaller rotations about the same axis.

So you can think of S and T as “phase knobs” that rotate the state around the $z$ axis by controlled angles.

Formal Description

The S gate

S is defined by

$$S|0\rangle = |0\rangle,\qquad S|1\rangle = i|1\rangle.$$

So it adds a phase of $\tfrac{\pi}{2}$ (a quarter-turn) to the $|1\rangle$ component.

Matrix form:

$$S = \begin{pmatrix} 1 & 0\\ 0 & i \end{pmatrix}.$$

Interpretation of the matrix:

• The top-left “1” means the $|0\rangle$ amplitude stays $\alpha$.
• The bottom-right “$i$” means the $|1\rangle$ amplitude becomes $i\beta$.

So for $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$,

$$S|\psi\rangle = \alpha|0\rangle + i\beta|1\rangle.$$

The T gate

T is defined by

$$T|0\rangle = |0\rangle,\qquad T|1\rangle = e^{i\pi/4}|1\rangle.$$

Matrix form:

$$T = \begin{pmatrix} 1 & 0\\ 0 & e^{i\pi/4} \end{pmatrix}.$$

Again, the meaning is direct:

• the $|0\rangle$ amplitude stays $\alpha$,
• the $|1\rangle$ amplitude becomes $e^{i\pi/4}\beta$.

So

$$T|\psi\rangle = \alpha|0\rangle + e^{i\pi/4}\beta|1\rangle.$$

Difference between S and T

Both are rotations about $z$. The only difference is the rotation angle:

• S applies a $\pi/2$ phase to $|1\rangle$.
• T applies a $\pi/4$ phase to $|1\rangle$.

You can view S as “two T steps”:

$$T^2 = S.$$

(We’re stating this relationship as a useful fact; the main point is that S is a bigger phase rotation than T.)

Worked Example

Take

$$|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle).$$

Apply S:

$$S|+\rangle = \tfrac{1}{\sqrt{2}}(S|0\rangle + S|1\rangle) = \tfrac{1}{\sqrt{2}}(|0\rangle + i|1\rangle).$$

Computational-basis probabilities are still $1/2$ and $1/2$. But the state has moved on the Bloch sphere:

• $|+\rangle$ points along +$x$.
• $\tfrac{1}{\sqrt{2}}(|0\rangle + i|1\rangle)$ points along +$y$.

So S rotates the equator direction by 90° about the $z$ axis.

Similarly, applying T to $|+\rangle$ rotates by 45° about $z$:

$$T|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + e^{i\pi/4}|1\rangle),$$

which lands halfway between +$x$ and +$y$ on the equator.

Turtle Tip

Common Pitfalls

Quick Check

What’s Next

We’ve learned named gates that rotate around axes by special angles. Next we generalize: instead of a fixed 180°, 90°, or 45° turn, we define rotation gates $R_x(\theta)$, $R_y(\theta)$, and $R_z(\theta)$ that rotate by an arbitrary angle $\theta$.