Phase Gates on |+⟩

A qubit in the equal superposition $|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ sits exactly on the equator of the Bloch sphere, pointing along the $+x$ axis. Phase gates leave the north and south poles fixed but rotate every equatorial state around the $z$-axis, moving it to a new point on the equator.

Phase gates as $z$-axis rotations

A phase gate $P(\lambda)$ multiplies the $|1\rangle$ amplitude by $e^{i\lambda}$ while leaving $|0\rangle$ unchanged:

The three standard phase gates are special cases of $P(\lambda)$:

| Gate | $\lambda$ | Effect on $|+\rangle$ | |------|-----------|----------------------| | $T$ | $\pi/4$ | $\tfrac{1}{\sqrt{2}}(|0\rangle + e^{i\pi/4}|1\rangle)$ — rotated 45° | | $S$ | $\pi/2$ | $\tfrac{1}{\sqrt{2}}(|0\rangle + i\,|1\rangle)$ — rotated 90° to $+y$ axis | | $Z$ | $\pi$ | $\tfrac{1}{\sqrt{2}}(|0\rangle - |1\rangle) = |-\rangle$ — rotated 180° to $-x$ axis |

Starting from $|+\rangle$ and applying $S$ yields

On the Bloch sphere this state points along $+y$, often written $|{+i}\rangle$ or $|R\rangle$.

Keeping track on the Bloch sphere

The Bloch sphere provides a clear picture: any state $\cos(\theta/2)|0\rangle + e^{i\phi}\sin(\theta/2)|1\rangle$ lives at polar angle $\theta$ and azimuthal angle $\phi$. For equatorial states ($\theta = \pi/2$), $P(\lambda)$ shifts $\phi$ by $\lambda$:

Starting from $|+\rangle$ with $\phi = 0$, applying $T$ gives $\phi = \pi/4$, applying $S$ gives $\phi = \pi/2$, and applying $Z$ gives $\phi = \pi$, landing at $|-\rangle$.

Try it

Prepare $|+\rangle$ by applying $H$ to $|0\rangle$, then apply the $S$ gate to rotate it 90° to the $+y$ axis. The grader checks the full state vector.

After running, the amplitudes should be $\tfrac{1}{\sqrt{2}} \approx 0.707$ for $|0\rangle$ and $\tfrac{i}{\sqrt{2}}$ for $|1\rangle$, confirming the 90° equatorial rotation.

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