X, Y, Z on Superpositions

The Pauli gates $X$, $Y$, and $Z$ do familiar things to the computational basis states $|0\rangle$ and $|1\rangle$ — but their effect on the superposition states $|+\rangle$ and $|-\rangle$ is just as important, and it reveals a beautiful symmetry.

Recall the two states:

What X does to |+⟩ and |−⟩

The matrix $X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ swaps the two amplitudes. Applying it to $|+\rangle$:

And for $|-\rangle$ with column vector $\tfrac{1}{\sqrt{2}}(1,-1)^T$:

So $|+\rangle$ and $|-\rangle$ are eigenstates of $X$ with eigenvalues $+1$ and $-1$ respectively. On the Bloch sphere, $X$ is a rotation by $\pi$ about the $x$-axis — which fixes the $\pm x$ poles.

What Z does to |+⟩ and |−⟩

The matrix $Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ flips the sign of the $|1\rangle$ amplitude. This swaps $|+\rangle$ and $|-\rangle$:

In other words, $Z$ acts like a bit-flip in the $X$ basis — it exchanges the two $X$-basis states, just as $X$ exchanges $|0\rangle$ and $|1\rangle$ in the $Z$ basis.

What Y does to |+⟩ and |−⟩

The matrix $Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$ introduces imaginary factors. Direct multiplication gives:

So $Y$ also swaps $|+\rangle$ and $|-\rangle$, but with an imaginary phase factor of $\mp i$.

Try it

Verify the prediction $Z|+\rangle = |-\rangle$ by building the circuit. Start in $|0\rangle$, apply $H$ to reach $|+\rangle$, then apply $Z$. The simulator will confirm the resulting state is $|-\rangle$.

After running, you should see amplitudes $\approx +0.707$ for $|0\rangle$ and $\approx -0.707$ for $|1\rangle$ — the characteristic signature of $|-\rangle$. The Bloch vector points along $-x$.

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