The Bloch sphere formula: \(|\psi\rangle = \cos\frac{\theta}{2}|0\rangle + e^{i\phi}\sin\frac{\theta}{2}|1\rangle\)
The 6 named states sit at the ends of the 3 axes:
| Name | Vector | Bloch position | How to remember |
|---|---|---|---|
| \(|0\rangle\) | \(\begin{pmatrix}1\\0\end{pmatrix}\) | North pole θ=0 |
All probability on first component |
| \(|1\rangle\) | \(\begin{pmatrix}0\\1\end{pmatrix}\) | South pole θ=π |
All probability on second component |
| \(|+\rangle\) | \(\frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}\) | +X (equator) θ=π/2, φ=0 |
Equal split, positive sign between them |
| \(|-\rangle\) | \(\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-1\end{pmatrix}\) | −X (equator) θ=π/2, φ=π |
Equal split, negative sign between them |
| \(|{+i}\rangle\) | \(\frac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix}\) | +Y (equator) θ=π/2, φ=π/2 |
Equal split, \(\beta\) is +i |
| \(|{-i}\rangle\) | \(\frac{1}{\sqrt{2}}\begin{pmatrix}1\\-i\end{pmatrix}\) | −Y (equator) θ=π/2, φ=−π/2 |
Equal split, \(\beta\) is −i |
So the name literally tells you the coefficient: \(|+\rangle\) has \(+1\), \(|-\rangle\) has \(-1\), \(|+i\rangle\) has \(+i\), \(|{-i}\rangle\) has \(-i\).
All 4 equator states give 50/50 measurement probabilities in the standard basis (\(|0\rangle\)/\(|1\rangle\)). You can't tell them apart by measurement alone! The difference is purely in the relative phase (φ), which gates can see but measurement in the Z-basis can't.
Plug in θ=π/2, φ=0:
$$\cos\frac{\pi/2}{2}|0\rangle + e^{i \cdot 0}\sin\frac{\pi/2}{2}|1\rangle = \cos\frac{\pi}{4}|0\rangle + \sin\frac{\pi}{4}|1\rangle = \frac{1}{\sqrt{2}}|0\rangle + \frac{1}{\sqrt{2}}|1\rangle \;\checkmark$$
Plug in θ=π/2, φ=−π/2:
$$\cos\frac{\pi}{4}|0\rangle + e^{-i\pi/2}\sin\frac{\pi}{4}|1\rangle = \frac{1}{\sqrt{2}}|0\rangle + (-i)\frac{1}{\sqrt{2}}|1\rangle = \frac{1}{\sqrt{2}}|0\rangle - \frac{i}{\sqrt{2}}|1\rangle \;\checkmark$$
\(e^{-i\pi/2} = \cos(-90°) + i\sin(-90°) = 0 + i(-1) = -i\) — unit circle reverse lookup!