The minus-i state $\lvert -i\rangle$ (also written $\lvert -y\rangle$) is an equal superposition of $\lvert 0\rangle$ and $\lvert 1\rangle$ with a relative phase of $-i$. It is one of the six cardinal states on the Bloch sphere, sitting at the negative $y$-axis at coordinates $(0, -1, 0)$.
$$\lvert -i\rangle = \frac{1}{\sqrt{2}}(\lvert 0\rangle - i\lvert 1\rangle) = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-i\end{pmatrix}$$
On the Bloch sphere, $\lvert -i\rangle$ is an eigenstate of the Pauli-Y gate with eigenvalue $-1$, meaning $Y\lvert -i\rangle = -\lvert -i\rangle$. It is prepared from $\lvert 0\rangle$ by $S^\dagger H\lvert 0\rangle$: the Hadamard rotates to the $+x$ equatorial point, then $S^\dagger$ rotates $-90°$ around $z$ to reach the $-y$ pole. Together with $\lvert +i\rangle$, it forms the Y eigenbasis.
# Prepare |−i⟩ = S†H|0⟩ — Hadamard to create superposition, then S† to rotate to the −y pole. from qiskit import QuantumCircuit from qiskit.quantum_info import Statevector qc = QuantumCircuit(1) qc.h(0) qc.sdg(0) print(Statevector(qc).data)
| Gate | Result | Comment |
|---|---|---|
| I gate | $I\lvert -i\rangle = \lvert -i\rangle$ | The identity gate leaves $\lvert -i\rangle$ unchanged. |
| X gate | $X\lvert -i\rangle = -i\lvert +i\rangle$ | Swaps the amplitudes; the result $\tfrac{1}{\sqrt{2}}(-i, 1)^T = -i\lvert +i\rangle$ is $\lvert +i\rangle$ up to global phase. |
| Y gate | $Y\lvert -i\rangle = -\lvert -i\rangle$ | $\lvert -i\rangle$ is an eigenstate of $Y$ with eigenvalue $-1$; the state acquires a global minus sign, unobservable in isolation. |
| Z gate | $Z\lvert -i\rangle = \lvert +i\rangle$ | Z negates the $\lvert 1\rangle$ component, flipping the phase from $-i$ to $+i$ and taking the $-y$ pole to the $+y$ pole. |
| Hadamard gate | $H\lvert -i\rangle = e^{-i\pi/4}\lvert +i\rangle$ | Produces $\lvert +i\rangle$ up to a global phase of $e^{-i\pi/4}$. |
| S gate | $S\lvert -i\rangle = \lvert +\rangle$ | Multiplying $-i$ by $i$ gives $-i^2 = +1$; the $-y$ equatorial point rotates to the $+x$ equatorial point. |
| T gate | $T\lvert -i\rangle = \tfrac{1}{\sqrt{2}}(1,\, e^{-i\pi/4})^T$ | Adds $\pi/4$ to the azimuthal angle; lands midway between $\lvert -i\rangle$ and $\lvert +\rangle$ at $315°$. |
| Rotation-X gate | $R_x(\theta)\lvert -i\rangle = \tfrac{1}{\sqrt{2}}(\cos\tfrac{\theta}{2} - \sin\tfrac{\theta}{2},\; -i(\cos\tfrac{\theta}{2} + \sin\tfrac{\theta}{2}))^T$ | Tilts off the $-y$ pole toward the $z$-axis; at $\theta=\pi/2$ the amplitudes become unequal. |
| Rotation-Y gate | $R_y(\theta)\lvert -i\rangle = e^{i\theta/2}\lvert -i\rangle$ | Global phase only; $\lvert -i\rangle$ is on the $y$-axis so a $y$-rotation has no observable effect. |
| Rotation-Z gate | $R_z(\theta)\lvert -i\rangle = \tfrac{e^{-i\theta/2}}{\sqrt{2}}(1,\, -ie^{i\theta})^T$ | Sweeps the azimuthal angle; at $\theta=\pi/2$ gives $\lvert +\rangle$, at $\theta=\pi$ gives $\lvert +i\rangle$, at $\theta=3\pi/2$ gives $\lvert -\rangle$. |
| Unitary gate | general rotation from $-y$ pole | Both $\phi$ and $\lambda$ contribute at all angles. |
| State | Gates | Comment |
|---|---|---|
| $\lvert 0\rangle$ | $HS\lvert -i\rangle = \lvert 0\rangle$ | S rotates to $\lvert +\rangle$, then H maps to $\lvert 0\rangle$. |
| $\lvert 1\rangle$ | $HS^\dagger\lvert -i\rangle = \lvert 1\rangle$ | $S^\dagger$ rotates to $\lvert -\rangle$, then H maps to $\lvert 1\rangle$. |
| $\lvert +\rangle$ | $S\lvert -i\rangle = \lvert +\rangle$ | S rotates $+90°$ around $z$, moving from the $-y$ pole to the $+x$ pole. |
| $\lvert -\rangle$ | $S^\dagger\lvert -i\rangle = \lvert -\rangle$ | $S^\dagger$ rotates $-90°$ around $z$, moving from the $-y$ pole to the $-x$ pole. |
| $\lvert +i\rangle$ | $Z\lvert -i\rangle = \lvert +i\rangle$ | Z flips the phase of the $\lvert 1\rangle$ component, taking the $-y$ pole to the $+y$ pole. |
| $\lvert -i\rangle$ | $I\lvert -i\rangle = \lvert -i\rangle$ | The identity gate leaves $\lvert -i\rangle$ unchanged. |