Table of Contents
$\lvert +i\rangle$ (Plus-i state)
The plus-i state $\lvert +i\rangle$ (also written $\lvert i\rangle$ or $\lvert y\rangle$) 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 positive $y$-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 $SH\lvert 0\rangle$: the Hadamard rotates to the $+x$ equatorial point, then the S gate rotates $90°$ around $z$ to reach the $+y$ pole. Together with $\lvert -i\rangle$, it forms the Y eigenbasis.
Qiskit
# Prepare |+i⟩ = SH|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.s(0) print(Statevector(qc).data)
Applying gates
| 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 gate returns the state unchanged. |
| 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 $-1$; the $+y$ equatorial point rotates to the $-x$ equatorial point. |
| T gate | $T\lvert +i\rangle = \tfrac{1}{\sqrt{2}}(1,\, e^{i3\pi/4})^T$ | Adds $\pi/4$ to the azimuthal angle; lands midway between $\lvert -\rangle$ and $\lvert -i\rangle$ at $135°$. |
| 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 $\theta=\pi$ gives $\lvert -i\rangle$ up to global phase, at $\theta=3\pi/2$ gives $\lvert +\rangle$. |
| Unitary gate | general rotation from $+y$ pole | Both $\phi$ and $\lambda$ contribute at all angles, unlike from $\lvert 0\rangle$ where $\lambda$ drops out. |
Reaching other states
| State | Gates | Comment |
|---|---|---|
| $\lvert 0\rangle$ | $HS^\dagger\lvert +i\rangle = \lvert 0\rangle$ | $S^\dagger$ rotates $-90°$ around $z$ to $\lvert +\rangle$, then H maps to $\lvert 0\rangle$. |
| $\lvert 1\rangle$ | $HS\lvert +i\rangle = \lvert 1\rangle$ | S rotates to $\lvert -\rangle$, then H maps to $\lvert 1\rangle$. |
| $\lvert +\rangle$ | $S^\dagger\lvert +i\rangle = \lvert +\rangle$ | $S^\dagger$ rotates $-90°$ around $z$, moving from the $+y$ pole to the $+x$ pole. |
| $\lvert -\rangle$ | $S\lvert +i\rangle = \lvert -\rangle$ | S rotates $+90°$ around $z$, moving from the $+y$ pole to the $-x$ pole. |
| $\lvert +i\rangle$ | $I\lvert +i\rangle = \lvert +i\rangle$ | The identity gate leaves $\lvert +i\rangle$ unchanged. |
| $\lvert -i\rangle$ | $Z\lvert +i\rangle = \lvert -i\rangle$ | Z flips the phase of the $\lvert 1\rangle$ component, taking the $+y$ pole to the $-y$ pole. |