ket-plus
Table of Contents
$\lvert +\rangle$ (Plus state)
The plus state $\lvert +\rangle$ is an equal superposition of $\lvert 0\rangle$ and $\lvert 1\rangle$ with equal positive amplitudes. It is one of the six cardinal states on the Bloch sphere, sitting at the positive $x$-axis at coordinates $(1, 0, 0)$.
$$\lvert +\rangle = \frac{1}{\sqrt{2}}(\lvert 0\rangle + \lvert 1\rangle) = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix}$$
On the Bloch sphere, $\lvert +\rangle$ is an eigenstate of the Pauli-X gate with eigenvalue $+1$, meaning $X\lvert +\rangle = \lvert +\rangle$. It is prepared from $\lvert 0\rangle$ by a single Hadamard: $H\lvert 0\rangle = \lvert +\rangle$. Together with $\lvert -\rangle$, it forms the X eigenbasis (Hadamard basis).
Qiskit
# Prepare |+⟩ = H|0⟩ — Hadamard creates equal superposition from |0⟩. from qiskit import QuantumCircuit from qiskit.quantum_info import Statevector qc = QuantumCircuit(1) qc.h(0) print(Statevector(qc).data)
Applying gates
| Gate | Result | Comment |
|---|---|---|
| I gate | $I\lvert +\rangle = \lvert +\rangle$ | The identity gate leaves $\lvert +\rangle$ unchanged. |
| X gate | $X\lvert +\rangle = \lvert +\rangle$ | $\lvert +\rangle$ is an eigenstate of $X$ with eigenvalue $+1$; both amplitudes are equal so swapping them changes nothing. |
| Y gate | $Y\lvert +\rangle = -i\lvert -\rangle$ | Swaps and phase-shifts; the result is $\lvert -\rangle$ up to global phase $-i$. |
| Z gate | $Z\lvert +\rangle = \lvert -\rangle$ | Z negates the $\lvert 1\rangle$ component, flipping the $+x$ pole to the $-x$ pole. |
| Hadamard gate | $H\lvert +\rangle = \lvert 0\rangle$ | The two amplitudes add constructively for $\lvert 0\rangle$ and cancel for $\lvert 1\rangle$; collapses to the north pole. |
| S gate | $S\lvert +\rangle = \lvert +i\rangle$ | Rotates $+90°$ around $z$, taking the $+x$ pole to the $+y$ pole. |
| T gate | $T\lvert +\rangle = \tfrac{1}{\sqrt{2}}(1,\, e^{i\pi/4})^T$ | Adds $\pi/4$ to the azimuthal angle; lands midway between $\lvert +\rangle$ and $\lvert +i\rangle$ at $45°$. |
| Rotation-X gate | $R_x(\theta)\lvert +\rangle = e^{-i\theta/2}\lvert +\rangle$ | Global phase only; $\lvert +\rangle$ is on the $x$-axis so an $x$-rotation has no observable effect. |
| Rotation-Y gate | $R_y(\theta)\lvert +\rangle = \tfrac{1}{\sqrt{2}}(\cos\tfrac{\theta}{2} - \sin\tfrac{\theta}{2},\; \cos\tfrac{\theta}{2} + \sin\tfrac{\theta}{2})^T$ | Tilts off the $+x$ pole toward the poles; at $\theta=\pi/2$ gives $\lvert 1\rangle$, at $\theta=-\pi/2$ gives $\lvert 0\rangle$. |
| Rotation-Z gate | $R_z(\theta)\lvert +\rangle = \tfrac{e^{-i\theta/2}}{\sqrt{2}}(1,\, e^{i\theta})^T$ | Sweeps the azimuthal angle; at $\theta=\pi/2$ gives $\lvert +i\rangle$, at $\theta=\pi$ gives $\lvert -\rangle$, at $\theta=3\pi/2$ gives $\lvert -i\rangle$. |
| Unitary gate | general rotation from $+x$ pole | $\lambda$ contributes here (unlike from $\lvert 0\rangle$); both $\phi$ and $\lambda$ affect the result. |
Reaching other states
| State | Gates | Comment |
|---|---|---|
| $\lvert 0\rangle$ | $H\lvert +\rangle = \lvert 0\rangle$ | Hadamard maps the $+x$ pole back to the north pole; constructive interference on $\lvert 0\rangle$. |
| $\lvert 1\rangle$ | $R_y(\pi/2)\lvert +\rangle = \lvert 1\rangle$ | Tilts the $+x$ equatorial point to the south pole with a real-amplitude rotation. |
| $\lvert +\rangle$ | $I\lvert +\rangle = \lvert +\rangle$ | The identity gate leaves $\lvert +\rangle$ unchanged. |
| $\lvert -\rangle$ | $Z\lvert +\rangle = \lvert -\rangle$ | Z flips the $\lvert 1\rangle$ phase, taking the $+x$ pole to the $-x$ pole. |
| $\lvert +i\rangle$ | $S\lvert +\rangle = \lvert +i\rangle$ | S rotates $+90°$ around $z$, moving from the $+x$ pole to the $+y$ pole. |
| $\lvert -i\rangle$ | $S^\dagger\lvert +\rangle = \lvert -i\rangle$ | $S^\dagger$ rotates $-90°$ around $z$, moving from the $+x$ pole to the $-y$ pole. |
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