The one state $\lvert 1\rangle$ is one of the two computational basis states of a qubit. It is the quantum analogue of a classical 1 bit. The other computational basis state is $\lvert 0\rangle$.
$$\lvert 1\rangle = \begin{pmatrix}0\\1\end{pmatrix}$$
On the Bloch sphere, $\lvert 1\rangle$ corresponds to the south pole at coordinates $(0, 0, -1)$. It is an eigenstate of the Pauli-Z gate with eigenvalue $-1$, meaning $Z\lvert 1\rangle = -\lvert 1\rangle$. Applying the Hadamard gate to $\lvert 1\rangle$ produces the equal superposition state $\lvert -\rangle = (\lvert 0\rangle - \lvert 1\rangle)/\sqrt{2}$.
# Prepare |1⟩ = X|0⟩ — flip the default |0⟩ with an X gate. from qiskit import QuantumCircuit from qiskit.quantum_info import Statevector qc = QuantumCircuit(1) qc.x(0) print(Statevector(qc).data)
| Gate | Result | Comment |
|---|---|---|
| I gate | $I\lvert 1\rangle = \lvert 1\rangle$ | The identity gate leaves $\lvert 1\rangle$ unchanged. |
| X gate | $X\lvert 1\rangle = \lvert 0\rangle$ | Flips $\lvert 1\rangle$ to $\lvert 0\rangle$; quantum analogue of classical NOT. |
| Y gate | $Y\lvert 1\rangle = -i\lvert 0\rangle$ | Bit flip with an imaginary phase factor of $-i$. |
| Z gate | $Z\lvert 1\rangle = -\lvert 1\rangle$ | $\lvert 1\rangle$ is an eigenstate of $Z$ with eigenvalue $-1$; the minus sign is a global phase in isolation but observable in superpositions. |
| Hadamard gate | $H\lvert 1\rangle = \lvert -\rangle$ | Rotates the south pole to the $-x$ equatorial point of the Bloch sphere. |
| S gate | $S\lvert 1\rangle = i\lvert 1\rangle$ | $S$ adds a phase of $i$ to the $\lvert 1\rangle$ component; acting on $\lvert 1\rangle$ alone this is a global phase. |
| T gate | $T\lvert 1\rangle = e^{i\pi/4}\lvert 1\rangle$ | $T$ adds a phase of $e^{i\pi/4}$ to the $\lvert 1\rangle$ component; global phase when acting on $\lvert 1\rangle$ alone. |
| Rotation-X gate | $R_x(\theta)\lvert 1\rangle = -i\sin\tfrac{\theta}{2}\lvert 0\rangle + \cos\tfrac{\theta}{2}\lvert 1\rangle$ | Tilts the state from $\lvert 1\rangle$ toward $\lvert 0\rangle$ with an imaginary phase on the $\lvert 0\rangle$ component. |
| Rotation-Y gate | $R_y(\theta)\lvert 1\rangle = -\sin\tfrac{\theta}{2}\lvert 0\rangle + \cos\tfrac{\theta}{2}\lvert 1\rangle$ | Real amplitudes; at $\theta=\pi/2$ gives $-\lvert -\rangle$, equivalent to $\lvert -\rangle$ up to global phase. |
| Rotation-Z gate | $R_z(\theta)\lvert 1\rangle = e^{i\theta/2}\lvert 1\rangle$ | Global phase only; $\lvert 1\rangle$ is on the $z$-axis so a $z$-rotation has no observable effect. |
| Unitary gate | $U\lvert 1\rangle = -e^{i\lambda}\sin\tfrac{\theta}{2}\lvert 0\rangle + e^{i(\phi+\lambda)}\cos\tfrac{\theta}{2}\lvert 1\rangle$ | Both $\phi$ and $\lambda$ contribute (unlike from $\lvert 0\rangle$ where $\lambda$ drops out). |
| State | Gates | Comment |
|---|---|---|
| $\lvert 0\rangle$ | $X\lvert 1\rangle = \lvert 0\rangle$ | X flips the qubit; the quantum analogue of classical NOT. $X$ is its own inverse: $X^2 = I$. |
| $\lvert 1\rangle$ | $I\lvert 1\rangle = \lvert 1\rangle$ | The identity gate leaves $\lvert 1\rangle$ unchanged. |
| $\lvert +\rangle$ | $ZH\lvert 1\rangle = \lvert +\rangle$ | H first produces $\lvert -\rangle$, then Z flips its relative phase to give $\tfrac{1}{\sqrt{2}}(\lvert 0\rangle + \lvert 1\rangle)$. |
| $\lvert -\rangle$ | $H\lvert 1\rangle = \lvert -\rangle$ | Hadamard rotates the south pole to the $-x$ equatorial point, creating the equal superposition $\tfrac{1}{\sqrt{2}}(\lvert 0\rangle - \lvert 1\rangle)$. |
| $\lvert +i\rangle$ | $S^\dagger H\lvert 1\rangle = \lvert +i\rangle$ | H rotates to the $-x$ equatorial point, then $S^\dagger$ rotates $-90°$ around $z$ to land on the $+y$ pole. |
| $\lvert -i\rangle$ | $SH\lvert 1\rangle = \lvert -i\rangle$ | H rotates to the $-x$ equatorial point, then S rotates $90°$ around $z$ to land on the $-y$ pole. |