ket-01
Table of Contents
$\lvert 01\rangle$
The $\lvert 01\rangle$ state is a two-qubit computational basis state with qubit 0 in $\lvert 0\rangle$ and qubit 1 in $\lvert 1\rangle$. It is the tensor product $\lvert 0\rangle\otimes\lvert 1\rangle$ — an unentangled product state. It is reached from $\lvert 00\rangle$ by a single $X$ gate on qubit 1.
$$\lvert 01\rangle = \lvert 0\rangle\otimes\lvert 1\rangle = \begin{pmatrix}0\\1\\0\\0\end{pmatrix}$$
Measuring qubit 0 yields $0$ with certainty; measuring qubit 1 yields $1$ with certainty. Applying $H\otimes I$ then CX maps $\lvert 01\rangle$ directly to $\lvert\Psi^+\rangle$, making it the natural starting state for the $\Psi$-type Bell states.
Qiskit
# Prepare |01⟩ — flip qubit 1 from the default |00⟩ state. from qiskit import QuantumCircuit from qiskit.quantum_info import Statevector qc = QuantumCircuit(2) qc.x(1) # |00⟩ → |01⟩ print(Statevector(qc).data)
Applying gates
Single-qubit gates act on qubit 0. Qubit 1 is in $\lvert 1\rangle$, so gates on qubit 1 follow the $\lvert 1\rangle$ single-qubit rules with qubit 0 unchanged.
| Gate | Result | Comment |
|---|---|---|
| I gate | $I_0\lvert 01\rangle = \lvert 01\rangle$ | Identity leaves the state unchanged. |
| X gate | $X_0\lvert 01\rangle = \lvert 11\rangle$ | Flips qubit 0 from $\lvert 0\rangle$ to $\lvert 1\rangle$; qubit 1 is unchanged. |
| Y gate | $Y_0\lvert 01\rangle = i\lvert 11\rangle$ | Bit flip with an imaginary phase factor on qubit 0. |
| Z gate | $Z_0\lvert 01\rangle = \lvert 01\rangle$ | Qubit 0 is in $\lvert 0\rangle$, an eigenstate of $Z$ with eigenvalue $+1$; no effect. |
| Hadamard gate | $H_0\lvert 01\rangle = \tfrac{1}{\sqrt{2}}(\lvert 01\rangle + \lvert 11\rangle)$ | Puts qubit 0 into equal superposition; qubits remain unentangled. |
| S gate | $S_0\lvert 01\rangle = \lvert 01\rangle$ | Phase gate only affects $\lvert 1\rangle$; qubit 0 is in $\lvert 0\rangle$ so nothing changes. |
| T gate | $T_0\lvert 01\rangle = \lvert 01\rangle$ | Same as $S$: no effect when the target qubit is in $\lvert 0\rangle$. |
| Rotation-X gate | $R_x(\theta)_0\lvert 01\rangle = \cos\tfrac{\theta}{2}\lvert 01\rangle - i\sin\tfrac{\theta}{2}\lvert 11\rangle$ | Tilts qubit 0 toward $\lvert 1\rangle$; qubit 1 is unchanged throughout. |
| Rotation-Y gate | $R_y(\theta)_0\lvert 01\rangle = \cos\tfrac{\theta}{2}\lvert 01\rangle + \sin\tfrac{\theta}{2}\lvert 11\rangle$ | Real amplitudes; at $\theta=\pi/2$ gives $\tfrac{1}{\sqrt{2}}(\lvert 01\rangle + \lvert 11\rangle)$, the same result as $H_0$. |
| Rotation-Z gate | $R_z(\theta)_0\lvert 01\rangle = e^{-i\theta/2}\lvert 01\rangle$ | Global phase only; qubit 0 is on the $z$-axis so a $z$-rotation has no observable effect. |
| CX (CNOT) gate | $\text{CX}_{0\to 1}\lvert 01\rangle = \lvert 01\rangle$ | Control is qubit 0 = $\lvert 0\rangle$; CX only fires on $\lvert 1\rangle$, so qubit 1 is unchanged. |
| SWAP gate | $\text{SWAP}\lvert 01\rangle = \lvert 10\rangle$ | Exchanges the two qubits. |
| iSWAP gate | $\text{iSWAP}\lvert 01\rangle = i\lvert 10\rangle$ | Exchanges $\lvert 01\rangle$ and $\lvert 10\rangle$ with a phase of $i$. |
Reaching other states
| State | Gates | Comment |
|---|---|---|
| $\lvert 01\rangle$ | $I\lvert 01\rangle = \lvert 01\rangle$ | The identity leaves $\lvert 01\rangle$ unchanged. |
| $\lvert 00\rangle$ | $X_1\lvert 01\rangle = \lvert 00\rangle$ | X on qubit 1 flips it from $\lvert 1\rangle$ back to $\lvert 0\rangle$. |
| $\lvert 11\rangle$ | $X_0\lvert 01\rangle = \lvert 11\rangle$ | X on qubit 0 flips the first bit. |
| $\lvert 10\rangle$ | $X_0 X_1\lvert 01\rangle = \lvert 10\rangle$ | X on both qubits; equivalent to SWAP. |
| $\lvert\Psi^+\rangle$ | $\text{CX}\cdot(H_0\otimes I)\lvert 01\rangle$ | Bell preparation: H on qubit 0, then CX. The standard starting state for $\lvert\Psi^+\rangle$. |
| $\lvert\Psi^-\rangle$ | $Z_0\cdot\text{CX}\cdot(H_0\otimes I)\lvert 01\rangle$ | Bell preparation then Z on qubit 0 flips the relative sign. |
| $\lvert\Phi^+\rangle$ | $\text{CX}\cdot(H_0\otimes I)\cdot X_1\lvert 01\rangle$ | Flip qubit 1 first to reach $\lvert 00\rangle$, then apply the standard Bell preparation. |
| $\lvert\Phi^-\rangle$ | $Z_0\cdot\text{CX}\cdot(H_0\otimes I)\cdot X_1\lvert 01\rangle$ | As $\lvert\Phi^+\rangle$ above, then Z on qubit 0. |
ket-01.txt · Last modified: (external edit)