Table of Contents
$\lvert 00\rangle$
The $\lvert 00\rangle$ state is the two-qubit computational basis state where both qubits are $\lvert 0\rangle$. It is the tensor product $\lvert 0\rangle\otimes\lvert 0\rangle$ — a product state with no entanglement between the qubits. It is the default initial state of every qubit register in a quantum circuit.
$$\lvert 00\rangle = \lvert 0\rangle\otimes\lvert 0\rangle = \begin{pmatrix}1\\0\\0\\0\end{pmatrix}$$
Measuring either qubit of $\lvert 00\rangle$ yields $0$ with certainty and leaves the state unchanged. Applying $H\otimes I$ then CX prepares $\lvert\Phi^+\rangle$ directly; all four Bell states are reachable from $\lvert 00\rangle$ by that circuit plus at most one additional Pauli gate.
Qiskit
# Prepare |00⟩ — default state; Qiskit initializes all qubits to |0⟩, no gates needed. from qiskit import QuantumCircuit from qiskit.quantum_info import Statevector qc = QuantumCircuit(2) print(Statevector(qc).data)
Applying gates
Single-qubit gates act on qubit 0. By symmetry of $\lvert 00\rangle$, applying the same gate to qubit 1 gives the same result with the qubit labels swapped.
| Gate | Result | Comment |
|---|---|---|
| I gate | $I_0\lvert 00\rangle = \lvert 00\rangle$ | Identity leaves the state unchanged. |
| X gate | $X_0\lvert 00\rangle = \lvert 10\rangle$ | Flips qubit 0; the quantum analogue of flipping the first bit of a classical 00. |
| Y gate | $Y_0\lvert 00\rangle = i\lvert 10\rangle$ | Bit flip with an imaginary phase factor on qubit 0. |
| Z gate | $Z_0\lvert 00\rangle = \lvert 00\rangle$ | $\lvert 0\rangle$ is an eigenstate of $Z$ with eigenvalue $+1$; no observable effect. |
| Hadamard gate | $H_0\lvert 00\rangle = \tfrac{1}{\sqrt{2}}(\lvert 00\rangle + \lvert 10\rangle)$ | Puts qubit 0 into equal superposition; the two qubits remain unentangled. A subsequent CX entangles them into $\lvert\Phi^+\rangle$. |
| S gate | $S_0\lvert 00\rangle = \lvert 00\rangle$ | Phase gate only affects the $\lvert 1\rangle$ component; qubit 0 is in $\lvert 0\rangle$ so nothing changes. |
| T gate | $T_0\lvert 00\rangle = \lvert 00\rangle$ | Same as $S$: no effect when the target qubit is in $\lvert 0\rangle$. |
| Rotation-X gate | $R_x(\theta)_0\lvert 00\rangle = \cos\tfrac{\theta}{2}\lvert 00\rangle - i\sin\tfrac{\theta}{2}\lvert 10\rangle$ | Tilts qubit 0 from $\lvert 0\rangle$ toward $\lvert 1\rangle$ with an imaginary phase on the $\lvert 10\rangle$ component. |
| Rotation-Y gate | $R_y(\theta)_0\lvert 00\rangle = \cos\tfrac{\theta}{2}\lvert 00\rangle + \sin\tfrac{\theta}{2}\lvert 10\rangle$ | Real amplitudes; at $\theta=\pi/2$ gives $\tfrac{1}{\sqrt{2}}(\lvert 00\rangle + \lvert 10\rangle)$, the same result as $H_0$. |
| Rotation-Z gate | $R_z(\theta)_0\lvert 00\rangle = e^{-i\theta/2}\lvert 00\rangle$ | Global phase only; qubit 0 is on the $z$-axis so a $z$-rotation has no observable effect. |
| CX (CNOT) gate | $\text{CX}_{0\to 1}\lvert 00\rangle = \lvert 00\rangle$ | Control is $\lvert 0\rangle$; CX only fires on $\lvert 1\rangle$, so the target qubit is unchanged. |
| SWAP gate | $\text{SWAP}\lvert 00\rangle = \lvert 00\rangle$ | Swapping two $\lvert 0\rangle$ qubits leaves the state unchanged. |
| iSWAP gate | $\text{iSWAP}\lvert 00\rangle = \lvert 00\rangle$ | No $\lvert 01\rangle$ or $\lvert 10\rangle$ terms to exchange; the state is unchanged. |
Reaching other states
| State | Gates | Comment |
|---|---|---|
| $\lvert 00\rangle$ | $I\lvert 00\rangle = \lvert 00\rangle$ | The identity leaves $\lvert 00\rangle$ unchanged. |
| $\lvert 01\rangle$ | $X_1\lvert 00\rangle = \lvert 01\rangle$ | X on qubit 1 flips the second bit. |
| $\lvert 10\rangle$ | $X_0\lvert 00\rangle = \lvert 10\rangle$ | X on qubit 0 flips the first bit. |
| $\lvert 11\rangle$ | $X_0 X_1\lvert 00\rangle = \lvert 11\rangle$ | X on both qubits; quantum analogue of classical NOT NOT 00 = 11. |
| $\lvert\Phi^+\rangle$ | $\text{CX}\cdot(H_0\otimes I)\lvert 00\rangle$ | The standard Bell state preparation circuit: H on qubit 0, then CX. |
| $\lvert\Phi^-\rangle$ | $Z_0\cdot\text{CX}\cdot(H_0\otimes I)\lvert 00\rangle$ | Bell preparation then Z on qubit 0 flips the relative sign. |
| $\lvert\Psi^+\rangle$ | $\text{CX}\cdot(H_0\otimes I)\cdot X_1\lvert 00\rangle$ | Flip qubit 1 first to get $\lvert 01\rangle$, then apply the Bell preparation circuit. |
| $\lvert\Psi^-\rangle$ | $Z_0\cdot\text{CX}\cdot(H_0\otimes I)\cdot X_1\lvert 00\rangle$ | As $\lvert\Psi^+\rangle$ above, then Z on qubit 0. |