The $\lvert 10\rangle$ state is a two-qubit computational basis state with qubit 0 in $\lvert 1\rangle$ and qubit 1 in $\lvert 0\rangle$. It is the tensor product $\lvert 1\rangle\otimes\lvert 0\rangle$ — an unentangled product state. It is the SWAP of $\lvert 01\rangle$, reached from $\lvert 00\rangle$ by a single $X$ gate on qubit 0.
$$\lvert 10\rangle = \lvert 1\rangle\otimes\lvert 0\rangle = \begin{pmatrix}0\\0\\1\\0\end{pmatrix}$$
Measuring qubit 0 yields $1$ with certainty; measuring qubit 1 yields $0$ with certainty. Applying $H\otimes I$ then CX maps $\lvert 10\rangle$ directly to $\lvert\Phi^-\rangle$, because $H\lvert 1\rangle = \lvert -\rangle$ and $\text{CX}\lvert -\rangle\lvert 0\rangle = \lvert\Phi^-\rangle$.
# Prepare |10⟩ — flip qubit 0 from the default |00⟩ state. from qiskit import QuantumCircuit from qiskit.quantum_info import Statevector qc = QuantumCircuit(2) qc.x(0) # |00⟩ → |10⟩ print(Statevector(qc).data)
Single-qubit gates act on qubit 0, which is in $\lvert 1\rangle$. Qubit 1 is in $\lvert 0\rangle$, so gates on qubit 1 follow the $\lvert 0\rangle$ single-qubit rules with qubit 0 unchanged.
| Gate | Result | Comment |
|---|---|---|
| I gate | $I_0\lvert 10\rangle = \lvert 10\rangle$ | Identity leaves the state unchanged. |
| X gate | $X_0\lvert 10\rangle = \lvert 00\rangle$ | Flips qubit 0 from $\lvert 1\rangle$ to $\lvert 0\rangle$; qubit 1 is unchanged. |
| Y gate | $Y_0\lvert 10\rangle = -i\lvert 00\rangle$ | Bit flip with a phase factor; $Y\lvert 1\rangle = -i\lvert 0\rangle$. |
| Z gate | $Z_0\lvert 10\rangle = -\lvert 10\rangle$ | Qubit 0 is in $\lvert 1\rangle$, an eigenstate of $Z$ with eigenvalue $-1$; the overall sign flips. |
| Hadamard gate | $H_0\lvert 10\rangle = \tfrac{1}{\sqrt{2}}(\lvert 00\rangle - \lvert 10\rangle)$ | $H\lvert 1\rangle = \lvert -\rangle$; qubit 0 enters the $\lvert -\rangle$ superposition. A subsequent CX produces $\lvert\Phi^-\rangle$. |
| S gate | $S_0\lvert 10\rangle = i\lvert 10\rangle$ | $S\lvert 1\rangle = i\lvert 1\rangle$; the result is $\lvert 10\rangle$ up to global phase $i$. |
| T gate | $T_0\lvert 10\rangle = e^{i\pi/4}\lvert 10\rangle$ | $T\lvert 1\rangle = e^{i\pi/4}\lvert 1\rangle$; global phase only. |
| Rotation-X gate | $R_x(\theta)_0\lvert 10\rangle = -i\sin\tfrac{\theta}{2}\lvert 00\rangle + \cos\tfrac{\theta}{2}\lvert 10\rangle$ | Tilts qubit 0 from $\lvert 1\rangle$ toward $\lvert 0\rangle$ with an imaginary phase on the $\lvert 00\rangle$ component. |
| Rotation-Y gate | $R_y(\theta)_0\lvert 10\rangle = -\sin\tfrac{\theta}{2}\lvert 00\rangle + \cos\tfrac{\theta}{2}\lvert 10\rangle$ | Real amplitudes; at $\theta=\pi/2$ gives $\tfrac{1}{\sqrt{2}}(-\lvert 00\rangle + \lvert 10\rangle)$, same as $H_0$ up to a sign. |
| Rotation-Z gate | $R_z(\theta)_0\lvert 10\rangle = e^{i\theta/2}\lvert 10\rangle$ | Global phase only; $R_z(\theta)\lvert 1\rangle = e^{i\theta/2}\lvert 1\rangle$. |
| CX (CNOT) gate | $\text{CX}_{0\to 1}\lvert 10\rangle = \lvert 11\rangle$ | Control is qubit 0 = $\lvert 1\rangle$; CX fires and flips qubit 1 from $\lvert 0\rangle$ to $\lvert 1\rangle$. |
| SWAP gate | $\text{SWAP}\lvert 10\rangle = \lvert 01\rangle$ | Exchanges the two qubits. |
| iSWAP gate | $\text{iSWAP}\lvert 10\rangle = i\lvert 01\rangle$ | Exchanges $\lvert 10\rangle$ and $\lvert 01\rangle$ with a phase of $i$. |
| State | Gates | Comment |
|---|---|---|
| $\lvert 10\rangle$ | $I\lvert 10\rangle = \lvert 10\rangle$ | The identity leaves $\lvert 10\rangle$ unchanged. |
| $\lvert 00\rangle$ | $X_0\lvert 10\rangle = \lvert 00\rangle$ | X on qubit 0 flips it from $\lvert 1\rangle$ back to $\lvert 0\rangle$. |
| $\lvert 11\rangle$ | $X_1\lvert 10\rangle = \lvert 11\rangle$ | X on qubit 1 flips the second bit. |
| $\lvert 01\rangle$ | $X_0 X_1\lvert 10\rangle = \lvert 01\rangle$ | X on both qubits; equivalent to SWAP. |
| $\lvert\Phi^-\rangle$ | $\text{CX}\cdot(H_0\otimes I)\lvert 10\rangle$ | Bell preparation: $H\lvert 1\rangle = \lvert -\rangle$, then CX produces $\lvert\Phi^-\rangle$ directly. |
| $\lvert\Phi^+\rangle$ | $Z_0\cdot\text{CX}\cdot(H_0\otimes I)\lvert 10\rangle$ | Bell preparation then Z on qubit 0 cancels the relative sign. |
| $\lvert\Psi^-\rangle$ | $\text{CX}\cdot(H_0\otimes I)\cdot X_1\lvert 10\rangle$ | Flip qubit 1 first to reach $\lvert 11\rangle$, then apply Bell preparation. |
| $\lvert\Psi^+\rangle$ | $Z_0\cdot\text{CX}\cdot(H_0\otimes I)\cdot X_1\lvert 10\rangle$ | As $\lvert\Psi^-\rangle$ above, then Z on qubit 0. |