Table of Contents
$\lvert\Psi^+\rangle$ (Psi-plus state)
The psi-plus state $\lvert\Psi^+\rangle$ is one of the four Bell states — the maximally entangled two-qubit states. It is an equal superposition of $\lvert 01\rangle$ and $\lvert 10\rangle$ with the same sign, meaning the two qubits always give opposite outcomes when measured in the computational basis. The other $\Psi$ Bell state is $\lvert\Psi^-\rangle$.
$$\lvert\Psi^+\rangle = \frac{1}{\sqrt{2}}(\lvert 01\rangle + \lvert 10\rangle) = \frac{1}{\sqrt{2}}\begin{pmatrix}0\\1\\1\\0\end{pmatrix}$$
It is prepared from $\lvert 01\rangle$ by applying $H\otimes I$ then CX, or equivalently from $\lvert 00\rangle$ by first flipping qubit 1 with $X$, then applying $H\otimes I$ and CX. Applying $X$ to either qubit of $\lvert\Phi^+\rangle$ also reaches $\lvert\Psi^+\rangle$. In the X basis, $\lvert\Psi^+\rangle = \tfrac{1}{\sqrt{2}}(\lvert ++\rangle - \lvert --\rangle)$, giving same-value X-basis correlations despite the anti-correlations in Z.
Qiskit
# Prepare |Ψ+⟩ = (|01⟩ + |10⟩)/√2 — X on qubit 1, then H on qubit 0, then CX(0→1). from qiskit import QuantumCircuit from qiskit.quantum_info import Statevector qc = QuantumCircuit(2) qc.x(1) # |00⟩ → |01⟩ qc.h(0) # |01⟩ → (|01⟩ + |11⟩)/√2 qc.cx(0, 1) # CX: → (|01⟩ + |10⟩)/√2 = |Ψ+⟩ print(Statevector(qc).data)
Applying gates
Single-qubit gates act on qubit 1. By symmetry of $\lvert\Psi^+\rangle$, applying $X$, $Y$, or $Z$ to qubit 2 gives the same result.
| Gate | Result | Comment |
|---|---|---|
| I gate | $I_1\lvert\Psi^+\rangle = \lvert\Psi^+\rangle$ | Identity leaves the state unchanged. |
| X gate | $X_1\lvert\Psi^+\rangle = \lvert\Phi^+\rangle$ | Flips qubit 1; $\lvert 01\rangle\to\lvert 11\rangle$ and $\lvert 10\rangle\to\lvert 00\rangle$, switching to same-value Z-basis correlations. |
| Y gate | $Y_1\lvert\Psi^+\rangle = i\lvert\Phi^-\rangle$ | Bit-flip with phase; the $-i$ on the $\lvert 00\rangle$ term introduces a relative minus sign, giving $\lvert\Phi^-\rangle$ up to global phase. |
| Z gate | $Z_1\lvert\Psi^+\rangle = \lvert\Psi^-\rangle$ | Negates the $\lvert 1\rangle$ component of qubit 1, which appears only in $\lvert 10\rangle$, flipping the relative sign to antisymmetric. |
| Hadamard gate | $H_1\lvert\Psi^+\rangle$ (not a Bell state) | Bell measurement circuit: CX then $H_1$ maps $\lvert\Psi^+\rangle\to\lvert 01\rangle$. |
| S gate | $S_1\lvert\Psi^+\rangle = \tfrac{1}{\sqrt{2}}(\lvert 01\rangle + i\lvert 10\rangle)$ | Adds a phase of $i$ to the $\lvert 10\rangle$ term; not a standard Bell state. |
| T gate | $T_1\lvert\Psi^+\rangle = \tfrac{1}{\sqrt{2}}(\lvert 01\rangle + e^{i\pi/4}\lvert 10\rangle)$ | Adds a phase of $e^{i\pi/4}$ to the $\lvert 10\rangle$ term; not a standard Bell state. |
| Rotation-X gate | Mixes $\lvert\Psi^+\rangle$ with $\lvert\Phi^+\rangle$ | X-type rotations couple the $\lvert\Psi^+\rangle$ and $\lvert\Phi^+\rangle$ Bell states. |
| Rotation-Y gate | Mixes $\lvert\Psi^+\rangle$ with $\lvert\Phi^-\rangle$ | Y-type rotations couple the $\lvert\Psi^+\rangle$ and $\lvert\Phi^-\rangle$ Bell states. |
| Rotation-Z gate | $R_z(\theta)_1\lvert\Psi^+\rangle = \tfrac{e^{-i\theta/2}}{\sqrt{2}}(\lvert 01\rangle + e^{i\theta}\lvert 10\rangle)$ | Modifies relative phase only; at $\theta=\pi$ gives $\lvert\Psi^-\rangle$ up to global phase. |
| CX (CNOT) gate | $\text{CX}\lvert\Psi^+\rangle = \lvert +\rangle\lvert 1\rangle$ | Disentangles $\lvert\Psi^+\rangle$ back to the product state used to prepare it. |
| SWAP gate | $\text{SWAP}\lvert\Psi^+\rangle = \lvert\Psi^+\rangle$ | Symmetric under qubit exchange; eigenstate of SWAP with eigenvalue $+1$. |
| iSWAP gate | $\text{iSWAP}\lvert\Psi^+\rangle = i\lvert\Psi^+\rangle$ | Both terms pick up a factor of $i$; eigenstate of iSWAP with eigenvalue $i$. |
Reaching other Bell states
| State | Gates | Comment |
|---|---|---|
| $\lvert\Psi^+\rangle$ | $I\lvert\Psi^+\rangle = \lvert\Psi^+\rangle$ | The identity gate leaves $\lvert\Psi^+\rangle$ unchanged. |
| $\lvert\Psi^-\rangle$ | $Z_1\lvert\Psi^+\rangle = \lvert\Psi^-\rangle$ | Z on qubit 1 flips the relative sign, making the state antisymmetric; the two qubits remain anti-correlated in Z. |
| $\lvert\Phi^+\rangle$ | $X_1\lvert\Psi^+\rangle = \lvert\Phi^+\rangle$ | X on either qubit switches from anti-correlated to same-value Z-basis correlations. |
| $\lvert\Phi^-\rangle$ | $Y_1\lvert\Psi^+\rangle = i\lvert\Phi^-\rangle$ | Y on qubit 1 reaches $\lvert\Phi^-\rangle$ up to global phase $i$. |