Table of Contents
$\lvert\Psi^-\rangle$ (Psi-minus state)
The psi-minus state $\lvert\Psi^-\rangle$ is one of the four Bell states — the maximally entangled two-qubit states. It is the only antisymmetric Bell state: swapping the two qubits picks up a minus sign, $\text{SWAP}\lvert\Psi^-\rangle = -\lvert\Psi^-\rangle$. It is also called the singlet state, because it is the spin-0 singlet of SU(2). 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 $\lvert\Psi^+\rangle$ by applying $Z$ to qubit 1, or directly as $\text{CX}\lvert -\rangle\lvert 1\rangle = \lvert\Psi^-\rangle$. The antisymmetry makes $\lvert\Psi^-\rangle$ rotationally invariant: in any orthonormal basis the state takes the form $\tfrac{1}{\sqrt{2}}(\lvert\uparrow\downarrow\rangle - \lvert\downarrow\uparrow\rangle)$, so measurements are anti-correlated in every basis.
Qiskit
# Prepare |Ψ−⟩ = (|01⟩ − |10⟩)/√2 — same as |Ψ+⟩ preparation, then Z on qubit 0. 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+⟩ qc.z(0) # Z on qubit 0: |Ψ+⟩ → |Ψ−⟩ print(Statevector(qc).data)
Applying gates
Single-qubit gates act on qubit 1. By antisymmetry, applying $X$, $Y$, or $Z$ to qubit 2 gives the same result up to a global phase.
| 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 $\lvert 10\rangle\to\lvert 00\rangle$ with the minus sign preserved. Equivalent to $\lvert\Phi^-\rangle$ up to global phase. |
| Y gate | $Y_1\lvert\Psi^-\rangle = i\lvert\Phi^+\rangle$ | The $-i$ on the $\lvert 00\rangle$ term cancels the minus sign in $\lvert\Psi^-\rangle$, leaving same-sign terms. Equivalent to $\lvert\Phi^+\rangle$ up to global phase. |
| Z gate | $Z_1\lvert\Psi^-\rangle = \lvert\Psi^+\rangle$ | Negates the $\lvert 10\rangle$ term, cancelling the relative minus sign and recovering the symmetric state. |
| Hadamard gate | $H_1\lvert\Psi^-\rangle$ (not a Bell state) | Bell measurement circuit: CX then $H_1$ maps $\lvert\Psi^-\rangle\to\lvert 11\rangle$. |
| S gate | $S_1\lvert\Psi^-\rangle = \tfrac{1}{\sqrt{2}}(\lvert 01\rangle - i\lvert 10\rangle)$ | S adds $i$ to the $\lvert 10\rangle$ term; combined with the minus sign gives $-i$. 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; combined with the minus sign. 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 | $\text{SWAP}\lvert\Psi^-\rangle = -\lvert\Psi^-\rangle$ | Antisymmetric under qubit exchange; the only Bell state with SWAP eigenvalue $-1$. |
| iSWAP gate | $\text{iSWAP}\lvert\Psi^-\rangle = -i\lvert\Psi^-\rangle$ | The two factors of $i$ from the exchange combine with the antisymmetry to give 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 cancels the relative minus sign, converting the antisymmetric state to the symmetric $\lvert\Psi^+\rangle$. |
| $\lvert\Phi^+\rangle$ | $Y_1\lvert\Psi^-\rangle = i\lvert\Phi^+\rangle$ | Y on qubit 1 reaches $\lvert\Phi^+\rangle$ up to global phase $i$. |
| $\lvert\Phi^-\rangle$ | $X_1\lvert\Psi^-\rangle = -\lvert\Phi^-\rangle$ | X on qubit 1 reaches $\lvert\Phi^-\rangle$ up to global phase $-1$. |