Singlet

Recall Bell States. Only ∣Ψ−⟩\ket{\Psi^-} is called a singlet. The rest are triplets. It is the only one which has a total spin of 0. The rest all have spin of 1

Note that

σAZσBZ∣Ψ−⟩=−∣Ψ−⟩\boxed{\sigma^Z_A \sigma^Z_B\ket{\Psi^-} = -\ket{\Psi^-}} σAXσBX∣Ψ−⟩=−∣Ψ−⟩\boxed{\sigma^X_A \sigma^X_B\ket{\Psi^-} = -\ket{\Psi^-}}

This is because if we measure on Z

σZ∣↑⟩=(100−1)(10)=(10)=+∣↑⟩\sigma^Z\ket{\uparrow} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} = +\ket{\uparrow}

+1 is an eigenvalue of σZ∣↑⟩\sigma^Z\ket{\uparrow} -1 is an eigenvalue of σZ∣↓⟩\sigma^Z\ket{\downarrow} acting on

σAZσBZ∣↑↓⟩=−∣↑↓⟩\sigma^Z_A \sigma^Z_B \ket{\uparrow\downarrow} = -\ket{\uparrow\downarrow} σAZσBZ∣↓↑⟩=−∣↓↑⟩\sigma^Z_A \sigma^Z_B \ket{\downarrow\uparrow} = -\ket{\downarrow\uparrow}

acting on the singlet

σAZσBZ∣Ψ−⟩=12(−∣↑↓⟩+∣↓↑⟩)=−∣Ψ−⟩\sigma^Z_A \sigma^Z_B \ket{\Psi^-} = \frac{1}{\sqrt{2}}(-\ket{\uparrow \downarrow}+\ket{\downarrow \uparrow}) =-\ket{\Psi^-}

combining we get

σAXσBZ∣Ψ−⟩=−∣Φ+⟩\sigma^X_A \sigma^Z_B\ket{\Psi^-} = -\ket{\Phi_+} σAZσBX∣Ψ−⟩=+∣Φ+⟩\sigma^Z_A \sigma^X_B\ket{\Psi^-} = +\ket{\Phi_+}

Getting the EV

⟨Ψ−∣σAZσBZ∣Ψ−⟩=−1,⟨Ψ−∣σAXσBX∣Ψ−⟩=−1\langle\Psi^-|\sigma^Z_A \sigma^Z_B|\Psi^-\rangle = -1, \quad \langle\Psi^-|\sigma^X_A \sigma^X_B|\Psi^-\rangle = -1 E[σAZσBZ]=−1E[σAXσBX]=−1\boxed{\mathbb{E}\left[\sigma^Z_A \sigma^Z_B\right] = -1\quad \mathbb{E}\left[\sigma^X_A \sigma^X_B\right]= -1}

These are perfect anti-correlations

Cross terms

We can also look at the cross terms

⟨Ψ−∣σAXσBZ∣Ψ−⟩=0,⟨Ψ−∣σAZσBX∣Ψ−⟩=0\langle\Psi^-|\sigma^X_A \sigma^Z_B|\Psi^-\rangle = 0, \quad \langle\Psi^-|\sigma^Z_A \sigma^X_B|\Psi^-\rangle = 0 E[σAXσBZ]=0,E[σAZσBX]=0\boxed{\mathbb{E}[\sigma^X_A \sigma^Z_B] = 0, \quad \mathbb{E}[\sigma^Z_A \sigma^X_B]= 0}

Cross terms vanish