I am trying to solve an exercise on deriving the equal-time anti-commutator of the Dirac fields. But I got stuck somewhere and couldn't get the desired result.
I would like to show that $$ \{\psi(x), \psi(y) \} = \gamma^0 \delta^{(3)}(\vec{x} - \vec{y}). $$ It has been shown in previous exercises that $$ \{\psi^\pm (x), \overline{\psi}^\mp(y) \} = i(i\gamma^\mu \partial_\mu + m) \Delta^\pm (x-y). $$
The following is what I got: $$ \begin{aligned} \{\psi(x), \overline{\psi}(y) \} &= \{\psi^+(x), \overline{\psi}^-(y) \} + \{\psi^-(x), \overline{\psi}^+(y) \} \\ &= i(i\gamma^\mu \partial_\mu + m) \Delta^+ (x-y) + i(i\gamma^\mu \partial_\mu - m) \Delta^- (x-y) \\ &= i(i\gamma^\mu \partial_\mu + m) \int\frac{d\vec{p}}{(2\pi)^3 2\omega(\vec{p})} e^{-ip(x-y)} + i(i\gamma^\mu \partial_\mu + m) \ \int\frac{d\vec{p}}{(2\pi)^3 2\omega(\vec{p})} e^{ip(x-y)}. \end{aligned} $$
I have no idea how to proceed after this.
The mode expansion of the Dirac fields is given by: $$ \psi(x) = \int\frac{d\vec{p}}{(2\pi)^3 2\omega(\vec{p})} a_s(p)u(p,s) e^{-ipx} + b^\dagger_s(p) v(p,s) e^{ipx}. $$
