Here are the two Feynman diagrams that corresponds to $e^- + e^+ \rightarrow e^- + e^+$ processes: electron positron scattering and electron positron annihilation respectively.
I can clearly see the final results should be in the form of
$$S_{fi} \propto (\bar{v}\gamma_{\nu}{v})(\bar{u}\gamma_{\mu}{u})$$ $$S_{fi} \propto (\bar{u}\gamma_{\mu}{v})(\bar{v}\gamma_{\nu}{u}).$$
however when I expand the field operators in terms of negative and positive energy fields (momenta and spin subscripts are omitted), I run into a conceptual problem.
$$S_{fi} = \langle{0 | d b : \bar{\psi_1}\gamma_{\mu}{\psi_1}\bar{\psi_2}\gamma_{\nu}{\psi_2}: \underbrace{A_{1}^{\mu}A_{2}^{\nu}} d^{\dagger}b^{\dagger}} |0 \rangle $$
$$S_{fi} = \langle{e^- e^+ | : (\bar{\psi^+_1}+\bar{\psi^-_1})\gamma_{\mu}({\psi^+_1}+{\psi^-_1})(\bar{\psi^+_2}+\bar{\psi^-_2})\gamma_{\nu}({\psi^+_2}+{\psi^-_2}): \underbrace{A_{1}^{\mu}A_{2}^{\nu}}} | e^- e^+ \rangle$$
$$ =\langle{e^- e^+ | (\bar{\psi^-_1})\gamma_{\mu}({\psi^-_1})(\bar{\psi^+_2})\gamma_{\nu}({\psi^+_2}) \underbrace{A_{1}^{\mu}A_{2}^{\nu}}} | e^- e^+ \rangle $$
which produces the correct result for the $s$-channel, while for the $t$-channel I am supposed to get the following based on what I expect from the first equation
$$\langle{e^- e^+ | (\bar{\psi^+_1})\gamma_{\nu}({\psi^-_1})(\bar{\psi^-_2})\gamma_{\mu}({\psi^+_2}) \underbrace{A_{1}^{\mu}A_{2}^{\nu}}} | e^- e^+ \rangle. $$
But I don't understand how $\bar{\psi^+_1}$ term with an annihilation operator can act on the final state after normal ordering.