In Peskin and Schroeder's QFT book, P.56 Eq.(3.95) mentions that
$$\begin{align} \langle 0|\bar\psi(y)_b\psi(x)_a|0\rangle = (\gamma \cdot p -m)_{ab}\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_p}Be^{ip(x-y)}\tag{3.95} \end{align}$$
When
$$\begin{align} \psi(x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}} \sum_s \big(a_p^su_p^s(p)e^{-ipx} + b_p ^{\dagger s} v^s(p)e^{ipx}\big) \end{align}$$ $$\begin{align} \bar\psi(y) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}} \sum_s \big(a_p^{\dagger s}\bar{u}_p^s(p)e^{ipy} + b_p^s\bar{v}^s(p)e^{-ipy}\big). \end{align}\tag{3.92}$$
The authors claim that letting $a_p^s|0\rangle = 0$ and $\langle 0|a_p^{\dagger s}$ and computing $\langle 0|\bar{\psi}_b(y)\psi_a(x)|0\rangle$ gives the term $\sum_s v^s\bar v^s = (\gamma \cdot p - m)$ in the middle of the calculation so the result is same as above.
But I found out that if I compute $\bar \psi(y) \psi(x)$ then the term with $\sum_s \bar v^s v^s$ emerge instead of $v^s \bar v^s$, which is equal to $2m$ from Eq(3.60) in the same book $\bar u^r u^s = 2m\delta^{rs}$.
And for another question I'm also curious about the index $_{ab}$ under the amplitude
$$\begin{align} \langle 0|\bar\psi(y)_b\psi(x)_a|0\rangle = (\gamma \cdot p -m)_{ab}\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_p}Be^{ip(x-y)}. \end{align}\tag{3.95}$$
Is it okay to write indices as $_{ab}$ instead of $_{ba}$? I know it's spinor index but not sure exactly.