# Error in Peskin-Schroeder calculation? ("The Dirac Propagator equation (3.115) )

I was trying to calculate $$\langle0|\bar{\psi}(y) \psi(x)|0 \rangle$$

where the wave-function operator is $$\psi(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{2E_P} \sum_{r=1}^{2} \left( a_p^r u^r(p) e^{-ipx} + b_p^{r,\dagger} v^r(p) e^{ipx}\right)$$

Peskin reports that this object is equal to: $$\int \frac{d^3p}{(2\pi)^3} \frac{1}{2E_P} \sum_{r=1}^{2} e^{-ip(y-x)} v^r(p) \bar{v}^r(p)$$

But I get a different order for my spinors: $$\int \frac{d^3p}{(2\pi)^3} \frac{1}{2E_P} \sum_{r=1}^{2} e^{-ip(y-x)} \bar{v}^r(p) v^r(p)$$

I get this because when I deal with $$\psi(x)|0 angle$$ the only term that survives is linked to the fact that $$b_p^{s,\dagger} |0\rangle$$ is different from zero. But when I act on this vector with $$\langle0| \bar{\psi}(y)$$ only the term that multiplies $$b_p^{s}$$ survives because now $$\langle0| b_p^{s}$$ is not zero. So I have $$\bar{v}^r(p) v^r(p)$$ and not $$v^r(p) \bar{v}^r(p)$$. This changes everything because the first one gives me -2m$$\delta_{rs}$$ and the second one $$\gamma^u p_u - m$$. What am I doing wrong?

• That's not what Peskin is computing. He's computing $\langle 0 | \bar{\psi}_b(y) \psi_a(x) | 0 \rangle$, so Peskin's result has $v^r(p)_a \bar{v}^r(p)_b$. Of course, these two quantities are just numbers, so you can exchange their order to get what you have. Commented Nov 4, 2023 at 19:37
• If this is hard to see, just think about how it works for ordinary vectors. Obviously, $\mathbf{a}^T \mathbf{b}$ is not the same thing as $\mathbf{b} \mathbf{a}^T$. But Peskin is computing the analogue of $a_i b_j$ which is of course equal to $b_j a_i$. Commented Nov 4, 2023 at 19:38
• Now that's clear, that was a silly mistake by me :c. Thank you so much! Commented Nov 4, 2023 at 19:42