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?

  • 2
    $\begingroup$ 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. $\endgroup$
    – knzhou
    Commented Nov 4, 2023 at 19:37
  • 2
    $\begingroup$ 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 $b_j a_i$. $\endgroup$
    – knzhou
    Commented Nov 4, 2023 at 19:38
  • $\begingroup$ Now that's clear, that was a silly mistake by me :c. Thank you so much! $\endgroup$ Commented Nov 4, 2023 at 19:42


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