3
$\begingroup$

In (3.87, page. 53) Peskin and Schroeder write $$\psi(\vec{x}) = \int\frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2E_{\vec{p}}}} e^{i\vec{p} \cdot \vec{x}} \sum_{s=1,2} (a_{\vec{p}}^{s}u^{s}(\vec{p}) + b_{-\vec{p}}^{s}v^{s}(-\vec{p})) $$ and then compute

$$[\psi(\vec{x}),\psi^{\dagger}(\vec{y})]=\int\frac{d^3p d^3q}{(2\pi)^{6}} \frac{1}{\sqrt{2E_{\vec{p}} 2E_{\vec{q}}}} e^{i(\vec{p} \cdot {\vec{x}}-\vec{q} \cdot {\vec{y}})} \sum_{r,s} ([a_{\vec{p}}^{r},a_{\vec{q}}^{s \dagger}] u^{r}(\vec{p}) \bar{u}^{s}(\vec{q}) + [b_{\vec{p}}^{r},b_{\vec{q}}^{s \dagger}] v^{r}(-\vec{p}) \bar{v}^{s}(-\vec{q})) \gamma^{0}$$

Where does the $ \gamma^{0} $ at the end come from?

Improve this question
$\endgroup$
1
  • 8
    $\begingroup$ At a glance: $\overline{\psi} =\psi^\dagger \gamma^0$, so since you have $\psi^\dagger$ in the LHS and $\overline{u}, \overline{v}$ in the RHS, a $\gamma^0$ has to show up to account for the difference between dagger and bar. $\endgroup$
    – Robin Ekman
    Commented Jul 12, 2014 at 3:53

1 Answer 1

Reset to default
3
$\begingroup$

$u^\dagger = \bar u \, \gamma^0$ and similarly for $v^\dagger$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.