I have troubles trying to prove a statement made by Peskin-Schroeder in page 61, section 3.5 where he says that the "spin" operator $J_z$ given by the non numbered equation

$$ J_z= \int d^3 x ~\psi^{\dagger} \frac{\Sigma^3}{2} \psi = \int d^3 x \int \frac{d^3 p ~ d^3q}{(2\pi)^6} \frac{1}{2\sqrt{E_p E_q}} e^{i x\cdot(p-q)} \\ \times \sum_{r,s} (a_{\bf q}^{s \dagger} u^{s}({\bf q})^{\dagger} + b_{-\bf q}^{s} v^{s \dagger}(-{\bf q})) \frac{\Sigma^3}{2} (a^r_{\bf p} u^{r}({\bf p} ) + b^{r \dagger}_{-{\bf p}} v^r (-{\bf p})) $$ annihilates the fermionic vacuum state $|0 \rangle$, so that $J_z |0 \rangle =0$. Where $E_p = \sqrt{m^2 +{\bf p}^2}$ and $u$ and $v$ are the basis of spinor solutions in momentum space of the Dirac equation, they are explicitely:


$$ u^{r}( {\bf p}) = \left( \begin{array}{c} \sqrt{p \cdot \sigma } \xi^s \\ \sqrt{ p \cdot \overline{\sigma}} \xi^s \end{array} \right), \quad v^{r}( {\bf p}) = \left( \begin{array}{c} \sqrt{p \cdot \sigma }\xi^s \\ -\sqrt{ p \cdot \overline{\sigma}}\xi^s \end{array} \right) $$ above $\xi^s$ is the canonical $\mathbb{C}^2$ basis and $\sigma = ({\bf 1}, \vec{\sigma}), ~~\overline{\sigma} = ({\bf 1}, \vec{\sigma})$. The vacuum state is annihilated by the ladder operators $a^r_{\bf p}|0\rangle = b^{r}_{\bf p}|0 \rangle = 0$ and the matrix $\Sigma^3$ is given by

$$ \Sigma^3 =\left( \begin{array}{cc} \sigma_3 & 0 \\ 0 & \sigma_3 \end{array} \right) $$ with Pauli matrix $\sigma_3$.


So far I have managed to show that all terms vanish except for this one (after performing the $x$ integration and then the $q$ integration), where I have failed miserably:

$$ \frac{1}{(2\pi)^3} \int \frac{d^3p}{2 E_p} \sum_{r,s} u^{s}({\bf p})^{\dagger}% \frac{\Sigma^3}{2}v^{r}(-{\bf p}) a_{\bf p}^{s \dagger}b_{-{\bf p}}^{r \dagger} |0\rangle =0?$$

How can I prove that this is vanishing? Furthermore, since the $z$ direction is not preferential, this should vanish if I replace $\sigma_3 \to \sigma_{1,2}$ also.


1 Answer 1


Take a look at equation (3.65) on page 48.

While the barred spinors $\bar{u}^s$ are properly orthogonal to $v^r$ \begin{align} \bar{u}^s(p)v^r(p) = u^{s\dagger}(p)\gamma^0v^r(p)=0 \end{align} the hermitian conjugated $u^{s\dagger}$ obey

$$ u^{s\dagger}(\mathbf{p})v^r(\mathbf{-p})=0$$ where $(-\mathbf{p})=(E_p,-\mathbf{p})$ is imho. a bit of an unfortunate notation. Once you have obtained this result, it's not hard to see that also $$ u^{s\dagger}(\mathbf{p})\Sigma^3 v^r(\mathbf{-p})=0 $$

  • $\begingroup$ Thanks, I checked by hand $\Sigma^1$ and $\Sigma^3$ but still no luck with $\Sigma^2$, is there any reason why all $\Sigma^i$ hold this property? I mean an explicit way to do it, without recoursing to rotational invariance of the original expression. $\endgroup$ Commented Mar 31, 2015 at 17:37
  • $\begingroup$ @RogelioMolina I'll come back to you tomorrow and expand my answer. $\endgroup$
    – Nephente
    Commented Mar 31, 2015 at 19:35

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