0
$\begingroup$

For two majorana field $\psi$ and $\chi$, which satisfy $\psi_{c}=\psi$ and $\chi_{c} = \chi$, where the charge-conjugation operation we define as $$ \Psi_{c} = C \Psi^{\ast} $$ where we work in the Weyl basis so that the matrix $C$ is given by $C = -i\gamma^2$.

It seems to me that the above implies $\bar{\psi}_{c} = ( C \psi^{\ast} )^{\dagger} \gamma^0 = \psi^{T} C^{\dagger} \gamma^0 = - \psi^{T} \gamma^0 C$, and so from this it follows that: $$ \bar{\psi} \chi = \psi^{\dagger} \gamma^0 \chi = \chi^{T} \gamma^0 \psi^{\ast} = \chi^{T} \gamma^0 C C \psi^{\ast} = - \left( - \chi^{T} \gamma^0 C \right) ( C \psi^{\ast} ) = - \bar{\chi}_c \psi_c = - \bar{\chi}\psi $$ where we have used $CC=I$, and the last equality uses the fact that these fields are Majorana fermions.

This implies that $\bar{\psi} \chi + \bar{\chi} \psi =0$ which seems wrong....where am I making a mistake?

$\endgroup$
  • 1
    $\begingroup$ For the record, the spinor flip in such bilinears is even for the scalar, pseudoscalar, and axial vector, and odd for the vector and tensor, unlike your expression. For $\chi=\psi$ you can sustain a Dirac mass term, unlike your expression. $\endgroup$ – Cosmas Zachos Feb 24 at 0:59
  • 1
    $\begingroup$ Also this near duplicate. $\endgroup$ – Cosmas Zachos Feb 24 at 14:56
  • 1
    $\begingroup$ ...and. $\endgroup$ – Cosmas Zachos Feb 24 at 15:08
1
$\begingroup$

I'm not adept in catching other people's mistakes, but I'll stick instead to the Bjorken & Drell v1 section 5.2 discussion and eliminate distractions. You are right in your instinct to look at the simplest bilinear, the scalar, and reject your answer, as it is odd, and $\psi=\chi$ would exclude a Dirac mass for Majorana spinors, which cannot be! So here is the trail map.

The conventions of that text, which are independent of the particular realizations and details all but guaranteed to bamboozle you, are $$ C=-C^{-1}= -C^T= -C^\dagger, $$ and imposing the Majorana condition yields $$\chi_c =C \bar{\chi}^{~T}=\chi, \qquad \psi_c= C \bar{\psi} ^{~T}=\psi ~~~~ \leadsto ~~~~\psi^T C =\bar{\psi}. $$

It then follows that $$ \bar{\psi} \chi = \psi^T C C \bar{\chi}^{~T} = -\psi^T \bar{\chi}^{~T}= \bar{\chi}\psi, $$ being a mere dot product of two vectors, and picking up a minus sign by anticommuting two Grassmann numbers at the very end.

From this point on, you may work out the evenness of the pseudoscalar and axial vector, as well, and the oddness of the vector and the tensor. It's all in the C properties of minus transposing gamma matrices.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you very much! I have been in a state of utter confusion for about a day now, and this resolves things. $\endgroup$ – QuantumEyedea Feb 24 at 16:48
  • $\begingroup$ I have a follow up question: why don't we care about dirac deltas when doing a swap in the order of the fields in the above? The equal time commutation relations $\{\psi_{a}(t,\mathbf{x}),\psi^{\dagger}_{b}(t,\mathbf{y})\} = \delta^{(3)}(x-y) \delta_{ab}$ would seem to indicate that there are some operators, for which you'd have to track a $\delta$-function. For example in doing an analogous calculation for $\bar{\psi}(x)\psi(x)$... $\endgroup$ – QuantumEyedea Feb 24 at 16:52
  • $\begingroup$ Yoou might have to give me an example of "not caring". In the above transposes, the two fermions are different. In a mass term, they are a symmetric expression, and Lagrangians don't care about quantum infinite constants. $\endgroup$ – Cosmas Zachos Feb 24 at 17:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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