I want to calculate the Dirac term from the canonical Kahler potential, \begin{equation} K = \Phi ^\ast \Phi \tag{1} \end{equation} but I'm coming across a pesky negative sign in the final result. I am finding (see derivation below), \begin{equation} - i \bar{\psi} \bar{\sigma} ^\mu \partial _\mu \psi \tag{2} \end{equation} This agrees with lecture notes by Matteo Bertolini (pg. 87) but is off by sign from lecture notes by Fernando Quevedo (pg. 50), while they both supposedly use the same conventions.

I would be okay with deciding Quevedo's notes have an error but my result also seems to contradict the regular Dirac Lagrangian, which in four-vector notation is \begin{equation} {\cal L} _D = + i\bar{\Psi} \gamma ^\mu \partial _\mu \Psi \tag{3} \end{equation} Any ideas where this problem is coming from?

Here are my steps:

The chiral and antichiral superfields take the form (I am using the (+---) metric), \begin{align} & \Phi = \phi + \sqrt{2} \theta \psi + \theta ^2 F + i \theta \sigma ^\mu \bar{\theta} \partial _\mu \phi - \frac{ i }{ \sqrt{2} } \theta ^2 \partial _\mu \psi \sigma ^\mu \bar{\theta} - \frac{1}{4} \theta ^2 \bar{\theta} ^2 \Box \phi \tag{4}\\ & \Phi ^\ast = \phi ^\ast + \sqrt{2} \bar{\psi} \bar{\theta} + \bar{\theta} ^2 F ^\ast - i \theta \sigma ^\mu \bar{\theta} \partial _\mu \phi ^\ast + \frac{ i }{ \sqrt{2} } \bar{\theta} ^2 \theta \sigma ^\mu \partial _\mu \bar{\psi} - \frac{1}{4} \theta ^2 \bar{\theta}^2 \Box \phi ^\ast \tag{5} \end{align} Calculating the Dirac term involves the product of the second term of $ \Phi ^\ast $ and the fifth term of $ \Phi $ and vice versa. I find: \begin{align} - i \bar{\psi} \bar{\theta} \theta ^2 \partial _mu\psi \sigma ^\mu \bar{\theta} & = - i \partial _\mu \psi ^\alpha \sigma ^\mu _{ \alpha \dot{\alpha} } \bar{\psi} _{\dot{\beta}} \bar{\theta} ^{\dot{\beta}} \bar{\theta} ^{\dot{\alpha}} \theta ^2 \tag{6}\\ & = \frac{ i }{ 2} \partial _\mu \psi ^\alpha \sigma ^\mu _{ \alpha \dot{\alpha} } \bar{\psi} ^{\dot{\alpha}} \bar{\theta} ^2 \theta ^2 \tag{7}\\ & = - \frac{ i }{ 2} \bar{\psi} \bar{\sigma} ^\mu \partial _\mu \psi \bar{\theta} ^2 \theta ^2 \tag{8} \end{align} where in the last step I used the spinor identity, $ \psi \sigma ^\mu \bar{\chi} = - \bar{\chi} \bar{\sigma} ^\mu \psi $. Repeating the calculation for the product of the fifth term of $ \Phi ^\ast $ and the second term of $ \Phi $ and summing the two results gives: \begin{equation} - i \bar{\psi} \bar{\sigma} ^\mu \partial _\mu \psi \bar{\theta} ^2 \theta ^2 \tag{9} \end{equation} which after stripping off the $\theta,\bar{\theta}$ is the result I quote above.

  • $\begingroup$ I apologize for asking a question about a silly minus sign but I can't seem to solve it myself. $\endgroup$
    – JeffDror
    Jan 9 '14 at 15:07
  • 5
    $\begingroup$ Good physicists make a even number of sign errors. $\endgroup$
    – Trimok
    Jan 9 '14 at 19:21
  • $\begingroup$ @Trimok: Indeed, or they pick the most convenient sign :) I often say to myself, "if this were a minus everything would cancel perfectly and look much simpler; I must have made a sign error!" $\endgroup$
    – JamalS
    Jun 10 '14 at 11:51

The final sign $- i \bar{\psi} \bar{\sigma} ^\mu \partial _\mu \psi$ seems correct if we look at Matteo Bertolini, formula $5.2$, page $72$, just notice that the order of $\psi$ and $\bar \psi$ is inverted in the formula and apply $\frac{i}{2}\partial_\mu \psi \sigma ^\mu \bar{\psi} = -\frac{i}{2} \bar{\psi} \bar{\sigma} ^\mu \partial_\mu \psi$.

  • $\begingroup$ Thanks for you response. I fixed the typos. 1) I don't see a reason why the equation you mentioned (now labelled equation 7) is wrong. I use the identity you mentioned above for the product of Grassman variables and then used the Levi Cevita symbol to raise the index of $\psi_{\dot{\beta}}$. 2)I agree with you that this result agrees with Bertolini's notes. However, I don't think it agrees with Quevedo's notes (bottom of pg 50) nor with the typical expression for the Dirac Lagrangian which both quote a '+' sign in the final expression (see Peskin chapter 3.5 eq 3.83 for Dirac Larangian). $\endgroup$
    – JeffDror
    Jan 9 '14 at 21:14
  • $\begingroup$ You have to integrate on the $\bar \theta$. I mean, it is not correct to say $ \bar{\theta} ^{\dot{\beta}} \bar \theta ^{\dot{\alpha}} =-\frac{1}{2}\epsilon^{\dot\alpha \dot\beta}$. Look at page $46$ in Quevedo's notes, you have the same expression that in Bertolini. $\endgroup$
    – Trimok
    Jan 9 '14 at 21:36
  • 1
    $\begingroup$ Thanks again. Sorry, I don't think I was clear. The identity I was using was $\bar{\theta}^{\dot{\alpha}}\bar{\theta}^{\dot{\beta}} = -\frac{1}{2}\epsilon ^{\dot{\beta}\dot{\alpha}}\bar{\theta}^2$. Looking at pg 46 I see the result does agree with Quevedo's on this page. Though he does contradict himself later... If my result is correct, I still don't understand though why this seems to contradict the typical Dirac Lagrangian though which I'm pretty sure has a positive sign, ${\cal L} _D=i \bar{\Psi} \gamma ^\mu \partial _\mu \Psi = i \psi _L \gamma ^\mu \partial _\mu\psi _L + ... $ $\endgroup$
    – JeffDror
    Jan 9 '14 at 23:08
  • $\begingroup$ @JeffDror : You are right, your expression is correct... I delete the part of my answer about this. For the sign, you compare with the Dirac equation which is using gamma matrices, not Pauli matrices. So, it is not a problem, I think. $\endgroup$
    – Trimok
    Jan 10 '14 at 9:39
  • $\begingroup$ Sorry, I had a typo is my last comment again. Though I think the gamma matrix and Pauli matrix Dirac equation have the same sign since The explicit Lagrangian in terms of Pauli matrices is given by (in the Weyl basis) $i \bar{\Psi} \partial _\mu \gamma ^\mu \Psi = i \bar{\psi} _L \partial _\mu \bar{\sigma} ^\mu \psi _L + i \bar{\psi} _R \partial _\mu \sigma ^\mu \psi _R $ (I denote the four-component Dirac equation with $\Psi $ and the two component with $\psi$), which unfortunately again has a positive sign. $\endgroup$
    – JeffDror
    Jan 10 '14 at 12:05

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