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In lectures on the standard model I recently saw, in writing down the SM Lagrangian the professor was careful to refer to fields like $e^c$ as a conjugate electron rather than an antielectron. He also mentioned that all the fields he wrote down were left-handed Weyl spinors.

First Question: What is the relationship between conjugate electrons and antielectrons?

Second Question: Are fields in the SM Lagrangian left-handed spinors by convention, or is this physically meaningful?

Third Question: Is there any connection between charge conjugation and handedness of a spinor?

Yly
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2 Answers2

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  1. Given a Dirac spinor $\psi$, its "charge conjugate" spinor is given by $\psi^c = C\psi^\ast$, where $C$ is a charge conjugation matrix defined by a certain convention (e.g. $C^\dagger \gamma^\mu C = -(\gamma^\mu)^\ast$). Meanwhile, the anti-particles are associated with the Dirac adjoint (or Driac conjugate) spinor $\bar{\psi} = \psi^\dagger \gamma^0$.

  2. Electrons are not left-handed, they are massive Dirac spinors with both left- and right-handed components. The SM neutrinos are massless and usually thought to be only left-handed, but there are plenty of ideas for sterile right-handed neutrinos in formulations with massive neutrinos beyond the SM.

  3. No. You can charge conjugate a Dirac spinor as well as just a Weyl spinor of either chirality. The charge conjugation matrix for Weyl spinors is just given by projecting $C$ down to the subspace of the given chirality.

ACuriousMind
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3, The charge conjugation $$ C: \Psi \to -i \gamma^2 \Psi^* = \Psi^c $$ exchanges the chirality. By the way, this is what $e^c$ means. To see this, note that a Dirac spinor can be regarded as a direct sum of two Weyl spinors $$ \Psi = \begin{pmatrix} \psi_L \\ \bar \chi_R \end{pmatrix}. $$ For the Weyl spinor, I wrote $\bar \chi = \chi^*$. Then $$ C: \Psi \to -i \gamma^2 \Psi^* = \begin{pmatrix} 0& -i \sigma^2 \\ i \sigma^2 & 0 \end{pmatrix} \begin{pmatrix} \bar \psi_L \\ \chi_R \end{pmatrix} = \begin{pmatrix} \chi_L \\ \bar \psi_R \end{pmatrix}. $$ You can verify this using van der Waerden indices, where $-i \sigma^2 = \epsilon$ is the antisymmetric tensor making the Lorentz spinor indices correct.

2, As a result, $C$ not only filps the chirality but also makes the field complex conjutate. Thus, a complex representation $\bf R$ became its complex conjugate $\overline{\bf R}$. Or, the $U(1)$ charge $q$ became $-q$.

They are equivalent in the sense that, you may use either $\Psi$ or $\Psi^c = -i \gamma^2 \Psi^*$ in your model, because only one of them has the desired chirality and gauge charge.

This means, we can always choose the chirality of the defining field. Equivalently, we can make all the chiral field into the left-handed. This is what your professor wanted to do. Especially this can be done equivalently to Weyl spinors to construct the holomopic superfield in the supersymmetry. In fact this is the genius of Howard Georgi to incorporate all the Standard Model fields into unified, left-handed multipliets.

1, $C$ exchanges a spin-up electron with a spin-down positron. You may verify it using the standard four solutions of the Dirac equation.

The above answer about $\Psi$ and $\overline \Psi$ exchange is about the time reversal $T$ or the combination of charge conjugation and parity $CP$. An electron flowing in the future is equivalent to a positron going back into the past.

user112002
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