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The Lagrangian of a fermion field is \begin{equation} \mathcal{L} = \overline{\psi} (i\gamma_{\mu} \partial^{\mu} - m)\psi \end{equation} It is said that the fermion field $\psi$ is necessarily complex because of the Dirac structure. I don't quite understand this. Why is the fermion field complex from a physical point of view? A complex field has two components, i.e., the real and imaginary components. Does this imply that all fermions are composite particles? For example, an electron is assumed to be a point particle that does not have structure. How can it have two components if it is structureless?

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    $\begingroup$ The Dirac fermion field does contain two components: the particle and the antiparticle, which are distinct. But this doesn't imply compositeness. $\endgroup$ Jul 16, 2018 at 13:22

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Fermions are not necessarily complex. Majorana spinors fulfill a reality condition $\Psi=\Psi^\mathcal{C}$, and in the Majorana basis a Majorana spinor is manifestly real. Clearly, only neutral fermions can be described by Majorana spinors, and neutrinos are candidates for such fermions, even though at present it is not clear whether they are Dirac or Majorana particles.

I think that there are some issues with your question. The answer above is on the title "Why are fermions complex". In your question, you suggest that all fermions need to be described by the Dirac equation, which is not the case. There is also the Weyl equation, and some refer to the field equation of Majorana spinors as Majorana equation. And I cannot follow the statement on the degrees of freedom vs. structure either, photons are described by a vector and do not have a "structure" either, at least according to our present knowledge.

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Any type of field can be complex, not only the fermions. The reason is the $U(1)_{EM}$ symmetry, i.e., the electromagnetic interactions.

The electric charge is the conserved quantity of the $U(1)_{EM}$ gauge symmetry in nature. A transformation of this symmetry is such that $$ \phi(x) \mapsto e^{iq\theta (x)} \phi (x) \\ \phi^{\dagger}(x) \mapsto e^{-iq\theta (x)} \phi^\dagger (x) $$ where $q$ is the electric charge of the field, and $\theta(x)$ is the gauge parameter. If $\phi(x)$ is a real-valued field, then the first and second equations should be identical, which implies $$ e^{iq\theta(x)} = e^{-iq\theta(x)} $$ This is only true if and only if $q=0$. For complex fields the charge would be opposite for their conjugates.

So, complex fields are charged and real fields are neutral. For example, after the electroweak breaking, the Higgs field is neutral therefore a real-valued boson, while W bosons, i.e., $W^\pm_\mu \equiv W^1_\mu \mp i W^2_\mu$, are charged, so complex-valued bosons. Neutrinos are neutral so they are real-valued fermions, but electrons are charged thus complex-valued fermions.

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Classical fermions in $3+1$ or higher dimensions are not complex. They are Grassmann number-valued. i.e. $\theta_{i}\theta_{j}+\theta_{j}\theta_{i}=0$ for any two fermions $\theta_{i}$ and $\theta_{j}$. There can be no difficulties regarding fermionic fields as complex numbers as long as you don't consider the product of two fields. If you consider them as complex numbers, then there is a thereom saying that fermions live in three representations: 1. complex. 2. real. 3. pseudo-real. For example, the Dirac fermion in 3+1 dimensions can be regarded as the $\mathbb{C}^{2}\oplus\mathbb{C}^{2}$ representation of $SL(2,\mathbb{C})$. But writing the Dirac Lagrangian requires it anti-commutes, otherwise you run into inconsistency.

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A charged particle requires a complex valued field. For a neutral particle it is believed that a real valued field suffices. For example the Schrödinger and Klein-Gordon current operator is zero for a real wave function.

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    $\begingroup$ Why does a charged particle require a complex-valued field instead of a real-valued field? $\endgroup$
    – Shen
    Jul 16, 2018 at 15:58
  • $\begingroup$ @Shen A conserved four-current (sometimes called the "Noether current") can be derived from the field using the the Noether procedure if the field is complex. Noether's theorem states that every continuous symmetry corresponds to a conserved quantity, and in the case of a complex field $\varphi$ (with a conserved current), the Lagrangian is invariant to the transformation $\varphi \rightarrow e^{i\alpha}\varphi$; this constitutes a symmetry and this symmetry is associated with a four-current, and integrating a four-current yields a charge. $\endgroup$ Mar 13, 2021 at 0:33

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