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The concept of the Grassmann number makes me confused. It is used to describe fermionic fields, especially path integral quantization. Also, it is used to deal with the classical field theory of fermions. To reduce my confusion, I want to ask the question only focusing on the classical field theory. Let us discuss the Dirac fermion. Its Lagrangian is given as $\mathcal L = i \bar\psi \gamma^\mu \partial_\mu \psi - m\bar \psi \psi$. Here $\bar \psi$ and $\psi$ are Grassman number. I understand that the action should be Grassmann-even for the typical theories. The Dirac Lagrangian satisfies this condition. However, I am not sure whether this action is an ordinary number. The Lagrangian mechanics is governed by the least action principle. If it is not real, how can we define the least action? Of course, I thought that we could give up the least action principle and just alternatively define the stationary condition compatible with Grassmann numbers (this is a rough description). Then what is that condition? Also, in terms of Hamiltonian mechanics, Hamiltonian $H$ is a physical observable that can be measured like $100$ GeV. If $H$ is built by Grassmann numbers, how can we relate it to the physical quantities that are real numbers? I understand that the classical field theory of the fermion field would be significantly different from the classical field theories that I knew. The deep aspect of it might not be my interest. Therefore, I want to ask a very simplified question.

Is the Dirac Lagrangian $\mathcal L$ (or action $S$) or Dirac Hamiltonian $H$ built with the Grassmann numbers field a real number with well-defined ordering?

Qmechanic
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Jaeok Yi
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    Possible duplicate: https://physics.stackexchange.com/q/40746/ – akhmeteli Feb 14 '24 at 07:09
  • The Dirac Lagrangian has many issues. – my2cts Feb 14 '24 at 09:22
  • I suggest this for a discussion of the classical Dorac field (that is sometimes considered complex and sometimes Grassmann): https://doi.org/10.1016/j.shpsb.2019.10.003 – Quillo Feb 14 '24 at 10:43
  • About "least action" and "stationary action". The version "least action" should never have been introduced in the first place. William Rowan Hamilton proposed several names, among them 'stationary action' and 'law of varying action'. With the benefit of hindsight: Hamilton was right. The true trajectory has the mathematical property: as you sweep out variation: the derivative of the action is zero. You write: "we could give up the least action principle and just alternatively define the stationary condition". I assert: it's the reverse of that: "least action" should never have been introduced. – Cleonis Feb 14 '24 at 21:18

1 Answers1

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The value of the Dirac action$^1$ $$ S~=~\int\! d^4x ~\bar{\psi}\left(\frac{i}{2}\gamma^{\mu}\underbrace{\stackrel{\leftrightarrow}{\partial}_{\!\mu}}_{=\stackrel{\rightarrow}{\partial}_{\mu}-\stackrel{\leftarrow}{\partial}_{\mu}}-m \right)\psi$$ is a real supernumber $S=S_B+S_S$ with a vanishing body $S_B=0$ and a soul $S_S$; it is not a real ordinary number, cf. e.g. this related Phys.SE post, which also answers many of OP's other questions.

More generally, the body part $S_B$ of an action with fermionic and bosonic fields is a real ordinary number, and hence endowed with an ordering, which (partially) explains why the least action principle still works.

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$^1$ Here we have included an appropriate boundary term, cf. this Phys.SE post.

Qmechanic
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  • Well, the soul summand $S_S$ is not an ordinary number. 2. I updated the answer.
    • – Qmechanic Feb 14 '24 at 10:52
  • I become more interested in these subjects. Could you recommend some textbooks or lecture notes that deal with supernumbers? – Jaeok Yi Feb 15 '24 at 05:38
  • Hi @Jaeok Yi. Maybe start with the references in the linked Phys.SE post. – Qmechanic Feb 15 '24 at 08:06