5
$\begingroup$

How many Grassmann generators are sufficient for the description of a Dirac spinor in 4 dimensions? i.e. The Dirac field is a map to $\Lambda_N$, the space of supernumbers with $N$ real Grassmann generators. What is $N$?

This is a follow up question to my previous question Grassmann Paradox Weirdness. I am following Prakash's book Mathemaical Perspectives on Theoretical Physics, where they say a supernumber $z\in\Lambda_N$ can be thought of the extension of complex numbers by the addition of $N$ Grassmann generators $\zeta^1,\, \zeta^2,\,\ldots \zeta^N$. The most general supernumber is written $$z = z_0+z_i\zeta^i+\textstyle\frac{1}{2!}z_{ij}\zeta^i\zeta^j+\ldots,$$ where $z_i$, $z_{ij}$, $\ldots$, are complex-valued and antisymmetric. The odd part of this is anticommuting and is used to describe Fermion fields. The book says that for finite $N$, it takes $2^{N-1}$ complex numbers to specify an anticommuting number.

How do I figure out how many Grassmann generators $\zeta^i$ I need to specify a Dirac spinor in 4 dimensions.

$\endgroup$
5
  • 2
    $\begingroup$ I would say that you need uncountable infinite number of Grassmann generators. One for each spacetime point and spinor component, then double everything up for the conjugate field. Then the Dirac field is an element of the subspace of the Grassmann algebra with only one generator. So the only product of Dirac fields which is zero is $\psi_\alpha(x)\psi_\alpha(x)=0$, where both are the same spinor component and at the same spacetime point. $\endgroup$
    – Heidar
    Oct 22, 2012 at 18:03
  • $\begingroup$ I don't agree with the answer. It's very likely because I don't fully understand your logic. Would you flesh it out in detail as an answer? $\endgroup$
    – QuantumDot
    Oct 25, 2012 at 1:31
  • $\begingroup$ @Heidar Now I agree with your answer. Please post your comment as an answer so that I may accept it. $\endgroup$
    – QuantumDot
    Apr 21, 2014 at 19:00
  • $\begingroup$ I had not seen your earlier comment, sorry for not replying. I think its always good to add @ to make sure people notice your comment. I have added an answer with a few more details. $\endgroup$
    – Heidar
    Apr 22, 2014 at 17:06
  • $\begingroup$ Related: physics.stackexchange.com/q/474625/2451 $\endgroup$
    – Qmechanic
    Apr 26, 2019 at 12:03

1 Answer 1

3
$\begingroup$

For the Dirac field you need an uncountable infinite number of Grassmann generators. What we want is the following property $$ \psi_\mu(x)\psi_\nu(x')=-\psi_\nu(x')\psi_\mu(x)$$ for any $x,x',\mu$ and $\nu$, where $\psi_\mu(x)\psi_\nu(x')=0$ only for $x=x'$ and $\mu=\nu$. You can clearly not achieve this using a finite number of Grassmann generators.

Generate a Grassmann algebra with a generator for each point $x$ and component $\mu$ $$ \Omega(\mathcal M) = \langle\zeta_\mu(x)|x\in\mathcal M,\mu=1,\dots n\rangle,$$ where the notation $\langle a,b,c,\dots\rangle$ stands for a Grassmann algebra generated by elements $a,b,c,\dots$ and $\mathcal M$ is the manifold your spinor lives on. A general element of this space is thus $z\in\Omega(\mathcal M)$: $$ z= z_0 + \int_{\mathcal M}\text dx\; z_\mu(x)\zeta_\mu(x) + \frac 1{2!}\int_{\mathcal M}\text dx\text dy\; z_{\mu,\nu}(x,y)\zeta_\mu(x)\zeta_\nu(y) +\dots.$$

The spinor $\psi_\mu(x)$ then lives in the subspace $\Omega_1(\mathcal M)\subset\Omega(\mathcal M)$ with only one generator. In other words $$ \psi_\mu(x) = z_\mu(x)\zeta_\mu(x) $$ for some complex number $z_\mu(x)\in\mathbb C$. This will ensure the property we want. For conjugate field $\bar{\psi}_{\mu}(x)$, you need to extend the algebra with a new set of generators $\bar \zeta_\mu(x)$ and similarly extend the algebra for each new set of fermions in the theory.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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