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I've been struggling with the use of Grassmann numbers in QFT e.g. Peskin and Schroeder. They are introduced as "numbers" whose product is antisymmetric, and associative (this isn't said, but used in following calculations), furthermore, they can be multiplied by complex numbers in the usual way. This all means that they are a complex algebra that alternates and is associative. My problem is that this isn't enough to specify what they are and how they behave.

Example: The exterior algebra $\Lambda \mathbb{C^2}$ has the property that the product of any three elements must vanish. The exterior algebra $\Lambda \mathbb{C^3}$ does not have this property. Both satisfy the conditions for Grassmann numbers.

So, it seems to me that the Grassmann numbers described in field theory are underspecified. Have I got something confused? If not, how should they be accurately specified?

Qmechanic
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Joshua Tilley
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  • Grassmann numbers are just assumed to obey no further relations besides anticommutativity. – knzhou Apr 23 '19 at 22:18
  • You can multiply them by complex numbers and add them, so they form a vector space, what is it's dimension? – Joshua Tilley Apr 23 '19 at 22:26
  • For quantum fields, always infinity, since there's basically an independent Grassmann algebra for every point. Field spaces are always infinite-dimensional. – knzhou Apr 23 '19 at 22:27
  • Ok, and if we take the Grassman algebra at a single point, what is the dimension of that? Also infinite? – Joshua Tilley Apr 23 '19 at 22:31
  • Related: https://physics.stackexchange.com/q/5005/2451 , https://physics.stackexchange.com/q/40746/2451 . More on Grassmann numbers. – Qmechanic Apr 23 '19 at 22:36
  • @JoshuaTilley Depends on the theory. For example, for one Dirac spinor, there are 4 independent Grassmann numbers, so the algebra is a $2^4 = 16$-dimensional vector space. – knzhou Apr 23 '19 at 22:58
  • I recommend checking out the book "Supermanifolds" by Bryce DeWitt. These stuff are spelled out in great detail there. – Bence Racskó Apr 23 '19 at 23:23

1 Answers1

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It is perhaps conceptionally simplest to define the algebra of supernumbers as (a copy of) an exterior algebra $$\Lambda_{\infty}~=~\bigwedge\!{}^{\Large\bullet} V_1 ,$$ where $V_1$ is an infinite-dimensional$^1$ $\mathbb{C}$-vector space. Then

  • $\bigwedge^{0} V_{1} \cong \mathbb{C}$ is called the body,

  • $\bigwedge^{>0} V_{1}$ is the soul,

  • $\mathbb{C}^{1|0} \cong \bigwedge^{\rm even} V_{1}$ is the vector space of Grassmann-even/bosonic supernumbers/$c$-numbers, and

  • $\mathbb{C}^{0|1} \cong \bigwedge^{\rm odd} V_{1}$ is the the vector space of Grassmann-odd/fermionic supernumbers/$a$-numbers,

cf. Refs. 1 & 2. See also this related Phys.SE post.

References:

  1. planetmath.org/supernumber.

  2. Bryce DeWitt, Supermanifolds, Cambridge Univ. Press, 1992.

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$^1$ We choose infinite dimensions to avoid the truncation problem that OP is alluding to. However, note that infinite dimensions leads to delicate issues of mathematical analysis.

Qmechanic
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