If I have a wavefunction of a fermion field $\Psi[\psi]$ I can expand it like so about some vacuum:

$$\Psi[\psi] = \Psi_0[\psi]( a + \int a(x)\psi(x)dx+\int a(x,y)\psi(x)\psi(y)dxdy+...)$$

Now all the coefficients are complex numbers as they are amplitudes (related to the amplitude for $N$-fermions). The Grassmann nature just ensures that the functions are anti-symmetric. e.g. $a(x,y)=-a(y,x)$ since $\psi(x)\psi(y)+\psi(y)\psi(x)=0$.

Now think of a superfield $\Phi(x,\theta)$ with Grassmann variables such that $\theta^\alpha \theta^\beta +\theta^\beta \theta^\alpha=0$ . We expand it like this:

$$\Phi(x,\theta)=\phi(x)+\psi^\alpha(x)\theta^\alpha + A^{\alpha \beta}(x) \theta^\alpha \theta^\beta +...$$

But in this case the coefficients such as the fermion field $\psi^\alpha(x)$ are grassmann valued are they not? But only when they occur with an odd number of grassmann variables.

This seems to be two ways to expand a super-field. One where the coefficients are all c-valued (non-Grassmann) and one where the coefficients are such that the whole expression is sort of c-valued.

Have I got something wrong here or is this really two ways functions of Grassmann variables can be expanded?

I am trying to see how we get from a super-field to the fact that the fermion fields are anti-commuting Grassmann fields. If I just take the wavefunction of a superfield $\Psi[\Phi]$ it doesn't seem to work.

Edit: I guess with superfields we must also impose (???)


This would ensure the coefficients Grassmann-valued for odd terms I think. But then what space does the superfield belong? It is not simply a member of $\mathbb{C}[\theta]$ because some of the coefficients are not in $\mathbb{C}$. How do we write this space? Is it something like $\mathbb{C}[\theta^{even}]\times G[\theta^{odd}]$?

For the wavefunction case I think we can say it is a member of $\mathbb{C}[G^{\mathbb{R}}]$, where $G$ are grassmann numbers parameterised by a real number (or $D$ real numbers in $D$-dimensional space.) And when expanded out this would be isomophic to fermionic Fock Space.

Again these seem like two very different ways of having Grassmann fields.

  • $\begingroup$ Consider to elaborate what "the whole expression is sort of c-valued" is supposed to mean. $\endgroup$
    – Qmechanic
    Commented Jul 15, 2019 at 20:35
  • $\begingroup$ In this case I guess it means the superfields commute with each other. (Probably not the correct usage of c-valued.) $\endgroup$
    – user84158
    Commented Jul 15, 2019 at 21:02

1 Answer 1

  1. A superfield $\Phi(x,\theta)$ is assumed to have definite Grassmann parity [which we will denote as $|\Phi|$ in this answer]. Here $x^{\mu}$ and $\theta^{\alpha}$ are Grassmann-even and Grassmann-odd coordinates of superspace $\mathbb{R}^{n|m}$, respectively$^1$.

  2. The superfield components $\phi^n_{\alpha_1\ldots \alpha_n}(x)$ in a superfield $$\Phi(x,\theta)~=~\sum_n \frac{1}{n!}\phi^n_{\alpha_1\ldots \alpha_n}(x)\theta^{\alpha_n}\ldots \theta^{\alpha_1}$$ are supernumbers$^2$ of Grassmann parity $|\Phi|+n$ $(\text{mod } 2)$.

  3. The superfield $\Phi(x,\theta)$ may be further constrained [e.g. chiral, real, $\ldots$], which imposes further conditions on the superfield components $\phi^n_{\alpha_1\ldots \alpha_n}(x)$ [such as, e.g., being a real supernumber].

  4. For more details, see e.g. this related Phys.SE post.


$^1$ In this answer the notation $\theta^{\alpha}$ is a shorthand that in practice includes complex conjugate variables too.

$^2$ Warning: The Grassmann-odd generators of a supernumber $\in \Lambda_{\infty}$ should not be conflated with the Grassmann-odd superspace coordinates $\theta^{\alpha}$.

  • $\begingroup$ Yes, that seems about right. But I wonder if there is a way to denote the (mathematical) field that the superfield belongs to? i.e it isn't just the field extension of real numbers by the Grassmann variables, it's something more complicated. $\endgroup$
    – user84158
    Commented Jul 15, 2019 at 21:44
  • $\begingroup$ Well, indeed, they are different. 1. For a start there's a continuum of them rather than a finite set. 2. But anwyay, according to Wikipedia the superfield cannot be a member of the Grassmann algebra because that is defined with complex coefficients. 3. So I just wondered if there was a name for the algebra that the superfield belongs to? $\endgroup$
    – user84158
    Commented Jul 16, 2019 at 3:59
  • $\begingroup$ 1. Usually taken to be a countable number rather than a continuum. $\endgroup$
    – Qmechanic
    Commented Jul 16, 2019 at 7:46
  • $\begingroup$ 1. What I mean is if $\psi(x)$ is grassmann-odd. It is a continuum of grassmann-odd numbers parametrised by the continuous index $x$. $\endgroup$
    – user84158
    Commented Jul 16, 2019 at 14:08
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Commented Jul 16, 2019 at 20:06

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