# Expanding superfields: inconsistency of notation?

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 (???)

$$[\Phi(x,\theta),\Phi(y,\theta')]=0$$

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.

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

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 $$\theta^{\alpha}$$.

• 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.
– user84158
Commented Jul 15, 2019 at 21:44
• 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?
– user84158
Commented Jul 16, 2019 at 3:59
• 1. Usually taken to be a countable number rather than a continuum. Commented Jul 16, 2019 at 7:46
• 1. What I mean is if $\psi(x)$ is grassmann-odd. It is a continuum of grassmann-odd numbers parametrised by the continuous index $x$.
– user84158
Commented Jul 16, 2019 at 14:08
• I updated the answer. Commented Jul 16, 2019 at 20:06