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I find the definition of a function $\phi$ on a superspace $(z,\theta)$ confusing for the following reason. $\phi(z,\theta)$ can be expanded as $$\phi(z,\theta) = \phi_0(z) + \theta \psi(z)\tag{1}.$$ Now, it is usually said that $\phi_0(z)$ is a smooth function. However, $z$ has a body $z_b$ (which is a complex number) and a soul $z_s$ (which is a sum of product of even number of Grassmann numbers). Perhaps what is meant is following. Let $x \mapsto f(x)$ be any smooth function. Then $$\phi_0(z) = \phi_0(z_b+z_s) \equiv f(z_b)+\sum_{n=1}^\infty \frac{1}{n!} f^{(n)}(z_b)z_s^n\tag{2}$$ seems a possible definition. Is this what is meant when one considers a function on a superspace?

Qmechanic
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Bronsteinx
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1 Answers1

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The Taylor expansion (1) of a superfield $\phi(\theta)$ in component fields should not be conflated with the Taylor expansion (2) of a component field/function $\phi_0(z)\equiv f(z)$ at the body $z_b$.

Note in particular that not all functions of supernumbers are superfields; and that the Grassmann numbers in the soul $z_s$ have nothing to do with Grassmann number $\theta$ in the superfield (1).

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

Qmechanic
  • 201,751
  • Hi. I am trying to understand what kind of object $\phi_0(z)$ is. $z$ can't be just the regular spacetime because under a supersymmetry it mixes with grassmannian numbers. Therefore, the domain of $\phi_0(z)$ must include even numbers. I don't know how such a thing can be defined if not as above. – Bronsteinx Nov 21 '22 at 15:04
  • A related and perhaps equivalent question is how $\partial_{z} \phi_0(z)$ is defined which is used frequently in field theory books. – Bronsteinx Nov 21 '22 at 15:25