Let's consider a free Wess-Zumino Lagrangian given by $$\mathcal{L} = \partial^{\mu}\overline{\phi}\partial_{\mu}\phi + i\psi^{\dagger}\overline{\sigma}^{\mu}\partial_{\mu}\psi\tag{1}$$

Whose supersymmetry transformations are

$$\delta\phi = \epsilon\psi, \space \delta\overline{\phi} = \epsilon^{\dagger}\psi^{\dagger}, \space \delta\psi_{\alpha} = -i(\sigma^{\mu}\epsilon^{\dagger})_{\alpha}\partial_{\mu}\phi, \space \delta\overline{\psi}_{\dot{\alpha}} = i(\epsilon\sigma^{\mu})_{\dot{\alpha}}\partial_{\mu}\overline{\phi}\tag{2}$$

Where $\phi$ is a bosonic scalar field, $\psi$ is a fermionic field, and $\{\alpha, \dot{\alpha}\}$ are spinor indices.

In the process of defining supersymmetry transformations, why is an infinitesimal, anti-commuting Grassmann parameter $\epsilon$ necessary?


2 Answers 2


TL;DR: A Grassmann-odd infinitesimal parameter $\epsilon$ is not necessary: A supersymmetry transformation can alternatively be defined via a Grassmann-odd (appropriately generalized) vector field/graded linear derivation.

Nevertheless, it is often introduced for the following reason:

  1. On one hand, an infinitesimal variation $\phi\to\phi+\delta\phi$ of a field $\phi$ should not change the Grassmann-parity of the field $\phi$, so $\delta$ itself must be Grassmann-even.

  2. On the other hand, a supersymmetry transformation (such as a Poincare supersymmetry transformation or a BRST transformation) is a transformation between fields of opposite Grassmann-parity.

It is therefore natural to introduce a Grassmann-odd infinitesimal parameter $\epsilon$ to describe the infinitesimal supersymmetry transformation (2).


You can define an infinitesimal transformation without the Grassmann parameter $\epsilon$, but the variation $\delta_F$ obtained this way will be an odd-graded derivation, and therefore its algebraic properties are slightly more complicated than the even-graded derivation $\delta_B = \epsilon\delta_F$. Also, the $\epsilon$ usually carries spinor indices. To deal with those, you can consider the transformation with a commuting variable $\epsilon$ carrying the spinor indices, or you can omit it completely, in which case the variation $\delta$ itself will carry the indices.

Explicitely, let's work in $1+0$ dimensions , so that we don't have to deal with spinor indices. Let us take the example used in Hori's book on Mirror symmetry : we take a real even variable $x(t) $ and a complex odd variable $\psi(t)$ with complex conjugate $\psi^\dagger =\bar \psi$. The Lagrangian is : $$L = \frac 12 \dot x^2 - \frac 12 (h'(x))^2 + i\bar \psi\dot\psi-h''(x) \bar \psi\psi$$ for some arbitrary function $h(x)$.

First, the usual way to do the calculations. This is invariant under a supersymmetric transformation, with parameter $\epsilon$ a complex Grassmann variable (with complex conjugate $\bar \psi$) : \begin{align} \delta_0 x &= \epsilon\psi - \bar \epsilon \bar\psi\\ \delta_0 \psi &= \epsilon (i\dot x + h'(x) ) \\ \delta_0 \bar \psi &= \epsilon \bar (- i\dot x + h'(x)) \end{align}

This preserves the parity of fields. It acts on any function of the fields as a derivation : it is linear over complex numbers and if $\mathcal A,\mathcal B$ are any two functions of the fields, we have : $$\delta_0 (\mathcal A \mathcal B ) = (\delta_0 \mathcal A) \mathcal B + \mathcal A (\delta_0 \mathcal B)$$

Using this formula, we can compute : \begin{align} \delta L &= \dot x(\delta_0\dot x ) - h'(x)h''(x)\delta_0 x+i\big[(\delta_0\bar \psi)\psi + \bar \psi\delta_0\psi\big] -h'''(x) (\delta_0 x)\bar \psi\psi - h''(x) \big[ \bar \psi \delta_0 \psi + (\delta_0 \bar \psi )\psi\big] \end{align} Replacing the variations of fields by the corresponding expressions and collecting terms, we find $\delta_0 L =0$ (up to integration by part, or more precisely $\delta_0 L$ equal some total derivative).

Now, we can also do this without the Grassman parameter $\epsilon$. We define two transformations (corresponding to $\epsilon$ and $\bar \epsilon$) : \begin{align} \delta x &=\bar \psi & \bar \delta \psi &= - \psi\\ \delta\psi &= i\dot x + h'(x) & \bar \delta \psi &= 0\\ \delta \bar \psi &= 0 & \bar \delta\bar \psi &= -i\dot x + h'(x) \end{align} These transformations reverse the parity of fields. This is not a problem in itself. It just means that we have to treat the symbol $\delta$ as an odd "variable" (meaning that it will pick up a minus sign when we commute it with another odd variable). In particular, it means that we have to adapt the Leibniz rule. If $\mathcal A$ has parity $\deg \mathcal A$, then we have : $$\delta (\mathcal A \mathcal B ) = (\delta \mathcal A) \mathcal B + (-1)^{\deg\mathcal A} \mathcal A (\delta \mathcal B)$$ and similarly for $\bar \delta$.

We have to do this to recover the usual Leibniz rule for $\delta_0 = \epsilon \delta + \bar \epsilon \bar \delta$. With this in mind, we can compute : \begin{align} \delta L = \dot x \delta \dot x - h'(x) h''(x) \delta x- i\bar \psi\delta \dot \psi - h'''(x) \delta x \bar \psi\psi - h''(x) \bar \psi\delta \psi \end{align} Here, we have omitted the in $\delta \bar \psi$ which vanish and picked up some extra minus signs from (anti)commuting $\delta$ with $\bar \psi$. Then : \begin{align} \delta L &= \dot x \dot{\bar \psi}- h'(x) h''(x)\bar \psi - i\bar \psi(i\ddot x + h''(x)\dot x)- h'''(x)\bar \psi \bar \psi\psi + h''(x) \bar \psi(i\dot x +h'(x)) \\ &=\dot x \dot{\bar \psi} + \ddot x\bar \psi - h'(x) h''(x)\bar \psi +h'(x) h''(x) \bar \psi -i\dot x h''(x)\bar \psi + i \dot x h''(x) \bar \psi \\ &= 0 \end{align} Repeating the same calculation, we would find $\bar \delta L =0$.

In short : the odd parameter $\epsilon$ is not strictly necessary. It has two roles : it carries some spinor indices and it makes the transformation $\delta$ into an even derivation, which is simpler to handle in calculations.


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