Grassmann variables and their calculus

A supersymmetry variation contains a product of an odd parameter and an odd operation. To calculate with it, we need more than the instruction “fermions anticommute”: we need a reliable way to differentiate products and integrate expressions without losing their signs. The smallest useful setting has only two anticommuting variables. It already contains the essential rules.

After What a supercharge does, the required preparation is ordinary differentiation and polynomial algebra. We will calculate both derivatives of one polynomial, integrate it in two different orders, and explain why those answers have opposite signs. The construction is algebraic and applies in any spacetime dimension or signature. No gauge theory, particular supersymmetry algebra, or spinor-index convention is being chosen here.

Two variables that cannot be numbers

Introduce symbols $\theta_1,\theta_2$ with

$$\theta_i\theta_j=-\theta_j\theta_i.$$

We work over the real or complex numbers, so setting $i=j$ gives $2\theta_i^2=0$, hence $\theta_i^2=0$. This does not imply $\theta_i=0$. A Grassmann generator is an element of a different algebra, not an unusual real number. The product $\theta_1\theta_2$ is nonzero, but its square vanishes:

$$(\theta_1\theta_2)^2 =-\theta_1\theta_1\theta_2\theta_2=0.$$

Choose the order $\theta_1$ before $\theta_2$. Every polynomial reduces uniquely to

$$p=a+b\theta_1+c\theta_2+d\theta_1\theta_2.$$

For the moment $a,b,c,d$ are ordinary commuting coefficients. There are four basis elements, $1,\theta_1,\theta_2,\theta_1\theta_2$. Any word of length three repeats a generator and vanishes after reordering. With $n$ independent generators, the basis consists of their ordered products, each generator appearing zero or one times; it has $2^n$ elements.

This is the exterior algebra of the generators. Its product resembles the wedge product of differential forms, but its generators here will represent odd coordinates or fermionic quantities. The subscripts $1,2$ are labels; they are not spinor indices to be raised and lowered.

An element is even or odd when all its terms contain an even or odd number of generators. An element with a definite parity is called homogeneous in parity. This parity, written $|u|=0$ or $1$, determines the exchange sign:

$$uv=(-1)^{|u||v|}vu.$$

Thus even elements commute with everything in this algebra. “Even” does not mean “ordinary nonzero number”: $\theta_1\theta_2$ is even and nilpotent. A general sum can have both parities and must be separated into its even and odd parts before applying a parity formula.

The derivative must know which side it acts from

An ordinary product rule would give

$$\partial_2(\theta_1\theta_2+\theta_2\theta_1) =\theta_1+\theta_1.$$

The expression being differentiated is zero, so this cannot be the correct rule.

We use left derivatives, denoted $\partial_i^L$. Define them by replacing $\theta_i$ by $\theta_i+\eta$, where $\eta$ is a new odd generator, and placing $\eta$ at the far left:

$$f(\ldots,\theta_i+\eta,\ldots) =f(\ldots,\theta_i,\ldots)+\eta\,\partial_i^L f.$$

The parameter $\eta$ anticommutes with every odd quantity, and the coefficients of $f$ are independent of $\eta$. For a single generator this expansion is exact, since $\eta^2=0$. In particular, $\partial_i^L\theta_j=\delta_{ij}$ and $\partial_i^L1=0$.

Now shift the second variable in a product:

$$\theta_1(\theta_2+\eta) =\theta_1\theta_2-\eta\theta_1.$$

The coefficient to the right of $\eta$ is $-\theta_1$. Shifting the first variable gives the other sign:

$$\partial_2^L(\theta_1\theta_2)=-\theta_1, \qquad \partial_1^L(\theta_1\theta_2)=\theta_2.$$

For a homogeneous element $u$, shifting a product $uv$ gives

$$\begin{aligned} (u+\eta\partial_i^L u)(v+\eta\partial_i^L v) &=uv+\eta(\partial_i^L u)v+u\eta(\partial_i^L v)\\ &=uv+\eta\left[(\partial_i^L u)v +(-1)^{|u|}u(\partial_i^L v)\right]. \end{aligned}$$

The term containing two $\eta$‘s vanishes. Comparing with the definition produces the graded product rule:

$$\partial_i^L(uv) =(\partial_i^L u)v+(-1)^{|u|}u(\partial_i^L v).$$

The minus sign appears when the derivative passes an odd factor. Applied to $\theta_1\theta_2+\theta_2\theta_1$, the two derivatives now cancel, as they must.

For the concrete polynomial

$$p=2+3\theta_1-5\theta_2+7\theta_1\theta_2,$$

the result is

$$\partial_1^L p=3+7\theta_2, \qquad \partial_2^L p=-5-7\theta_1.$$

Differentiating again yields

$$\partial_2^L\partial_1^L p=7, \qquad \partial_1^L\partial_2^L p=-7.$$

Operators act from the rightmost one first. The derivatives themselves are odd and obey

$$(\partial_i^L)^2=0, \qquad \partial_i^L\partial_j^L+\partial_j^L\partial_i^L=0.$$

One can verify these identities on the four basis elements; linearity then establishes them on every two-variable polynomial. Right derivatives are another consistent convention, but switching sides midway through a calculation changes signs. We will retain the superscript $L$ here to keep the choice visible.

Berezin integration extracts a coefficient

Integration over an odd variable is an algebraic operation, not an area under a curve. We want it to be invariant under an odd translation. If $I$ denotes the one-variable integral, translation invariance requires

$$I(\theta+\eta)=I(\theta).$$

Since the extra term is proportional to $I(1)$, this requires $I(1)=0$. Choose the normalization $I(\theta)=1$. The resulting Berezin integral is

$$\int d\theta\,(a+\theta b)=b, \qquad \int d\theta\,1=0, \qquad \int d\theta\,\theta=1.$$

The integrated variable is written to the left of its coefficient. With this convention the integral acts just like its left derivative. For example, if $\xi$ is an independent odd coefficient,

$$\int d\theta\,\xi\theta =-\int d\theta\,\theta\xi=-\xi.$$

Moving an odd coefficient through the integral therefore requires attention to its sign. Ordinary real and complex coefficients introduce no such sign.

For several variables, the ordered measure is part of the definition. We choose

$$\int d\theta_2\,d\theta_1\,f :=\partial_2^L\partial_1^L f.$$

The integral nearest the polynomial, $d\theta_1$, acts first. Consequently,

$$\int d\theta_2\,d\theta_1\,\theta_1\theta_2=1, \qquad \int d\theta_1\,d\theta_2\,\theta_1\theta_2=-1.$$

The first ordering extracts the coefficient of $\theta_1\theta_2$. The second extracts the coefficient of $\theta_2\theta_1$. Our worked polynomial therefore has integral $7$ in the first orientation and $-7$ in the second. Nothing is ambiguous once both the product order and the measure order are specified.

Bilal introduces the same single-generator normalization and opposite ordering between an iterated measure and its normalized monomial in section 4.1. His compact spinor measures include additional factors fixed by index contractions. Our explicit $d\theta_2\,d\theta_1$ should not be silently identified with his $d^2\theta$. See the text preceding equation (4.3) and equations (4.3)–(4.4) of Bilal’s lectures, version 1.

Two useful checks on an integral

First, a full Berezin integral annihilates a derivative with respect to any of its integration variables: differentiating removes that generator, so the top ordered product is absent. With $D\theta=d\theta_2\,d\theta_1$, the graded product rule gives

$$0=\int D\theta\,\partial_i^L(uv) =\int D\theta\,(\partial_i^L u)v +(-1)^{|u|}\int D\theta\,u(\partial_i^L v).$$

Thus integration by parts reads

$$\int D\theta\,(\partial_i^L u)v =-(-1)^{|u|}\int D\theta\,u(\partial_i^L v).$$

For odd $u$, the sign is plus. Test this with $u=\theta_2$, $v=\theta_1\theta_2$, and $i=2$. The left side is $\int D\theta\,\theta_1\theta_2=1$. The right side is $\int D\theta\,\theta_2(-\theta_1)=1$. Copying the ordinary integration-by-parts sign would give the wrong answer.

Second, consider a rescaling $\eta=\lambda\theta$, where $\lambda$ is a nonzero commuting number. Normalization demands

$$1=\int d\eta\,\eta =\int (\lambda^{-1}d\theta)(\lambda\theta).$$

The odd measure transforms with the inverse scaling factor:

$$d\eta=\lambda^{-1}d\theta.$$

This differs from an ordinary commuting integration variable. Exercise 2 checks the corresponding determinant for two generators.

A finite exponential offers another quick calculation. For an even commuting parameter $m$,

$$e^{m\theta_1\theta_2}=1+m\theta_1\theta_2, \qquad \int D\theta\,e^{m\theta_1\theta_2}=m.$$

The series truncation is exact, not a small-$m$ approximation. Changing the sign of the exponent changes the integral to $-m$; changing the measure orientation does the same. This is why a fermionic Gaussian calculation must state its exponent and measure together.

Coordinates and fermionic coefficients

In a field theory, coefficients can themselves be Grassmann-valued. For example, consider the even expression

$$\mathcal P =a+\theta_1\psi_1+\theta_2\psi_2+\theta_1\theta_2 B,$$

where $a,B$ are even and $\psi_1,\psi_2$ are odd, independent of the $\theta_i$. Every term has even total parity. The left derivatives are

$$\partial_1^L\mathcal P=\psi_1+\theta_2 B, \qquad \partial_2^L\mathcal P=\psi_2-\theta_1 B, \qquad \int D\theta\,\mathcal P=B.$$

The $\psi_i$ anticommute with the coordinates as well as with each other. A two-variable polynomial with ordinary coefficients is useful for learning the calculus; it does not automatically describe an even superfield. In a superfield, the component parities ensure that all terms have the same total parity.

These algebraic generators also differ from quantum fermion creation and annihilation operators: the latter act on states and can have a nonzero anticommutator. Grassmann calculus supplies the classical variables used in fermionic actions and path integrals. It does not replace the operator algebra on the Hilbert space.

Exercises

1. Find a derivative in two ways

Let $u=\theta_1+2\theta_2$ and $v=3\theta_1-\theta_2$. Reduce $uv$ to the ordered basis. Compute $\partial_2^L(uv)$ both by differentiating the result and by the graded product rule. Finally evaluate $\int D\theta\,uv$.

First hint

Both $u$ and $v$ are odd. Terms containing a squared generator vanish.

Second hint

Reorder $6\theta_2\theta_1$ before combining terms. In the product rule, the second term has a minus sign.

Complete solution

Multiplication gives

$$uv=-\theta_1\theta_2+6\theta_2\theta_1=-7\theta_1\theta_2.$$

Hence $\partial_2^L(uv)=7\theta_1$. Independently,

$$\begin{aligned} (\partial_2^L u)v-u(\partial_2^L v) &=2(3\theta_1-\theta_2) -(\theta_1+2\theta_2)(-1)\\ &=7\theta_1. \end{aligned}$$

Agreement checks the product-rule sign. The ordered integral is the top coefficient, $-7$.

2. Change the generators

Define $\eta_1=2\theta_1+\theta_2$ and $\eta_2=3\theta_2$. Find $\eta_1\eta_2$ and the factor relating $d\eta_2\,d\eta_1$ to $D\theta$. Use them to evaluate

$$\int d\eta_2\,d\eta_1\,e^{m\eta_1\eta_2}$$

in the $\theta$ variables. Explain why the result must agree with the calculation directly in the $\eta$ variables.

First hint

Require $\int d\eta_2\,d\eta_1\,\eta_1\eta_2=1$.

Second hint

The coefficient of $\theta_1\theta_2$ in $\eta_1\eta_2$ is the determinant of the linear transformation.

Complete solution

Nilpotence gives

$$\eta_1\eta_2=(2\theta_1+\theta_2)(3\theta_2) =6\theta_1\theta_2.$$

The normalization therefore fixes

$$d\eta_2\,d\eta_1=\frac16D\theta.$$

Substituting into the integral gives

$$\frac16\int D\theta\,(1+6m\theta_1\theta_2)=m.$$

Direct coefficient extraction in the $\eta$ basis also gives $m$. We have changed coordinates on the same algebra, not changed the quantity being calculated. More generally, for an invertible commuting matrix $A$ with $\eta_i=A_{ij}\theta_j$, the ordered top product acquires $\det A$, so the Berezin measure acquires $(\det A)^{-1}$. If $\det A=0$, the new generators are not independent and this change-of-variables formula is unavailable.

3. Extract a component without losing a fermion sign

For the even polynomial $\mathcal P$ above, calculate $\int D\theta\,\mathcal P^2$. Explain why treating $\psi_1,\psi_2$ as commuting coefficients would change the answer and would also change the parity of the terms.

First hint

Only products containing exactly one $\theta_1$ and one $\theta_2$ can contribute.

Second hint

Move $\psi_1$ past $\theta_2$ in $(\theta_1\psi_1)(\theta_2\psi_2)$ before reading its coefficient.

Complete solution

The terms involving $a$ and $B$ give $2a\theta_1\theta_2B$. The other contributing pair is

$$(\theta_1\psi_1)(\theta_2\psi_2) =-\theta_1\theta_2\psi_1\psi_2.$$

Since $\theta_1\psi_1$ and $\theta_2\psi_2$ are both even, the product in the reverse order gives the same answer. Consequently,

$$\int D\theta\,\mathcal P^2=2aB-2\psi_1\psi_2.$$

Terms with a repeated $\theta_i$ vanish. As a check, setting both $\psi_i$ to zero leaves $2aB$. If the coefficients were commuting, the two linear terms would instead be odd, their cross products would cancel, and $\mathcal P$ would no longer be an even expression. The different answer would describe a different algebraic object.

Checkpoint and next calculation

You should now be able to explain the minus sign in $\partial_2^L(\theta_1\theta_2)$, predict which coefficient an ordered measure extracts, and calculate that coefficient when it contains fermion fields. These are the tools needed to turn a supersymmetry transformation into a dependable calculation.

The next new ingredient is spinor geometry: how the components transform under rotations and how their indices are contracted. Continue with Spinors in three dimensions for the curved-space route. The Grassmann rules remain the same in the three-dimensional and four-dimensional entrances. Only after specifying those index conventions will a compact symbol such as $d^2\theta$ have a definite normalization.

For a direct application of the integration rules, try a finite-dimensional localization example. Two ordinary variables and two Grassmann generators are enough to see how a symmetry can preserve an integral while changing which configurations contribute.