When I first came across the presentation of linear algebra in terms of Grassmann’s exterior products, I was struck by its elegance. An introduction to linear algebra in terms of Grassmann’s ideas can be found here. The Grassmann approach is so much more intuitive that, once learned, there is no going back to the old way of thinking. For example, although in college I found determinants and permanents relatively easy to understand using the conventional approach, I only really came to understand the meaning of Pfaffians after learning exterior algebra (a.k.a Grassmann algebra).
Curiously, Grassmann did not have a university education in mathematics. Rather, he was actually trained as a linguist. Yet, his contributions to mathematics are widely recognized today. Terms such as Grassmann numbers, Grassmann algebra and Grassmanian manifold are all named in his honor.
Also fascinating is how Grassmann numbers make their appearance very naturally in quantum field theory. Specifically, they are used in the path integral formulaton for fermionic fields. Readers interested in finding out more about the connection with fermionic path integrals should refer to standard textbooks, for instance the books by Weinberg or the one by Ryder.
Some years ago I derived, for my own convenience, a few of the basic identities satisfied by Grassmann variables, in the context of differentiation and integration. Integration of Grassmann variables is known as Berezin integration. This article is based on my old study notes.
2. Grassmann numbers
Consider the algebra generated by grassmann numbers , that anticommute according to
Such elements are nilpotent with degree 2 due to the antisymmetry property: .
What is the dimension of this algebra (as a vector space)? Consider the set of all monomials. There are generators which are each nilpotent with degree 2. Hence, a general nonzero monomial can have a generator as factor only once. Hence, there are exactly monomials. It is then easy to check that the most general function of the generators can be expressed as a linear combination of these monomials. Indeed, all power series (e.g., Taylor series) terminate, i.e. the most general function is a polynomial in the generators. The dimension of the Grassmann algebra is thus equal to the number of linearly independent monomials, .
It is worth calling attention to an antisymmetry property of the coefficients of monomials. Consider the monomial term . By definition, it equals . So the representation of a function as a linear combination of monomials is not unique. However, if we define the coefficients to be antisymmetric, then we recover uniqueness. For instance, if the coefficients satisfy .
Because Grassmann variables do not commute, we can define derivatives acting from the right and from the left. Here, I consider only derivatives acting to the right. We define the derivative as follows for a single generator:
To extend the derivative to a monomial, we must first bring the matching generator all the way to the left, multiplying by where is the number of generators to the left of in the original monomial. Then, the derivative is obtained by dropping . For example,
The derivative then extends to all functions via linearity, i.e. differentiation is a linear operator.
The chain rule holds in the usual manner.
The product rule holds as usual if the first factor is of even degree in the generators:
However, if the first factor has odd degree, then there is a sign change in the second term. Since a general funtion need not be homogeneous and may have terms of both odd and even degree, I consider it safer to assume that the product rule does not hold in general, and instead to calculate term by term explicitly, unless you know what you are doing.
4. Berezin integration
Now we come to the most interesting part of this article! Is it possible to define an integral for Grassmann numbers? The usual antiderivative of a variable
would be zero if were a Grassmann variable, so it does not make sense to define integration in this manner. However, one can define the equivalent of a definite Riemann integral. The definite integral
has the following properties:
(i) Translation invariance:
Hence, we will require that the integral of a Grassmann number also have these two properties. Let and now denote Grassmann numbers.
Then first we require translation invariance.
Given that is at most a linear function of , let us write
where and are complex numbers. Substituting, and invoking linearity, we get
so that we are forced to assume
Hence, we get
Berezin chose the convention that
although other conventions are possible for the constant. Below I use Berezin’s convention.
From these observations, we define for Grassmann numbers,
(If this is too difficult or “wierd” to accept, try to imagine that supposed integrated quantity vanishes at the boundary.)
Next we define
Note that the integral is independent of quantities such as which is what you would expect for a Riemann integral for normal (i.e., non-Grassmann) variables, since that would be zero due to the nilpotent property. Instead, Berezin integration is similar to standard differentiation: the usual derivative of is 1 and the usual derivative of is 0.
Moreover, the differential anticommmutes with . In fact, the anticommutation property holds generally:
Multiple integrals are defined, in analogy with Fubini’s theorem, as iterated integrals:
Note that there are other sign conventions for Berezin integrals. Physicists usually use the convention
In what follows I use the former sign convention.
Next, we come to one of the most interesting and unexpected properties of the Berezin integral. Let represent a function of all the generators. Recall that the most general function of the Grassmann algebra generators is a polynomial. Hence, the most general function can be written
Indeed, every element of the Grassmann algebra can be written as such.
Now consider the multiple Berezin integral
Note that all monomial terms of degree with will vanish, because each of the iterated integrals for the variables not appearing in the monomial vanish separately. Only monomials of degree survive:
5. Change of variables
Riemann integrals satisfy
We will show that Berezin integrals satisfy intead
The reason for this opposite behavior is that Berezin integration is actually similar to (standard) differentiation.
Let for Grassmann variables and and consider that by definition
which means that
In standard calculus, we would instead have , so Berezin differentials scale opposite to what one would expect from standard calculus.
Now let us generalize to generators. Let Those with some familiarity with exterior products will recogize that the product
corresponds to the exterior product of maximal grade, so that we naturally expect the determinant to make an appearance:
Moreover, because the differentials scale inversely to the generators, we have
6. Gaussian integrals
Consider the following Riemann integrals:
We will next derive an analog of the above for the Berezin integral. The analog of the first integral is zero, due to the nilpotent property. We thus look at the Berezin analog of the second:
Note that the is in the numerator rather than the denominator. Morever, there is no more factor of . (In fact, there are conventions that I do not discuss here, as previously mentioned.)
Let and be generators, so that there are generators total. Moreover, let
Now consider the multiple Gaussian Berezin integral:
Let us change basis in order to diagonalize the matrix , via a unitary transformation. In the new variables and , the transformed matrix is diagonal, so that the exponential factors into a product of terms such as . Hence, the full integral can be written as a product of simple Gaussian integrals and the value of the full Gaussian integral will simply be
Here we have used the fact that unitary transformations leave the determinant invariant.
In contrast, for a Riemann integral the correct expression is
So, besides the factors of , the determinant of appears in the numerator instead of in the denominator for the Berezin integral.
7. A Gaussian integral in terms of a Pfaffian
Let us use a change of variable and define
Note that the reason the Jacobian matrix is inverted is due to the strange way that Grassmann variables behave under change of variables, as explained above.
Let be an antisymmetric matrix of dimension with even. Consider
Since is antisymmetric, the cross terms will cancel, so that
Integrating the exponential of the above and substituting into (32). and remembering that is even we get,
Recall the following identity for the Pfaffian of an antisymmetric even dimensional matrix:
We thus obtain
for any even dimensional antisymmetric matrix .
8. Berezin integration as a contraction
My favorite way of thinking about Berezin integration is in terms of interior products in Grassmann algebras. (Note: interior products are not the same as inner products.) In fact, interior products are how I explicitly calculate more difficult Berezin integrals in practice. If time permits, I may write something up in the future on this topic. This idea is of course not new. It is known that Berezin integrals are a type of contraction, see here.