One variable

The Berezin integral over the sole Grassmann variable ${\displaystyle \theta =\theta _{1}}$ is defined to be a linear functional

${\displaystyle \int [af(\theta )+bg(\theta )]\,d\theta =a\int f(\theta )\,d\theta +b\int g(\theta )\,d\theta ,\quad a,b\in \mathbb {C} }$

where we define

${\displaystyle \int \theta \,d\theta =1,\qquad \int \,d\theta =0}$

so that :

${\displaystyle \int {\frac {\partial }{\partial \theta }}f(\theta )\,d\theta =0.}$

These properties define the integral uniquely and imply

${\displaystyle \int (a\theta +b)\,d\theta =a,\quad a,b\in \mathbb {C} .}$

Take note that ${\displaystyle f(\theta )=a\theta +b}$ is the most general function of ${\displaystyle \theta }$ because Grassmann variables square to zero, so ${\displaystyle f(\theta )}$ cannot have non-zero terms beyond linear order.

Multiple variables

The Berezin integral on ${\displaystyle \Lambda ^{n}}$ is defined to be the unique linear functional ${\displaystyle \int _{\Lambda ^{n}}\cdot {\textrm {d}}\theta }$ with the following properties:

${\displaystyle \int _{\Lambda ^{n}}\theta _{n}\cdots \theta _{1}\,\mathrm {d} \theta =1,}$

${\displaystyle \int _{\Lambda ^{n}}{\frac {\partial f}{\partial \theta _{i}}}\,\mathrm {d} \theta =0,\ i=1,\dots ,n}$

for any ${\displaystyle f\in \Lambda ^{n},}$ where ${\displaystyle \partial /\partial \theta _{i}}$ means the left or the right partial derivative. These properties define the integral uniquely.

Notice that different conventions exist in the literature: Some authors define instead^[1]

${\displaystyle \int _{\Lambda ^{n}}\theta _{1}\cdots \theta _{n}\,\mathrm {d} \theta$ :=1.}

The formula

${\displaystyle \int _{\Lambda ^{n}}f(\theta )\,\mathrm {d} \theta =\int _{\Lambda ^{1}}\left(\cdots \int _{\Lambda ^{1}}\left(\int _{\Lambda ^{1}}f(\theta )\,\mathrm {d} \theta _{1}\right)\,\mathrm {d} \theta _{2}\cdots \right)\mathrm {d} \theta _{n}}$

expresses the Fubini law. On the right-hand side, the interior integral of a monomial ${\displaystyle f=g(\theta ')\theta _{1}}$ is set to be ${\displaystyle g(\theta '),}$ where ${\displaystyle \theta '=\left(\theta _{2},\ldots ,\theta _{n}\right)}$; the integral of ${\displaystyle f=g(\theta ')}$ vanishes. The integral with respect to ${\displaystyle \theta _{2}}$ is calculated in the similar way and so on.

Change of Grassmann variables

Let ${\displaystyle \theta _{i}=\theta _{i}\left(\xi _{1},\ldots ,\xi _{n}\right),\ i=1,\ldots ,n,}$ be odd polynomials in some antisymmetric variables ${\displaystyle \xi _{1},\ldots ,\xi _{n}}$. The Jacobian is the matrix

${\displaystyle D=\left\{{\frac {\partial \theta _{i}}{\partial \xi _{j}}},\ i,j=1,\ldots ,n\right\},}$

where ${\displaystyle \partial /\partial \xi _{j}}$ refers to the right derivative (${\displaystyle \partial (\theta _{1}\theta _{2})/\partial \theta _{2}=\theta _{1},\;\partial (\theta _{1}\theta _{2})/\partial \theta _{1}=-\theta _{2}}$). The formula for the coordinate change reads

${\displaystyle \int f(\theta )\,\mathrm {d} \theta =\int f(\theta (\xi ))(\det D)^{-1}\,\mathrm {d} \xi .}$

Definition

Consider now the algebra ${\displaystyle \Lambda ^{m\mid n}}$ of functions of real commuting variables ${\displaystyle x=x_{1},\ldots ,x_{m}}$ and of anticommuting variables ${\displaystyle \theta _{1},\ldots ,\theta _{n}}$ (which is called the free superalgebra of dimension ${\displaystyle (m|n)}$). Intuitively, a function ${\displaystyle f=f(x,\theta )\in \Lambda ^{m\mid n}}$ is a function of m even (bosonic, commuting) variables and of n odd (fermionic, anti-commuting) variables. More formally, an element ${\displaystyle f=f(x,\theta )\in \Lambda ^{m\mid n}}$ is a function of the argument ${\displaystyle x}$ that varies in an open set ${\displaystyle X\subset \mathbb {R} ^{m}}$ with values in the algebra ${\displaystyle \Lambda ^{n}.}$ Suppose that this function is continuous and vanishes in the complement of a compact set ${\displaystyle K\subset \mathbb {R} ^{m}.}$ The Berezin integral is the number

${\displaystyle \int _{\Lambda ^{m\mid n}}f(x,\theta )\,\mathrm {d} \theta \,\mathrm {d} x=\int _{\mathbb {R} ^{m}}\,\mathrm {d} x\int _{\Lambda ^{n}}f(x,\theta )\,\mathrm {d} \theta .}$

Change of even and odd variables

Let a coordinate transformation be given by ${\displaystyle x_{i}=x_{i}(y,\xi ),\ i=1,\ldots ,m;\ \theta _{j}=\theta _{j}(y,\xi ),j=1,\ldots ,n,}$ where ${\displaystyle x_{i}}$ are even and ${\displaystyle \theta _{j}}$ are odd polynomials of ${\displaystyle \xi }$ depending on even variables ${\displaystyle y.}$ The Jacobian matrix of this transformation has the block form:

${\displaystyle \mathrm {J} ={\frac {\partial (x,\theta )}{\partial (y,\xi )}}={\begin{pmatrix}A&B\\C&D\end{pmatrix}},}$

where each even derivative ${\displaystyle \partial /\partial y_{j}}$ commutes with all elements of the algebra ${\displaystyle \Lambda ^{m\mid n}}$; the odd derivatives commute with even elements and anticommute with odd elements. The entries of the diagonal blocks ${\displaystyle A=\partial x/\partial y}$ and ${\displaystyle D=\partial \theta /\partial \xi }$ are even and the entries of the off-diagonal blocks ${\displaystyle B=\partial x/\partial \xi ,\ C=\partial \theta /\partial y}$ are odd functions, where ${\displaystyle \partial /\partial \xi _{j}}$ again mean right derivatives.

When the function ${\displaystyle D}$ is invertible in ${\displaystyle \Lambda ^{m\mid n},}$

${\displaystyle \mathrm {J} ={\frac {\partial (x,\theta )}{\partial (y,\xi )}}={\begin{pmatrix}A&B\\C&D\end{pmatrix}}={\begin{pmatrix}I&B\\0&D\end{pmatrix}}{\begin{pmatrix}A-BD^{-1}C&0\\D^{-1}C&I\end{pmatrix}}}$

So we have the Berezinian (or superdeterminant) of the matrix ${\displaystyle \mathrm {J} }$, which is the even function

${\displaystyle \operatorname {Ber} \mathrm {J} =\det \left(A-BD^{-1}C\right)(\det D)^{-1}}$

Suppose that the real functions ${\displaystyle x_{i}=x_{i}(y,0)}$ define a smooth invertible map ${\displaystyle F:Y\to X}$ of open sets ${\displaystyle X,Y}$ in ${\displaystyle \mathbb {R} ^{m}}$ and the linear part of the map ${\displaystyle \xi \mapsto \theta =\theta (y,\xi )}$ is invertible for each ${\displaystyle y\in Y.}$ The general transformation law for the Berezin integral reads

${\displaystyle {\begin{aligned}&\int _{\Lambda ^{m\mid n}}f(x,\theta )\,\mathrm {d} \theta \,\mathrm {d} x=\int _{\Lambda ^{m\mid n}}f(x(y,\xi ),\theta (y,\xi ))\varepsilon \operatorname {Ber} \mathrm {J} \,\mathrm {d} \xi \,\mathrm {d} y\\[6pt]={}&\int _{\Lambda ^{m\mid n}}f(x(y,\xi ),\theta (y,\xi ))\varepsilon {\frac {\det \left(A-BD^{-1}C\right)}{\det D}}\,\mathrm {d} \xi \,\mathrm {d} y,\end{aligned}}}$

where ${\displaystyle \varepsilon =\mathrm {sgn} (\det \partial x(y,0)/\partial y}$) is the sign of the orientation of the map ${\displaystyle F.}$ The superposition ${\displaystyle f(x(y,\xi ),\theta (y,\xi ))}$ is defined in the obvious way, if the functions ${\displaystyle x_{i}(y,\xi )}$ do not depend on ${\displaystyle \xi .}$ In the general case, we write ${\displaystyle x_{i}(y,\xi )=x_{i}(y,0)+\delta _{i},}$ where ${\displaystyle \delta _{i},i=1,\ldots ,m}$ are even nilpotent elements of ${\displaystyle \Lambda ^{m\mid n}}$ and set

${\displaystyle f(x(y,\xi ),\theta )=f(x(y,0),\theta )+\sum _{i}{\frac {\partial f}{\partial x_{i}}}(x(y,0),\theta )\delta _{i}+{\frac {1}{2}}\sum _{i,j}{\frac {\partial ^{2}f}{\partial x_{i}\,\partial x_{j}}}(x(y,0),\theta )\delta _{i}\delta _{j}+\cdots ,}$

where the Taylor series is finite.