Grassmann Numbers and Berezin Integration

Hello! Welcome to the next lesson in our journey through the path integral formulation of quantum field theory.

In our last lesson, we saw the power of the generating functional, $Z_{0} [J]$ , for scalar fields. By taking functional derivatives, we could systematically compute any correlation function, and in doing so, we derived Wick's theorem as a natural consequence of Gaussian integration. The entire formalism was built on ordinary, commuting numbers, which is appropriate for describing bosons.

However, the universe also contains fermions—particles like electrons and quarks. From your study of advanced quantum mechanics, you know that fermionic creation and annihilation operators fundamentally differ from their bosonic counterparts: they anticommute. For example, ${ψ_{a}, ψ_{b}^{†}} \neq = 0$ . This raises a critical question: how can we represent anticommuting fields within a path integral, which is a sum over classical field configurations? A classical field is just a function that returns a number for each point in spacetime, and ordinary numbers commute.

The solution is to invent a new kind of number that has the required anticommuting property built in. Today, our learning outcome is to define Grassmann numbers and the Berezin integral for their integration. We will develop the essential mathematical tools that allow us to build path integrals for fermions.

What are Grassmann Numbers?

To handle anticommuting fields, we need anticommuting numbers. These are called Grassmann numbers, and they are defined by a single, simple axiom.

A set of Grassmann numbers, ${θ_{1}, θ_{2}, \dots, θ_{N}}$ , are objects that anticommute with each other:
$θ_{i} θ_{j} = - θ_{j} θ_{i}$
This is their defining property. They can be multiplied by ordinary complex numbers and added together as usual.

A striking consequence of this rule emerges when we consider the product of a Grassmann number with itself. For any $θ_{i}$ :
$θ_{i} θ_{i} = - θ_{i} θ_{i}$
The only way this equation can be true is if the product is zero:
$θ_{i}^{2} = 0$
This property, which seems trivial, makes the calculus of Grassmann numbers vastly simpler than that of ordinary numbers.

The following video provides a clear introduction to this concept.

QFT2 Lecture 5d: introduction to Grassman numbers

In this lecture, Dr. Nabil Iqbal introduces the motivation for Grassmann numbers and defines their fundamental properties.

Please watch the segment from 01:36 to 04:36. This part defines Grassmann numbers, their anticommutation relation, and the crucial consequence that their square is zero.

Functions of Grassmann Numbers

The property $θ^{2} = 0$ has a profound implication for functions. If you try to Taylor expand a function of a single Grassmann number, $f (θ)$ , the series terminates immediately.
$f (θ) = c_{0} + c_{1} θ + c_{2} θ^{2} + c_{3} θ^{3} + \dots$
Since $θ^{2} = 0$ , all terms of order two and higher vanish. Therefore, the most general function of a single Grassmann number is just a linear expression:
$f (θ) = A + B θ$
where $A$ and $B$ are ordinary (commuting) numbers or other Grassmann-valued expressions.

For a function of two Grassmann numbers, $θ_{1}$ and $θ_{2}$ , the expansion stops at the $θ_{1} θ_{2}$ term:
$f (θ_{1}, θ_{2}) = A + B θ_{1} + C θ_{2} + D θ_{1} θ_{2}$
Note that $θ_{2} θ_{1} = - θ_{1} θ_{2}$ , so we only need one of the quadratic terms. This finite nature is a general feature.

For a more formal algebraic perspective, these notes discuss the structure of the algebra generated by these numbers.

Fermionic Functional Integrals

These notes on 'Fermionic Functional Integrals' by Vadim Kaplunovsky provide a slightly more formal introduction to Grassmann algebras.

Read the first two pages, up to the section 'Next, consider functions of Grassmann numbers.' This will introduce the concept of a Z2-graded algebra, distinguishing between 'even' (bosonic) and 'odd' (fermionic) elements, which is a useful way to think about these structures.

Integration of Grassmann Numbers: The Berezin Integral

Having defined functions, we now need to define integration. We can't use the standard Riemann or Lebesgue definitions because the notion of a continuous range for $θ$ doesn't make sense. Instead, we define integration by requiring it to have certain desirable properties.

The most important property we demand is translational invariance, just as we did for path integrals over ordinary fields. We require that shifting the integration variable by another Grassmann number $η$ does not change the value of the integral:
$\int d θ f (θ) = \int d θ f (θ + η)$
Let's see what this implies for our general function $f (θ) = A + B θ$ .
$f (θ + η) = A + B (θ + η) = (A + B η) + B θ$
The "constant" part of the function changes, but the coefficient of the linear term, $B$ , does not. For the integral to be invariant, it must depend only on the part of the function that is invariant under the shift. This uniquely selects the coefficient of the linear term as the result of the integration.

This leads to the Berezin integration rules:

$\int d θ A = 0$
$\int d θ B θ = B$ (by convention, we set the normalization to 1)

In essence, Grassmann integration is the same as differentiation: $\int d θ \equiv \frac{\partial}{\partial θ}$ .

The following video segment walks through this derivation and establishes the rules for integration.

QFT2 Lecture 5d: introduction to Grassman numbers

Dr. Iqbal now derives the rules for Berezin integration by demanding shift invariance, a key physical principle.

Watch from 05:39 to 13:00. This part covers functions of Grassmann numbers, introduces left/right derivatives (which we won't heavily use but is good context), and then derives the Berezin integration rules. Pay close attention to the argument based on shift invariance.

For multiple variables, we define the integral as an iterated integral. However, since the variables anticommute, the order of integration matters. The standard convention is:
$\int d θ_{N} \dots \int d θ_{2} \int d θ_{1} \dots$
This means the innermost differential, $d θ_{1}$ , corresponds to the first integration performed. For example:
$\int d θ_{2} \int d θ_{1} (θ_{1} θ_{2}) = \int d θ_{2} (θ_{2}) = 1$
But if we switch the order of the variables in the integrand:
$\int d θ_{2} \int d θ_{1} (θ_{2} θ_{1}) = \int d θ_{2} \int d θ_{1} (- θ_{1} θ_{2}) = \int d θ_{2} (- θ_{2}) = - 1$
The general rule is that the integral picks out the coefficient of the top-degree term, $θ_{N} \dots θ_{1}$ .

Gaussian Integrals: The Main Result

Now we arrive at the most important application for our purposes: Gaussian integrals. Recall from our study of scalar fields that the path integral was a giant Gaussian integral, and its solution involved the inverse of the matrix in the exponent, which gave rise to the propagator. For ordinary numbers $z_{i}$ , we have the famous formula:
$\int \prod_{i} \frac{d z _{i}^{*} d z _{i}}{2 π i} exp (- z^{†} A z) = \frac{1}{d e t ( A )}$
Let's see what happens with Grassmann numbers. We need to evaluate:
$I (A) = \int \prod_{i = 1}^{N} d θ ˉ_{i} d θ_{i} exp (- θ ˉ_{i} A_{ij} θ_{j})$
where $θ_{i}$ and $θ ˉ_{i}$ are independent complex Grassmann variables.

Let's start with the simplest case, $N = 1$ :
$\int d θ ˉ d θ e^{- b θ ˉ θ}$
The exponential expands to $1 - b θ ˉ θ$ . We use $θ ˉ θ = - θ θ ˉ$ and the integration rules:
$\int d θ ˉ \int d θ (1 - b θ ˉ θ) = \int d θ ˉ \int d θ (1 + b θ θ ˉ) = \int d θ ˉ (b θ ˉ) = b$
The result is $b$ , not $1/ b$ !

This generalizes to the N-dimensional case. The result, which is fundamental to all of QFT for fermions, is:
$\int (\prod_{i = 1}^{N} d θ ˉ_{i} d θ_{i}) exp (- \sum_{j, k} θ ˉ_{j} A_{j k} θ_{k}) = det (A)$
The determinant appears in the numerator, not the denominator.

This is a crucial difference between bosonic and fermionic path integrals. The following reading provides a clear, step-by-step proof of this amazing result.

Fermionic Functional Integrals

The notes by Vadim Kaplunovsky provide an excellent derivation of the Gaussian integral formula for Grassmann numbers.

Please read the section 'Gaussian Integrals over Odd Grassmann Numbers' (pages 4-6). It starts with the N=1 case, works through N=2 explicitly to show how the determinant emerges from the combinatorics, and then presents the proof for general N.

Test your understanding!

For a $2 \times 2$ matrix $A = (a c b d)$ , the quadratic term in the exponent is $- (θ ˉ_{1} a θ_{1} + θ ˉ_{1} b θ_{2} + θ ˉ_{2} c θ_{1} + θ ˉ_{2} d θ_{2})$ . The integral $\int d θ ˉ_{1} d θ_{1} d θ ˉ_{2} d θ_{2} exp (- Θ ˉ^{T} A Θ)$ picks out the coefficient of the $θ_{1} θ ˉ_{1} θ_{2} θ ˉ_{2}$ term in the expansion of the exponential. What is this term? Show that its coefficient is indeed $det (A)$ .

(Hint: Only the $\frac{1}{2 !} (Θ ˉ^{T} A Θ)^{2}$ term in the Taylor series of the exponential can contribute. You will need to carefully anticommute the variables to get them into the correct order for integration.)

Show answer

The term in the expansion of the exponential that contains all four variables is $\frac{1}{2} (θ ˉ_{i} A_{ij} θ_{j}) (θ ˉ_{k} A_{k l} θ_{l})$ . To get a non-zero contribution, all indices must be different, which means we must have terms like $(θ ˉ_{1} A_{12} θ_{2}) (θ ˉ_{2} A_{21} θ_{1})$ and $(θ ˉ_{1} A_{11} θ_{1}) (θ ˉ_{2} A_{22} θ_{2})$ .

Let's expand the quadratic term:
$\frac{1}{2} (θ ˉ_{1} a θ_{1} + θ ˉ_{1} b θ_{2} + θ ˉ_{2} c θ_{1} + θ ˉ_{2} d θ_{2})^{2}$

The only terms that survive are cross terms that don't repeat a variable. The two such terms are:

$(θ ˉ_{1} a θ_{1}) (θ ˉ_{2} d θ_{2}) = a d (θ ˉ_{1} θ_{1} θ ˉ_{2} θ_{2}) = a d (θ_{1} θ ˉ_{1} θ_{2} θ ˉ_{2})$
$(θ ˉ_{1} b θ_{2}) (θ ˉ_{2} c θ_{1}) = b c (θ ˉ_{1} θ_{2} θ ˉ_{2} θ_{1}) = - b c (θ ˉ_{1} θ_{1} θ_{2} θ ˉ_{2}) = - b c (θ_{1} θ ˉ_{1} θ_{2} θ ˉ_{2})$

The total coefficient of the $θ_{1} θ ˉ_{1} θ_{2} θ ˉ_{2}$ term in the expansion of $exp (- Θ ˉ^{T} A Θ)$ is therefore $(a d - b c) = det (A)$ . The integral picks up exactly this coefficient.

Finally, just as we did for the scalar field, we can add source terms to the Gaussian integral. This allows us to define the generating functional for fermions, from which we can derive propagators and correlation functions.

QFT2 Lecture 5d: introduction to Grassman numbers

To conclude, Dr. Iqbal demonstrates the Gaussian integral and extends it to include source terms, which directly mirrors what we did for the scalar field and sets the stage for defining the fermionic generating functional.

Watch from 14:12 to 21:05 and then from 29:32 to 31:40. The first part covers the Gaussian integral for multiple variables. The second part (after a discussion of change of variables) introduces source terms and derives the final, all-important formula for the generating functional.

Conclusion

Today, we have assembled a new mathematical toolkit. We addressed the challenge of representing anticommuting fermionic fields in a path integral by introducing a new type of number that satisfies this property by definition.

Key Takeaways:

Grassmann numbers are anticommuting objects: $θ_{i} θ_{j} = - θ_{j} θ_{i}$ , which implies $θ_{i}^{2} = 0$ .
Functions of Grassmann numbers are simple, finite polynomials.
Berezin integration is defined by requiring translational invariance and is equivalent to differentiation: $\int d θ (A + B θ) = B$ .
The fundamental result for fermionic path integrals is the Gaussian integral formula: $\int d θ ˉ d θ e^{- θ ˉ^{T} A θ} = det (A)$ . The determinant appears in the numerator, in stark contrast to the bosonic case.

Preview of the Next Lesson:
We are now fully equipped to tackle fermionic fields. In our next lesson, we will use Grassmann numbers and the Berezin integral to construct the path integral for the free Dirac field. We will define its generating functional and, by applying the results from today's lesson, derive the Feynman propagator for fermions directly from the path integral.

Recommended resources

What are Grassmann Numbers?

Functions of Grassmann Numbers

Integration of Grassmann Numbers: The Berezin Integral

Gaussian Integrals: The Main Result

Conclusion