Appendix C: Gaussian Integrals and Grassmann Integrals¶
Story so far:
In Appendix B, we organized the spinor representation of the Lorentz group and the Dirac equation, seeing how spin-1/2 fermions are incorporated into quantum field theory.
Goals of this chapter
- Organize the Gaussian integrals (single-variable and multi-variable) and the algebra of Grassmann numbers along with Berezin integration, which repeatedly appear in quantum field theory calculations
- Gaussian integrals form the mathematical foundation for bosonic path integrals (Chapters 10–11), while Grassmann integrals form the foundation for fermionic path integrals (Ch. 12)
- Once you grasp the contrast between "bosonic \((\det A)^{-1/2}\)" and "fermionic \(\det A\)," the difference in how the two types of particles are treated in path integrals becomes clear
- This Appendix focuses on the basics of Gaussian and Grassmann integrals
- Practical tools for loop calculations—dimensional analysis, Feynman parameters, Wick rotation, and loop integral formulas—are collected in the following Appendix D
C.1 Single-Variable Gaussian Integral¶
🟡 Lina: This Appendix is a "toolbox." You'll reference it many times in the path integral chapters (Chapters 10–12) and loop calculation chapters (Chapters 13–14), so let's start from the very beginning—the Gaussian integral.
C.1.1 Basic Formula¶
🟡 Lina: Calculations in quantum field theory, when you boil them down, all reduce to Gaussian integrals—that's barely an exaggeration. Here's the basic formula.
🔵 Kai: Can the parameter \(a\) be something other than a real number?
🟡 Lina: Good question. \(a\) can be complex. However, the condition \(\mathrm{Re}(a) > 0\)—meaning the real part of \(a\) is positive—is required. The reason is that if you write \(a = \alpha + i\beta\), then
The second factor \(e^{-i\beta q^2/2}\) merely oscillates and stays at magnitude 1. For the integral to converge, the first factor \(e^{-\alpha q^2/2}\) needs to approach zero as \(q \to \pm\infty\), which requires \(\alpha > 0\).
⚪ Mei: So \(\mathrm{Re}(a) > 0\) is the condition for "the integral not to diverge."
✅ Comprehension Check: For the Gaussian integral \(\int_{-\infty}^{\infty} dq\; e^{-aq^2/2}\) where the parameter \(a\) is complex, what condition must be imposed on \(a\) for the integral to converge? Explain the reason as well.
Answer
\(\mathrm{Re}(a) > 0\) is required. Writing \(a = \alpha + i\beta\), we get \(e^{-aq^2/2} = e^{-\alpha q^2/2} \cdot e^{-i\beta q^2/2}\), where the second factor merely oscillates with magnitude 1. For the integral to converge as \(q \to \pm\infty\), the first factor \(e^{-\alpha q^2/2}\) must approach zero, which requires \(\alpha = \mathrm{Re}(a) > 0\).
C.1.2 Proof—The Squaring Trick¶
🟡 Lina: The proof of this formula is beautiful, so let's go through it carefully. Let \(I\) denote the value of the integral.
Computing this directly is difficult, so we consider \(I^2\).
🔵 Kai: So you use two independent variables \(q_1, q_2\) to turn it into a two-dimensional integral.
🟡 Lina: Exactly. Now we convert to 2D polar coordinates. Setting \(q_1 = r\cos\theta\), \(q_2 = r\sin\theta\), we have \(q_1^2 + q_2^2 = r^2\). The area element becomes \(dq_1\,dq_2 = r\,dr\,d\theta\). Intuitively, at radius \(r\), sweeping through angle \(d\theta\) gives an arc length of \(r\,d\theta\), so the infinitesimal area is "\(dr\) × \(r\,d\theta\)" = \(r\,dr\,d\theta\). Recall the arc length formula \(r\theta\) from high school—for an infinitesimal angle \(d\theta\), the arc length is \(r\,d\theta\), so the area of the thin "strip" is width \(dr\) × length \(r\,d\theta\). Close to the origin (small \(r\)), the strips are short; far away (large \(r\)), they're long—that's why the area element picks up a factor of \(r\). In general, how area elements transform under variable changes is organized by the concept of the "Jacobian"—I'll explain this properly in C.2.1. For now, just remember that "in polar coordinates, an extra factor of \(r\) appears."
Since the integrand doesn't contain \(\theta\), the \(\theta\) integration simply gives \(2\pi\).
⚪ Mei: Since there's no angular dependence, the \(\theta\) integral immediately gives \(2\pi\) and drops out.
🟡 Lina: The \(r\) integral can be handled with the substitution \(u = \frac{a}{2}r^2\). Since \(du = ar\,dr\), we have \(r\,dr = du/a\). Substituting:
🔵 Kai: Oh, the substitution absorbs the \(r\) factor perfectly, leaving just a simple exponential integral.
🟡 Lina: Exactly. Putting it together:
This proves equation (C.1).
🔵 Kai: When \(a\) is complex, how do we handle the sign of the square root?
🟡 Lina: Sharp question. For complex square roots, we need to be careful about "which sign to take." Write \(a = |a|e^{i\theta}\) in polar form. Since \(\mathrm{Re}(a) = |a|\cos\theta\), the condition \(\mathrm{Re}(a) > 0\) means \(\cos\theta > 0\), i.e., \(-\pi/2 < \theta < \pi/2\). If we define \(\sqrt{a} = \sqrt{|a|}\,e^{i\theta/2}\) (since \(\theta/2\) ranges over \(-\pi/4 < \theta/2 < \pi/4\), we're guaranteed \(\mathrm{Re}(\sqrt{a}) = \sqrt{|a|}\cos(\theta/2) > 0\)), then writing \(\sqrt{2\pi/a} = \sqrt{2\pi}/\sqrt{a}\):
When \(\theta = 0\) (\(a\) is a positive real number), \(I\) is clearly a positive real number, confirming this sign convention is correct. We'll use this when \(a\) is close to pure imaginary (before Wick rotation) in the path integral (Ch. 10).
🔵 Kai: So within the range \(\mathrm{Re}(a) > 0\), the sign choice of the square root is uniquely determined.
C.1.3 Extended Formula with a Linear Term (Gaussian Integral with Source)¶
🟡 Lina: Next, let's derive the formula when a linear term in \(q\) is added to the exponent. This is always used when introducing a source \(J\) in path integrals.
(Note on signs: some references define it with \(e^{-aq^2/2 + Jq}\), i.e., \(+Jq\). In that case, the substitution \(J \to -J\) reduces it to equation (C.3). In this book, we use the same sign convention in the generating functional of Ch. 11.)
🔵 Kai: So \(e^{J^2/(2a)}\) appears as a "bonus" on the right-hand side. How do you derive this?
🟡 Lina: The key technique is completing the square—the same thing you used in high school to find the vertex of a quadratic function.
Let's rearrange the exponent:
⚪ Mei: Let me verify. Expanding the right-hand side:
Indeed it returns to the original.
🟡 Lina: Then we just substitute \(z \equiv q + J/a\). Since \(dz = dq\) and the integration range remains \((-\infty, +\infty)\):
🔵 Kai: Oh, it comes out cleanly! Since \(e^{J^2/(2a)}\) is a constant independent of \(z\), it can be pulled out of the integral.
🟡 Lina: This pattern—"complete the square → change variables → reduce to a Gaussian integral"—is used dozens of times in path integrals, so make it second nature.
✅ Comprehension Check: In deriving the Gaussian integral with source \(\int dq\; e^{-aq^2/2 - Jq}\), what specifically is the "completing the square" operation? Also, why does the integration range remain unchanged after the variable substitution?
Answer
Completing the square is the operation of rewriting the exponent \(-\frac{a}{2}q^2 - Jq\) as \(-\frac{a}{2}(q + J/a)^2 + J^2/(2a)\). Under the variable substitution \(z = q + J/a\), we have \(dz = dq\), and as \(q\) ranges from \(-\infty\) to \(+\infty\), \(z\) ranges over the same interval, so the integration limits are unchanged. Pulling out the constant factor \(e^{J^2/(2a)}\), the remainder reduces to the basic Gaussian integral \(\sqrt{2\pi/a}\).
Comprehension Check C.1
Evaluate \(\displaystyle\int_{-\infty}^{\infty} dq\; e^{-\frac{a}{2}q^2 + bq}\) (with \(\mathrm{Re}(a) > 0\)).
Answer: The exponent in equation (C.3) is \(-\frac{a}{2}q^2 - Jq\), and the exponent in this problem is \(-\frac{a}{2}q^2 + bq\). Comparing the two: \(-Jq = +bq\), so \(J = -b\). Substituting \(J = -b\) into equation (C.3) gives \(\sqrt{2\pi/a}\;e^{J^2/(2a)} = \sqrt{2\pi/a}\;e^{(-b)^2/(2a)} = \sqrt{2\pi/a}\;e^{b^2/(2a)}\). Alternatively, completing the square directly gives the same result: \(-\frac{a}{2}q^2 + bq = -\frac{a}{2}(q - b/a)^2 + b^2/(2a)\).
C.1.4 Gaussian Integral with \(q^n\) Insertion¶
🟡 Lina: Let me introduce another frequently used formula—the case where the Gaussian integrand is multiplied by \(q^n\).
Result:
Here \((2m-1)!! = (2m-1)(2m-3)\cdots 3 \cdot 1\) is called the double factorial. You multiply every other number.
🔵 Kai: I've never seen the double factorial before. What values does it take concretely?
🟡 Lina: For example, \(m = 1\) gives \((2 \cdot 1 - 1)!! = 1!! = 1\); \(m = 2\) gives \(3!! = 3 \cdot 1 = 3\); \(m = 3\) gives \(5!! = 5 \cdot 3 \cdot 1 = 15\). While the ordinary factorial \(n!\) multiplies all integers, the double factorial multiplies every other one.
By the way, let's check that formula (C.6) works correctly for \(m = 0\) (i.e., \(n = 0\)). Substituting \(m = 0\) gives \(\sqrt{2\pi/a} \cdot (-1)!!/a^0 = \sqrt{2\pi/a} \cdot (-1)!!\). On the other hand, \(I_0 = \int dq\; e^{-aq^2/2} = \sqrt{2\pi/a}\) is already known (equation (C.1) itself). For these to match, we must have \((-1)!! = 1\). In other words, we define \((-1)!! = 1\) so that "the formula works correctly even at \(m = 0\)." This is the same reasoning as \(0! = 1\). Recall that \(0!\) is also determined by consistency: substituting \(n = 1\) into \(n! = n \cdot (n-1)!\) gives \(1! = 1 \cdot 0!\), so \(0! = 1\). Intuitively, think of it as the convention that "when there's nothing to multiply, the product equals 1"—if you multiply nothing together, the result stays at 1. Similarly, \((2m-1)!!\) is "the product of \(m\) odd numbers," so for \(m = 0\) it's "the product of 0 numbers" = 1.
🔵 Kai: The result being zero for odd \(n\) is because the integrand is an odd function, right?
🟡 Lina: Exactly. \(e^{-aq^2/2}\) is an even function, and \(q^n\) (for odd \(n\)) is an odd function. Their product is an odd function, so integrating from \(-\infty\) to \(+\infty\), the positive and negative parts cancel to give zero.
⚪ Mei: How do you derive the even case?
🟡 Lina: The easiest way is to use a recurrence relation. Here's the relation, which can be shown by integration by parts:
Proof: Split the integrand of \(I_n(a)\) as \(q^{n-1} \cdot q\,e^{-aq^2/2}\). Since \(\frac{d}{dq}e^{-aq^2/2} = -aq\,e^{-aq^2/2}\) (chain rule), we can write \(q\,e^{-aq^2/2} = -\frac{1}{a}\frac{d}{dq}e^{-aq^2/2}\). Using integration by parts \(\int u\,dv = [uv] - \int v\,du\) with
- \(u = q^{n-1}\) → \(du = (n-1)q^{n-2}\,dq\)
- \(v = -\frac{1}{a}e^{-aq^2/2}\) → \(dv = q\,e^{-aq^2/2}\,dq\)
we get
Since \(\mathrm{Re}(a) > 0\), the boundary term vanishes. The remaining integral is simply \(I_{n-2}(a)\).
🔵 Kai: I see—starting from \(I_0 = \sqrt{2\pi/a}\), you can chain together \(I_2 = (1/a)\sqrt{2\pi/a}\), \(I_4 = (3/a^2)\sqrt{2\pi/a}\), and so on. By the way, is there any significance to the coefficient \(3\) in \(I_4\)? Like, the number of ways to pair up 4 \(q\)'s into pairs of 2...?
🟡 Lina: Exactly right! The number of ways to partition 4 \(q\)'s into pairs of 2 is \((q_1 q_2)(q_3 q_4)\), \((q_1 q_3)(q_2 q_4)\), \((q_1 q_4)(q_2 q_3)\)—3 ways—which matches precisely \((2m-1)!! = 3!! = 3\). This is no coincidence; it's directly connected to Wick's theorem (Ch. 7).
⚪ Mei: The number of pairings equals the double factorial—that's why it's directly linked to Wick's theorem.
🟡 Lina: Let me summarize the concrete values in a table.
Table C.1: Concrete values of the Gaussian integral \(I_n(a)\)
| \(n\) | \(I_n(a)\) |
|---|---|
| 0 | \(\sqrt{2\pi/a}\) |
| 2 | \(\sqrt{2\pi/a}\cdot a^{-1}\) |
| 4 | \(\sqrt{2\pi/a}\cdot 3a^{-2}\) |
| 6 | \(\sqrt{2\pi/a}\cdot 15a^{-3}\) |
Verify that the coefficients \(1, 3, 15\) correspond to \((2m-1)!!\).
✅ Comprehension Check: Explain why the Gaussian integral \(I_n(a) = \int dq\; q^n e^{-aq^2/2}\) vanishes when \(n\) is odd. Also, what technique is used to show the recurrence relation \(I_n = \frac{n-1}{a} I_{n-2}\) for even \(n\)?
Answer
When \(n\) is odd, \(q^n\) is an odd function and \(e^{-aq^2/2}\) is an even function, so their product is an odd function, and integrating over the symmetric interval \((-\infty, +\infty)\) gives zero. The recurrence relation is shown by integration by parts. Split the integrand as \(q^{n-1} \cdot q e^{-aq^2/2}\), use \(q e^{-aq^2/2} = -\frac{1}{a}\frac{d}{dq}e^{-aq^2/2}\), and integrate by parts; the boundary term vanishes and the result reduces to \(I_{n-2}\).
📝 Exercises:
- Computing \(I_6(a)\) using the Gaussian integral recurrence relation → Problem B-3. Gaussian Integral Containing \(q^n\) (Application of Recurrence Relation)
C.2 Multi-Variable Gaussian Integral¶
🟡 Lina: In quantum field theory, there isn't just one integration variable—there are infinitely many. So the extension to multiple variables is essential.
C.2.1 Basic Formula¶
🟡 Lina: Let's collectively write \(n\) real variables \(q_1, q_2, \ldots, q_n\) as the vector \(\mathbf{q}\). Let \(A\) be an \(n \times n\) real symmetric positive-definite matrix. "Symmetric" means \(A_{ij} = A_{ji}\) (unchanged under transposition). The formal definition of "positive-definite" is "for any \(\mathbf{q} \neq \mathbf{0}\), \(\mathbf{q}^T A\,\mathbf{q} > 0\) holds," but intuitively it's the same condition as needing \(a > 0\) so that \(aq^2 > 0\) in the single-variable case—"no matter which direction you pull, the quadratic form is positive."
🔵 Kai: I've heard it also means "all eigenvalues are positive"—but what are eigenvalues?
🟡 Lina: An eigenvalue is a scalar \(\lambda\) such that there exists a nonzero vector \(\mathbf{v}\) satisfying \(A\mathbf{v} = \lambda\mathbf{v}\)—the matrix merely "stretches" a particular direction vector \(\mathbf{v}\) (called an eigenvector) by a factor \(\lambda\). For example, the \(2 \times 2\) diagonal matrix \(\begin{pmatrix} 3 & 0 \\ 0 & 5 \end{pmatrix}\) stretches vectors in the \(x\)-direction by 3 and in the \(y\)-direction by 5, so the eigenvalues are 3 and 5.
🔵 Kai: So an eigenvector is a vector whose direction doesn't change when the matrix acts on it, and the eigenvalue is the "stretch factor."
🟡 Lina: Exactly. Being positive-definite is equivalent to "all eigenvalues of \(A\) are positive." The reason for this equivalence is that a real symmetric matrix can always be diagonalized by an orthogonal matrix—that is, there exists an orthogonal matrix \(O\) such that \(O^T A\,O = \mathrm{diag}(\lambda_1, \ldots, \lambda_n)\) (this is called the spectral theorem in linear algebra. You don't need to memorize the name, but we'll use the fact that "real symmetric matrices can always be diagonalized by orthogonal matrices" as a tool in this chapter. I'll leave the proof to linear algebra textbooks). I'll show you shortly why diagonalization is useful.
🔵 Kai: Can matrices that aren't "real symmetric" fail to be diagonalizable?
🟡 Lina: In general, yes. But in physics, the \(A\) appearing in quadratic forms \(\mathbf{q}^T A\,\mathbf{q}\) can always be taken to be symmetric (the antisymmetric part doesn't contribute to \(\mathbf{q}^T A\,\mathbf{q}\)), so in this chapter we always deal with diagonalizable situations. In the diagonalized coordinate system, \(\mathbf{q}^T A\,\mathbf{q} = \sum_i \lambda_i z_i^2\). For this to be positive for all \(\mathbf{z} \neq \mathbf{0}\), each \(\lambda_i > 0\) is necessary and sufficient—the same logic as \(a > 0\) in the single-variable case.
⚪ Mei: So "diagonalize and separate into each direction, and it ultimately reduces to the single-variable condition."
🟡 Lina: Exactly. Then:
Here \(d^n q = dq_1\,dq_2\cdots dq_n\). The expression \(\mathbf{q}^T A\,\mathbf{q}\) is the product "row vector × matrix × column vector," which in components reads \(\mathbf{q}^T A\,\mathbf{q} = \sum_{i,j} q_i A_{ij} q_j\). For example, with \(n = 2\):
If \(A\) is symmetric (\(A_{12} = A_{21}\)), the cross term is \(2A_{12}q_1 q_2\).
🔵 Kai: So in the single-variable case it was \(\sqrt{2\pi/a}\), and going to \(n\) variables, \(a\) gets replaced by the matrix \(A\), and \(1/\sqrt{a}\) becomes \(1/\sqrt{\det A}\).
🟡 Lina: Precisely. Intuitively, when \(A\) can be diagonalized, you can perform independent Gaussian integrals for each eigenvalue \(\lambda_i\):
⚪ Mei: Since symmetric matrices can be diagonalized by orthogonal matrices, each variable separates into an independent single-variable Gaussian integral.
🔵 Kai: Wait, when you change variables... doesn't the area element pick up an extra factor, like with polar coordinates?
🟡 Lina: Good point. In multi-variable changes of variables, a factor called the Jacobian multiplies the volume element. In general, when transforming from variables \(\mathbf{q}\) to new variables \(\mathbf{z}\), the volume element changes as
This \(\left|\det\frac{\partial q_i}{\partial z_j}\right|\) is the Jacobian. Here \(\frac{\partial q_i}{\partial z_j}\) is a partial derivative—"the rate at which \(q_i\) changes when you vary \(z_j\) slightly while holding all other variables \(z_k\) (\(k \neq j\)) fixed." It's the multi-variable version of the single-variable derivative \(dq/dz\). Arranging these into a matrix and taking the determinant gives the Jacobian. In the polar coordinate example from C.1.2, with \(q_1 = r\cos\theta\), \(q_2 = r\sin\theta\):
This is the reason for the extra factor of \(r\) in polar coordinates.
🔵 Kai: I see—the intuitive explanation earlier about "the strip length being proportional to \(r\)" matches the Jacobian calculation exactly.
🟡 Lina: Now, in our problem we're considering the variable change \(\mathbf{q} = O\mathbf{z}\) with an orthogonal matrix \(O\). In this case \(\partial q_i/\partial z_j = O_{ij}\), so the Jacobian is \(|\det O|\). An orthogonal matrix satisfies \(O^T O = I\) (the transpose equals the inverse), and geometrically represents "rotations and reflections." Since \(\det(O^T O) = (\det O)^2 = \det I = 1\), we have \(\det O = \pm 1\), meaning \(|\det O| = 1\). Rotations and reflections don't change areas or volumes, right? That's why the Jacobian is 1.
🔵 Kai: I see—since rotations don't change area, no extra factor appears. So we can safely diagonalize.
✅ Comprehension Check: In the multi-variable Gaussian integral result \((2\pi)^{n/2}/(\det A)^{1/2}\), explain the physical/mathematical reason why the determinant \(\det A\) appears.
Answer
A real symmetric positive-definite matrix \(A\) can be diagonalized by an orthogonal transformation. With eigenvalues \(\lambda_1, \ldots, \lambda_n\), each variable allows an independent single-variable Gaussian integral \(\sqrt{2\pi/\lambda_i}\). Their product is \(\prod_i \sqrt{2\pi/\lambda_i} = (2\pi)^{n/2}/\sqrt{\lambda_1 \cdots \lambda_n} = (2\pi)^{n/2}/\sqrt{\det A}\). Since the Jacobian of an orthogonal transformation is 1, no additional factor arises from the change of variables.
C.2.2 Multi-Variable Gaussian Integral with Source¶
🟡 Lina: When a linear term (source \(\mathbf{J}\)) is added, the result is:
🔵 Kai: So in the single-variable case it was \(e^{J^2/(2a)}\), and now it's replaced by \(e^{\frac{1}{2}\mathbf{J}^T A^{-1}\mathbf{J}}\). The \(1/a\) has become the inverse matrix \(A^{-1}\). But can you do completing the square in the matrix case too?
🟡 Lina: Good question. Yes, you can. The derivation is the same as the single-variable case. Complete the square:
The variable substitution \(\mathbf{z} = \mathbf{q} + A^{-1}\mathbf{J}\) has Jacobian 1, so it reduces to equation (C.8).
⚪ Mei: In the single-variable case it was "shift \(q\) by \(q + J/a\)," and in the multi-variable case it's "shift \(\mathbf{q}\) by \(\mathbf{q} + A^{-1}\mathbf{J}\)"—the structure is exactly the same.
🟡 Lina: This formula becomes the heart of computing the generating functional \(Z[J]\) in the field path integral (Ch. 11). In the \(n \to \infty\) limit, the discrete sum is replaced by a continuous integral, and \(A^{-1}\) becomes the Feynman propagator.
⚪ Mei: So the structure where \(1/a\) in the single-variable case becomes \(A^{-1}\) in the multi-variable case carries over directly to the infinite-variable field theory.
✅ Comprehension Check: What does the \(A^{-1}\) appearing in the result of the multi-variable Gaussian integral with source correspond to in the path integral of quantum field theory?
Answer
In the \(n \to \infty\) limit where discrete variables are replaced by continuous fields, the inverse matrix \(A^{-1}\) corresponds to the Feynman propagator. The exponential part \(\frac{1}{2}\mathbf{J}^T A^{-1} \mathbf{J}\) of the formula with source provides the structure of the generating functional \(Z[J]\), and each component of \(A^{-1}\) describes propagation between two points.
Comprehension Check C.2
Verify that setting \(n = 1\), \(A = (a)\), \(\mathbf{J} = (J)\) in equation (C.9) reproduces equation (C.3).
Answer: \(\det A = a\), \(A^{-1} = 1/a\), \(\mathbf{J}^T A^{-1}\mathbf{J} = J^2/a\). Substituting gives \(\sqrt{2\pi/a}\;e^{J^2/(2a)}\), which matches equation (C.3).
C.3 Algebra and Integration of Grassmann Numbers¶
🟡 Lina: From here on, we discuss Grassmann numbers, which are indispensable for the fermionic path integral (Ch. 12). While the bosonic path integral could use "ordinary numbers" as integration variables, for fermions we need anticommuting numbers.
C.3.1 Why Grassmann Numbers Are Needed¶
🔵 Kai: Why can't we use ordinary numbers?
🟡 Lina: Fermionic field operators satisfy anticommutation relations:
In path integrals, the advantage is not using operators but integrating over "classical field values." However, if the integration variables are ordinary commuting numbers, this anticommutation property cannot be reproduced. That's why we introduce a new type of number that anticommutes—Grassmann numbers.
✅ Comprehension Check: Why must Grassmann numbers, rather than ordinary real or complex numbers, be used as integration variables in the fermionic path integral?
Answer
Fermionic field operators satisfy the anticommutation relation \(\hat{\psi}(x)\hat{\psi}(y) = -\hat{\psi}(y)\hat{\psi}(x)\). In path integrals, operators are replaced by "classical field values" (integration variables), but ordinary commuting numbers cannot reproduce this anticommutation property. Therefore, Grassmann numbers—which change sign upon exchange—must be introduced as integration variables.
C.3.2 Basic Anticommutativity and \(\eta^2 = 0\)¶
🟡 Lina: The definition is surprisingly simple. Two Grassmann numbers \(\eta\), \(\zeta\) satisfy
This is anticommutativity.
From this, immediately setting \(\zeta = \eta\):
Adding \(\eta\eta\) to both sides gives \(2\eta^2 = 0\). Since the coefficient is an ordinary real or complex number, we can divide by 2:
🔵 Kai: \(\eta^2 = 0\) but \(\eta \neq 0\)! That's unthinkable for ordinary numbers.
🟡 Lina: Think of this as the mathematical expression of the Pauli exclusion principle. When you try to occupy the same fermionic state twice, you get zero—that's precisely the exclusion principle, right?
And since \(\eta^2 = 0\), Taylor expansions of functions of Grassmann numbers terminate after finitely many terms. For a single variable:
is the most general form. All terms of degree 2 or higher vanish. For example, \(e^\eta = 1 + \eta + \eta^2/2! + \cdots = 1 + \eta\) (truncated at \(\eta^2 = 0\)). Similarly, \(\sin\eta = \eta - \eta^3/3! + \cdots = \eta\), \(\cos\eta = 1 - \eta^2/2! + \cdots = 1\).
⚪ Mei: Since the Taylor expansion ends at just 2 terms, any function fits into the form \(a + b\eta\). Very manageable.
✅ Comprehension Check: The property \(\eta^2 = 0\) for a Grassmann number \(\eta\) corresponds physically to what principle? Also, what constraint does this property impose on functions of Grassmann numbers?
Answer
\(\eta^2 = 0\) corresponds to the mathematical expression of the Pauli exclusion principle. It reflects the fact that attempting to occupy the same fermionic state twice gives zero. Due to this property, the Taylor expansion of a function of a single Grassmann variable truncates to \(f(\eta) = a + b\eta\) with just 2 terms, and all terms of degree 2 or higher vanish.
C.3.3 Differentiation of Grassmann Numbers¶
🟡 Lina: The definition of differentiation is natural:
So differentiating \(f(\eta) = a + b\eta\) gives
🔵 Kai: So far this is just like normal.
🟡 Lina: The important point is that the differentiation operator itself behaves in a Grassmann-like (odd) manner. Specifically, Grassmann differentiation is defined as a left derivative. \(\frac{\partial}{\partial\zeta}\) is the operation of "bringing \(\zeta\) to the leftmost position in the product, then removing it." When there are two variables \(\eta, \zeta\), try differentiating \(\eta\zeta\) with respect to \(\zeta\).
First, bring \(\zeta\) to the left end: \(\eta\zeta = -\zeta\eta\) (anticommutativity). Then remove the leftmost \(\zeta\): \(-\eta\). That is:
🔵 Kai: A minus sign appeared! It's because swapping \(\eta\) and \(\zeta\) when bringing \(\zeta\) to the left changes the sign.
🟡 Lina: Exactly. In general, to apply \(\frac{\partial}{\partial\zeta}\), you move the variable \(\zeta\) you want to differentiate all the way to the left. Each time it passes over another Grassmann variable along the way, the sign changes once.
⚪ Mei: So in that example, it jumped over one \(\eta\), so the sign changed once to give \(-\eta\). Even with more variables, you just repeat the same rule.
C.3.4 Berezin Integration¶
🟡 Lina: Here's the most surprising part of Grassmann numbers. We define integration as follows:
This is called Berezin integration.
🔵 Kai: Wait, this is completely different from ordinary integration! Isn't \(\int dx\; 1 = x + C\)?
🟡 Lina: Grassmann integration is a different beast from ordinary Riemann integration. This definition is demanded by translation invariance. The ordinary definite integral \(\int_{-\infty}^{\infty} dx\; f(x)\) is unchanged under \(x \to x + a\) (when the integrand decays sufficiently fast). Let's demand the same property for Grassmann integration.
The most general function is \(f(\eta) = \alpha + \beta\eta\), so we also demand linearity for the integral, just as with ordinary integration. That is, \(\int d\eta\; f(\eta) = \alpha\int d\eta\; 1 + \beta\int d\eta\; \eta = c_0 \alpha + c_1 \beta\). Here \(c_0 = \int d\eta\; 1\) and \(c_1 = \int d\eta\; \eta\) are the constants to be determined.
Let's require invariance under the translation \(\eta \to \eta + \xi\). Here \(\xi\) is another Grassmann number. The reason we translate by a Grassmann number rather than an ordinary number is that since \(\eta\) is an anticommuting variable, \(\eta + \xi\) after the translation needs to maintain its properties as a Grassmann number. Let's see this concretely. If \(\xi\) were an ordinary number \(c\), then \((\eta + c)^2 = \eta^2 + 2c\eta + c^2 = 2c\eta + c^2\), which doesn't vanish for \(c \neq 0\). This means \(\eta + c\) loses the fundamental Grassmann property \(\theta^2 = 0\). On the other hand, if \(\xi\) is a Grassmann number, then \((\eta + \xi)^2 = \eta^2 + \eta\xi + \xi\eta + \xi^2 = 0 + \eta\xi - \eta\xi + 0 = 0\), properly remaining a Grassmann number. It's natural to translate by a number of the same kind.
🔵 Kai: I see—shifting by an ordinary number breaks the "square equals zero" property, so you must shift by a Grassmann number.
🟡 Lina: Substituting \(\eta \to \eta + \xi\) into \(f(\eta) = \alpha + \beta\eta\):
Here \(\alpha, \beta\) are ordinary numbers (not Grassmann numbers), so the ordering of \(\beta\xi\) and \(\beta\eta\) is not an issue. Looking at this expression, in terms of \(\eta\) it separates into "the part not containing \(\eta\) (a constant with respect to \(\eta\)) \(= \alpha + \beta\xi\)" and "the coefficient of \(\eta\) \(= \beta\)." Although \(\beta\xi\) is a Grassmann number, it's a different variable from \(\eta\), so when integrating over \(\eta\) it's treated as a constant. Using linearity to integrate over \(\eta\):
For this to equal \(\int d\eta\; f(\eta) = c_0\alpha + c_1\beta\), we need \(c_0\beta\xi = 0\) for arbitrary \(\beta, \xi\). Therefore \(c_0 = 0\), meaning \(\int d\eta\; 1 = 0\). We set \(c_1 = 1\) as a normalization convention.
🔵 Kai: I see—\(\int d\eta\; 1 = 0\) comes from the requirement that "the value doesn't change under translation."
⚪ Mei: Let me verify. Replacing \(\eta \to \eta + \xi\) in \(f(\eta) = \alpha + \beta\eta\) gives \(f(\eta + \xi) = (\alpha + \beta\xi) + \beta\eta\). Integrating this:
Integrating the original function also gives \(\int d\eta\; f(\eta) = \beta\). They indeed match.
🟡 Lina: And now, compare with equation (C.14):
In the world of Grassmann numbers, differentiation and integration are the same operation.
🔵 Kai: Whoa, it's true. In the world of ordinary numbers, differentiation and integration are inverse operations...
🟡 Lina: Since the only function of \(\eta\) is \(f(\eta) = a + b\eta\), "removing \(\eta\) and extracting the coefficient \(b\)" is a unique operation. Both differentiation and integration perform that same operation.
✅ Comprehension Check: In the definition of Berezin integration \(\int d\eta\; \eta = 1\), \(\int d\eta\; 1 = 0\), from what requirement is the seemingly strange rule \(\int d\eta\; 1 = 0\) derived? Also, what is the relationship between Grassmann differentiation and integration?
Answer
\(\int d\eta\; 1 = 0\) is derived from the requirement of "translation invariance." Requiring that the integral value doesn't change under \(\eta \to \eta + \xi\) forces the integral of a constant to be zero. Also, in the Grassmann world, differentiation and integration are the same operation. For \(f(\eta) = a + b\eta\), we have \(\partial f/\partial\eta = b\) and \(\int d\eta\; f(\eta) = b\)—both are the same operation of "extracting the coefficient of \(\eta\)."
C.3.5 Grassmann Gaussian Integral¶
🟡 Lina: Let me derive the formula that is central to the fermionic path integral. In the bosonic Gaussian integral, the square of the variable appeared in the exponent as \(e^{-aq^2/2}\). But for Grassmann numbers, \(\eta^2 = 0\), so "\(e^{-a\eta^2/2}\)" using the same variable squared just gives \(e^0 = 1\), which is meaningless. So to construct a quadratic form, we need a different Grassmann variable. In the fermionic path integral (Ch. 12), the Dirac field \(\psi\) and \(\bar{\psi} = \psi^\dagger\gamma^0\) are treated as independent integration variables. Correspondingly, here we consider two independent Grassmann variables \(\bar{\eta}, \eta\).
🔵 Kai: Is \(\bar{\eta}\) the complex conjugate of \(\eta\)?
🟡 Lina: Good question, but no. The bar on \(\bar{\eta}\) is a notational convention to correspond to \(\bar{\psi}\) of the Dirac field—it's not the complex conjugate of \(\eta\). The variables \(\bar{\eta}\) and \(\eta\) are completely independent, separate Grassmann variables. Think of them as two variables with no mathematical relationship, just with a bar on the name. And importantly, all Grassmann variables anticommute with each other. Between \(\bar{\eta}\) and \(\eta\) we also have \(\bar{\eta}\eta = -\eta\bar{\eta}\), and in the multi-variable case, any combination of \(\bar{\eta}_i\) and \(\eta_j\), \(\bar{\eta}_i\) and \(\bar{\eta}_j\), \(\eta_i\) and \(\eta_j\) all anticommute.
Let's compute this. Since \(\eta^2 = 0\), expanding the exponential:
Terms of degree 2 or higher vanish due to \(\eta^2 = 0\) or \(\bar{\eta}^2 = 0\).
🔵 Kai: The Taylor expansion ending at 2 terms is uniquely Grassmann.
🟡 Lina: Now we just use the definition of Berezin integration. We adopt the convention that \(\int d\bar{\eta}\,d\eta\) means "first integrate over \(\eta\), then integrate over \(\bar{\eta}\)." This is the same reading as ordinary multiple integrals—execute starting from the integration symbol closest to the integrand. That is, a nested structure \(\int d\bar{\eta}\bigl(\int d\eta\;(\cdots)\bigr)\). This convention remains the same throughout this chapter—in multi-variable and source-added cases as well, execute starting from the rightmost \(d\eta\). In Grassmann numbers, swapping the order of integration changes the sign. This is because, as we saw in C.3.4, Berezin integration is the same operation as differentiation. Since the differentiation operator \(\frac{\partial}{\partial\eta}\) behaves in a Grassmann-like (odd) manner (recall from C.3.3 how signs changed with left derivatives), the integration measures \(d\bar{\eta}\) and \(d\eta\) also anticommute—meaning \(d\bar{\eta}\,d\eta = -d\eta\,d\bar{\eta}\). Swapping the order introduces one minus sign. Respecting this convention is important.
⚪ Mei: Just swapping the order of integration changes the sign—something that doesn't happen with ordinary integration is a natural consequence in the Grassmann world.
🟡 Lina: Now let's compute.
First term: Since the inner \(\int d\eta\; 1 = 0\) already vanishes, we don't even need to perform the outer \(\bar{\eta}\) integration—this entire term is zero.
Second term: As we confirmed at the end of C.3.4, in the Grassmann world differentiation and integration are the same operation. That is, the Berezin integral \(\int d\eta\) performs the same thing as the left derivative defined in C.3.3. In other words, \(\int d\eta\) is the operation of "bringing \(\eta\) to the leftmost position, then removing it." To integrate \(\bar{\eta}\eta\) over \(\eta\), we first need to move \(\eta\) to the left end. Since \(\bar{\eta}\eta = -\eta\bar{\eta}\) (anticommutativity):
Here \(\bar{\eta}\) is a Grassmann variable independent of \(\eta\), so with respect to integration over \(\eta\), it can be treated as "a factor not containing \(\eta\)." However, there's a subtlety. For an ordinary constant \(c\), you can freely pull it out: \(\int dx\; x \cdot c = c\int dx\; x\). But a Grassmann "constant" \(\bar{\eta}\) anticommutes with other Grassmann variables, so moving it could change the sign. In the present case \(\int d\eta\;\eta\bar{\eta}\), \(\eta\) is already at the left end. By linearity of Berezin integration, factors independent of \(\eta\) can be pulled out of the integral. Since \(\bar{\eta}\) is a different variable from \(\eta\), it's treated as a "constant" with respect to \(\eta\). Here the Berezin integral \(\int d\eta\) is the operation "remove the leftmost \(\eta\) and return what remains." In \(\int d\eta\;\eta\bar{\eta}\), since \(\eta\) is already at the left end, removing it leaves just \(\bar{\eta}\) on the right. So \(\int d\eta\;\eta\bar{\eta} = 1 \cdot \bar{\eta} = \bar{\eta}\). Since \(\bar{\eta}\) is to the right of \(\eta\) and hasn't jumped over \(\eta\), no additional sign change occurs. Therefore:
Next, integrating over \(\bar{\eta}\): \(\int d\bar{\eta}\;(-\bar{\eta}) = -1\).
Altogether, \(\int d\bar{\eta}\,d\eta\;\bar{\eta}\eta = -1\).
🔵 Kai: So bringing \(\eta\) to the left end means jumping over \(\bar{\eta}\) once, changing the sign once—exactly the left derivative rule from C.3.3!
🟡 Lina: Exactly. If you remember that Berezin integration and left differentiation are the same operation, you won't get lost with signs.
So the second term is \(-a \cdot (-1) = a\). The first term was zero, so the total is:
🔵 Kai: The result is simply \(a\)! That's a completely different form from the bosonic \(\sqrt{2\pi/a}\).
🟡 Lina: What I want you to notice is that in the bosonic Gaussian integral (C.1), we got \(\sqrt{2\pi/a}\) with \(a\) in the denominator, whereas in the Grassmann case the result is \(a\) itself—in the numerator. This is the essential difference.
⚪ Mei: It's inverted. For bosons, larger \(a\) makes the integral value smaller, while for fermions it makes it larger—completely opposite behavior.
🟡 Lina: Exactly. Extending to multiple variables:
Here \(\bar{\boldsymbol{\eta}}^T A\,\boldsymbol{\eta} = \sum_{i,j} \bar{\eta}_i A_{ij} \eta_j\) (with \(\bar{\eta}_i\) on the left, \(\eta_j\) on the right—since order matters for Grassmann numbers, I'm being explicit). The ordering convention for the integration measure is \(\prod_i d\bar{\eta}_i\,d\eta_i \equiv d\bar{\eta}_1\,d\eta_1\,d\bar{\eta}_2\,d\eta_2\cdots d\bar{\eta}_n\,d\eta_n\) (the \(i = 1\) pair leftmost). The execution order is from the right—that is, \(d\eta_n \to d\bar{\eta}_n \to \cdots \to d\eta_1 \to d\bar{\eta}_1\). The matrix \(A\) is \(n \times n\) and can be either real or complex. Actually, when \(\det A = 0\) both sides are zero and the equality still holds, but for the version with sources (C.21) we need \(A^{-1}\), so we assume \(\det A \neq 0\) (non-singular). No symmetry or positive-definiteness is required—since Grassmann integrals have Taylor expansions that terminate after finitely many terms, there's no concern about convergence. While for bosons we had \((\det A)^{-1/2}\), for fermions \(\det A\) appears in the numerator.
🔵 Kai: For the bosonic case, you explained it by "diagonalizing and integrating independently for each eigenvalue." How should we understand the Grassmann case?
🟡 Lina: Same idea. Let me explain the strategy of the proof. Since the Taylor expansion of the Grassmann integral terminates after finitely many terms, we expand \(e^{-\bar{\boldsymbol{\eta}}^T A\boldsymbol{\eta}}\) and use the fact that only "the coefficient of \(\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\cdots\bar{\eta}_n\eta_n\)" survives the integration. Why? Recall the Berezin integration definition \(\int d\eta_i\;1 = 0\), \(\int d\eta_i\;\eta_i = 1\). If some term in the expansion doesn't contain \(\eta_i\), it vanishes the moment \(\int d\eta_i\) is applied. Conversely, if \(\eta_i\) appears twice or more, that term itself is zero by \(\eta_i^2 = 0\). So only terms where each \(\eta_i\) and each \(\bar{\eta}_i\) appear exactly once survive.
🔵 Kai: Each variable exactly once—too many or too few and it's zero. Just like the Pauli exclusion principle.
🟡 Lina: Precisely. (As an alternative, you could also diagonalize as in the bosonic case. Write \(A = P\,\mathrm{diag}(\lambda_1, \ldots, \lambda_n)\,P^{-1}\) with an invertible matrix \(P\) and change variables. In Grassmann variable changes, the Jacobian appears in the opposite way from usual—the determinant of the transformation matrix multiplies directly (not its inverse). I'll explain this using a single-variable example shortly. The result is that equation (C.18) can be applied independently for each pair, giving \(\lambda_1\lambda_2\cdots\lambda_n = \det A\). But for now let's use the expansion method.)
I'll now verify two things with an explicit \(n = 2\) example: (1) rearranging Grassmann variables into standard order within the expansion naturally produces the determinant's signs, and (2) performing the Berezin integration indeed yields \(\det A\).
🔵 Kai: Concretely, how do you expand and pick out the coefficient?
🟡 Lina: Let's first recall what the determinant "contains." For a \(2 \times 2\) matrix, \(\det A = A_{11}A_{22} - A_{12}A_{21}\). The "\(-\)" sign comes from swapping the column indices \((1,2) \to (2,1)\). For a general \(n \times n\) matrix, the determinant is a sum over all permutations of columns, taking one entry from each row, multiplied together. This is called the Leibniz formula:
Here \(\sigma\) is a permutation of \(1, 2, \ldots, n\)—specifying how to rearrange the \(n\) numbers—and there are \(n!\) of them in total.
🔵 Kai: So a permutation is just "rearranging \(1, 2, \ldots, n\) into a different order"?
🟡 Lina: Exactly. For example, with \(n = 3\), rearranging \((1,2,3)\) to \((2,3,1)\) is one permutation, and to \((2,1,3)\) is another. There are \(3! = 6\) in total. \(\mathrm{sgn}(\sigma)\) is the sign: "\(+1\) if it can be realized by an even number of adjacent transpositions, \(-1\) if odd." While there are multiple ways to decompose into transpositions, whether the number is even or odd is uniquely determined for each permutation (the proof is left to linear algebra textbooks). Checking with the \(n = 2\) example: the minus in \(\det A = A_{11}A_{22} - A_{12}A_{21}\) comes from the column indices \((2,1)\) being obtained by 1 swap, so \(\mathrm{sgn} = -1\). For \(n = 3\), \((1,2,3) \to (2,1,3)\) is "swap 1 and 2"—1 swap, so \(\mathrm{sgn} = -1\). \((1,2,3) \to (2,3,1)\) is "first swap 1 and 2 to get \((2,1,3)\), then swap 1 and 3 to get \((2,3,1)\)"—2 adjacent swaps, so \(\mathrm{sgn} = +1\).
🔵 Kai: I see—the minus sign in the \(2 \times 2\) determinant generalizes to the "sign of the permutation" in the general case.
🟡 Lina: Let's see the correspondence with Grassmann variables concretely for \(n = 2\). Expanding \(e^{-(\bar{\eta}_1 A_{11}\eta_1 + \bar{\eta}_1 A_{12}\eta_2 + \bar{\eta}_2 A_{21}\eta_1 + \bar{\eta}_2 A_{22}\eta_2)}\) and picking out terms where \(\bar{\eta}_1, \eta_1, \bar{\eta}_2, \eta_2\) all appear exactly once (everything else vanishes under Berezin integration). There are 2 such terms:
The first is \((-\bar{\eta}_1 A_{11}\eta_1)(-\bar{\eta}_2 A_{22}\eta_2) = A_{11}A_{22}\,\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\).
The second is \((-\bar{\eta}_1 A_{12}\eta_2)(-\bar{\eta}_2 A_{21}\eta_1) = A_{12}A_{21}\,\bar{\eta}_1\eta_2\bar{\eta}_2\eta_1\).
🔵 Kai: The first is already in standard order \(\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\), but the second is \(\bar{\eta}_1\eta_2\bar{\eta}_2\eta_1\) with a different ordering. Rearranging it will change the sign?
🟡 Lina: Exactly. Let's rearrange the second into standard order \(\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\). Starting from \(\bar{\eta}_1\eta_2\bar{\eta}_2\eta_1\). At each step, swapping two adjacent Grassmann variables changes the sign once. Let me track the cumulative sign.
- Swap \(\eta_2\) and \(\bar{\eta}_2\) (adjacent swap #1): \(\bar{\eta}_1\eta_2\bar{\eta}_2\eta_1 \to -\bar{\eta}_1\bar{\eta}_2\eta_2\eta_1\) (cumulative sign \(-1\))
- Swap \(\eta_2\) and \(\eta_1\) (adjacent swap #2): \(-\bar{\eta}_1\bar{\eta}_2\eta_2\eta_1 \to +\bar{\eta}_1\bar{\eta}_2\eta_1\eta_2\) (cumulative sign \(+1\))
- Swap \(\bar{\eta}_2\) and \(\eta_1\) (adjacent swap #3): \(+\bar{\eta}_1\bar{\eta}_2\eta_1\eta_2 \to -\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\) (cumulative sign \(-1\))
Let me verify step 3. Looking at \(\bar{\eta}_1\bar{\eta}_2\eta_1\eta_2\), the 2nd element \(\bar{\eta}_2\) and 3rd element \(\eta_1\) are adjacent. Swapping these two adjacent elements changes the sign once, giving \(\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\)—that is, \(\bar{\eta}_1\bar{\eta}_2\eta_1\eta_2 = -\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\).
A total of 3 adjacent swaps gives sign \((-1)^3 = -1\). So the original \(\bar{\eta}_1\eta_2\bar{\eta}_2\eta_1 = -\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\). Therefore the second term becomes \(A_{12}A_{21} \cdot (-\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2) = -A_{12}A_{21}\,\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\).
⚪ Mei: Three adjacent swaps giving \((-1)^3 = -1\)—the rearrangement of Grassmann variables automatically produces the minus sign.
🟡 Lina: Combined, we get \((A_{11}A_{22} - A_{12}A_{21})\,\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\). Each anticommutation produces a minus, which automatically reproduces the permutation sign \(\mathrm{sgn}(\sigma) = -1\). That is, the anticommutativity of Grassmann variables naturally generates the sign structure of the determinant. Now if we can show that \(\int d\bar{\eta}_1 d\eta_1 d\bar{\eta}_2 d\eta_2\;\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 = 1\) (which I'll verify right below), then the integral value is \(A_{11}A_{22} - A_{12}A_{21} = \det A\).
⚪ Mei: I see—the signs arising from rearranging Grassmann variables correspond perfectly to \(\mathrm{sgn}(\sigma)\) in the Leibniz formula for the determinant.
🔵 Kai: How do you concretely compute \(\int d\bar{\eta}_1 d\eta_1 d\bar{\eta}_2 d\eta_2\;\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 = 1\)? With 4 integrations, I might get confused about the order...
🟡 Lina: Let me explain the execution order. Just as the ordinary multiple integral \(\int dx\int dy\; f(x,y)\) "executes the inner \(\int dy\) first," we execute starting from the integration symbol closest to the integrand (rightmost). So the symbol closest to the integrand \(\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\) is \(d\eta_2\), so we first integrate over \(\eta_2\), then \(d\bar{\eta}_2\), \(d\eta_1\), \(d\bar{\eta}_1\) proceeding outward.
As confirmed in C.3.4, Berezin integration is the same operation as left differentiation. So \(\int d\eta_2\) is the operation of "bringing \(\eta_2\) to the leftmost position in the integrand, then removing it." I'll track signs cumulatively at each step.
Step 1: Compute \(\int d\eta_2\;\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\). To bring \(\eta_2\) to the left end, it must jump over \(\bar{\eta}_1, \eta_1, \bar{\eta}_2\)—3 variables—giving \((-1)^3 = -1\), so \(\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 = -\eta_2\bar{\eta}_1\eta_1\bar{\eta}_2\). Since \(\int d\eta_2\;\eta_2 = 1\), what remains is \((-1)\bar{\eta}_1\eta_1\bar{\eta}_2 = -\bar{\eta}_1\eta_1\bar{\eta}_2\).
Step 2: Compute \(\int d\bar{\eta}_2\;(-\bar{\eta}_1\eta_1\bar{\eta}_2)\). The integrand is \(-\bar{\eta}_1\eta_1\bar{\eta}_2\) (the result from Step 1 with its minus sign). First, bring \(\bar{\eta}_2\) to the left end within \(\bar{\eta}_1\eta_1\bar{\eta}_2\). As stated in C.3.2, all Grassmann variables anticommute with each other—even \(\bar{\eta}\)'s among themselves: \(\bar{\eta}_1\bar{\eta}_2 = -\bar{\eta}_2\bar{\eta}_1\). Whether barred or not, they're all Grassmann variables, so swapping any two adjacent ones always changes the sign once. Let me do it step by step:
- Swap #1: swap \(\eta_1\) and \(\bar{\eta}_2\) → \(\bar{\eta}_1\eta_1\bar{\eta}_2 = -\bar{\eta}_1\bar{\eta}_2\eta_1\)
- Swap #2: swap \(\bar{\eta}_1\) and \(\bar{\eta}_2\) → \(-\bar{\eta}_1\bar{\eta}_2\eta_1 = +\bar{\eta}_2\bar{\eta}_1\eta_1\)
Two adjacent swaps total, giving \((-1)^2 = +1\). So \(\bar{\eta}_1\eta_1\bar{\eta}_2 = +\bar{\eta}_2\bar{\eta}_1\eta_1\). Since \(\int d\bar{\eta}_2\;\bar{\eta}_2 = 1\), we're left with \(\bar{\eta}_1\eta_1\). Including the minus sign in front of the integrand, Step 2's result is \(-\bar{\eta}_1\eta_1\).
🔵 Kai: Count the jumps, track the sign—I'm getting the rhythm.
🟡 Lina: Good pace.
Step 3: Compute \(\int d\eta_1\;(-\bar{\eta}_1\eta_1)\). The integrand is \(-\bar{\eta}_1\eta_1\). Bringing \(\eta_1\) to the left end within \(\bar{\eta}_1\eta_1\): jumping over \(\bar{\eta}_1\) once gives \(\bar{\eta}_1\eta_1 = -\eta_1\bar{\eta}_1\). So \(-\bar{\eta}_1\eta_1 = -(-\eta_1\bar{\eta}_1) = +\eta_1\bar{\eta}_1\). Then \(\int d\eta_1\;\eta_1\bar{\eta}_1 = \bar{\eta}_1\) (\(\bar{\eta}_1\) is a different variable from \(\eta_1\), so it's treated as a constant).
Step 4: \(\int d\bar{\eta}_1\;\bar{\eta}_1 = 1\).
Indeed the total is \(1\). So at each step, it's just "bring to the left → sign changes by the number of jumps → remove." Summarizing:
- \(\int d\eta_2\): \(\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 \to -\bar{\eta}_1\eta_1\bar{\eta}_2\)
- \(\int d\bar{\eta}_2\): \(-\bar{\eta}_1\eta_1\bar{\eta}_2 \to -\bar{\eta}_1\eta_1\)
- \(\int d\eta_1\): \(-\bar{\eta}_1\eta_1 \to \bar{\eta}_1\)
- \(\int d\bar{\eta}_1\): \(\bar{\eta}_1 \to 1\)
⚪ Mei: I can see the pattern. Even with more variables, it's just the same mechanical operation repeated.
🟡 Lina: The anticommutativity of Grassmann variables naturally produces the permutation signs, which is why the expansion gives \(\det A\). I'll leave the detailed proof for exercises, but to state the result: for the case when \(A\) is a diagonal matrix, equation (C.18) can be applied independently for each pair \(\bar{\eta}_i, \eta_i\), giving \(\lambda_1\lambda_2\cdots\lambda_n = \det A\). For a general non-singular matrix, we diagonalize by a change of variables—but note that for Grassmann variable changes, the Jacobian appears in the opposite way from ordinary integration. Let me explain this using a single-variable example below. In the bosonic case, \(A\) needed to be real symmetric positive-definite for the integral to converge, but Grassmann integration has no convergence issues since Taylor expansions terminate after finitely many terms. Actually, when \(\det A = 0\) both sides of (C.19) are zero and the equality still holds. But for the source version (C.21) we need \(A^{-1}\), so we assume \(\det A \neq 0\) (non-singular).
🔵 Kai: With ordinary integration, changing variables divides by the absolute value of the Jacobian, right? Is it different for Grassmann?
🟡 Lina: Good catch. In ordinary integration, \(dx = |\partial x/\partial y|\,dy\), so the inverse of the transformation matrix's determinant appears. Let's check what happens for Grassmann with a simple example. Under the change of variables \(\eta' = c\eta\) (\(c\) is an ordinary nonzero number), we have \(\eta = \eta'/c\).
For ordinary integration: \(d\eta = d\eta'/c\), so \(\int d\eta\;\eta = \int (d\eta'/c)\;(\eta'/c) = (1/c^2)\int d\eta'\;\eta'\).
But in Berezin integration, by definition \(\int d\eta\;\eta = 1\), and for the variable \(\eta'\) also \(\int d\eta'\;\eta' = 1\). Let's check consistency. Substituting \(\eta = \eta'/c\): \(\int d\eta\;\eta = \int d\eta\;(\eta'/c) = (1/c)\int d\eta\;\eta'\). For this to equal 1, we need \(\int d\eta\;\eta' = c\). Since \(\int d\eta'\;\eta' = 1\), we need the relation \(\int d\eta = c\int d\eta'\). That is, for Grassmann \(d\eta = c\,d\eta'\), which is the opposite of the ordinary \(d\eta = d\eta'/c\).
🔵 Kai: The opposite! The variable change rule is reversed between ordinary and Grassmann integration...
🟡 Lina: In general, the determinant of the transformation matrix multiplies directly (not its inverse). This is a consequence of "Berezin integration being the same operation as differentiation"—it follows the same transformation rule as the chain rule for differentiation: \(\partial/\partial\eta' = (\partial\eta/\partial\eta')\,\partial/\partial\eta = (1/c)\,\partial/\partial\eta\).
🔵 Kai: So when changing variables with a non-singular matrix, what about the Jacobian's effect?
🟡 Lina: The same principle works for multiple variables. To state just the conclusion: combining the transformations \(\boldsymbol{\eta}' = P\boldsymbol{\eta}\) and \(\bar{\boldsymbol{\eta}}' = \bar{\boldsymbol{\eta}}\,(P^T)^{-1}\) preserves \(\bar{\boldsymbol{\eta}}'^T\boldsymbol{\eta}' = \bar{\boldsymbol{\eta}}^T\boldsymbol{\eta}\). Applying the Grassmann Jacobian rule (the determinant of the transformation matrix multiplies directly, not as its inverse) to both \(\boldsymbol{\eta}\) and \(\bar{\boldsymbol{\eta}}\): for \(\boldsymbol{\eta}' = P\boldsymbol{\eta}\), the transformation matrix is \(P\), so \(\prod_i d\eta'_i\) picks up \(\det P\). For \(\bar{\boldsymbol{\eta}}' = \bar{\boldsymbol{\eta}}(P^T)^{-1}\), the transformation matrix is \((P^T)^{-1}\), so its determinant \(\det(P^T)^{-1} = (\det P)^{-1}\) multiplies directly. Combined: \(\det P \cdot (\det P)^{-1} = 1\), so no additional factor appears. Thus you can change variables without worry. Verify the details in the exercises.
⚪ Mei: The Grassmann Jacobian is the opposite of usual, yet applying transformations to both \(\eta\) and \(\bar{\eta}\) makes them cancel to 1—it's elegantly designed.
🔵 Kai: The determinant being in the numerator for fermions vs. the denominator for bosons... does this have physical meaning?
🟡 Lina: Absolutely! Wonderful intuition. Intuitively, since the bosonic \(\log Z \propto -\tfrac{1}{2}\log\det A\) and the fermionic \(\log Z \propto +\log\det A\) have opposite signs in the logarithm, a relative minus sign arises per loop contribution. We'll discuss in detail why fermion loops in Feynman diagrams carry a minus sign in Ch. 13, but its origin lies in this equation (C.19).
🔵 Kai: Just from the difference in integration definitions, a sign difference appears in loop calculations... The mathematical structure determines the physics. So concretely, what kind of calculation produces "one minus per loop"?
🟡 Lina: We'll do that in detail when deriving the Feynman rules in Ch. 13. For now, just hold on to the correspondence that "the determinant \(\det A\) appearing in the numerator" is the root of the sign.
🔵 Kai: Got it, I'll look forward to Chapter 13. ...Oh, but that raises another question. If the signs are opposite, then if bosons and fermions have the same matrix \(A\), wouldn't they cancel each other when added together? Just a simple question.
🟡 Lina: Sharp observation. In fact, if bosons and fermions had the same mass and the same number of degrees of freedom, since the signs are opposite, quantum corrections would cancel dramatically. This is precisely the starting point of an idea called supersymmetry. It goes beyond the scope of this book so I won't go deeper, but the idea that "if there's a symmetry between bosons and fermions, divergence problems are alleviated" comes precisely from this sign difference in \(\det A\).
🔵 Kai: A single sign difference in one integral formula connects to such a big story...
⚪ Mei: So the bosonic \(-\frac{1}{2}\log\det A\) and the fermionic \(+\log\det A\) cancel for the same \(A\)—if such a symmetry exists, divergence problems are dramatically alleviated. A single sign in an integral formula can fundamentally change the ultraviolet structure of the theory.
✅ Comprehension Check: The bosonic multi-variable Gaussian integral gives \((\det A)^{-1/2}\), while the fermionic (Grassmann) case gives \(\det A\). How is this difference of "the determinant in the denominator vs. numerator" related to phenomena in quantum field theory?
Answer
This difference is the root cause of the additional minus sign that fermion loops carry in Feynman diagrams. For bosons \(\log Z \propto -\frac{1}{2}\log\det A = -\frac{1}{2}\mathrm{Tr}\log A\), while for fermions \(\log Z \propto +\log\det A = +\mathrm{Tr}\log A\), with opposite signs. This produces the relative minus sign for fermion loops in one-loop calculations.
Comprehension Check C.3
Compute \(\displaystyle\int d\bar{\eta}\,d\eta\; (\bar{\eta}\eta)\,e^{-\bar{\eta}\,a\,\eta}\).
Answer: Since \(e^{-\bar{\eta}\,a\,\eta} = 1 - a\bar{\eta}\eta\), we have \(\bar{\eta}\eta \cdot (1 - a\bar{\eta}\eta) = \bar{\eta}\eta - a(\bar{\eta}\eta)(\bar{\eta}\eta)\). Computing \((\bar{\eta}\eta)^2 = \bar{\eta}\eta\bar{\eta}\eta\): swap the middle \(\eta\bar{\eta}\) using anticommutativity \(\eta\bar{\eta} = -\bar{\eta}\eta\), so \(\bar{\eta}(\eta\bar{\eta})\eta = \bar{\eta}(-\bar{\eta}\eta)\eta = -\bar{\eta}\bar{\eta}\eta\eta = -\bar{\eta}^2\eta^2 = 0\) (using \(\bar{\eta}^2 = 0\)), hence \(\bar{\eta}\eta \cdot (1 - a\bar{\eta}\eta) = \bar{\eta}\eta\). Computing \(\int d\bar{\eta}\,d\eta\;\bar{\eta}\eta\): since \(\bar{\eta}\eta = -\eta\bar{\eta}\), we get \(\int d\eta\;(-\eta\bar{\eta}) = -\bar{\eta}\), then \(\int d\bar{\eta}\;(-\bar{\eta}) = -1\). So \(\int d\bar{\eta}\,d\eta\;\bar{\eta}\eta = -1\), and the answer is \(-1\). Alternatively, differentiating both sides of \(\int d\bar{\eta}\,d\eta\; e^{-\bar{\eta}a\eta} = a\) with respect to \(a\) gives \(\int d\bar{\eta}\,d\eta\;(-\bar{\eta}\eta)e^{-\bar{\eta}a\eta} = 1\), hence \(\int d\bar{\eta}\,d\eta\;\bar{\eta}\eta\,e^{-\bar{\eta}a\eta} = -1\).
C.3.6 Grassmann Gaussian Integral with Source¶
🟡 Lina: Let's derive the fermionic version corresponding to the bosonic formula with source (C.3). We introduce Grassmann sources \(\bar{\xi}, \xi\). These are also Grassmann numbers, so they anticommute with all Grassmann variables including \(\bar{\eta}, \eta\) (e.g., \(\bar{\xi}\eta = -\eta\bar{\xi}\)). We want to compute:
We attack this with completing the square, just like the bosonic case. Rearranging the exponent:
🔵 Kai: You can complete the square with Grassmann numbers too!
🟡 Lina: Try verifying it. The key point is that \(a^{-1}\) is an ordinary number, so it commutes freely with Grassmann variables—you only need to worry about ordering between Grassmann variables themselves.
🔵 Kai: Um, so I need to expand \(-a(\bar{\eta} - \bar{\xi}a^{-1})(\eta - a^{-1}\xi)\), right? Four terms come out... the last term is \(-a \cdot (-\bar{\xi}a^{-1})\cdot(-a^{-1}\xi)\), with two minuses making it positive... wait, is the ordering of \(\bar{\xi}\) and \(\xi\) okay?
🟡 Lina: \(a^{-1}\) is an ordinary number, so it passes through Grassmann variables freely. So \(\bar{\xi}a^{-1} \cdot a^{-1}\xi = a^{-2}\bar{\xi}\xi\), multiplied by \(-a\) gives \(-a^{-1}\bar{\xi}\xi\). Combining everything: \(-a\bar{\eta}\eta + \bar{\eta}\xi + \bar{\xi}\eta - a^{-1}\bar{\xi}\xi\).
⚪ Mei: Since \(a^{-1}\) is an ordinary number that commutes freely with Grassmann variables, the signs of each term can be tracked straightforwardly.
🔵 Kai: Let me verify the four terms one by one. Expanding \(-a(\bar{\eta} - \bar{\xi}a^{-1})(\eta - a^{-1}\xi)\)... Term 1: \(-a \cdot \bar{\eta} \cdot \eta = -a\bar{\eta}\eta\). Term 2: \(-a \cdot \bar{\eta} \cdot (-a^{-1}\xi) = +\bar{\eta}\xi\). Term 3: \(-a \cdot (-\bar{\xi}a^{-1}) \cdot \eta = +\bar{\xi}\eta\). Term 4: \(-a \cdot (-\bar{\xi}a^{-1}) \cdot (-a^{-1}\xi) = -a \cdot a^{-2}\bar{\xi}\xi = -a^{-1}\bar{\xi}\xi\). In that last \(\bar{\xi}a^{-1} \cdot a^{-1}\xi\) part, is the Grassmann ordering okay?
🟡 Lina: Since \(a^{-1}\) is an ordinary number (not a Grassmann number), it commutes freely with Grassmann variables. So \(\bar{\xi}a^{-1} \cdot a^{-1}\xi = a^{-1} \cdot a^{-1} \cdot \bar{\xi}\xi = a^{-2}\bar{\xi}\xi\). Multiplied by \(-a\) gives \(-a^{-1}\bar{\xi}\xi\). The ordering of the Grassmann variables is maintained as \(\bar{\xi}\xi\), so there's no problem.
🔵 Kai: I see—ordinary numbers pass through Grassmann variables freely so ordering isn't a concern. But conversely, if you mix up the ordering of Grassmann variables with each other, the sign flips, so the ordering of the source terms \(\bar{\xi}\eta + \bar{\eta}\xi\) matters too, right?
🟡 Lina: Right. Now add the \(+\bar{\xi}\,a^{-1}\,\xi\) from the right-hand side of the completing the square formula to Kai's expansion. Since \(a^{-1}\) is an ordinary number, it commutes freely with the Grassmann variables \(\bar{\xi}, \xi\): \(\bar{\xi}\,a^{-1}\,\xi = a^{-1}\bar{\xi}\xi\). Adding this to the last term \(-a^{-1}\bar{\xi}\xi\) from the expansion gives \(+a^{-1}\bar{\xi}\xi - a^{-1}\bar{\xi}\xi = 0\)—they cancel, and what remains is \(-a\bar{\eta}\eta + \bar{\eta}\xi + \bar{\xi}\eta\). The source terms match the original expression with ordering \(\bar{\xi}\eta + \bar{\eta}\xi\).
⚪ Mei: So expanding and adding everything up returns to the original—confirming the completing the square is correct.
🟡 Lina: Exactly. Substituting \(\bar{\eta}' = \bar{\eta} - \bar{\xi}a^{-1}\), \(\eta' = \eta - a^{-1}\xi\), by translation invariance of Berezin integration the integral value doesn't change. Pull out the constant factor \(e^{\bar{\xi}\,a^{-1}\,\xi}\), and the rest reduces to equation (C.18):
🔵 Kai: The same 3-step process of "complete the square → change variables → reduce to basic formula" works cleanly in Grassmann too.
🟡 Lina: The multi-variable version is:
🟡 Lina: Let's compare side by side with the bosonic version (C.9).
Table C.2: Comparison of bosonic and fermionic Gaussian integrals
| Boson | Fermion | |
|---|---|---|
| Prefactor | \((\det A)^{-1/2}\) | \(\det A\) |
| Source factor | \(e^{\frac{1}{2}\mathbf{J}^T A^{-1}\mathbf{J}}\) | \(e^{\bar{\boldsymbol{\xi}}^T A^{-1}\boldsymbol{\xi}}\) |
⚪ Mei: For bosons, \(\det A\) appears to the \(-1/2\) power in the denominator; for fermions, to the \(+1\) power in the numerator—the same structural difference Prof. Lina mentioned earlier appears here too.
📝 Exercises:
- Direct verification that the multi-variable Grassmann Gaussian integral \(= \det A\) → Problem M-2. Derivation of the Multi-variable Grassmann Gaussian Integral
Summary: Core Formulas of Gaussian and Grassmann Integrals¶
🟡 Lina: Let me compile the main formulas of this Appendix. The contrast between bosons and fermions should be clear at a glance.
Table C.3: Appendix C Main Formula Summary
| Number | Formula | Condition/Application |
|---|---|---|
| (C.1) | \(\int dq\;e^{-aq^2/2} = \sqrt{2\pi/a}\) | Bosonic starting point |
| (C.3) | \(\int dq\;e^{-aq^2/2 - Jq} = \sqrt{2\pi/a}\;e^{J^2/(2a)}\) | With source / completing the square |
| (C.6) | \(I_n(a)\): for even \(n=2m\), \(\sqrt{2\pi/a}\cdot(2m-1)!!/a^m\); zero for odd | Directly linked to Wick's theorem |
| (C.8) | \(\int d^nq\;e^{-\mathbf{q}^T A\mathbf{q}/2} = (2\pi)^{n/2}/\sqrt{\det A}\) | Multi-variable, \(A\) real symmetric positive-definite |
| (C.9) | With source, multi-variable: \(\times\,e^{\mathbf{J}^T A^{-1}\mathbf{J}/2}\) | Generating functional \(Z[J]\) |
| (C.18) | \(\int d\bar{\eta}\,d\eta\;e^{-\bar{\eta}a\eta} = a\) | Grassmann |
| (C.19) | Multi-variable Grassmann: \(= \det A\) | \(A\) non-singular (\(\det A \neq 0\)) / Opposite of boson! |
| (C.21) | Grassmann with source | Fermionic \(Z[\bar{\xi},\xi]\) |
🔵 Kai: Bosons give \((\det A)^{-1/2}\), fermions give \(\det A\). From this single point alone, you get the key to understanding why fermion loops carry a minus sign at one loop.
⚪ Mei: The contrast between Gaussian integrals (bosons) and Grassmann Gaussian integrals (fermions) forms the backbone of quantum field theory.
Preview of Next Chapter¶
Appendix D: Loop Calculation Toolbox — Dimensional Analysis, Feynman Parameters, Wick Rotation — With the foundations of "bosonic and fermionic integration" now in place, the next Appendix D will organize practical techniques for executing concrete loop integrals—natural units and mass dimensions, the technique of combining denominators using Feynman parameters, the procedure of Wick rotation from Minkowski to Euclidean space, and the momentum integral formulas needed for dimensional regularization—all at once. Let's hone these techniques to the level where "your hands can move" in the renormalization calculations of Chapters 13–14.
Exercises¶
📝 Exercises:
- Computing \(I_6(a)\) using the Gaussian integral recurrence relation → Problem B-3. Gaussian Integral Containing \(q^n\) (Application of Recurrence Relation)
- Direct verification that the multi-variable Grassmann Gaussian integral \(= \det A\) → Problem M-2. Derivation of the Multi-variable Grassmann Gaussian Integral
References¶
- Sakamoto Mahito, "Quantum Field Theory II — Focusing on Feynman Graphs and Renormalization" Chapter 6, "Gaussian Integrals and Fresnel Integrals"
- Lancaster & Blundell, "Quantum Field Theory for the Gifted Amateur" Chapter 28, "Grassmann numbers"
- Peskin & Schroeder, "An Introduction to Quantum Field Theory" Appendix A (Conventions and formula summary)
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