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Appendix C Solutions

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Table of Contents

Basic

Medium

Advanced

Basic

B-1. Calculation of a Basic Gaussian Integral

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Solution strategy: Compare with equation (C.1) \(\int_{-\infty}^{\infty} dq \; e^{-\frac{a}{2}q^2} = \sqrt{\frac{2\pi}{a}}\).

Calculation:

\[ e^{-3q^2} = e^{-\frac{a}{2}q^2} \quad \Longrightarrow \quad \frac{a}{2} = 3 \quad \Longrightarrow \quad a = 6 \]

Substituting into equation (C.1),

\[ \int_{-\infty}^{\infty} dq \; e^{-3q^2} = \sqrt{\frac{2\pi}{6}} = \sqrt{\frac{\pi}{3}} \]

Final answer:

\[ \boxed{\int_{-\infty}^{\infty} dq \; e^{-3q^2} = \sqrt{\frac{\pi}{3}}} \]

Verification: As a dimensional check, \(a = 6\) is a positive real number, so it satisfies the convergence condition \(\mathrm{Re}(a) > 0\). Also, if \(a = 2\), we would get \(\sqrt{\pi}\), which is consistent with the known result \(\int e^{-q^2} dq = \sqrt{\pi}\).

B-2. Completing the Square in a Gaussian Integral with a Source

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Solution Strategy: Organize the exponent into the form \(-\frac{a}{2}q^2 + bq\) and apply the variant of equation (C.3).

Calculation:

Organizing the exponent:

\[ -2q^2 + 6q = -\frac{a}{2}q^2 + bq \]

By comparison, \(\frac{a}{2} = 2\) gives \(a = 4\), and \(b = 6\).

From the result of Comprehension Check C.1 (equation (C.3) with \(J \to -b\)):

\[ \int_{-\infty}^{\infty} dq \; e^{-\frac{a}{2}q^2 + bq} = \sqrt{\frac{2\pi}{a}} \; e^{\frac{b^2}{2a}} \]

Substituting:

\[ \int_{-\infty}^{\infty} dq \; e^{-2q^2 + 6q} = \sqrt{\frac{2\pi}{4}} \; e^{\frac{36}{8}} = \sqrt{\frac{\pi}{2}} \; e^{9/2} \]

Final Answer:

\[ \boxed{\int_{-\infty}^{\infty} dq \; e^{-2q^2 + 6q} = \sqrt{\frac{\pi}{2}} \; e^{9/2}} \]

Verification: Directly confirm by completing the square. \(-2q^2 + 6q = -2(q^2 - 3q) = -2\left(q - \frac{3}{2}\right)^2 + \frac{9}{2}\). With the substitution \(z = q - 3/2\):

\[ e^{9/2} \int_{-\infty}^{\infty} dz \; e^{-2z^2} = e^{9/2} \sqrt{\frac{\pi}{2}} \]

This agrees.

B-3. Gaussian Integral Containing \(q^n\) (Application of Recurrence Relation)

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Solution strategy: Set \(a = 1\) and repeatedly apply the recurrence relation (C.7) \(I_n(a) = \frac{n-1}{a} I_{n-2}(a)\).

Calculation:

\[ I_6(1) = \frac{5}{1} \cdot I_4(1) = 5 \cdot I_4(1) \]
\[ I_4(1) = \frac{3}{1} \cdot I_2(1) = 3 \cdot I_2(1) \]
\[ I_2(1) = \frac{1}{1} \cdot I_0(1) = I_0(1) = \sqrt{2\pi} \]

Therefore:

\[ I_6(1) = 5 \cdot 3 \cdot \sqrt{2\pi} = 15\sqrt{2\pi} \]

Final answer:

\[ \boxed{\int_{-\infty}^{\infty} dq \; q^6 \, e^{-\frac{1}{2}q^2} = 15\sqrt{2\pi}} \]

Verification: From equation (C.6), with \(n = 6 = 2m\) so \(m = 3\), we get \(I_6(1) = \sqrt{2\pi} \cdot (2 \cdot 3 - 1)!! / 1^3 = \sqrt{2\pi} \cdot 5!! = \sqrt{2\pi} \cdot 15\). This agrees.

B-4. Confirming that odd-order Gaussian integrals vanish

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Solution strategy: Perform the variable substitution \(q \to -q\) and examine the symmetry of the integrand.

Calculation:

Let \(f(q) = q^3 \, e^{-\frac{a}{2}q^2}\). Under the transformation \(q \to -q\):

\[ f(-q) = (-q)^3 \, e^{-\frac{a}{2}(-q)^2} = -q^3 \, e^{-\frac{a}{2}q^2} = -f(q) \]

Therefore \(f(q)\) is an odd function. Integrating an odd function over the symmetric interval \((-\infty, +\infty)\) gives zero:

\[ \int_{-\infty}^{\infty} dq \; f(q) = \int_{-\infty}^{0} f(q)\,dq + \int_{0}^{\infty} f(q)\,dq = -\int_{0}^{\infty} f(q)\,dq + \int_{0}^{\infty} f(q)\,dq = 0 \]

Final answer:

\[ \boxed{\int_{-\infty}^{\infty} dq \; q^3 \, e^{-\frac{a}{2}q^2} = 0} \]

Verification: In general, \(q^n e^{-aq^2/2}\) is an odd function when \(n\) is odd and an even function when \(n\) is even. This is consistent with the result of Eq. (C.6).

B-5. Two-Variable Gaussian Integral

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Solution strategy: Substitute \(n = 2\) into equation (C.8) \(\int d^n q \; e^{-\frac{1}{2}\mathbf{q}^T A\,\mathbf{q}} = \frac{(2\pi)^{n/2}}{(\det A)^{1/2}}\).

Calculation:

Compute the determinant:

\[ \det A = 2 \times 3 - 1 \times 1 = 5 \]

The eigenvalues of \(A\) are positive (since \(\mathrm{tr}\,A = 5 > 0\) and \(\det A = 5 > 0\), both eigenvalues are positive), so the positive-definite condition is satisfied.

Substituting into equation (C.8):

\[ \int d^2 q \; e^{-\frac{1}{2}\mathbf{q}^T A\,\mathbf{q}} = \frac{(2\pi)^{2/2}}{(\det A)^{1/2}} = \frac{2\pi}{\sqrt{5}} \]

Final answer:

\[ \boxed{\int d^2 q \; e^{-\frac{1}{2}\mathbf{q}^T A\,\mathbf{q}} = \frac{2\pi}{\sqrt{5}}} \]

Consistency check: If \(A\) were the diagonal matrix \(\mathrm{diag}(2, 3)\), then \(\det A = 6\) and the result would be \(2\pi/\sqrt{6}\). Integrating each variable independently gives \(\sqrt{2\pi/2} \cdot \sqrt{2\pi/3} = \sqrt{\pi} \cdot \sqrt{2\pi/3} = 2\pi/\sqrt{6}\). This is consistent.

B-6. Expansion Using Anticommutation Relations of Grassmann Numbers

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Solution Strategy: Expand using the distributive law, then apply \(\eta_i^2 = 0\) and \(\eta_i \eta_j = -\eta_j \eta_i\).

Calculation:

\[ (\eta_1 + \eta_2)(\eta_2 + \eta_3) = \eta_1\eta_2 + \eta_1\eta_3 + \eta_2\eta_2 + \eta_2\eta_3 \]

Since \(\eta_2^2 = 0\), the third term vanishes:

\[ = \eta_1\eta_2 + \eta_1\eta_3 + \eta_2\eta_3 \]

Final Answer:

\[ \boxed{(\eta_1 + \eta_2)(\eta_2 + \eta_3) = \eta_1\eta_2 + \eta_1\eta_3 + \eta_2\eta_3} \]

Verification: This can also be written in a different order using the anticommutation relations. For example, since \(\eta_1\eta_2 = -\eta_2\eta_1\), it can also be written as \(-\eta_2\eta_1 + \eta_1\eta_3 + \eta_2\eta_3\). Setting \(\eta_1 = \eta_2\) in the original expression gives \(2\eta_1(\eta_1 + \eta_3) = 2\eta_1\eta_3\) (using \(\eta_1^2 = 0\)). On the other hand, substituting into the result gives \(\eta_1^2 + \eta_1\eta_3 + \eta_1\eta_3 = 0 + 2\eta_1\eta_3\). These agree.

B-7. Basic Calculation of Berezin Integration

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Solution Strategy: Apply the definition of Berezin integration (C.16) \(\int d\eta\;\eta = 1\), \(\int d\eta\;1 = 0\) to each term.

Calculation:

\[ \int d\eta \; (3 + 5\eta) = 3\int d\eta \; 1 + 5\int d\eta \; \eta = 3 \cdot 0 + 5 \cdot 1 = 5 \]

Final Answer:

\[ \boxed{\int d\eta \; (3 + 5\eta) = 5} \]

Verification: Confirm that the Grassmann integral is the same operation as differentiation. \(\frac{\partial}{\partial\eta}(3 + 5\eta) = 5\). This agrees.

B-8. Single-Variable Grassmann Gaussian Integral

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Solution strategy: Use \(\bar{\eta}^2 = \eta^2 = 0\) to expand the exponential function into finite terms, then perform the Berezin integration.

Calculation:

Since \(\bar{\eta}\eta\) is a product of Grassmann numbers, \((\bar{\eta}\eta)^2 = \bar{\eta}\eta\bar{\eta}\eta = -\bar{\eta}\bar{\eta}\eta\eta = -\bar{\eta}^2\eta^2 = 0\). Therefore:

\[ e^{-5\bar{\eta}\eta} = 1 + (-5\bar{\eta}\eta) + \frac{1}{2!}(-5\bar{\eta}\eta)^2 + \cdots = 1 - 5\bar{\eta}\eta \]

We perform the Berezin integration. The integration order is \(\int d\bar{\eta}\,d\eta\), integrating over \(\eta\) on the inside and over \(\bar{\eta}\) on the outside:

\[ \int d\bar{\eta}\,d\eta \; (1 - 5\bar{\eta}\eta) \]

First, integrate over \(\eta\):

\[ \int d\eta \; (1 - 5\bar{\eta}\eta) = \int d\eta \; 1 - 5\bar{\eta}\int d\eta \; \eta = 0 - 5\bar{\eta} \cdot 1 = -5\bar{\eta} \]

Then integrate over \(\bar{\eta}\):

\[ \int d\bar{\eta} \; (-5\bar{\eta}) = -5 \int d\bar{\eta} \; \bar{\eta} = -5 \cdot 1 = -5 \]

Wait, let us recheck the sign. When computing \(\int d\eta\;(\bar{\eta}\eta)\) in the \(-5\bar{\eta}\eta\) term, \(\bar{\eta}\) is a Grassmann constant with respect to \(\eta\), so:

\[ \int d\eta \; \bar{\eta}\eta = \bar{\eta} \int d\eta \; \eta = \bar{\eta} \cdot 1 = \bar{\eta} \]

(When pulling \(\bar{\eta}\) outside the integral sign, does a sign arise from exchanging \(d\eta\) and \(\bar{\eta}\)? In the Berezin integration convention, \(\int d\eta\) acts as a left derivative, so in \(\int d\eta\;(\bar{\eta}\eta)\) we need to bring \(\bar{\eta}\) past \(\eta\) from the left. Since \(\bar{\eta}\eta = -\eta\bar{\eta}\), we get \(\int d\eta\;(-\eta\bar{\eta}) = -\bar{\eta}\)...)

Let us carefully verify the convention here. In the standard convention:

\[ \int d\eta \; \bar{\eta}\eta \]

We compute \(\frac{\partial}{\partial\eta}(\bar{\eta}\eta)\). Using the left derivative convention, we bring \(\eta\) to the leftmost position: since \(\bar{\eta}\eta = -\eta\bar{\eta}\),

\[ \frac{\partial}{\partial\eta}(\bar{\eta}\eta) = \frac{\partial}{\partial\eta}(-\eta\bar{\eta}) = -\bar{\eta} \]

Therefore \(\int d\eta\;\bar{\eta}\eta = -\bar{\eta}\).

Recalculating:

\[ \int d\eta \; (1 - 5\bar{\eta}\eta) = 0 - 5\int d\eta\;\bar{\eta}\eta = -5 \cdot (-\bar{\eta}) = 5\bar{\eta} \]

Then:

\[ \int d\bar{\eta} \; 5\bar{\eta} = 5 \]

Final answer:

\[ \boxed{\int d\bar{\eta}\,d\eta \; e^{-5\bar{\eta}\eta} = 5} \]

Verification: According to Eq. (C.18), \(\int d\bar{\eta}\,d\eta \; e^{-a\bar{\eta}\eta} = a\). Substituting \(a = 5\) gives \(5\). This matches. This is also consistent with \(\det A = 5\) for the \(1 \times 1\) matrix \(A = (5)\).

B-9. Sign of Grassmann Differentiation

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Solution strategy: First, rearrange \(\theta\phi\theta\) using the anticommutation relations, then differentiate with respect to \(\phi\).

Calculation:

\[ \theta\,\phi\,\theta = -\theta\,\theta\,\phi = -\theta^2\,\phi = 0 \]

(A sign change occurs when exchanging \(\phi\) and \(\theta\), and since \(\theta^2 = 0\), the entire expression vanishes.)

Therefore:

\[ \frac{\partial}{\partial\phi}(\theta\,\phi\,\theta) = \frac{\partial}{\partial\phi}(0) = 0 \]

Final answer:

\[ \boxed{\frac{\partial}{\partial\phi}(\theta\,\phi\,\theta) = 0} \]

Verification: We confirm using an alternative method. To apply \(\frac{\partial}{\partial\phi}\) under the left-derivative convention, we bring \(\phi\) to the leftmost position. \(\theta\phi\theta = -\phi\theta\theta = -\phi\theta^2 = 0\). Since the expression is already zero before differentiation, the result is also zero.

B-10. Multivariable Gaussian Integral with Source

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Solution strategy: Apply equation (C.9).

Calculation:

Given data:

\[ A = \begin{pmatrix}4 & 0\\0 & 4\end{pmatrix}, \quad \mathbf{J} = \begin{pmatrix}2\\0\end{pmatrix} \]

Compute the necessary quantities:

\[ \det A = 4 \times 4 - 0 = 16 \]
\[ A^{-1} = \frac{1}{4}\begin{pmatrix}1 & 0\\0 & 1\end{pmatrix} \]
\[ \mathbf{J}^T A^{-1}\mathbf{J} = (2, 0)\frac{1}{4}\begin{pmatrix}2\\0\end{pmatrix} = (2, 0)\begin{pmatrix}1/2\\0\end{pmatrix} = 1 \]

Substituting into equation (C.9) (with \(n = 2\)):

\[ \int d^2 q \; e^{-\frac{1}{2}\mathbf{q}^T A\,\mathbf{q} - \mathbf{J}^T\mathbf{q}} = \frac{(2\pi)^{2/2}}{(\det A)^{1/2}} \; e^{\frac{1}{2}\mathbf{J}^T A^{-1}\mathbf{J}} = \frac{2\pi}{\sqrt{16}} \; e^{1/2} = \frac{2\pi}{4} \; e^{1/2} = \frac{\pi}{2}\,e^{1/2} \]

Final answer:

\[ \boxed{\int d^2 q \; e^{-\frac{1}{2}\mathbf{q}^T A\,\mathbf{q} - \mathbf{J}^T\mathbf{q}} = \frac{\pi}{2}\,e^{1/2}} \]

Verification: Since \(A\) is diagonal, we can integrate each variable independently. The \(q_1\) integral: \(\int dq_1\;e^{-2q_1^2 - 2q_1} = \sqrt{\pi/2}\,e^{1/2}\) (using equation (C.3) with \(a = 4\), \(J = 2\)). The \(q_2\) integral: \(\int dq_2\;e^{-2q_2^2} = \sqrt{\pi/2}\). The product is \((\pi/2)\,e^{1/2}\). This agrees.

Medium

M-1. Generating Correlation Functions via Gaussian Integrals with Sources

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Solution strategy: Differentiate \(Z(J) = \sqrt{2\pi/a}\;e^{J^2/(2a)}\) with respect to \(J\) and evaluate at \(J = 0\).

(a) \(\langle q^2 \rangle = 1/a\)

Calculation:

\[ Z(J) = \sqrt{\frac{2\pi}{a}} \; e^{J^2/(2a)} \]
\[ \frac{\partial Z}{\partial J} = \sqrt{\frac{2\pi}{a}} \; \frac{J}{a} \; e^{J^2/(2a)} \]
\[ \frac{\partial^2 Z}{\partial J^2} = \sqrt{\frac{2\pi}{a}} \; e^{J^2/(2a)} \left[\frac{1}{a} + \frac{J^2}{a^2}\right] \]

Evaluating at \(J = 0\):

\[ \left.\frac{\partial^2 Z}{\partial J^2}\right|_{J=0} = \sqrt{\frac{2\pi}{a}} \cdot \frac{1}{a} \]

Since \(Z(0) = \sqrt{2\pi/a}\):

\[ \langle q^2 \rangle = \frac{1}{Z(0)}\left.\frac{\partial^2 Z}{\partial J^2}\right|_{J=0} = \frac{1}{a} \]
\[ \boxed{\langle q^2 \rangle = \frac{1}{a}} \]

(b) \(\langle q^4 \rangle = 3/a^2\)

Calculation:

Expanding \(e^{J^2/(2a)}\) in a power series:

\[ e^{J^2/(2a)} = \sum_{k=0}^{\infty} \frac{1}{k!}\left(\frac{J^2}{2a}\right)^k = 1 + \frac{J^2}{2a} + \frac{J^4}{8a^2} + \cdots \]

Applying \(\frac{\partial^4}{\partial J^4}\), only the \(J^4\) term contributes at \(J = 0\):

\[ \frac{\partial^4}{\partial J^4}\left(\frac{J^4}{8a^2}\right) = \frac{4!}{8a^2} = \frac{24}{8a^2} = \frac{3}{a^2} \]

Therefore:

\[ \left.\frac{\partial^4 Z}{\partial J^4}\right|_{J=0} = Z(0) \cdot \frac{3}{a^2} \]
\[ \langle q^4 \rangle = \frac{1}{Z(0)}\left.\frac{\partial^4 Z}{\partial J^4}\right|_{J=0} = \frac{3}{a^2} \]
\[ \boxed{\langle q^4 \rangle = \frac{3}{a^2}} \]

(c) Correspondence with Wick's theorem

Calculation:

\[ \langle q^4 \rangle = \frac{3}{a^2} = 3 \cdot \left(\frac{1}{a}\right)^2 = 3\langle q^2 \rangle^2 \]

According to Wick's theorem, \(\langle q^4 \rangle\) is given by the sum over all ways of pairing (contracting) the four \(q\)'s into pairs of two. Labeling the four \(q\)'s as \(q_1, q_2, q_3, q_4\), the possible pairings are:

  1. \((q_1 q_2)(q_3 q_4)\)
  2. \((q_1 q_3)(q_2 q_4)\)
  3. \((q_1 q_4)(q_2 q_3)\)

giving 3 ways. Each pair contributes \(\langle q^2 \rangle = 1/a\), so:

\[ \langle q^4 \rangle = 3 \times \langle q^2 \rangle^2 = \frac{3}{a^2} \]

The number of pairings is given in general by \(\frac{(2m)!}{2^m \cdot m!}\), and for \(m = 2\) this gives \(\frac{4!}{2^2 \cdot 2!} = \frac{24}{4 \cdot 2} = 3\).

Verification: From the recurrence relation (C.7), one can directly confirm that \(I_4(a) = 3/a^2 \cdot I_0(a)\), which is consistent with \(\langle q^4 \rangle = I_4/I_0 = 3/a^2\).

M-2. Derivation of the Multi-variable Grassmann Gaussian Integral

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Solution strategy: Expand the exponential in Grassmann variables and pick out the non-zero terms via Berezin integration.

Calculation:

Writing \(A = \begin{pmatrix}a & b \\ c & d\end{pmatrix}\), we expand the exponent:

\[ \bar{\boldsymbol{\eta}}^T A \boldsymbol{\eta} = a\bar{\eta}_1\eta_1 + b\bar{\eta}_1\eta_2 + c\bar{\eta}_2\eta_1 + d\bar{\eta}_2\eta_2 \]

Denoting this as \(X\), we expand \(e^{-X}\):

\[ e^{-X} = 1 - X + \frac{1}{2}X^2 - \cdots \]

For the Berezin integral \(\int d\bar{\eta}_1\,d\eta_1\,d\bar{\eta}_2\,d\eta_2\) to be non-zero, the integrand must contain all four Grassmann variables \(\bar{\eta}_1, \eta_1, \bar{\eta}_2, \eta_2\) each appearing exactly once.

  • The 1 term: Contains no Grassmann variables, so the integral is zero.
  • The \(-X\) term: Each term contains only 2 Grassmann variables, so it gives zero.
  • The \(\frac{1}{2}X^2\) term: Terms containing all 4 Grassmann variables survive.

Computing \(X^2\):

\[ X^2 = (a\bar{\eta}_1\eta_1 + b\bar{\eta}_1\eta_2 + c\bar{\eta}_2\eta_1 + d\bar{\eta}_2\eta_2)^2 \]

The combinations in which all four variables appear are:

  • \((a\bar{\eta}_1\eta_1)(d\bar{\eta}_2\eta_2) = ad\,\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\)
  • \((d\bar{\eta}_2\eta_2)(a\bar{\eta}_1\eta_1) = ad\,\bar{\eta}_2\eta_2\bar{\eta}_1\eta_1\)
  • \((b\bar{\eta}_1\eta_2)(c\bar{\eta}_2\eta_1) = bc\,\bar{\eta}_1\eta_2\bar{\eta}_2\eta_1\)
  • \((c\bar{\eta}_2\eta_1)(b\bar{\eta}_1\eta_2) = bc\,\bar{\eta}_2\eta_1\bar{\eta}_1\eta_2\)

We reorder to a standard form. Taking \(\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\) as the reference.

Term 1: \(\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\) (already in standard order).

Term 2: \(\bar{\eta}_2\eta_2\bar{\eta}_1\eta_1\). Moving \(\bar{\eta}_2\eta_2\) from the left to the right of \(\bar{\eta}_1\eta_1\): since \(\bar{\eta}_2\eta_2\) is a product of an even number of Grassmann numbers, it commutes with other Grassmann numbers without sign change. Therefore \(\bar{\eta}_2\eta_2\bar{\eta}_1\eta_1 = \bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\).

Term 3: \(\bar{\eta}_1\eta_2\bar{\eta}_2\eta_1\). Rearranging to the order \(\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\):

\[ \bar{\eta}_1\eta_2\bar{\eta}_2\eta_1 = \bar{\eta}_1(-\bar{\eta}_2\eta_2)\eta_1 = -\bar{\eta}_1\bar{\eta}_2\eta_2\eta_1 = -\bar{\eta}_1\bar{\eta}_2(-\eta_1\eta_2) = \bar{\eta}_1\bar{\eta}_2\eta_1\eta_2 \]

Then for \(\bar{\eta}_1\bar{\eta}_2\eta_1\eta_2\): exchanging \(\bar{\eta}_2\) and \(\eta_1\) gives \(\bar{\eta}_1(-\eta_1\bar{\eta}_2)\eta_2 = -\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\).

Therefore Term 3 \(= -\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\).

Term 4: \(\bar{\eta}_2\eta_1\bar{\eta}_1\eta_2\). Rearranging similarly:

\[ \bar{\eta}_2\eta_1\bar{\eta}_1\eta_2 \]

Exchanging \(\eta_1\) and \(\bar{\eta}_1\): \(\bar{\eta}_2(-\bar{\eta}_1\eta_1)\eta_2 = -\bar{\eta}_2\bar{\eta}_1\eta_1\eta_2\).

Exchanging \(\bar{\eta}_2\) and \(\bar{\eta}_1\): \(-(-\bar{\eta}_1\bar{\eta}_2)\eta_1\eta_2 = \bar{\eta}_1\bar{\eta}_2\eta_1\eta_2\).

Exchanging \(\bar{\eta}_2\) and \(\eta_1\): \(\bar{\eta}_1(-\eta_1\bar{\eta}_2)\eta_2 = -\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\).

Therefore Term 4 \(= -\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\).

Combining:

\[ X^2 \big|_{\text{4-variable}} = (ad + ad - bc - bc)\,\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 = 2(ad - bc)\,\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 \]

Therefore:

\[ \frac{1}{2}X^2\big|_{\text{4-variable}} = (ad - bc)\,\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 \]

Now we perform the Berezin integration. Using the convention \(\int d\bar{\eta}_1\,d\eta_1\,d\bar{\eta}_2\,d\eta_2\):

\[ \int d\bar{\eta}_1\,d\eta_1\,d\bar{\eta}_2\,d\eta_2 \; \bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 \]

Starting from the inside outward:

\[ \int d\eta_2 \; \eta_2 = 1 \]
\[ \int d\bar{\eta}_2 \; \bar{\eta}_2 = 1 \]
\[ \int d\eta_1 \; \eta_1 = 1 \]
\[ \int d\bar{\eta}_1 \; \bar{\eta}_1 = 1 \]

However, we must be careful about signs. For \(\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\), when integrating sequentially from the right:

\[ \int d\eta_2\;(\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2) \]

Bringing \(\eta_2\) to the left end: \(\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 = -\bar{\eta}_1\eta_1\eta_2\bar{\eta}_2\)...

Let us reorganize using a different approach. As a standard convention:

\[ \int d\bar{\eta}_1\,d\eta_1\,d\bar{\eta}_2\,d\eta_2 \; \bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 \]

is evaluated by applying the rightmost \(d\eta_2\) first. \(\int d\eta_2\;\eta_2 = 1\) leaves \(\bar{\eta}_1\eta_1\bar{\eta}_2\). Then \(\int d\bar{\eta}_2\;\bar{\eta}_2 = 1\) leaves \(\bar{\eta}_1\eta_1\). Then \(\int d\eta_1\;\eta_1 = 1\) leaves \(\bar{\eta}_1\). Finally \(\int d\bar{\eta}_1\;\bar{\eta}_1 = 1\).

However, signs arising from bringing variables to the left end at each step must be taken into account.

More systematically: using the left-derivative convention, where each Berezin integral acts as a left derivative.

Acting from the right sequentially:

\[ \int d\eta_2 \; (\bar{\eta}_1\eta_1\bar{\eta}_2\eta_2) \]

Bringing \(\eta_2\) to the left end: passing through 3 Grassmann variables:

\[ \bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 = (-1)^3 \eta_2\bar{\eta}_1\eta_1\bar{\eta}_2 = -\eta_2\bar{\eta}_1\eta_1\bar{\eta}_2 \]

Therefore \(\frac{\partial}{\partial\eta_2}(-\eta_2\bar{\eta}_1\eta_1\bar{\eta}_2) = -\bar{\eta}_1\eta_1\bar{\eta}_2\).

Next \(\int d\bar{\eta}_2\): \(\frac{\partial}{\partial\bar{\eta}_2}(-\bar{\eta}_1\eta_1\bar{\eta}_2)\). Bringing \(\bar{\eta}_2\) to the left end:

\[ -\bar{\eta}_1\eta_1\bar{\eta}_2 = -(-1)^2\bar{\eta}_2\bar{\eta}_1\eta_1 = -\bar{\eta}_2\bar{\eta}_1\eta_1 \]

\(\frac{\partial}{\partial\bar{\eta}_2}(-\bar{\eta}_2\bar{\eta}_1\eta_1) = -\bar{\eta}_1\eta_1\).

Next \(\int d\eta_1\): \(\frac{\partial}{\partial\eta_1}(-\bar{\eta}_1\eta_1)\). Bringing \(\eta_1\) to the left end:

\[ -\bar{\eta}_1\eta_1 = -(-1)^1\eta_1\bar{\eta}_1 = \eta_1\bar{\eta}_1 \]

\(\frac{\partial}{\partial\eta_1}(\eta_1\bar{\eta}_1) = \bar{\eta}_1\).

Finally \(\int d\bar{\eta}_1\): \(\frac{\partial}{\partial\bar{\eta}_1}(\bar{\eta}_1) = 1\).

Therefore:

\[ \int d\bar{\eta}_1\,d\eta_1\,d\bar{\eta}_2\,d\eta_2 \; \bar{\eta}_1\eta_1\bar{\eta}_2\eta_2 = 1 \]

Thus:

\[ \int d\bar{\eta}_1\,d\eta_1\,d\bar{\eta}_2\,d\eta_2 \; e^{-\bar{\boldsymbol{\eta}}^T A\boldsymbol{\eta}} = ad - bc = \det A \]

Final answer:

\[ \boxed{\int \prod_{i=1}^{2} d\bar{\eta}_i\,d\eta_i \; e^{-\bar{\boldsymbol{\eta}}^T A\boldsymbol{\eta}} = ad - bc = \det A} \]

Verification: When \(A = \mathbf{1}\) (identity matrix), \(\det A = 1\). With \(X = \bar{\eta}_1\eta_1 + \bar{\eta}_2\eta_2\), we get \(\frac{1}{2}X^2 = \bar{\eta}_1\eta_1\bar{\eta}_2\eta_2\). The integral gives 1. This is consistent. In the bosonic case \((\det A)^{-1/2}\) appears in the denominator, whereas for fermions \(\det A\) appears in the numerator—this is the essential characteristic of Grassmann integration.

M-3. Source Terms and Inverse Matrix in Grassmann Gaussian Integrals

→ Back to problem

Solution strategy: Perform completing the square for Grassmann variables, then reduce to the source-free integral after a change of variables.

Calculation:

Organize the exponent:

\[ -\bar{\boldsymbol{\eta}}^T A\,\boldsymbol{\eta} + \bar{\boldsymbol{\xi}}^T\boldsymbol{\eta} + \bar{\boldsymbol{\eta}}^T\boldsymbol{\xi} \]

Perform completing the square. Define new variables:

\[ \boldsymbol{\eta}' = \boldsymbol{\eta} - A^{-1}\boldsymbol{\xi}, \qquad \bar{\boldsymbol{\eta}}' = \bar{\boldsymbol{\eta}} - \bar{\boldsymbol{\xi}}(A^{-1})^T = \bar{\boldsymbol{\eta}} - \bar{\boldsymbol{\xi}}(A^T)^{-1} \]

(Note the position of the transpose here: from the form \(\bar{\boldsymbol{\eta}}^T A\boldsymbol{\eta}\), we have \(\bar{\boldsymbol{\eta}}'^T = \bar{\boldsymbol{\eta}}^T - \bar{\boldsymbol{\xi}}^T A^{-1}\))

Verify the completing-the-square identity:

\[ \bar{\boldsymbol{\eta}}^T A\,\boldsymbol{\eta} - \bar{\boldsymbol{\xi}}^T\boldsymbol{\eta} - \bar{\boldsymbol{\eta}}^T\boldsymbol{\xi} = (\bar{\boldsymbol{\eta}}^T - \bar{\boldsymbol{\xi}}^T A^{-1})\,A\,(\boldsymbol{\eta} - A^{-1}\boldsymbol{\xi}) - \bar{\boldsymbol{\xi}}^T A^{-1}\boldsymbol{\xi} \]

Expand the right-hand side to verify:

\[ (\bar{\boldsymbol{\eta}}^T - \bar{\boldsymbol{\xi}}^T A^{-1})\,A\,(\boldsymbol{\eta} - A^{-1}\boldsymbol{\xi}) \]
\[ = \bar{\boldsymbol{\eta}}^T A\boldsymbol{\eta} - \bar{\boldsymbol{\eta}}^T A \cdot A^{-1}\boldsymbol{\xi} - \bar{\boldsymbol{\xi}}^T A^{-1} A\boldsymbol{\eta} + \bar{\boldsymbol{\xi}}^T A^{-1} A \cdot A^{-1}\boldsymbol{\xi} \]
\[ = \bar{\boldsymbol{\eta}}^T A\boldsymbol{\eta} - \bar{\boldsymbol{\eta}}^T\boldsymbol{\xi} - \bar{\boldsymbol{\xi}}^T\boldsymbol{\eta} + \bar{\boldsymbol{\xi}}^T A^{-1}\boldsymbol{\xi} \]

Adding \(-\bar{\boldsymbol{\xi}}^T A^{-1}\boldsymbol{\xi}\):

\[ = \bar{\boldsymbol{\eta}}^T A\boldsymbol{\eta} - \bar{\boldsymbol{\eta}}^T\boldsymbol{\xi} - \bar{\boldsymbol{\xi}}^T\boldsymbol{\eta} \]

This indeed matches the original expression. ✓

Perform the change of variables \(\boldsymbol{\eta}' = \boldsymbol{\eta} - A^{-1}\boldsymbol{\xi}\), \(\bar{\boldsymbol{\eta}}' = \bar{\boldsymbol{\eta}} - (A^{-1})^T\bar{\boldsymbol{\xi}}\).

Checking the Jacobian: For a linear transformation of Grassmann variables \(\eta_i' = \eta_i + c_i\) (where \(c_i\) are Grassmann constants), the Berezin integration measure is invariant:

\[ d\eta_i' = d\eta_i \]

This follows from the translation invariance of Berezin integration. In \(\int d\eta'\;f(\eta') = \int d\eta\;f(\eta + c)\), we have \(f(\eta + c) = f_0 + f_1(\eta + c) = (f_0 + f_1 c) + f_1\eta\), and the coefficient of \(\eta\) remains \(f_1\).

More generally, for a transformation \(\eta_i' = M_{ij}\eta_j + c_i\), we have \(\prod_i d\eta_i' = (\det M)^{-1}\prod_i d\eta_i\) (opposite to the bosonic case!). In the present case \(M = \mathbf{1}\) (identity matrix), so the Jacobian is 1.

Similarly, the Jacobian for \(\bar{\boldsymbol{\eta}}'\) is also 1.

Therefore:

\[ \prod_i d\bar{\eta}_i'\,d\eta_i' = \prod_i d\bar{\eta}_i\,d\eta_i \]

Evaluate the integral:

\[ \int \prod_i d\bar{\eta}_i\,d\eta_i \; e^{-\bar{\boldsymbol{\eta}}^T A\boldsymbol{\eta} + \bar{\boldsymbol{\xi}}^T\boldsymbol{\eta} + \bar{\boldsymbol{\eta}}^T\boldsymbol{\xi}} \]
\[ = \int \prod_i d\bar{\eta}_i'\,d\eta_i' \; e^{-\bar{\boldsymbol{\eta}}'^T A\boldsymbol{\eta}' + \bar{\boldsymbol{\xi}}^T A^{-1}\boldsymbol{\xi}} \]
\[ = e^{\bar{\boldsymbol{\xi}}^T A^{-1}\boldsymbol{\xi}} \int \prod_i d\bar{\eta}_i'\,d\eta_i' \; e^{-\bar{\boldsymbol{\eta}}'^T A\boldsymbol{\eta}'} \]
\[ = e^{\bar{\boldsymbol{\xi}}^T A^{-1}\boldsymbol{\xi}} \cdot \det A \]

Final answer:

\[ \boxed{\int \prod_i d\bar{\eta}_i\,d\eta_i \; e^{-\bar{\boldsymbol{\eta}}^T A\,\boldsymbol{\eta} + \bar{\boldsymbol{\xi}}^T\boldsymbol{\eta} + \bar{\boldsymbol{\eta}}^T\boldsymbol{\xi}} = \det A \; e^{\bar{\boldsymbol{\xi}}^T A^{-1}\boldsymbol{\xi}}} \]

Consistency check: Setting \(\boldsymbol{\xi} = \bar{\boldsymbol{\xi}} = 0\) recovers \(\det A\). ✓ Also, comparing with the bosonic formula (C.9), where the result was \((\det A)^{-1/2} e^{\frac{1}{2}\mathbf{J}^T A^{-1}\mathbf{J}}\), the fermionic case gives \(\det A \cdot e^{\bar{\boldsymbol{\xi}}^T A^{-1}\boldsymbol{\xi}}\). The determinant moves from the denominator to the numerator, and the absence of the \(1/2\) factor in the exponent corresponds to the fact that \(\bar{\eta}\) and \(\eta\) are independent variables.

M-4. Gaussian Integral as a Fresnel Integral

→ Back to problem

Solution strategy: Apply equation (C.2) for \(a = -i\alpha\).

Calculation:

The integral we seek is:

\[ \int_{-\infty}^{\infty} dq \; e^{\frac{i\alpha}{2}q^2} \]

Comparing this with the form \(\int dq\;e^{-\frac{a}{2}q^2}\) from equation (C.1):

\[ -\frac{a}{2} = \frac{i\alpha}{2} \quad \Longrightarrow \quad a = -i\alpha \]

Writing \(a\) in polar form:

\[ a = -i\alpha = \alpha \cdot e^{-i\pi/2} \]

Here \(|a| = \alpha\), \(\theta = -\pi/2\). Since \(\mathrm{Re}(a) = 0\), this lies strictly on the boundary of the convergence condition \(\mathrm{Re}(a) > 0\). This is understood as an analytic continuation from the region \(\mathrm{Re}(a) > 0\).

Applying equation (C.2):

\[ I = \sqrt{\frac{2\pi}{|a|}} \; e^{-i\theta/2} = \sqrt{\frac{2\pi}{\alpha}} \; e^{-i(-\pi/2)/2} = \sqrt{\frac{2\pi}{\alpha}} \; e^{i\pi/4} \]

Final answer:

\[ \boxed{\int_{-\infty}^{\infty} dq \; e^{\frac{i\alpha}{2}q^2} = \sqrt{\frac{2\pi}{\alpha}}\;e^{i\pi/4}} \]

Relation to the path integral in Minkowski space:

In the Minkowski-space path integral, the weight takes the form \(e^{iS}\). The action for a free scalar field is

\[ S = \int d^4x \; \frac{1}{2}(\partial_\mu\phi)^2 - \frac{1}{2}m^2\phi^2 \]

In momentum space, \(S \sim \int \frac{d^4k}{(2\pi)^4}\;\frac{1}{2}\phi(-k)(k^2 - m^2)\phi(k)\), and the path integral becomes

\[ \int \mathcal{D}\phi \; e^{iS} \sim \prod_k \int d\phi_k \; e^{\frac{i}{2}(k^2 - m^2)\phi_k^2} \]

The integral for each mode is precisely of the Fresnel type \(\int dq\;e^{i\alpha q^2/2}\).

This integral does not strictly converge at \(\mathrm{Re}(a) = 0\). The actual prescriptions are:

  1. \(i\epsilon\) prescription: Setting \(m^2 \to m^2 - i\epsilon\) gives \(a\) a small positive real part, ensuring convergence. This is the origin of the \(i\epsilon\) in the Feynman propagator.

  2. Wick rotation: Performing \(t \to -i\tau\) to move to Euclidean space gives \(e^{iS} \to e^{-S_E}\), so that the Gaussian integral takes the standard convergent form. After the calculation, one analytically continues back to Minkowski space.

The phase \(e^{i\pi/4}\) of the Fresnel integral corresponds to half the rotation angle \(\pi/2\) of the Wick rotation, and determines the overall phase factor of the path integral (which is absorbed into the normalization).

Verification: Since \(e^{i\pi/4} = \frac{1+i}{\sqrt{2}}\), we have \(I = \sqrt{\frac{2\pi}{\alpha}} \cdot \frac{1+i}{\sqrt{2}} = \sqrt{\frac{\pi}{\alpha}}(1+i)\). Then \(|I|^2 = \frac{\pi}{\alpha} \cdot 2 = \frac{2\pi}{\alpha}\). On the other hand, \(|I|^2 = |\int dq\;e^{i\alpha q^2/2}|^2 = \int dq_1\int dq_2\;e^{i\alpha(q_1^2 - q_2^2)/2}\). Substituting \(u = q_1 + q_2\), \(v = q_1 - q_2\) gives \(\frac{1}{2}\int du\int dv\;e^{i\alpha uv/2}\). This equals \(\frac{1}{2} \cdot 2\pi \cdot \frac{2}{\alpha} = \frac{2\pi}{\alpha}\) (using the delta function representation). The results agree.

Advanced

A-1. Ratio of Boson and Fermion Determinants and Supersymmetry

→ Back to problem

(a) Calculation of \(Z\)

Strategy: The bosonic and fermionic parts are independent, so they can be integrated separately.

Calculation:

Bosonic part: From equation (C.8)

\[ \int d^n q \; e^{-\frac{1}{2}\mathbf{q}^T A\,\mathbf{q}} = \frac{(2\pi)^{n/2}}{(\det A)^{1/2}} \]

Fermionic part: From equation (C.19)

\[ \int \prod_i d\bar{\eta}_i\,d\eta_i \; e^{-\bar{\boldsymbol{\eta}}^T A\,\boldsymbol{\eta}} = \det A \]

Therefore:

\[ \boxed{Z = \frac{(2\pi)^{n/2}}{(\det A)^{1/2}} \cdot \det A = (2\pi)^{n/2} \cdot (\det A)^{1/2}} \]

(b) Case \(A = m^2 \mathbf{1}\)

Calculation:

\[ \det A = \det(m^2 \mathbf{1}) = (m^2)^n = m^{2n} \]
\[ (\det A)^{1/2} = m^n \]
\[ Z = (2\pi)^{n/2} \cdot m^n \]

...this depends on \(m\). Let us re-examine the problem statement.

Indeed, the bosonic integral gives \((\det A)^{-1/2}\) and the fermionic part gives \(\det A\), so:

\[ Z = \frac{(2\pi)^{n/2}}{(\det A)^{1/2}} \cdot \det A = (2\pi)^{n/2} (\det A)^{1/2} \]

For \(A = m^2\mathbf{1}\), we get \(Z = (2\pi)^{n/2} m^n\). This depends on \(m\).

Let us reconsider the intent of the problem. In the context of supersymmetry, when the bosonic and fermionic degrees of freedom are equal, the partition function (or the determinant part of the effective action) becomes mass-independent. However here, the bosons are \(n\) real scalars (degrees of freedom \(n\)), while the fermions are \(n\) pairs of complex Grassmann variables (degrees of freedom \(2n\)), so the degrees of freedom do not match.

For supersymmetry to hold exactly, the bosonic and fermionic determinants must cancel completely. That is, we need a situation where \((\det A)^{-1/2}\) and \((\det A)^{1/2}\) cancel, not \((\det A)^{-1/2}\) and \(\det A\).

Looking at the hint, it says "(b) substitute \((\det A)^{1/2} = m^n\)", and does not explicitly state that \(Z\) is \(m\)-independent...

Re-reading the problem: "When \(A\) is a constant times the identity matrix \(A = m^2 \mathbf{1}\), show that \(Z\) is independent of \(m\)."

For this to hold, the bosonic and fermionic contributions must cancel completely. For \(n\) real bosonic variables \((\det A)^{-1/2} = m^{-n}\), and for \(n\) Grassmann pairs \(\det A = m^{2n}\). The product is \(m^{-n} \cdot m^{2n} = m^n\). This depends on \(m\).

Presumably the problem intends either complex bosonic variables (\(2n\) real degrees of freedom), or a setup where the fermionic part gives \((\det A)^{1/2}\).

Alternatively, with \(n\) complex bosonic variables \(z_i = (q_{2i-1} + iq_{2i})/\sqrt{2}\):

\[ \int d^n z\,d^n\bar{z}\;e^{-\bar{\mathbf{z}}^T A\mathbf{z}} = \frac{\pi^n}{\det A} \]

In this case, combined with the fermionic contribution \(\det A\), we get \(Z = \pi^n\). This is independent of \(m\)!

However, the problem statement says "\(n\) bosonic variables \(q_i\)", suggesting real variables.

Most consistent interpretation: The problem actually intends \(2n\) real bosonic variables (\(n\) complex bosons), in which case:

\[ \int d^{2n}q\;e^{-\frac{1}{2}\mathbf{q}^T(A\otimes\mathbf{1}_2)\mathbf{q}} = \frac{(2\pi)^n}{(\det A)^1} \]

Combined with the fermionic contribution \(\det A\), we get \(Z = (2\pi)^n\). This is independent of \(m\).

Here we answer by interpreting the problem literally and discussing the conditions under which "determinant cancellation" occurs.

Solution

(a)

Bosonic part:

\[ Z_B = \int d^n q \; e^{-\frac{1}{2}\mathbf{q}^T A\,\mathbf{q}} = \frac{(2\pi)^{n/2}}{(\det A)^{1/2}} \]

Fermionic part:

\[ Z_F = \int \prod_i d\bar{\eta}_i\,d\eta_i \; e^{-\bar{\boldsymbol{\eta}}^T A\,\boldsymbol{\eta}} = \det A \]

Total:

\[ \boxed{Z = Z_B \cdot Z_F = (2\pi)^{n/2} \cdot (\det A)^{1/2}} \]

(b)

For \(A = m^2\mathbf{1}\):

\[ (\det A)^{1/2} = (m^{2n})^{1/2} = m^n \]
\[ Z = (2\pi)^{n/2} \cdot m^n \]

At first glance this appears to depend on \(m\), but the physically meaningful quantity is the \(m\)-dependence of the determinant part of \(\ln Z\). Comparing the bosonic contribution \(-\frac{1}{2}\ln\det A = -n\ln m\) with the fermionic contribution \(+\ln\det A = +2n\ln m\), the sum is \(+n\ln m\).

However, for exact cancellation from supersymmetry, the bosonic and fermionic degrees of freedom must match. For \(n\) real bosonic degrees of freedom, the \(n\) Grassmann pairs have \(2n\) Grassmann degrees of freedom.

In the correct supersymmetric setup, one pairs \(n\) complex bosons (\(2n\) real degrees of freedom) with \(n\) Grassmann pairs. In this case:

\[ Z_B = \frac{(2\pi)^n}{\det A}, \qquad Z_F = \det A \]
\[ Z = (2\pi)^n \]

This is completely independent of \(m\). The \(\det A\) cancels completely between bosons and fermions.

In the problem's setup (\(n\) real bosons), the cancellation between \((\det A)^{-1/2}\) and \((\det A)^{+1}\) is incomplete, leaving \((\det A)^{1/2}\). This reflects the mismatch in degrees of freedom.

\[ \boxed{Z = (2\pi)^{n/2} m^n \quad \text{(for $n$ real bosons)}} \]

Note: In the setup where supersymmetry holds exactly (\(n\) complex bosons + \(n\) Grassmann pairs), \(Z = (2\pi)^n\) (independent of \(m\)).

(c) Correspondence with the one-loop effective potential

The one-loop effective potential in quantum field theory takes the following form (Ch. 14):

\[ V_{\text{1-loop}} = \frac{1}{2}\int \frac{d^4k}{(2\pi)^4}\left[\ln(k^2 + m_B^2) - 2\ln(k^2 + m_F^2)\right] \]

(For one bosonic degree of freedom and a Dirac fermion with 4 degrees of freedom. The coefficients depend on the degrees of freedom.)

In general, using the supertrace:

\[ V_{\text{1-loop}} = \frac{1}{2}\mathrm{STr}\int\frac{d^4k}{(2\pi)^4}\ln(k^2 + M^2) \]

where \(\mathrm{STr}\) is the trace with \(+\) sign for bosons and \(-\) sign for fermions:

\[ \mathrm{STr}\,f(M^2) = \sum_{\text{bosons}} f(m_B^2) - \sum_{\text{fermions}} f(m_F^2) \]

When supersymmetry holds, for each boson there exists a fermionic partner with the same mass, and the degrees of freedom also match. Therefore:

\[ \mathrm{STr}\,M^0 = n_B - n_F = 0 \]
\[ \mathrm{STr}\,M^2 = \sum m_B^2 - \sum m_F^2 = 0 \]

This ensures that the ultraviolet divergences (\(\Lambda^4\) and \(\Lambda^2\) terms) of \(V_{\text{1-loop}}\) automatically cancel. Furthermore, if \(\mathrm{STr}\,M^4\ln(M^2/\mu^2) = 0\) holds, the logarithmic divergence also vanishes.

In our result \(Z_B \cdot Z_F \propto (\det A)^{-1/2} \cdot \det A\), the complete cancellation \((\det A)^0\) occurs when the bosonic determinant is \((\det A)^{-1}\) (complex bosons), which corresponds precisely to the case where bosonic and fermionic degrees of freedom are equal (supersymmetry holds).

\[ \boxed{\text{SUSY: } \frac{\det A_F}{(\det A_B)^{1}} = 1 \quad \Longleftrightarrow \quad V_{\text{1-loop}} = 0} \]

Verification: In the supersymmetric Wess–Zumino model, one complex scalar (2 real degrees of freedom) and one Weyl fermion (2 real degrees of freedom) share the same mass, and the zero-point energies of the one-loop effective potential cancel. This corresponds to the \(n = 1\) case (complex boson setup) of the present problem.

A-2. Faddeev–Popov Ghost Representation of Determinants via Grassmann Integration

→ Back to problem

(a) Grassmann Integral Representation of \(\det M\)

Calculation:

This is equation (C.19) itself. For an \(n \times n\) matrix \(M\):

\[ \det M = \int \prod_{a=1}^n d\bar{c}^a\,dc^a \; e^{-\bar{c}^a M_{ab}\,c^b} \]

Verification: As shown in S2, the Grassmann Gaussian integral gives the determinant in the numerator. This is in contrast to the bosonic Gaussian integral, which gives \((\det A)^{-1/2}\) (in the denominator).

\[ \boxed{\det M = \int \prod_a d\bar{c}^a\,dc^a \; e^{-\bar{c}^a M_{ab}\,c^b}} \]

(b) Derivation of the Faddeev–Popov Operator

Calculation:

In \(SU(N)\) Yang–Mills theory, the infinitesimal gauge transformation of the gauge field \(A_\mu^a\) is:

\[ \delta A_\mu^a = \frac{1}{g}D_\mu^{ab}\alpha^b = \frac{1}{g}(\delta^{ab}\partial_\mu + g f^{acb}A_\mu^c)\alpha^b \]

where \(D_\mu^{ab} = \delta^{ab}\partial_\mu + g f^{acb}A_\mu^c\) is the covariant derivative in the adjoint representation.

(Note: Depending on the convention, one may write \(\delta A_\mu^a = D_\mu^{ab}\alpha^b/g\) or \(\delta A_\mu^a = D_\mu^{ab}\alpha^b\). Here we adopt the convention \(\delta A_\mu^a = D_\mu^{ab}\alpha^b\).)

Lorenz gauge condition:

\[ G^a[A] = \partial^\mu A_\mu^a \]

Variation of the gauge condition under a gauge transformation:

\[ \delta G^a = \partial^\mu(\delta A_\mu^a) = \partial^\mu(D_\mu^{ab}\alpha^b) = \partial^\mu(\delta^{ab}\partial_\mu + g f^{acb}A_\mu^c)\alpha^b \]

The Faddeev–Popov operator is:

\[ M^{ab}(x, y) = \frac{\delta G^a(x)}{\delta\alpha^b(y)} \]

From the form \(\delta G^a(x) = \int d^4y\;M^{ab}(x,y)\alpha^b(y)\):

\[ \delta G^a(x) = \partial^\mu_x D_\mu^{ab}(x)\alpha^b(x) = \int d^4y\;\partial^\mu_x D_\mu^{ab}(x)\delta^4(x-y)\alpha^b(y) \]

Therefore:

\[ \boxed{M^{ab}(x, y) = \partial^\mu D_\mu^{ab}\,\delta^4(x-y) = (\delta^{ab}\partial^2 + g f^{acb}\partial^\mu A_\mu^c + g f^{acb}A^{c\mu}\partial_\mu)\delta^4(x-y)} \]

Alternatively, in the convention using \(-M^{ab}\):

\[ M^{ab}(x,y) = -\partial_\mu D^{\mu\,ab}\,\delta^4(x-y) \]

where \(D^{\mu\,ab} = \delta^{ab}\partial^\mu + gf^{acb}A^{c\mu}\).

(c) Feynman Rules for Ghost Fields

Calculation:

The ghost action is:

\[ S_{\text{ghost}} = \int d^4x \; \bar{c}^a(-\partial_\mu D^{\mu\,ab})c^b \]

Expanding this:

\[ S_{\text{ghost}} = \int d^4x \left[-\bar{c}^a\partial_\mu(\delta^{ab}\partial^\mu + gf^{acb}A^{c\mu})c^b\right] \]
\[ = \int d^4x \left[-\bar{c}^a\partial^2 c^a - g f^{abc}\bar{c}^a\partial_\mu(A^{b\mu}c^c)\right] \]

Using integration by parts (dropping surface terms):

\[ = \int d^4x \left[-\bar{c}^a\partial^2 c^a - g f^{abc}(\partial^\mu\bar{c}^a)A_\mu^b c^c\right] \]

Propagator: From the free part \(-\bar{c}^a\partial^2 c^a\), we read off the momentum-space propagator:

\[ \langle c^a(k)\bar{c}^b(-k)\rangle = \frac{\delta^{ab}}{k^2} \]

This has the same form as the propagator of a massless scalar field.

Vertex: From the interaction term \(-gf^{abc}(\partial^\mu\bar{c}^a)A_\mu^b c^c\), we read off the ghost–gauge field vertex:

\[ V^{abc}_\mu = -gf^{abc}p^\mu \]

where \(p^\mu\) is the momentum of the outgoing \(\bar{c}\) (arising from \(\partial^\mu\bar{c}\)).

Reason why ghosts obey Fermi statistics:

  1. From the properties of Grassmann integration: The ghost fields \(c^a, \bar{c}^a\) are introduced as Grassmann variables. The Grassmann integral gives the determinant in the numerator (equation (C.19)). This corresponds to the \((-1)\) sign associated with fermion loops in the path integral.

  2. Sign of loops: In Feynman diagrams, closed ghost loops arise from traces over Grassmann variables and therefore carry an additional factor of \((-1)\). This is precisely the hallmark of Fermi statistics.

  3. Relation to the spin-statistics theorem: Ghosts have the propagator of a scalar field (spin 0) yet obey Fermi statistics. This violates the spin-statistics theorem. However, since ghosts are not physical particles (they do not appear on external lines) and arise as auxiliary fields from the gauge-fixing procedure, this violation is permissible. Ghosts are excluded from the physical state space by BRST symmetry.

  4. Physical role: Ghost loops cancel the contributions of unphysical polarizations (longitudinal and scalar components) of the gauge field in the path integral. The fact that \(\det M\) appears in the numerator (a property of Grassmann integration) is what makes this cancellation possible.

\[ \boxed{\text{Ghost propagator: } \frac{\delta^{ab}}{k^2}, \quad \text{Vertex: } -gf^{abc}p^\mu, \quad \text{Loop factor: } (-1)} \]

Consistency checks:

  • Gauge invariance: The ghost action \(\bar{c}(-\partial_\mu D^\mu)c\) is invariant under BRST transformations, which guarantees unitarity of the theory.
  • Dimensional analysis: \([c] = [\bar{c}] = 1\) (mass dimension), \([g] = 0\) (in 4 dimensions), \([A_\mu] = 1\). The dimension of the vertex \(gf^{abc}\partial^\mu\bar{c}^a A_\mu^b c^c\) is \(0 + 1 + 1 + 1 + 1 = 4\). This matches the dimension 4 of the Lagrangian density. ✓
  • Abelian limit: When \(f^{abc} = 0\) (QED), ghosts do not couple to the gauge field and propagate only as free fields. In Lorenz gauge QED, ghosts do not contribute to physics (since there are no vertices for loop diagrams). This is consistent with the well-known fact that ghosts can be ignored in QED. ✓