Saddle-Point Approximation and Functional Determinants

Source: Perimeter Scholars International, Quantum Field Theory II, Tutorial 3; 15 worked parts.

Question prompts paraphrased from the official Perimeter PSI Online tutorial sheet; the equations and notation follow the source.

1 Saddle-Point Approximation

Approximation methods for ordinary integrals are among the reasons that path integrals are useful. For example, consider \[I(z)=\int dx\,e^{if(x,z)}.\tag{1.1}\] The corresponding ideas extend to functional integrals such as \[Z[J]=\int\mathcal D\phi\,e^{iS[\phi,J]}.\tag{1.2}\] The saddle-point method is one especially useful example.

Restore $\hbar$ in the Euclidean partition function: \[Z[J] =\int\mathcal D\phi\, e^{-\frac{1}{\hbar}S[\phi,J]}.\tag{1.3}\] This makes it possible to organise a semiclassical expansion by treating $\hbar$ as small. In Euclidean signature, configurations $\phi_c$ minimising the action dominate because configurations with larger action are exponentially suppressed. For a Minkowski integral, all stationary configurations of the action must be considered rather than only its minima.

The following questions first develop this approximation for ordinary $n$-dimensional integrals.

Let $S(x^\mu;z)$ be a real function of $n$ spacetime variables $\mu=1,\ldots,n$ and a coupling $z$, and define \[I(z) =\int_{-\infty}^{\infty}d^n x\, e^{-\frac{1}{\hbar}S(x;z)}.\tag{1.4}\] Expand $S(x;z)$ around its extrema $x_c$ and show that the sum over its minima gives \[I(z) \approx \sum_{\operatorname{Min}[x_c]} e^{-\frac{1}{\hbar}S(x_c;z)} \left[ \frac{(2\pi\hbar)^n}{\det S''(x_c)} \right]^{1/2} +\mathcal O(\hbar^a),\tag{1.5}\] where \[\left[S''(x_c)\right]_{\mu\nu} =\left. \frac{\partial^2S}{\partial x^\mu\,\partial x^\nu} \right|_{x=x_c}.\tag{1.6}\]

Expanding $S(x;z)$ around its minima, we find \[S(x;z) = S(x_c;z) + \frac{1}{2}(x-x_c)^\mu (x-x_c)^\nu [S''(x_c;z)]_{\mu\nu} + \cdots.\tag{1.7}\] In the semi-classical regime, the field configurations around the minima dominate, giving \[\begin{aligned} I(z) &\approx \sum_{\operatorname{Min}[x_c]}\int d^nx e^{-\frac{1}{\hbar}(S(x_c;z) + \frac{1}{2}(x-x_c)^\mu (x-x_c)^\nu [S''(x_c;z)]_{\mu\nu})}\\ &=\sum_{\operatorname{Min}[x_c]} e^{-\frac{1}{\hbar}S(x_c;z)}\left[ \frac{(2\pi\hbar)^n}{\det S''(x_c)} \right]^{1/2}, \end{aligned}\tag{1.8}\] where we have used the results from conventional Gaussian integration in the last equality.
For a generic function $S(x;z)$, determine the exponent $a$.

The important fluctuations should be $x-x_c \sim \sqrt{\hbar}$. The next contribution comes from $S''''$ ($S'''$ is odd, and hence vanishes). Therefore, $a = n/2 +2 - 1 = n/2 +1$.
State the condition under which the approximation becomes exact.

The approximation is exact if all higher order derivative vanishes.
Explain how the saddle-point expression should change if the starting point is the Minkowski partition function \[Z=\int e^{iS/\hbar}.\tag{1.9}\]

In real time, $-S \to iS$. Besides that, we also have to add the $i\epsilon$ prescription to ensure the path integral converges. Therefore, we have \[I(z)\approx \sum_{\operatorname{Min}[x_c]} e^{\frac{i}{\hbar}S(x_c;z)} \left[ \frac{(2\pi i\hbar)^n}{\det (S''(x_c) + i\epsilon)} \right]^{1/2} +\mathcal O(\hbar^a).\tag{1.10}\]

2 Functional Determinants

Functional determinants occur throughout the path-integral treatment of QFT, including the semiclassical result above. Their formal definition as a product over operator eigenvalues is generally divergent and requires regularisation. For this exercise, set those analytic subtleties aside and manipulate the operators by analogy with matrices.

Define \[F[f] =\log\!\left( \det\!\left[-\frac{d^2}{dy^2}+f(y)\right] \right).\tag{2.1}\] Thus $F$ is the logarithm of the determinant of the operator \[\mathcal A[f(y)] =-\frac{d^2}{dy^2}+f(y).\tag{2.2}\] If $M(y,z)$ is the integral kernel of an operator $\mathcal M$, then \[\mathcal M g(y) =\int M(y,z)g(z)\,dz\tag{2.3}\] for every suitable function $g$.

The matrix-product analogy used here is \[\sum_y M_{xy}M_{yz} \quad\Longleftrightarrow\quad \int dy\,M(x,y)M(y,z).\tag{2.4}\]

Show that the integral kernel of $\mathcal A$ is \[A(y,z) =\left[-\frac{d^2}{dy^2}+f(y)\right]\delta(y-z).\tag{2.5}\]

\[\begin{aligned} \mathcal{A}[f(y)]g(y) = -\frac{d^2g}{dy^2} + fg = \int dz \, A(y,z) g(z). \end{aligned}\tag{2.6}\]

Suppose that $O(x,y)$ is the integral kernel of \[\mathcal O =\left(-\frac{d^2}{dy^2}+f(y)\right)^{-1}.\tag{2.7}\]
The operators satisfy $\mathcal A\mathcal O=1$. Rewrite this identity entirely in terms of their integral kernels.

\[\begin{aligned} \int dz A(x,z)O(z,y) = \delta(x-y). \end{aligned}\tag{2.8}\]
Using functional differentiation, prove \[\frac{\delta F[f]}{\delta f(x)}=O(x,x).\tag{2.9}\] Hint: Recall the matrix identity \[\det M=\exp(\operatorname{tr}\log M).\tag{2.10}\]

\[\begin{aligned} \frac{\delta F[f]}{\delta f(x)} &= \frac{\delta}{\delta f(x)} \operatorname{tr} \log \mathcal{A} = \frac{\delta}{\delta f(x)}\operatorname{tr} \mathcal{O} \delta\mathcal{A}= \frac{\delta}{\delta f(z)} \int dydz O(x,y)\delta f(y) \delta(y-z) =O(x,x). \end{aligned}\tag{2.11}\]
Next show \[\frac{\delta^2F[f]}{\delta f(x)\delta f(y)} =-O(x,y)O(y,x).\tag{2.12}\] Hint: To differentiate an inverse, begin with \[M^{-1}M=1.\tag{2.13}\]

\[\frac{\delta^2F[f]}{\delta f(x)\delta f(y)} = \frac{\delta}{\delta f(y)}\operatorname{tr}\mathcal{A}^{-1} = -\operatorname{tr}\mathcal{A}^{-1}\frac{\delta\mathcal{A}}{\delta f(y)}\mathcal{A}^{-1} = - O(x,y) O(y,x).\tag{2.14}\]
Finally derive \[\begin{aligned}\frac{\delta^3F[f]} {\delta f(x)\delta f(y)\delta f(z)} &=O(x,y)O(y,z)O(z,x) \notag\\ &\quad+O(x,z)O(z,y)O(y,x).\end{aligned}\tag{2.15}\]

Follows trivially from above.

3 Optional: The Gel'fand--Yaglom Formalism

The Gel'fand--Yaglom method can determine functional determinants without explicitly computing every eigenvalue. In favourable cases it is exact, and it can also be implemented numerically. Begin with the Riemann zeta function \[\zeta(s) \equiv \sum_{k=1}^{\infty}k^{-s}.\tag{3.1}\]

The contour required below is fixed by the argument principle. It is positively oriented and surrounds the positive zeros $\lambda_n$ of $F(\lambda)$, while avoiding the branch cut of $\lambda^{-s}$ on the negative real axis.

Figure 1. Schematic reconstruction of contour $\gamma$ in the $\lambda$-plane.

Let $A$ have eigenvalues $\lambda_n$ and eigenfunctions $\phi_n$: \[A\phi_n=\lambda_n\phi_n.\tag{3.2}\] Associate to it the zeta function \[\begin{aligned}\zeta_A(s) &\equiv \operatorname{Tr}\!\left[\frac{1}{A^s}\right] \notag\\ &=\sum_{k=1}^{\infty}\lambda_k^{-s}.\end{aligned}\tag{3.3}\]

Show that zeta regularisation gives \[\det A=\exp\!\left[-\zeta_A'(0)\right].\tag{3.4}\] Because $\zeta_A$ and its derivative need not converge near $s=0$, analytic continuation may be required before this expression is meaningful.

\[\det A =\prod_{k=1}^\infty \lambda_k,\tag{3.5}\] while \[\zeta'_A = \sum_{k=1}^\infty-\log \lambda_k \lambda^{-s}.\tag{3.6}\] Therefore, \[\det A=\exp\!\left[-\zeta_A'(0)\right].\tag{3.7}\]
Take \[A=-\frac{d^2}{d\tau^2}\tag{3.8}\] on $0\leq\tau\leq1$, with Dirichlet conditions \[\phi_n(0)=\phi_n(1)=0.\tag{3.9}\] Find the eigenvalues $\lambda_n$, then evaluate $\det A$ using \[\zeta(0)=-\frac{1}{2}, \qquad \zeta'(0)=-\frac{1}{2}\log(2\pi).\tag{3.10}\]

It can be easily found that $\lambda_n = n^2\pi^2$, $n \in \mathbb{Z}^{+}$. Then, \[\zeta_A (s) = \sum_{k=1}^\infty(n\pi)^{-2s},\tag{3.11}\] and \[\zeta_A'(0) = \sum_{k=1}^\infty-2\log k - 2\log\pi = 2\zeta'(0) -2\log \pi\zeta(0) = -\log2.\tag{3.12}\] Hence, we have \[\det A = 2.\tag{3.13}\]

In cases where the eigenvalues cannot be obtained explicitly, suppose one can construct $F(\lambda)$ whose zeros are precisely the eigenvalues of $A$.
Assume that the $\lambda_n>0$ are discrete and nondegenerate. Show that \[\zeta_A(s) =\frac{1}{2\pi i} \int_\gamma d\lambda\, \lambda^{-s} \frac{d\ln F(\lambda)}{d\lambda},\tag{3.14}\] where $\gamma$ is the contour specified in Figure 1 and the branch cut lies on the negative real axis.

The RHS is given by \[\begin{aligned} &\frac{1}{2\pi i} \int_\gamma d\lambda\, \lambda^{-s} \frac{d\ln F(\lambda)}{d\lambda}\\ &= \sum_{n=1}^\infty \lim_{\lambda\to\lambda_n} \frac{\lambda-\lambda_n}{\lambda^s}\frac{1}{F(\lambda)}\frac{d F(\lambda)}{d\lambda}\\ &=\sum_{n=1}^\infty \lim_{\lambda\to\lambda_n}\left(F'(\lambda_n) + F''(\lambda_n)(\lambda-\lambda_n) + \cdots \right)\left(F'(\lambda_n)(\lambda-\lambda_n) + \frac12F''(\lambda_n)(\lambda-\lambda_n)^2 + \cdots \right)^{-1}\\ &=\sum_{n=1}^\infty \lim_{\lambda\to\lambda_n}\frac{1}{\lambda^s}\left(1+ \frac{F''(\lambda_n)}{F'(\lambda_n)}(\lambda-\lambda_n) +\cdots\right)\left(1- \frac12\frac{F''(\lambda_n)}{F'(\lambda_n)}(\lambda-\lambda_n) +\cdots\right)\\ &=\sum_{n=1}^\infty \lambda_n^{-s}\\ &= \zeta_A(s). \end{aligned}\tag{3.15}\]
Deform $\gamma$ into the contour $\gamma_-$, which hugs the negative real axis as indicated in Figure 2, and derive \[\zeta_A(s) =\frac{\sin(\pi s)}{\pi} \int_0^{-\infty}d\lambda\, \lambda^{-s} \frac{d\ln F(\lambda)}{d\lambda}.\tag{3.16}\]

\[\begin{aligned} \zeta_A(s) &= \frac{1}{2\pi i} \int_{\gamma_-} d\lambda\, \lambda^{-s} \frac{d\ln F(\lambda)}{d\lambda}\\ &= \lim_{\epsilon \to 0^+}\frac{1}{2\pi i} \int_{-\infty}^0 d\lambda\, (\lambda+i\epsilon)^{-s} \frac{d\ln F(\lambda+i\epsilon)}{d\lambda} - \frac{1}{2\pi i} \int_{-\infty}^0 d\lambda\, (\lambda-i\epsilon)^{-s} \frac{d\ln F(\lambda-i\epsilon)}{d\lambda}\\ &=\frac{1}{2\pi i} \int_{-\infty}*0 d\lambda\, (e^{-s\log \lambda-is\pi}-e^{-s\log \lambda+ is\pi}) \frac{d\ln F(\lambda)}{d\lambda}\\ &=-\frac{\sin \pi s}{\pi}\int_{-\infty}^0 d\lambda\, \lambda^{-s} \frac{d\ln F(\lambda)}{d\lambda}. \end{aligned}\tag{3.17}\]

Figure 2. Finite contour deformation.
Apply the preceding results to the one-dimensional Schrödinger operator \[A=-\frac{d^2}{dx^2}+V(x)\tag{3.18}\] on $0\leq x\leq1$, with Dirichlet boundary conditions: \[\begin{aligned} A\phi_n&=\lambda_n\phi_n,\\ \phi(0)&=0,\\ \phi(1)&=0. \end{aligned}\tag{3.19}\] Assume that the $\lambda_n$ are positive, discrete, and nondegenerate. Instead of the Dirichlet eigenfunctions, solve the same eigenvalue equation with initial-value conditions \[\begin{aligned} Au_\lambda&=\lambda u_\lambda,\\ u_\lambda(0)&=0,\\ u_\lambda'(0)&=1. \end{aligned}\tag{3.20}\] Choose \[F(\lambda)=u_\lambda(1)\tag{3.21}\] and use the previous parts to establish the Gel'fand--Yaglom theorem \[\frac{\det A}{\det A_{\mathrm{free}}} =\frac{u_0(1)}{u_{\mathrm{free},0}(1)},\tag{3.22}\] where \[A_{\mathrm{free}}=-\frac{d^2}{dx^2}.\tag{3.23}\]

From the results above, we have \[\zeta_A'(0) = \int^{-\infty}_0d\lambda \frac{d\log F(\lambda)}{d\lambda} = \log\frac{F(-\infty)}{F(0)} \implies \det A = \exp [-\zeta_A'(0)] = \frac{F(0)}{F(-\infty)}.\tag{3.24}\] Note that $F(-\infty) = u_{-\infty}(1)$ is formally infinite. Intuitively, one might guess \[\lim_{\lambda\to -\infty}\frac{F(\lambda)}{F_\mathrm{free}(\lambda)} = 1,\tag{3.25}\] which follows \[\frac{\det A}{\det A_{\mathrm{free}}} =\frac{u_0(1)}{u_{\mathrm{free},0}(1)}.\tag{3.26}\] The guess is true, as long as $V(x)$ is bounded, when $\lambda \to -\infty$, $V(x)$ becomes negligible.
Use the Gel'fand--Yaglom theorem to compute $\det A$ for \[A=-\frac{d^2}{dx^2}+\omega^2\tag{3.27}\] on the unit interval with Dirichlet boundary conditions. Verify the result independently by finding the eigenvalues and using \[\sin(\pi z) =\pi z\prod_{n=1}^{\infty} \left(1-\frac{z^2}{n^2}\right).\tag{3.28}\]

\[u_{\mathrm{free},0}(x) = x, \quad u_0(x) = \frac{\sinh \omega x}{\omega}, \quad \det A_\mathrm{free} = 2 \implies \det A =2\frac{\sinh \omega}{\omega}.\tag{3.29}\] Meanwhile, \[\lambda_n = \pi^2n^2 + \omega^2 \implies \det A = \prod_{n=1}^\infty \pi^2n^2\left(1 +\frac{\omega^2}{\pi^2 n^2}\right),\tag{3.30}\] which looks disastrous. However, remember, we can regulate it with \[\det A_\mathrm{free} = \prod_{n=1}^\infty \pi^2n^2.\tag{3.31}\] Therefore, we find \[\frac{\det A}{\det A_\mathrm{free}} =\frac{\sinh \omega}{\omega},\tag{3.32}\] which recovers the Fel'danf-Yaglom theorem result by regulating $\det A_\mathrm{free}$ through the analytic continuation of the Reimann function.