Gelfand-Yaglom theorem for functional determinants

Question

What is the 'Gelfand-Yaglom' Theorem? I have heard that it is used to calculate Functional determinants by solving an initial value problem of the form

$Hy(x)-zy(x)=0$ with $y(0)=0$ and $y'(0)=1$. Here $H$ is the Hamiltonian and $z$ is a real parameter.

Is it that simple? If $H$ is a Hamiltonian, could I use the WKB approximation to solve the initial value problem and to be valid for $z$ big?

Answer 1

User Simon has already given a good answer. Here we sketch a derivation of the Gelfand-Yaglom formula.

1. Let there be given a self-adjoint Hamiltonian operator $$H~=~H^{(0)}+V, \tag{1}$$ with non-degenerate discrete energy levels $(\lambda_n)_{n\in\mathbb{N}}$, bounded from below, and not zero. Similarly, the free Hamiltonian $H^{(0)}$ has non-degenerate discrete energy levels $(\lambda^{(0)}_n)_{n\in\mathbb{N}}$, bounded from below, and not zero. (A zero-eigenvalue must be excluded to have a useful notion of determinant.) Let an entire function $f:\mathbb{C}\to \mathbb{C}$ have simple zeros at $(\lambda_n)_{n\in\mathbb{N}}$, i.e. it is of the form $$\begin{align}f(\lambda)~=~&(\lambda-\lambda_n)g_n(\lambda), \cr g_n(\lambda_n)~\neq~& 0.\end{align}\tag{2}$$ We shall later see how one in practice can construct such $f$-function, cf. eqs. (16) & (26) below. The function$^1$ $$\begin{align}({\rm Ln} f)^{\prime}(\lambda)~=~&\frac{f^{\prime}(\lambda)}{f(\lambda)}\cr ~\sim~&\frac{1}{\lambda-\lambda_n}+ \text{regular terms}\end{align}\tag{3}$$ has unit residue $${\rm Res}(({\rm Ln} f)^{\prime},\lambda=\lambda_n)~\stackrel{(3)}{=}~1\tag{4}$$ at $\lambda=\lambda_n$.

2. Now use zeta-function regularization $$\begin{align} \zeta_H(s)~=~&\sum_{n\in\mathbb{N}} \lambda_n^{-s}\cr ~\stackrel{(4)}{=}~& \int_{\gamma_+}\!\frac{d\lambda}{2\pi i} \exp\left(-s{\rm Ln}\lambda\right)~({\rm Ln} f)^{\prime}(\lambda) ,\tag{5}\cr -\zeta^{\prime}_H(s)~\stackrel{(5)}{=}~& \sum_{n\in\mathbb{N}} \lambda_n^{-s}~{\rm Ln}\lambda_n ,\tag{6}\end{align}$$ where the contour $\gamma_+$ is depicted in Fig. 1.

   

   $\uparrow$ Fig. 1: Original integration contour $\gamma_+$ in the complex $\lambda$ plane. The black dots represent the non-zero discrete energy levels $(\lambda_n)_{n\in\mathbb{N}}$. (Fig. taken from Ref. 2.)

3. For the 1D Sturm-Liouville problems that we have in mind, $$\lambda_n~\sim~ {\cal O}(n^2)\quad\text{for}\quad n~\to~ \infty,\tag{7} $$ so that the eqs. (5) & (6) are typically only valid for ${\rm Re}(s)>\frac{1}{2}$. This is not good enough since the zeta-function-regularized determinant is defined via analytic continuation to the point $s=0$: $$\begin{align} {\rm Ln} {\rm Det} H ~=~&{\rm Ln} \prod_{n\in\mathbb{N}}\lambda_n\cr ~=~&\sum_{n\in\mathbb{N}} {\rm Ln} \lambda_n\cr ~\stackrel{(6)}{=}~& -\zeta^{\prime}_H(s=0) .\end{align}\tag{8} $$ For large energies $\lambda \to \infty$, the potential $V$ should not matter, so that $$\frac{f(\lambda)}{f^{(0)}(\lambda)}~\longrightarrow~ 1 \quad\text{for}\quad |\lambda|~\to~ \infty.\tag{9}$$ The idea is to instead study the difference between the full and free theory: $$\begin{align} \zeta_H(s)&-\zeta_{H^{(0)}}(s)\cr ~\stackrel{(5)}{=}~&\int_{\gamma_+}\!\frac{d\lambda}{2\pi i} \exp\left(-s{\rm Ln}\lambda\right)~({\rm Ln} \frac{f}{f^{(0)}})^{\prime}(\lambda).\end{align}\tag{10}$$

   

   $\uparrow$ Fig. 2: Deformed integration contour $\gamma_-$ in the complex $\lambda$ plane. The black half-line at an angle $\theta$ in the upper half-plane denotes the branch cut of the complex logarithm. The black dots represent the non-zero discrete energy levels $(\lambda_n)_{n\in\mathbb{N}}$ and $(\lambda^{(0)}_n)_{n\in\mathbb{N}}$.

4. We next deform the integration contour $\gamma_+$ into $\gamma_-$, cf. Fig. 2. $$\begin{align} \zeta_H(s)&-\zeta_{H^{(0)}}(s) \cr ~\stackrel{(10)}{=}~&\int_{\gamma_-}\!\frac{d\lambda}{2\pi i} \exp\left(-s{\rm Ln}\lambda\right)~({\rm Ln} \frac{f}{f^{(0)}})^{\prime}(\lambda) \cr ~=~&\left(\int_{e^{i\theta}\infty}^0\!e^{-i\theta s}+\int_0^{e^{i\theta}\infty}\!e^{-i(\theta-2\pi) s} \right)\cr &|\lambda|^{-s}~({\rm Ln} \frac{f}{f^{(0)}})^{\prime}(\lambda) \frac{d\lambda}{2\pi i} \cr ~=~&e^{i(\pi -\theta) s} \frac{\sin(\pi s)}{\pi}\cr &\int_{e^{i\theta}\mathbb{R}_+}\!d\lambda~ |\lambda|^{-s}~({\rm Ln} \frac{f}{f^{(0)}})^{\prime}(\lambda) .\end{align}\tag{11}$$ Differentiation wrt. $s$ yields: $$\begin{align} \zeta^{\prime}_H(s)&-\zeta^{\prime}_{H^{(0)}}(s)\cr~\stackrel{(11)}{=}~& e^{i(\pi -\theta) s}\cos(\pi s)\cr &\int_{e^{i\theta}\mathbb{R}_+}\!d\lambda~ |\lambda|^{-s}~({\rm Ln} \frac{f}{f^{(0)}})^{\prime}(\lambda)\cr &+o(s).\end{align}\tag{12}$$ The zeta-function-regularized determinant is $$\begin{align}{\rm Ln}&\frac{{\rm Det} H}{{\rm Det} H^{(0)}}\cr ~\stackrel{(8)+(12)}{=}&~ -\int_{e^{i\theta}\mathbb{R}_+}\!d\lambda~ ({\rm Ln} \frac{f}{f^{(0)}})^{\prime}(\lambda)~\cr \stackrel{(9)}{=}~~& {\rm Ln} \frac{f(\lambda=0)}{f^{(0)}(\lambda=0)} , \end{align}\tag{13}$$ which is the Gelfand-Yaglom formula

   $$ \frac{{\rm Det} H}{{\rm Det} H^{(0)}}~\stackrel{(13)}{=}~ \frac{f(\lambda=0)}{f^{(0)}(\lambda=0)}. \tag{14}$$

   Since the requirements (2) to the $f$-function are scale-invariant, a relative result (14) is the best we could hope for.

5. Main application: Consider the 1D TISE on the finite interval $a\leq x\leq b $ with Dirichlet boundary conditions, with free$^2$ Hamiltonian $$H^{(0)} ~=~-\frac{\hbar^2}{2}\frac{d}{dx}m(x)^{-1}\frac{d}{dx}. \tag{15}$$ The $f$-function is chosen as $$ f(\lambda)~=~\psi_{\lambda}(x=b),\tag{16}$$ where $\psi_{\lambda}(x)$ is the unique solution to the initial value problem $$\begin{align} H\psi_{\lambda}~=~&\lambda\psi_{\lambda}, \cr \psi_{\lambda}(x=a)~=~&0,\cr \psi^{\prime}_{\lambda}(x=a)~=~&C\cr ~=~&\text{some fixed constant}.\end{align}\tag{17}$$

6. Example: Constant potential $V(x)=V_0$ and constant mass $m(x)=m_0$. The discrete energy eigenvalues for the infinite square well are $$\begin{align} \lambda_n~=~&\lambda^{(0)}_n+V_0, \cr \lambda^{(0)}_n~=~&\frac{(\pi\hbar n)^2}{2m_0(b-a)^2}, \cr n~\in~&\mathbb{N}.\end{align}\tag{18}$$ The zeta-function-regularized determinant becomes$^3$ $$\begin{align} {\rm Det} H~=~&\frac{2}{\sqrt{V_0}}\sinh\left(\frac{\sqrt{2m_0V_0}}{\hbar}(b-a)\right), \cr {\rm Det} H^{(0)}~=~&\frac{2\sqrt{2m_0}}{\hbar}(b-a).\end{align}\tag{19}$$ On the other hand $$\begin{align}\psi_{\lambda}(x)~=~&C\frac{\hbar }{\sqrt{2m_0(\lambda-V_0)}}\cr &\sin\left(\frac{\sqrt{2m_0(\lambda-V_0)}}{\hbar}(x-a)\right),\end{align}\tag{20}$$ so that $$\begin{align}\psi_{\lambda=0}(x=b)~=~&C\frac{\hbar}{\sqrt{2m_0V_0}}\cr &\sinh\left(\frac{\sqrt{2m_0V_0}}{\hbar}(b-a)\right), \cr \psi^{(0)}_{\lambda=0}(x=b)~=~&C(b-a) .\end{align}\tag{21}$$ Eqs. (19) & (21) should be compared with the Gelfand-Yaglom formula (14).

7. Modified main application. Consider again the free Hamiltonian (15). Let $\phi_{\lambda}(x)$ be an eigenfunction to the full Hamiltonian (1): $$\begin{align} H\phi_{\lambda}~=~&\lambda\phi_{\lambda}, \cr \phi_{\lambda}(x=a)~\neq~&0.\end{align}\tag{22}$$ Define $$\psi_{\lambda}(x)~:=~\phi_{\lambda}(x)\int_a^x\! dx^{\prime} \frac{m(x^{\prime})}{\phi_{\lambda}(x^{\prime})^2}. \tag{23}$$ Then one may show that (23) is an independent eigenfunction $$\begin{align} H\psi_{\lambda}~=~&\lambda\psi_{\lambda}, \cr \psi_{\lambda}(x=a)~=~&0.\end{align}\tag{24}$$ The Wronskian is $$\begin{align} W(\phi_{\lambda},\psi_{\lambda})~=~&\phi_{\lambda}\psi^{\prime}_{\lambda}-\phi^{\prime}_{\lambda}\psi_{\lambda}\cr~=~&m(x).\end{align} \tag{25}$$ The $f$-function is now instead chosen as $$\begin{align} f(\lambda)~=~~&\phi_{\lambda}(a)\frac{m(x)}{W(\phi_{\lambda},\psi_{\lambda})}\psi_{\lambda}(b)\cr ~\stackrel{(23)+(25)}{=}&~ \phi_{\lambda}(a)\phi_{\lambda}(b)\int_a^b\! dx \frac{m(x)}{\phi_{\lambda}(x)^2}.\end{align}\tag{26}$$ The middle formula in eq. (26) is independent of $\phi_{\lambda}$ and $\psi_{\lambda}$ satisfying eqs. (22) & (24).

References:

1. G.V. Dunne, Functional Determinants in QFT, lecture notes, 2009; Chap. 5. PDF & PDF.

2. K. Kirsten & A.J. McKane, J.Phys. A37 (2004) 4649, arXiv:math-ph/0403050.

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$^1$ ${\rm Ln}$ denotes the complex $\ln$ function: ${\rm Ln}(\lambda)=\ln|\lambda|+i{\rm Arg}(\lambda)$. We choose the branch ${\rm Arg}(\lambda)\in]\theta\!-\!2\pi,\theta[$, where the branch-cut $\theta\in]0,\pi[$ lies in the upper half-plane.

$^2$ The Hamiltonian (15) in this answer is for semantic reasons called free even if the particle is strictly speaking not free when the mass $m(x)$ is allowed to depend on the position $x$.

$^3$ Use the well-known regularization formulas $$\begin{align} \prod_{n\in \mathbb{N}} a~=~&a^{\zeta(0)}~=~\frac{1}{\sqrt{a}}, \cr \prod_{n\in \mathbb{N}} n~=~&e^{-\zeta^{\prime}(0)}~=~\sqrt{2\pi}, \end{align}\tag{27} $$ $$\begin{align} \prod_{n\in \mathbb{N}} \left[1-\left(\frac{a}{n}\right)^2 \right]~=~&\frac{\sin \pi a}{\pi a}, \cr \prod_{n\in \mathbb{N}} \left[1+\left(\frac{n}{a}\right)^2 \right]~=~&2\sinh \pi a, \end{align}\tag{28} $$ via analytic continuation of the Riemann zeta function $$\begin{align}\zeta(s)~=~&\sum_{n\in \mathbb{N}}n^{-s}, \cr {\rm Re}(s) ~>~&1.\end{align}\tag{29}$$

Answer 2

I was at a talk a while back by Gerald Dunne where he talked about the Gelfand-Yaglom theorem. He used it for calculating some Euler-Heisenberg type effective actions. A paper of his with Hyunsoo Min on the subject is A comment on the Gelfand–Yaglom theorem, zeta functions and heat kernels for PT-symmetric Hamiltonians and he's got some nice lecture notes: Functional Determinants in Quantum Field Theory (also see a wider spanning set of lectures of the same name).

Basically, it's a way of calculating the determinant of a 1-dimensional operator $\det(H)=\prod_i \lambda_i$ with out calculating, let alone multiplying, any of its eigenvalues $H \psi_i = \lambda_i \psi_i$.

To state the original theorem: assume that you have a Schrodinger operator (or Hamiltonian) $ H = -\frac{d^2}{d x^2} + V(x) $ on the interval $x\in[0,L]$ with Dirichlet boundary conditions: $$ H \psi_i(x) = \lambda_i \psi_i(x) \,, \quad \psi(0)=\psi(L)=0 \ . $$ Then we can compute its determinant by solving the related initial value problem $$ H \phi(x) = 0\,, \quad \phi(0)=0\,,\quad \phi'(0) = 1 \ ,$$ so that $$ \det H \approx \phi(L) \,,$$ where the final result is only $\approx$ as we can only really calculate the ratio of two determinants.

This basic result can be generalised to more general boundary conditions, coupled systems of ODEs and higher order linear ODEs.

Answer 3

1. In this answer, we would like to compare the Gelfand-Yaglom formula with a path integral evaluation of a functional determinant, cf. e.g. Ref. 1. Consider the action $$\begin{align} S~=~& \int_{t_i}^{t_f}\! dt~L, \cr L~=~&\frac{m(t)}{2}\dot{q}^2-V, \cr V~=~&\frac{k(t)}{2}q^2 , \end{align} \tag{1}$$ for a 1D harmonic oscillator where the mass $m(t)$ and the spring constant $k(t)$ may depend explicitly on time $t$. The Feynman amplitude/kernel/path integral
   $$\begin{align}\langle q_f\!=\!0,& t_f | q_i\!=\!0,t_i \rangle \cr ~=~& \int_{q(t_i)=0}^{q(t_f)=0} \! {\cal D}q~\exp\left(\frac{i}{\hbar} S\right),\cr\cr &\qquad\qquad\qquad {\cal D}q~\sim~\prod_{t_i <t< t_f} dq(t) , \cr~\stackrel{\begin{array}{c}\text{int. by} \cr\text{parts}\end{array}}{=}&~ \int_{q(t_i)=0}^{q(t_f)=0} \! {\cal D}q~\exp\left(-\frac{i}{\hbar}\int\! dt~ q(t) ~\hat{H} q(t)\right)\cr ~\stackrel{\begin{array}{c}\text{Wick.} \cr\text{rot.}\end{array}}{=}&~ \int \! {\cal D}q\exp\left[-\frac{1}{2\hbar}\iint_{[\tau_i,\tau_f]^2} d\tau~d\tau^{\prime} ~q(\tau)H(\tau,\tau^{\prime})q(\tau^{\prime}) \right]\cr ~\stackrel{\begin{array}{c}\text{Gauss.} \cr\text{ int.}\end{array}}{=}&~ {\rm Det}\hat{H}^{-1/2}\end{align}\tag{2}$$ becomes a functional determinant via Gaussian integration. We may in principle Wick rotate to Euclidean time $$ \tau ~=~it \tag{3}$$ to make the Hessian operator $$\begin{align}\hat{H}~:=~& \underbrace{\frac{d}{dt}m(t)\frac{d}{dt}}_{~=:~\hat{H}^{(0)}}+k(t)\cr ~\stackrel{(3)}{=}~&-\frac{d}{d\tau}m(\tau)\frac{d}{d\tau}+k(\tau)\cr ~>~&0\end{align}\tag{4}$$ positive definite. However, we shall mostly work with Minkowski time $t$. In eq. (2) the matrix elements of the Euclidean Hessian read $$H(\tau,\tau^{\prime}) ~:=~\hat{H}\delta(\tau-\tau^{\prime}).\tag{5}$$

2. Let $\phi_0(t)$ be a zero-mode solution to the homogeneous 2nd-order ODE $$\begin{align}\hat{H}\phi_0~=~&0, \cr \phi_0(t=t_i)~\neq~& 0.\end{align} \tag{6}$$ Introduce for later convenience the shorthand notation $$\begin{align} \Phi_0~:=~&{\rm Ln}\phi_0, \cr \dot{\Phi}_0~=~&\frac{\dot{\phi}_0}{\phi_0}. \end{align}\tag{7}$$
   Then the potential term (1) can be integrated by parts: $$\begin{align} V~\stackrel{(1)}{=}~&\frac{k(t)}{2}q^2\cr ~\stackrel{(6)}{=}~&-\frac{q^2}{2\phi_0} \frac{d(m(t)\dot{\phi}_0)}{dt}\cr ~\stackrel{(7)}{=}~&m(t)\dot{\Phi}_0q\dot{q}- \frac{m(t)}{2}\dot{\Phi}_0^2q^2\cr &-\frac{d}{dt}\left(\frac{m(t)}{2}\dot{\Phi}_0q^2\right).\end{align} \tag{8}$$ Note that the total derivative term (8) vanishes due to the Dirichlet boundary conditions (BCs). The action (1) becomes $$\begin{align} S~\stackrel{(1)+(8)}{=}&~\int_{t_i}^{t_f}\! dt~L^{\prime} ,\cr L^{\prime}~=~~&\frac{m(t)}{2} \left(\dot{q}- \dot{\Phi}_0q\right)^2. \end{align}\tag{9}$$

3. Now perform a non-local coordinate transformation $$\begin{align}Q(t)~=~&q(t)\cr &-\int_{t_i}^{t_f}\! dt^{\prime} ~\theta(t-t^{\prime})~ \dot{\Phi}_0(t^{\prime})q(t^{\prime}), \end{align}\tag{10}$$ so that $$\begin{align} \dot{Q} ~\stackrel{(10)}{=}~& \dot{q} - \dot{\Phi}_0q\cr ~\stackrel{(7)}{=}~& \phi_0\frac{d}{dt}\left(\frac{q}{\phi_0}\right)\end{align} \tag{11}$$ in order to turn the Lagrangian (9) into a free Lagrangian
   $$L^{\prime}~\stackrel{(9)+(11)}{=}~\frac{m(t)}{2} \dot{Q}^2. \tag{12}$$

4. The Jacobian matrix becomes $$\begin{align}\frac{\delta Q(t)}{\delta q(t^{\prime})} ~\stackrel{(10)}{=}~&\delta(t-t^{\prime}) - B(t,t^{\prime}), \cr B(t,t^{\prime}) ~:=~&\theta(t-t^{\prime})~ \dot{\Phi}_0(t^{\prime}), \end{align}\tag{13}$$ via functional differentiation $$ \frac{\delta q(t)}{\delta q(t^{\prime})}~=~\delta(t-t^{\prime}). \tag{14} $$ The trace is $$\begin{align}{\rm Tr} (B) ~=~&\iint_{[t_i,t_f]^2}\!dt~dt^{\prime}~\delta(t-t^{\prime}) B(t,t^{\prime}) \cr ~=~&\int_{[t_i,t_f]}\!dt~ B(t,t) \cr ~\stackrel{(13)}{=}~&\frac{1}{2}(\Phi_0(t_f)-\Phi_0(t_i))\cr ~\stackrel{(7)}{=}~&\frac{1}{2}{\rm Ln} \frac{\phi_0(t_f)}{\phi_0(t_i)}. \end{align}\tag{15}$$ The higher traces vanish $$\begin{align}{\rm Tr} (B^2)~=~&\iiint_{[t_i,t_f]^3}\!dt~dt^{\prime}~dt^{\prime\prime}~\delta(t-t^{\prime\prime}) B(t,t^{\prime})B(t^{\prime},t^{\prime\prime}) \cr ~=~&\iint_{[t_i,t_f]^2}\!dt~dt^{\prime} ~B(t,t^{\prime})B(t^{\prime},t)\cr ~\stackrel{(13)}{=}~&\frac{1}{4}\iint_{[t_i,t_f]^2}\!dt~dt^{\prime}\delta_{t,t^{\prime}} \dot{\Phi}_0(t^{\prime}) \dot{\Phi}_0(t)\cr ~=~&0, \end{align}\tag{16}$$ $$ {\rm Tr} (B^{n\geq 2})~=~0, \tag{17}$$ because the Kronecker delta function $\delta_{t,t^{\prime}}$ vanishes almost everywhere. So the Jacobian factor is $$\begin{align} J~:=~& {\rm Det} \left(\frac{\delta q}{\delta Q}\right)\cr ~=~&{\rm Det} \left(\frac{\delta Q}{\delta q}\right)^{-1}\cr ~\stackrel{(13)}{=}~&{\rm Det}(1-B)^{-1}\cr ~=~&\exp\left(-{\rm Tr}{\rm Ln}(1-B)\right) \cr ~=~&\exp\sum_{n=1}^{\infty} \frac{{\rm Tr} (B^n)}{n}\cr ~\stackrel{(17)}{=}~&\exp{\rm Tr} (B)\cr ~\stackrel{(15)}{=}~&\sqrt{\frac{\phi_0(t_f)}{\phi_0(t_i)}}. \end{align}\tag{18}$$

5. The inverse coordinate transformation is $$ \frac{q(t)}{\phi_0(t)} ~\stackrel{(11)}{=}~\int_{t_i}^{t_f}\! dt^{\prime} ~\theta(t-t^{\prime})~\frac{\dot{Q}(t^{\prime})}{\phi_0(t^{\prime})}. \tag{19}$$ Let us implement the final Dirichlet BC $$\begin{align}0~\approx~&q(t_f)\cr ~\stackrel{(19)}{=}~&\phi_0(t_f)\int_{t_i}^{t_f}\! dt ~\frac{\dot{Q}(t)}{\phi_0(t)} \end{align}\tag{20}$$ with a Lagrange multiplier $\lambda$. The new action becomes $$\begin{align} S^{\prime}~=~~&S+\lambda q(t_f)\cr ~\stackrel{(12)+(20)}{=}&~ \int_{t_i}^{t_f}\! dt~L^{\prime\prime} ,\cr L^{\prime\prime}~=~&\frac{m(t)}{2}\dot{Q}^2 + \lambda \phi_0(t_f) \frac{\dot{Q}}{\phi_0},\end{align} \tag{21}$$ and the Feynman amplitude/kernel/path integral becomes $$\begin{align}\langle q_f\!=\!0,& t_f | q_i\!=\!0,t_i \rangle \cr ~=~& \int_{q(t_i)=0} \! {\cal D}q~\frac{d\lambda}{2\pi\hbar}\exp\left(\frac{i}{\hbar} S^{\prime}\right)\cr ~=~& J\int_{Q(t_i)=0} \! {\cal D}Q~\frac{d\lambda}{2\pi\hbar}\exp\left(\frac{i}{\hbar} S^{\prime}\right).\end{align} \tag{22} $$

6. Next perform a second coordinate transformation $$\begin{align}\tilde{q}(t)~=~&Q(t)\cr &+ \lambda\phi_0(t_f) \int_{t_i}^{t_f}\! dt^{\prime} ~\theta(t-t^{\prime})~ \frac{1}{m(t^{\prime})\phi_0(t^{\prime})} , \end{align}\tag{23}$$ so that $$ \dot{\tilde{q}}~\stackrel{(23)}{=}\dot{Q} + \frac{\lambda\phi_0(t_f)}{m(t)\phi_0} \tag{24}$$ in order to simplify the action $$\begin{align} S^{\prime}~\stackrel{(21)+(24)}{=}&~ \int_{t_i}^{t_f}\! dt~L^{\prime\prime\prime} \cr &-\frac{\lambda^2\phi_0(t_f)^2}{2} \int_{t_i}^{t_f}\! \frac{dt}{m(t)\phi_0(t)^2}, \cr L^{\prime\prime\prime}~=~&\frac{m(t)}{2}\dot{\tilde{q}}^2 .\end{align}\tag{25}$$ Note that both coordinate transformations (10) and (23) do not change the initial Dirichlet BC $$\begin{align}q(t_i)~\approx~&0\cr \quad\stackrel{(10)}{\Leftrightarrow}\quad Q(t_i)~\approx~&0 \cr \quad\stackrel{(23)}{\Leftrightarrow}\quad {\tilde{q}}(t_i)~\approx~&0,\end{align} \tag{26}$$ and the Jacobian for the second coordinate transformations (23) is trivial. (The second transformation (23) is a pure shift/translation.)

7. The Gaussian integration over the Lagrange multiplier $\lambda$ yields $$\begin{align}&\langle q_f\!=\!0, t_f | q_i\!=\!0,t_i \rangle\cr ~\stackrel{(22)}{=}~&J\int_{\tilde{q}(t_i)=0} \! {\cal D}{\tilde{q}}~\frac{d\lambda}{2\pi\hbar}~\exp\left(\frac{i}{\hbar} S^{\prime}\right)\cr ~\stackrel{(25)}{=}~&J\left( 2\pi i\hbar ~\phi_0(t_f)^2 \int_{t_i}^{t_f}\! \frac{dt}{m(t)\phi_0(t)^2} \right)^{-1/2}\cr &\int_{\tilde{q}(t_i)=0} \! {\cal D}\tilde{q}~\exp\left(\frac{i}{\hbar} \int_{t_i}^{t_f}\! dt~L^{\prime\prime\prime}\right)\cr ~\stackrel{(18)}{=}~&\left( 2\pi i\hbar ~\phi_0(t_i)\phi_0(t_f)\int_{t_i}^{t_f}\! \frac{dt}{m(t)\phi_0(t)^2} \right)^{-1/2}\cr &\underbrace{ \int \! d\tilde{q}_f~\langle \tilde{q}_f, t_f | \tilde{q}_i\!=\!0,t_i \rangle^{(0)}}_{~=~1.} .\end{align} \tag{27} $$ Recall that the absolute square of the latter factor in eq. (27) has a physical interpretation in QM as the probability (=100%) that a free particle that starts at position $\tilde{q}_i\!=\!0$ ends somewhere, cf. e.g. this. (Alternatively, it is not difficult to perform the path integral for the free particle directly $$\begin{align}\langle q_f, t_f |q_i,t_i \rangle^{(0)} ~=~& \left(2\pi i \hbar\int_{t_i}^{t_f}\! \frac{dt}{m(t)}\right)^{-1/2} \cr &\exp\left( \frac{i}{2\hbar} \frac{(\Delta q)^2}{\int_{t_i}^{t_f}\! \frac{dt}{m(t)}} \right), \cr \Delta q~:=~&q_f-q_i,\end{align}\tag{28}$$ and a Gaussian integration of eq. (28) over $q_f$ clearly produce 1.) Altogether, the path integral evaluation yields the functional determinant

   $$ {\rm Det}\hat{H}~\stackrel{(2)+(3)+(27)}{=}~ 2\pi i \hbar ~\phi_0(t_i)\underbrace{\phi_0(t_f)\int_{t_i}^{t_f}\! \frac{dt}{m(t)\phi_0(t)^2}}_{~=:~\psi_0(t_f)}. \tag{29} $$

   The final expression (29) agrees with Gelfand-Yaglom formula, cf. eqs. (14) & (26) in my other answer in this thread. The corresponding free theory has a constant zero-eigenmode $\phi^{(0)}_0(t)\equiv 1$, so that the free overlap is given by the formula $$\begin{align}\langle q_f\!=\!0, t_f | q_i\!=\!0,t_i \rangle^{(0)}~=~&{\rm Det}(\hat{H}^{(0)})^{-1/2}, \cr {\rm Det}\hat{H}^{(0)} ~=~&2\pi i\hbar \int_{t_i}^{t_f}\! \frac{dt}{m(t)}.\end{align} \tag{30} $$ Eq. (30) is consistent with eq. (28) and well-known Feynman amplitude/kernel for a free particle.

References:

1. R. Rajaraman, Solitons and Instantons: An Intro to Solitons and Instantons in QFT, 1987; Appendix A.

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Answer 4

1. Another application of the Gelfand-Yaglom formula is the van Vleck determinant:

   $$\begin{align}\langle q_f,& t_f | q_i,t_i \rangle \cr ~=~& \int_{q(t_i)=q_i}^{q(t_f)=q_f} \! {\cal D}q~\exp\left(\frac{i}{\hbar} S[q]\right)\cr ~\sim~&\sqrt{\det\left(\frac{-1}{2\pi i \hbar}\frac{\partial^2 S_{\rm cl}}{\partial q_f \partial q_i} \right)} \exp\left(\frac{i}{\hbar} S_{\rm cl}\right) \cr &\quad\text{for}\quad \hbar~\to~0, \end{align}\tag{1}$$

   where $$S[q]~:=~ \int_{t_i}^{t_f}\! dt ~ L(q(t),\dot{q}(t),t) \tag{2}$$ is the off-shell action functional, and $$ S_{\rm cl}~:=~S[q_{\rm cl}] \tag{3}$$ is the Dirichlet on-shell action function for a classical path $q_{\rm cl}:[t_i,t_f]\to \mathbb{R}$. (In this answer, we assume for simplicity that the classical path exists and is unique, i.e. no instantons.)

2. Example: The harmonic oscillator $$ L~=~\frac{m}{2}\dot{q}^2 -\frac{m}{2}\omega^2 q^2 \tag{4}$$ has classical path $$\begin{align} q_{\rm cl}(t)~=~&\frac{q_f\sin \omega (t-t_i)+q_i\sin \omega (t_f-t)}{\sin (\omega \Delta t)}, \cr \Delta t~:=~&t_f-t_i, \end{align}\tag{5} $$ on-shell action $$ S_{\rm cl}~\stackrel{(4)+(5)}{=}~m\omega\frac{(q_f^2+q_i^2)\cos(\omega\Delta t)-2q_fq_i}{2\sin(\omega\Delta t)}, \tag{6}$$ and Feynman amplitude/kernel $$\begin{align}\langle q_f,& t_f | q_i,t_i \rangle\cr ~\stackrel{(1)+(6)}{=}&~\sqrt{\frac{m\omega}{2\pi i \hbar\sin(\omega\Delta t)}} \exp\left(\frac{i}{\hbar} S_{\rm cl}\right).\end{align}\tag{7}$$ It is remarkable that the full quantum amplitude (7) can be derived from the classical on-shell action (6) alone!

3. Proof of eq. (1) for 1D. Firstly, expand the Lagrangian to quadratic order in fluctuations $q=q_{\rm cl}+y$: $$\begin{align}L(q,\dot{q},t)~=~&L(q_{\rm cl},\dot{q}_{\rm cl},t) + L_1 \cr &+ L_2 + {\cal O}(y^3),\end{align}\tag{8}$$ $$\begin{align} L_1~:=~& p_{\rm cl}(t)\dot{y}+F_{\rm cl}(t)y~\stackrel{\begin{array}{c}\text{int. by} \cr\text{parts}\end{array}}{\sim}~0, \cr p_{\rm cl}(t)~:=~&\left. \frac{\partial L}{\partial \dot{q}}\right|_{q=q_{\rm cl}(t)}, \cr F_{\rm cl}(t)~:=~&\left. \frac{\partial L}{\partial q}\right|_{q=q_{\rm cl}(t)},\end{align}\tag{9} $$ $$\begin{align} L_2~:=~~&\frac{m(t)}{2}\dot{y}^2+ b(t)y\dot{y} - \frac{k(t)}{2}y^2\cr ~\stackrel{\begin{array}{c}\text{int. by} \cr\text{parts}\end{array}}{\sim}&~ \frac{m(t)}{2}\dot{y}^2 - \frac{k(t)+\dot{b}(t)}{2}y^2 ,\end{align}\tag{10}$$ $$\begin{align} m(t)~:=~&\left. \frac{\partial^2 L}{\partial \dot{q}^2}\right|_{q=q_{\rm cl}(t)}, \cr b(t)~:=~&\left.\frac{\partial^2 L}{\partial q~\partial \dot{q}}\right|_{q=q_{\rm cl}(t)}, \cr k(t)~:=~&-\left.\frac{\partial^2 L}{\partial q^2}\right|_{q=q_{\rm cl}(t)}. \end{align}\tag{11}$$ In eq. (10) the $b$-term is integrated by parts. The boundary terms vanish because of Dirichlet boundary conditions (BCs) $y(t_i)=0=y(t_f)$. Secondly, expand the momentum to linear order in fluctuations $q=q_{\rm cl}+y$: $$\begin{align} p~:=~&\frac{\partial L}{\partial \dot{q}}\cr ~=~&p_{\rm cl}(t) +b(t)y+m(t)\dot{y} + {\cal O}(y^2).\end{align}\tag{12}$$

4. Next use the WKB/stationary phase approximation for $\hbar \to 0$: $$\begin{align}\langle q_f,& t_f | q_i,t_i \rangle \cr ~=~& \int_{q(t_i)=q_i}^{q(t_f)=q_f} \! {\cal D}q~\exp\left(\frac{i}{\hbar} S[q]\right)\cr ~\stackrel{\text{WKB}}{\sim}&~ {\rm Det}\hat{H}^{-1/2} \exp\left(\frac{i}{\hbar} S_{\rm cl}\right) \cr ~\stackrel{(29)}{=}&~\left(2\pi i\hbar \phi_0(t_i)\psi_0(t_f) \right)^{-1/2}\exp\left(\frac{i}{\hbar} S_{\rm cl}\right),\end{align}\tag{13}$$ where the Hessian operator reads $$ \hat{H}~:=~\frac{d}{dt}m(t)\frac{d}{dt}+k(t) +\dot{b}(t).\tag{14} $$ In the last equality of eq. (13) was used eq. (29) from my other answer in this thread. Here $\phi_0$ is a zero-mode with $\phi_0(t_i)\neq 0$, and $$\begin{align} \psi_0(t)~:=~&\phi_0(t)\int_{t_i}^t\! \frac{dt^{\prime}}{m(t)\phi_0(t^{\prime})^2},\cr \psi_0(t)~=~&0,\end{align} \tag{15}$$ is an independent zero-mode, cf. the Gelfand-Yaglom formula. Note for later that the Wronskian is $$\begin{align} W(\phi_0,\psi_0) ~:=~&\phi_0\dot{\psi}_0-\dot{\phi}_0\psi_0\cr~=~&\frac{1}{m(t)}.\end{align} \tag{16}$$

5. On the other hand, the final momentum $p_f$ can be found from the on-shell formula $$ p_f ~=~ \frac{\partial S_{\rm cl}}{\partial q_f},\tag{17} $$ see e.g. eq. (11) in my Phys.SE answer here. Therefore the $1\times 1$ van Vleck matrix can be found $$ \frac{\partial^2 S_{\rm cl}}{\partial q_f \partial q_i} ~\stackrel{(17)}{=}~\frac{\partial p_f}{ \partial q_i}\tag{18}$$ by varying infinitesimally the initial position $\delta q_i= y(t_i)$ for fixed final position $\delta q_f= y(t_f)=0$, and such that the new path $q=q_{\rm cl}+y$ is also a classical solution. The EL eq. for the new path $q=q_{\rm cl}+y$ (i.e. the linearized EL eq. for $y$) implies that the infinitesimal variation $y$ is a zero-mode $\hat{H}y=0$, i.e. a linear combination $$ y(t)~=~A\phi_0(t)+ B\psi_0(t),\tag{19} $$ where $A$ & $B$ are 2 infinitesimal constants determined by the Dirichlet BCs: $$\begin{align}\delta q_i~=~~~& y(t_i)\cr ~\stackrel{(15)+(19)}{=}&~A\phi_0(t_i)\cr\qquad\Downarrow&\qquad\cr A~=~&\frac{\delta q_i}{\phi_0(t_i)} ,\end{align}\tag{20} $$ $$\begin{align}0~=~&\delta q_f\cr ~=~ &y(t_f)\cr ~\stackrel{(19)}{=}~& A\phi_0(t_f)+B\psi_0(t_f) \cr\qquad\Downarrow&\qquad\cr B~=~&-A\frac{\phi_0(t_f)}{\psi_0(t_f)}\cr ~\stackrel{(20)}{=}~& -\frac{\delta q_i}{\phi_0(t_i)}\frac{\phi_0(t_f)}{\psi_0(t_f)} .\end{align}\tag{21} $$ The change in the final momentum is $$\begin{align}\delta p_f ~\stackrel{(12)}{=}~&m(t_f) \dot{y}(t_f)\cr ~\stackrel{(19)}{=}~&m(t_f)\left(A\dot{\phi}_0(t_f)+B\dot{\psi}_0(t_f)\right)\cr ~\stackrel{(21)}{=}~&m(t_f)A\left(\dot{\phi}_0(t_f)-\frac{\phi_0(t_f)}{\psi_0(t_f)}\dot{\psi}_0(t_f)\right)\cr ~\stackrel{(16)}{=}~&-\frac{A}{\psi_0(t_f)}\cr ~\stackrel{(20)}{=}~&-\frac{\delta q_i}{\phi_0(t_i) \psi_0(t_f)}.\end{align}\tag{22}$$ Therefore $$ \frac{\partial^2 S_{\rm cl}}{\partial q_f \partial q_i} ~\stackrel{(18)+(22)}{=}~ -\frac{1}{\phi_0(t_i) \psi_0(t_f)}.\tag{23}$$ Comparing eqs. (13) & (23) yields the sought-for van Vleck formula (1). $\Box$

References:

1. B.S. DeWitt, The Global Approach to QFT, Vol 1, 2003; Chapter 14.

2. H. Kleinert, Path Integrals in QM, Statistics, Polymer Physics, and Financial Markets, 5th ed.; Section 2.4.

3. M. Blau, Notes for (semi-)advanced QM: The Path Integral Approach to QM; App. C.

4. R. Rattazzi, Lecture notes for QM IV: The Path Integral approach to QM; Section 3.1.