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?


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 $$f(\lambda)~=~(\lambda-\lambda_n)g_n(\lambda), \qquad g_n(\lambda_n)~\neq~ 0.\tag{2}$$ We shall later see how one in practice can construct such $f$-function, cf. eqs. (16) & (26) below. The function$^1$ $$({\rm Ln} f)^{\prime}(\lambda)~=~\frac{f^{\prime}(\lambda)}{f(\lambda)}~\sim~\frac{1}{\lambda-\lambda_n}+ \text{regular terms}\tag{3}$$ 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 $$ \zeta_H(s)~=~\sum_{n\in\mathbb{N}} \lambda_n^{-s} ~\stackrel{(4)}{=}~\int_{\gamma_+}\!\frac{d\lambda}{2\pi i} \exp\left(-s{\rm Ln}\lambda\right)~({\rm Ln} f)^{\prime}(\lambda) ,\tag{5}$$ $$ -\zeta^{\prime}_H(s)~\stackrel{(5)}{=}~ \sum_{n\in\mathbb{N}} \lambda_n^{-s}~{\rm Ln}\lambda_n ,\tag{6}$$ 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.)

  1. 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 $s=0$: $${\rm Ln} {\rm Det} H~=~{\rm Ln} \prod_{n\in\mathbb{N}}\lambda_n ~=~\sum_{n\in\mathbb{N}} {\rm Ln} \lambda_n ~\stackrel{(6)}{=}~ -\zeta^{\prime}_H(s=0) .\tag{8} $$ 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: $$ \zeta_H(s)-\zeta_{H^{(0)}}(s) ~\stackrel{(5)}{=}~\int_{\gamma_+}\!\frac{d\lambda}{2\pi i} \exp\left(-s{\rm Ln}\lambda\right)~({\rm Ln} \frac{f}{f^{(0)}})^{\prime}(\lambda).\tag{10}$$

$\uparrow$ 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 $(\lambda^{(0)}_n)_{n\in\mathbb{N}}$.

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

  2. 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 $$ H\psi_{\lambda}~=~\lambda\psi_{\lambda}, \qquad \psi_{\lambda}(x=a)~=~0,$$ $$\qquad \psi^{\prime}_{\lambda}(x=a)~=~C~=~\text{some fixed constant}.\tag{17}$$

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

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


  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.


$^1$ ${\rm Ln}$ 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 $$ \prod_{n\in \mathbb{N}} a~=~a^{\zeta(0)}~=~\frac{1}{\sqrt{a}}, \qquad \prod_{n\in \mathbb{N}} n~=~e^{-\zeta^{\prime}(0)}~=~\sqrt{2\pi}, \tag{27} $$ $$ \prod_{n\in \mathbb{N}} \left[1-\left(\frac{a}{n}\right)^2 \right]~=~\frac{\sin \pi a}{\pi a}, \qquad \prod_{n\in \mathbb{N}} \left[1+\left(\frac{n}{a}\right)^2 \right]~=~2\sinh \pi a, \tag{28} $$ via analytic continuation of the Riemann zeta function $$\zeta(s)~=~\sum_{n\in \mathbb{N}}n^{-s}, \qquad {\rm Re}(s) ~>~1.\tag{29}$$

| cite | improve this answer | |

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.

| cite | improve this answer | |
  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 $$ S~=~ \int_{t_i}^{t_f}\! dt~L, \qquad L~=~\frac{m(t)}{2}\dot{q}^2-V, \qquad V~=~\frac{k(t)}{2}q^2 , \tag{1}$$ 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
    $$\begin{align}\langle q_f\!=\!0, t_f | q_i\!=\!0,t_i \rangle ~=~& \int_{q(t_i)=0}^{q(t_f)=0} \! {\cal D}q~\exp\left(\frac{i}{\hbar} S\right), \qquad\qquad {\cal D}q~\sim~\prod_{t_i <t< t_f} dq(t) , \cr~\stackrel{\begin{array}{c}\text{int. by} \cr\text{parts}\end{array}}{=}&~ \int_{q(t_i)=0}^{q(t_f)=0} \! {\cal D}q~\exp\left(-\frac{i}{\hbar}\int\! dt~ q(t) ~\hat{H} q(t)\right)\cr ~\stackrel{\begin{array}{c}\text{Wick.} \cr\text{rot.}\end{array}}{=}&~ \int \! {\cal D}q\exp\left[-\frac{1}{2\hbar}\iint_{[\tau_i,\tau_f]^2} d\tau~d\tau^{\prime} ~q(\tau)H(\tau,\tau^{\prime})q(\tau^{\prime}) \right]\cr ~\stackrel{\begin{array}{c}\text{Gauss.} \cr\text{ int.}\end{array}}{=}&~ {\rm Det}\hat{H}^{-1/2}\end{align}\tag{2}$$ 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 $$\hat{H}~:=~ \underbrace{\frac{d}{dt}m(t)\frac{d}{dt}}_{~=:~\hat{H}^{(0)}}+k(t)~\stackrel{(3)}{=}~-\frac{d}{d\tau}m(\tau)\frac{d}{d\tau}+k(\tau)~>~0\tag{4}$$ 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 $$\hat{H}\phi_0~=~0, \qquad \phi_0(t=t_i)~\neq~ 0. \tag{6}$$ Introduce for later convenience the shorthand notation $$ \Phi_0~:=~{\rm Ln}\phi_0, \qquad \dot{\Phi}_0~=~\frac{\dot{\phi}_0}{\phi_0}. \tag{7}$$
    Then the potential term (1) can be integrated by parts: $$\begin{align} V~\stackrel{(1)}{=}~&\frac{k(t)}{2}q^2\cr ~\stackrel{(6)}{=}~&-\frac{q^2}{2\phi_0} \frac{d(m(t)\dot{\phi}_0)}{dt}\cr ~\stackrel{(7)}{=}~&m(t)\dot{\Phi}_0q\dot{q}- \frac{m(t)}{2}\dot{\Phi}_0^2q^2 -\frac{d}{dt}\left(\frac{m(t)}{2}\dot{\Phi}_0q^2\right).\end{align} \tag{8}$$ Note that the total derivative term (8) vanishes due to the Dirichlet boundary conditions (BCs). The action (1) becomes $$ S~\stackrel{(1)+(8)}{=}~\int_{t_i}^{t_f}\! dt~L^{\prime} ,\qquad L^{\prime}~=~\frac{m(t)}{2} \left(\dot{q}- \dot{\Phi}_0q\right)^2. \tag{9}$$

  3. Now perform a non-local coordinate transformation $$Q(t)~=~q(t)-\int_{t_i}^{t_f}\! dt^{\prime} ~\theta(t-t^{\prime})~ \dot{\Phi}_0(t^{\prime})q(t^{\prime}), \tag{10}$$ so that $$ \dot{Q}~\stackrel{(10)}{=}~\dot{q} - \dot{\Phi}_0q ~\stackrel{(7)}{=}~ \phi_0\frac{d}{dt}\left(\frac{q}{\phi_0}\right) \tag{11}$$ 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 $$\frac{\delta Q(t)}{\delta q(t^{\prime})} ~\stackrel{(10)}{=}~\delta(t-t^{\prime}) - B(t,t^{\prime}), \qquad B(t,t^{\prime}) ~:=~\theta(t-t^{\prime})~ \dot{\Phi}_0(t^{\prime}), \tag{13}$$ 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 $$0~\approx~q(t_f)~\stackrel{(19)}{=}~\phi_0(t_f)\int_{t_i}^{t_f}\! dt ~\frac{\dot{Q}(t)}{\phi_0(t)} \tag{20}$$ with a Lagrange multiplier $\lambda$. The new action becomes $$ S^{\prime}~=~S+\lambda q(t_f)~\stackrel{(12)+(20)}{=}~ \int_{t_i}^{t_f}\! dt~L^{\prime\prime} ,\qquad L^{\prime\prime}~=~\frac{m(t)}{2}\dot{Q}^2 + \lambda \phi_0(t_f) \frac{\dot{Q}}{\phi_0}, \tag{21}$$ and the Feynman amplitude/kernel/path integral becomes $$\begin{align}\langle q_f\!=\!0, t_f | q_i\!=\!0,t_i \rangle ~=~& \int_{q(t_i)=0} \! {\cal D}q~\frac{d\lambda}{2\pi\hbar}\exp\left(\frac{i}{\hbar} S^{\prime}\right)\cr ~=~& J\int_{Q(t_i)=0} \! {\cal D}Q~\frac{d\lambda}{2\pi\hbar}\exp\left(\frac{i}{\hbar} S^{\prime}\right).\end{align} \tag{22} $$

  6. Next perform a second coordinate transformation $$\tilde{q}(t)~=~Q(t)+ \lambda\phi_0(t_f) \int_{t_i}^{t_f}\! dt^{\prime} ~\theta(t-t^{\prime})~ \frac{1}{m(t^{\prime})\phi_0(t^{\prime})} , \tag{23}$$ 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 $$ S^{\prime}~\stackrel{(21)+(24)}{=}~ \int_{t_i}^{t_f}\! dt~L^{\prime\prime\prime} -\frac{\lambda^2\phi_0(t_f)^2}{2} \int_{t_i}^{t_f}\! \frac{dt}{m(t)\phi_0(t)^2}, \qquad L^{\prime\prime\prime}~=~\frac{m(t)}{2}\dot{\tilde{q}}^2 .\tag{25}$$ Note that both coordinate transformations (10) and (23) do not change the initial Dirichlet BC $$q(t_i)~\approx~0 \quad\stackrel{(10)}{\Leftrightarrow}\quad Q(t_i)~\approx~0 \quad\stackrel{(23)}{\Leftrightarrow}\quad {\tilde{q}}(t_i)~\approx~0, \tag{26}$$ 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 $$\begin{align}&\langle q_f\!=\!0, t_f | q_i\!=\!0,t_i \rangle\cr ~\stackrel{(22)}{=}~&J\int_{\tilde{q}(t_i)=0} \! {\cal D}{\tilde{q}}~\frac{d\lambda}{2\pi\hbar}~\exp\left(\frac{i}{\hbar} S^{\prime}\right)\cr ~\stackrel{(25)}{=}~&J\left( 2\pi i\hbar ~\phi_0(t_f)^2 \int_{t_i}^{t_f}\! \frac{dt}{m(t)\phi_0(t)^2} \right)^{-1/2} \int_{\tilde{q}(t_i)=0} \! {\cal D}\tilde{q}~\exp\left(\frac{i}{\hbar} \int_{t_i}^{t_f}\! dt~L^{\prime\prime\prime}\right)\cr ~\stackrel{(18)}{=}~&\left( 2\pi i\hbar ~\phi_0(t_i)\phi_0(t_f)\int_{t_i}^{t_f}\! \frac{dt}{m(t)\phi_0(t)^2} \right)^{-1/2} \underbrace{ \int \! d\tilde{q}_f~\langle \tilde{q}_f, t_f | \tilde{q}_i\!=\!0,t_i \rangle^{(0)}}_{~=~1.} .\end{align} \tag{27} $$ 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 $$\begin{align}\langle q_f, t_f |q_i,t_i \rangle^{(0)} ~=~& \left(2\pi i \hbar\int_{t_i}^{t_f}\! \frac{dt}{m(t)}\right)^{-1/2} \exp\left( \frac{i}{2\hbar} \frac{(\Delta q)^2}{\int_{t_i}^{t_f}\! \frac{dt}{m(t)}} \right), \cr \Delta q~:=~&q_f-q_i,\end{align}\tag{28}$$ 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 $$\langle q_f\!=\!0, t_f | q_i\!=\!0,t_i \rangle^{(0)}~=~{\rm Det}(\hat{H}^{(0)})^{-1/2}, \qquad {\rm Det}\hat{H}^{(0)} ~=~2\pi i\hbar \int_{t_i}^{t_f}\! \frac{dt}{m(t)}. \tag{30} $$ Eq. (30) is consistent with eq. (28) and well-known Feynman amplitude/kernel for a free particle.


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


| cite | improve this answer | |
  1. Another application of the Gelfand-Yaglom formula is the van Vleck determinant:

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

    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 $$ q_{\rm cl}(t)~=~\frac{q_f\sin \omega (t-t_i)+q_i\sin \omega (t_f-t)}{\sin (\omega \Delta t)}, \qquad \Delta t~:=~t_f-t_i, \tag{5} $$ 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 $$\langle q_f, t_f | q_i,t_i \rangle ~\stackrel{(1)+(6)}{=}~\sqrt{\frac{m\omega}{2\pi i \hbar\sin(\omega\Delta t)}} \exp\left(\frac{i}{\hbar} S_{\rm cl}\right).\tag{7}$$ 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 $q=q_{\rm cl}+y$: $$L(q,\dot{q},t)~=~L(q_{\rm cl},\dot{q}_{\rm cl},t) + L_1 + L_2 + {\cal O}(y^3),\tag{8}$$ $$ L_1~:=~ p_{\rm cl}(t)\dot{y}+F_{\rm cl}(t)y~\stackrel{\begin{array}{c}\text{int. by} \cr\text{parts}\end{array}}{\sim}~0, \quad p_{\rm cl}(t)~:=~\left. \frac{\partial L}{\partial \dot{q}}\right|_{q=q_{\rm cl}(t)}, \quad F_{\rm cl}(t)~:=~\left. \frac{\partial L}{\partial q}\right|_{q=q_{\rm cl}(t)},\tag{9} $$ $$ L_2~:=~\frac{m(t)}{2}\dot{y}^2+ b(t)y\dot{y} - \frac{k(t)}{2}y^2 ~\stackrel{\begin{array}{c}\text{int. by} \cr\text{parts}\end{array}}{\sim}~ \frac{m(t)}{2}\dot{y}^2 - \frac{k(t)+\dot{b}(t)}{2}y^2 ,\tag{10}$$ $$ m(t)~:=~\left. \frac{\partial^2 L}{\partial \dot{q}^2}\right|_{q=q_{\rm cl}(t)}, \qquad b(t)~:=~\left.\frac{\partial^2 L}{\partial q~\partial \dot{q}}\right|_{q=q_{\rm cl}(t)}, \qquad k(t)~:=~-\left.\frac{\partial^2 L}{\partial q^2}\right|_{q=q_{\rm cl}(t)}. \tag{11}$$ 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 $q=q_{\rm cl}+y$: $$ p~:=~\frac{\partial L}{\partial \dot{q}}~=~p_{\rm cl}(t) +b(t)y+m(t)\dot{y} + {\cal O}(y^2).\tag{12}$$

  4. Next use the WKB/stationary phase approximation for $\hbar \to 0$: $$\begin{align}\langle q_f, t_f | q_i,t_i \rangle ~=~& \int_{q(t_i)=q_i}^{q(t_f)=q_f} \! {\cal D}q~\exp\left(\frac{i}{\hbar} S[q]\right)\cr ~\stackrel{\text{WKB}}{\sim}&~ {\rm Det}\hat{H}^{-1/2} \exp\left(\frac{i}{\hbar} S_{\rm cl}\right) \cr ~\stackrel{(29)}{=}&~\left(2\pi i\hbar \phi_0(t_i)\psi_0(t_f) \right)^{-1/2}\exp\left(\frac{i}{\hbar} S_{\rm cl}\right),\end{align}\tag{13}$$ 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 $$ \psi_0(t)~:=~\phi_0(t)\int_{t_i}^t\! \frac{dt^{\prime}}{m(t)\phi_0(t^{\prime})^2},\qquad \psi_0(t)~=~0, \tag{15}$$ is an independent zero-mode, cf. the Gelfand-Yaglom formula. Note for later that the Wronskian is $$ W(\phi_0,\psi_0)~:=~\phi_0\dot{\psi}_0-\dot{\phi}_0\psi_0~=~\frac{1}{m(t)}. \tag{16}$$

  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: $$\delta q_i~=~ y(t_i)~\stackrel{(15)+(19)}{=}~A\phi_0(t_i) \qquad\Rightarrow\qquad A~=~\frac{\delta q_i}{\phi_0(t_i)} ,\tag{20} $$ $$0~=~\delta q_f~=~ y(t_f)~\stackrel{(19)}{=}~A\phi_0(t_f)+B\psi_0(t_f) $$ $$ \qquad\Rightarrow\qquad B~=~-A\frac{\phi_0(t_f)}{\psi_0(t_f)} ~\stackrel{(20)}{=}~-\frac{\delta q_i}{\phi_0(t_i)}\frac{\phi_0(t_f)}{\psi_0(t_f)} .\tag{21} $$ 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). $\Box$


  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.

| cite | improve this answer | |
  • $\begingroup$ Notes for later: $\frac{\partial^2 S_{\rm cl}}{\partial t_f \partial t_i}~=~-\dot{q}_f\frac{\partial^2 S_{\rm cl}}{\partial q_f \partial t_i}~=~-\frac{\partial^2 S_{\rm cl}}{\partial t_f \partial q_i}\dot{q}_i~=~\dot{q}_f\frac{\partial^2 S_{\rm cl}}{\partial q_f \partial q_i}\dot{q}_i$ $\endgroup$ – Qmechanic May 25 '18 at 17:55
  • $\begingroup$ Notes for later: Initial value problem: $\quad q_{\rm cl}(t)~=~\frac{v_i}{\omega}\sin \omega (t-t_i)+q_i\cos \omega (t-t_i)$; $\quad \dot{q}_{\rm cl}(t)~=~v_i\cos \omega (t-t_i)-q_i\omega\sin\omega (t-t_i)$; $\quad E~=~\frac{m}{2}(v_i^2 +\omega^2 q_i^2)$; $\quad L~=~\frac{m}{2}(v_i^2 -\omega^2 q_i^2)\cos 2\omega (t-t_i) -m\omega q_iv_i\sin 2\omega (t-t_i)$; $\quad S~=~\frac{m}{4\omega}(v_i^2 -\omega^2 q_i^2)\sin 2\omega \Delta t -m q_iv_i\sin^2\omega \Delta t$; $\endgroup$ – Qmechanic Aug 5 '18 at 17:31
  • $\begingroup$ Notes for later: $\quad q_f~=~\frac{v_i}{\omega}\sin \omega \Delta t + q_i\cos \omega \Delta t $; $\quad \Delta t~:=~t_f-t_i$; Non-unique path/Eigenmode only possible for $\Delta t=\frac{nT}{2}$, $n\in\mathbb{Z}$, and $q_f=(-1)^n q_i$. This corresponds to the $n$'th energy eigenmode, and is a zero-mode for the action. $\endgroup$ – Qmechanic Aug 6 '18 at 9:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.