operators - Gelfand-Yaglom theorem for functional determinants

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

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.)

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 $s=0$: $${\rm Ln} {\rm Det} H~=~{\rm Ln} \prod_{n\in\mathbb{N}}\lambda_n ~=~\sum_{n\in\mathbb{N}} {\rm Ln} \lambda_n ~\stackrel{(6)}{=}~ -\zeta^{\prime}_H(s=0) .\tag{8}$$ 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 $(\lambda^{(0)}_n)_{n\in\mathbb{N}}$.

4. 
   

   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.

   
   



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


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

   $$\lambda_n~=~\lambda^{(0)}_n+V_0, \qquad\lambda^{(0)}_n~=~\frac{(\pi\hbar n)^2}{2m_0(b-a)^2}, \qquad n~\in~\mathbb{N}.\tag{18}$$ 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).
   
   


7. 
   

   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).
   
   

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

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