A Stuck on evaluating this functional determinant

[TroyElliott]

I am trying to show that given the following stochastic differential equation: $\dot{x} = W(x(\tau))+\eta(\tau),$ we have

$det|\frac{d\eta(\tau)}{dx(\tau')}| = exp^{\int_{0}^{T}d\tau \,Tr \ln([\frac{d}{d\tau}-W'(x(\tau))]\delta (\tau - \tau'))} = exp^{\frac{1}{2}\int_{0}^{T}d\tau W'(x(\tau))}.$

Attempt at the solution: We can write the differential equation as $\eta(\tau) = \dot{x} - W(x(\tau)),$ and then take the function derivative with respect to $x(\tau')$ to get $\frac{d\eta(\tau)}{dx(\tau')} = (\frac{d}{d\tau}-W')\delta(\tau - \tau')$. Next we can use the identity $det(exp^{M}) = exp^{Tr(M)}$ to write $det|\frac{d\eta(\tau)}{dx(\tau')}| = exp^{Tr \, \ln(\frac{d\eta(\tau)}{dx(\tau')})} = exp^{Tr \, \ln[(\frac{d}{d\tau}-W')\delta(\tau - \tau')]}.$

This is where I am stuck at. I don't see how the integral arises in the first equality in the above equation, and furthermore I do not see how to evaluate this integral to end up with the second equality in the above equation. Any insight would be greatly appreciated. Thanks!

 

[samalkhaiat]

I don't see how the integral arises in the first equality in the above equation, and furthermore I do not see how to evaluate this integral to end up with the second equality in the above equation. Any insight would be greatly appreciated. Thanks!

The integral arises from the definition of the trace of infinite-dimensional matrices. A function A( \tau , \sigma) of two real variables can be regarded as \infty \times \infty matrix A. The trace of such matrix is defined by \mbox{Tr}(A) = \int d \tau \ A(\tau , \tau) . And the trace of a product of two such matrices, say A^{-1}B, is given by \mbox{Tr}(A^{-1}B) = \int d \tau d \sigma \ A^{-1}(\tau , \sigma)B (\sigma , \tau ) , \ \ \ \ \ \ \ \ (1) where the inverse matrix A^{-1} is defined by \int d \lambda \ A( \tau , \lambda )A^{-1} (\lambda , \sigma ) = \delta ( \tau - \sigma ) . \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
Okay, I must say that I am not an expert on stochastic processes, but let’s see what we can do about your problem. I write your functional derivative as \frac{\delta \eta (\tau)}{\delta x (\sigma)} = \frac{\partial}{\partial \tau} \delta ( \tau - \sigma ) - \frac{\partial W}{\partial x} \delta ( \tau - \sigma ) . \ \ \ \ \ (3) Now, if we define the matrices M(\tau , \sigma ) = \frac{\delta \eta (\tau)}{\delta x (\sigma)} ,A(\tau , \sigma ) = \frac{\partial}{\partial \tau} \delta ( \tau - \sigma ) , \ \ \ \ \ (4)B( \tau , \sigma ) = \frac{\partial W}{\partial x} \delta ( \tau - \sigma ) , \ \ \ \ \ \ (5) then we can write (3) as M = A ( 1 - A^{-1}B ). Thus \mbox{det}(M) = \mbox{det}(A) \ \mbox{det} ( 1 - A^{-1}B ) . Notice that all dynamical information is contained in the matrix A^{-1}B. And since A does not depend on trajectory x(\tau), we may include \mbox{det}(A) in the definition of an overall normalisation constant (if you have done path integral, you know what I am talking about). So, we may write \mbox{det}(M) = C \ \mbox{det} ( 1 - A^{-1}B ) = C \exp \left( \mbox{Tr} \ln (1 - A^{-1}B)\right) . Expanding the function \ln (1 - y), we get \mbox{det}(M) = C \exp \left( - \sum_{k = 1} \frac{1}{k} \mbox{Tr}(A^{-1}B)^{k} \right) , or \mbox{det}(M) = C \exp \left( - \mbox{Tr}(A^{-1}B) - \frac{1}{2} \mbox{Tr}(A^{-1}B)^{2} - \cdots \ \right) . \ \ \ \ (6) This is our final result. To make sense of (6), we need to use our definitions (4) and (5) and evaluate the all traces in (6). First, substituting (4) in (2), we find \frac{\partial}{\partial \tau} A^{-1}( \tau , \sigma ) = \delta ( \tau - \sigma ) . This means that A^{-1} is the step function [recall the identity \theta^{\prime}(x) = \delta (x)] A^{-1}( \tau , \sigma ) = \theta ( \tau - \sigma ) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (7) Now, by substituting (7) and (5) in (1), we find \mbox{Tr} (A^{-1}B) = \int d \tau d \sigma \ \theta ( \tau - \sigma) \delta (\sigma - \tau) \frac{\partial W}{\partial x} ( \sigma ) = \theta (0) \int d \tau \frac{\partial W}{\partial x} . Since the step function \theta ( \tau ) = +1 for \tau > 0, and \theta (\tau) = 0 for \tau < 0, it is reasonable to take \theta (0) = \frac{1}{2}. Indeed, this follows from the following regularized form of the step function \theta ( \tau ) = \lim_{\lambda \to 0} \left( \frac{1}{2} + \frac{1}{\pi} \tan^{-1} \left(\frac{\tau}{\lambda} \right)\right) . So, we have found our first trace \mbox{Tr} (A^{-1}B) = \frac{1}{2} \int d \tau \ \frac{\partial W}{\partial x} . \ \ \ \ (8) In fact, one can show that (8) is the only non-zero trace, i.e., \mbox{Tr}(A^{-1}B)^{n} = 0, for all n > 1. To make the equation shorter, I will prove that only for n = 2 case, because the general case is exactly the same:
\begin{align*}\mbox{Tr}(A^{-1}B)^{2} &= \int d \tau d \sigma \ (A^{-1}B)( \tau , \sigma ) \ (A^{-1}B) ( \sigma , \tau ) \\ &= \int d \tau d \sigma \left( \int d \lambda \ A^{-1}( \tau , \lambda ) B( \lambda , \sigma )\right) \left( \int d \rho \ A^{-1}( \sigma , \rho ) B( \rho , \tau )\right) . \end{align*} Now, if you substitute for B's their corresponding definition from (5), and do the delta’s integrations, you find \mbox{Tr}(A^{-1}B)^{2} = \int d \tau d \sigma \ A^{-1}( \tau , \sigma ) \ A^{-1}( \sigma , \tau ) \left( \frac{\partial W}{\partial x} \right)^{2} . Substituting for the A^{-1}’s their corresponding step functions from (7), we get \mbox{Tr}(A^{-1}B)^{2} = \int d \tau d \sigma \ \theta ( \tau - \sigma) \theta ( \sigma - \tau ) \left( \frac{\partial W}{\partial x}\right)^{2} . But the step functions are non-zero only if \tau > \sigma > \tau which is non-sense. Thus the integral must vanish. Technically speaking, when the integrand has support on a set of zero measure, the integral vanishes (because in this case there is no delta function to help us out). So, our final result (6) becomes [after using (8)]\mbox{det} \left( \frac{\delta \eta}{\delta x}\right) = C \ e^{- \frac{1}{2} \int d \tau \ \frac{\partial W}{\partial x}} .

 

[TroyElliott]

The integral arises from the definition of the trace of infinite-dimensional matrices. A function A( \tau , \sigma) of two real variables can be regarded as \infty \times \infty matrix A. The trace of such matrix is defined by \mbox{Tr}(A) = \int d \tau \ A(\tau , \tau) . And the trace of a product of two such matrices, say A^{-1}B, is given by \mbox{Tr}(A^{-1}B) = \int d \tau d \sigma \ A^{-1}(\tau , \sigma)B (\sigma , \tau ) , \ \ \ \ \ \ \ \ (1) where the inverse matrix A^{-1} is defined by \int d \lambda \ A( \tau , \lambda )A^{-1} (\lambda , \sigma ) = \delta ( \tau - \sigma ) . \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
Okay, I must say that I am not an expert on stochastic processes, but let’s see what we can do about your problem. I write your functional derivative as \frac{\delta \eta (\tau)}{\delta x (\sigma)} = \frac{\partial}{\partial \tau} \delta ( \tau - \sigma ) - \frac{\partial W}{\partial x} \delta ( \tau - \sigma ) . \ \ \ \ \ (3) Now, if we define the matrices M(\tau , \sigma ) = \frac{\delta \eta (\tau)}{\delta x (\sigma)} ,A(\tau , \sigma ) = \frac{\partial}{\partial \tau} \delta ( \tau - \sigma ) , \ \ \ \ \ (4)B( \tau , \sigma ) = \frac{\partial W}{\partial x} \delta ( \tau - \sigma ) , \ \ \ \ \ \ (5) then we can write (3) as M = A ( 1 - A^{-1}B ). Thus \mbox{det}(M) = \mbox{det}(A) \ \mbox{det} ( 1 - A^{-1}B ) . Notice that all dynamical information is contained in the matrix A^{-1}B. And since A does not depend on trajectory x(\tau), we may include \mbox{det}(A) in the definition of an overall normalisation constant (if you have done path integral, you know what I am talking about). So, we may write \mbox{det}(M) = C \ \mbox{det} ( 1 - A^{-1}B ) = C \exp \left( \mbox{Tr} \ln (1 - A^{-1}B)\right) . Expanding the function \ln (1 - y), we get \mbox{det}(M) = C \exp \left( - \sum_{k = 1} \frac{1}{k} \mbox{Tr}(A^{-1}B)^{k} \right) , or \mbox{det}(M) = C \exp \left( - \mbox{Tr}(A^{-1}B) - \frac{1}{2} \mbox{Tr}(A^{-1}B)^{2} - \cdots \ \right) . \ \ \ \ (6) This is our final result. To make sense of (6), we need to use our definitions (4) and (5) and evaluate the all traces in (6). First, substituting (4) in (2), we find \frac{\partial}{\partial \tau} A^{-1}( \tau , \sigma ) = \delta ( \tau - \sigma ) . This means that A^{-1} is the step function [recall the identity \theta^{\prime}(x) = \delta (x)] A^{-1}( \tau , \sigma ) = \theta ( \tau - \sigma ) . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (7) Now, by substituting (7) and (5) in (1), we find \mbox{Tr} (A^{-1}B) = \int d \tau d \sigma \ \theta ( \tau - \sigma) \delta (\sigma - \tau) \frac{\partial W}{\partial x} ( \sigma ) = \theta (0) \int d \tau \frac{\partial W}{\partial x} . Since the step function \theta ( \tau ) = +1 for \tau > 0, and \theta (\tau) = 0 for \tau < 0, it is reasonable to take \theta (0) = \frac{1}{2}. Indeed, this follows from the following regularized form of the step function \theta ( \tau ) = \lim_{\lambda \to 0} \left( \frac{1}{2} + \frac{1}{\pi} \tan^{-1} \left(\frac{\tau}{\lambda} \right)\right) . So, we have found our first trace \mbox{Tr} (A^{-1}B) = \frac{1}{2} \int d \tau \ \frac{\partial W}{\partial x} . \ \ \ \ (8) In fact, one can show that (8) is the only non-zero trace, i.e., \mbox{Tr}(A^{-1}B)^{n} = 0, for all n > 1. To make the equation shorter, I will prove that only for n = 2 case, because the general case is exactly the same:
\begin{align*}\mbox{Tr}(A^{-1}B)^{2} &= \int d \tau d \sigma \ (A^{-1}B)( \tau , \sigma ) \ (A^{-1}B) ( \sigma , \tau ) \\ &= \int d \tau d \sigma \left( \int d \lambda \ A^{-1}( \tau , \lambda ) B( \lambda , \sigma )\right) \left( \int d \rho \ A^{-1}( \sigma , \rho ) B( \rho , \tau )\right) . \end{align*} Now, if you substitute for B's their corresponding definition from (5), and do the delta’s integrations, you find \mbox{Tr}(A^{-1}B)^{2} = \int d \tau d \sigma \ A^{-1}( \tau , \sigma ) \ A^{-1}( \sigma , \tau ) \left( \frac{\partial W}{\partial x} \right)^{2} . Substituting for the A^{-1}’s their corresponding step functions from (7), we get \mbox{Tr}(A^{-1}B)^{2} = \int d \tau d \sigma \ \theta ( \tau - \sigma) \theta ( \sigma - \tau ) \left( \frac{\partial W}{\partial x}\right)^{2} . But the step functions are non-zero only if \tau > \sigma > \tau which is non-sense. Thus the integral must vanish. Technically speaking, when the integrand has support on a set of zero measure, the integral vanishes (because in this case there is no delta function to help us out). So, our final result (6) becomes [after using (8)]\mbox{det} \left( \frac{\delta \eta}{\delta x}\right) = C \ e^{- \frac{1}{2} \int d \tau \ \frac{\partial W}{\partial x}} .

Thank you. I really appreciate you taking your time to write out such a detailed solution!