5
$\begingroup$

Given a vector function $$f=(f_1,\ldots,f_n)\in L^2(\mathbb R,\mathbb R^n)$$ (for some $n\in\mathbb N$), let us define $$\Delta f:=(\Delta f_1,\ldots,\Delta f_n),$$ where $\Delta$ is the Laplacian operator, and let $Q:\mathbb R\to\mathbb R^{n\times n}$ be a potential taking values in symmetric $n\times n$ matrices. I'm interested in vector Schrödinger operators of the form $$Hf=-\Delta f+Qf,\qquad f\in L^2(\mathbb R,\mathbb R^n).$$

Question. Is there a Feynman-Kac type formula known for $H$'s semigroup in the case where $Q(x)$ is not necessarily diagonal?

(Note: I specify abote that I'm interested in the case where $Q$ is not diagonal; if $Q=\mathrm{diag}(Q_1,\ldots,Q_n)$, then $$Hf=(-\Delta f_1+Q_1 f_1,\ldots,-\Delta f_1+Q_1 f_1),$$ in which case we can simply apply the one-dimensional Feynman-Kac formula to each component.)

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
2
$\begingroup$

It's the usual formula with the exponential of $Q$ replaced by a time-ordered exponential --- needed in the integral over $t$ since $Q[x(t)]$ and $Q[x(t')]$ do not commute for $t\neq t'$. One reference where this "chronological'" integral is worked out is Equivalence of Two Definitions of a Chronological Integral.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ How would the fomula be modified for the more general case where $Hf=-\Delta f + R\nabla f + Qf$, where $R:\mathbb R\to\mathbb R^{n\times n}$? Specifically, I'm not sure how the Itô process would be modified since the $\textrm d t$ term now has a matrix coefficient. I can move this to a separate question if necessary. $\endgroup$
    – Dylan
    Commented Nov 30, 2019 at 13:21
  • $\begingroup$ I separate question would seem the way to go. $\endgroup$
    – Carlo Beenakker
    Commented Nov 30, 2019 at 14:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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