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.)


1 Answer 1


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.

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    $\begingroup$ 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
    Nov 30, 2019 at 13:21
  • $\begingroup$ I separate question would seem the way to go. $\endgroup$ Nov 30, 2019 at 14:31

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