Feynman–Kac Formulas for Dirichlet–Pauli–Fierz Operators with Singular Coefficients

Abstract

We derive Feynman–Kac formulas for Dirichlet realizations of Pauli–Fierz operators generating the dynamics of nonrelativistic quantum mechanical matter particles, which are minimally coupled to both classical and quantized radiation fields and confined to an arbitrary open subset of the Euclidean space. Thanks to a suitable interpretation of the involved Stratonovich integrals, we are able to retain familiar formulas for the Feynman–Kac integrands merely assuming local square-integrability of the classical vector potential and the coupling function in the quantized vector potential. Allowing for fairly general coupling functions becomes relevant when the matter-radiation system is confined to cavities with inward pointing boundary singularities.

Appendix A. A useful rule for vector-valued conditional expectations

The following lemma should be well-known, also in the infinite dimensional setting, but we could not find an appropriate reference. Therefore, we prove it for the convenience of the reader.

Lemma A.1

Let $(X,{\mathfrak {A}},P)$ be a probability space, ${\mathfrak {C}}$ be a sub-$\sigma$-algebra of ${\mathfrak {A}}$, and Y and Z separable Banach spaces equipped with their Borel $\sigma$-algebras ${\mathfrak {B}}(Y)$ and ${\mathfrak {B}}(Z)$, respectively. Let $f:X\times Y\rightarrow Z$ be a function such that $f(\cdot ,y):X\rightarrow Z$ is Bochner-Lebesgue integrable (in particular ${\mathfrak {A}}$-${\mathfrak {B}}(Z)$-measurable) and ${\mathfrak {C}}$-independent for every $y\in Y$, and such that $f(x,\cdot ):Y\rightarrow Z$ is continuous for every $x\in X$. (This implies that f is $({\mathfrak {A}}\otimes {\mathfrak {B}}(Y))$-${\mathfrak {B}}(Z)$-measurable.) Define

$$\begin{aligned} \phi (y):=E[f(\cdot ,y)]:=\int _X f(x,y)\mathrm{d}P(x),\quad y\in Y. \end{aligned}$$

(Then $\phi :Y\rightarrow Z$ is in any case Borel measurable.) Finally, let $g:X\rightarrow Y$ be ${\mathfrak {C}}$-${\mathfrak {B}}(Y)$-measurable and assume that

$$\begin{aligned} X\ni x\longmapsto h(x):=f(x,g(x))\in Z \end{aligned}$$

is Bochner-Lebesgue integrable. Then

$$\begin{aligned} E^{{\mathfrak {C}}}[h]=\phi (g),\quad P\text {-a.s.,} \end{aligned}$$

where $E^{{\mathfrak {C}}}$ denotes a version of the Z-valued conditional expectation with respect to P given the hypothesis ${\mathfrak {C}}$.

Proof

Let $\chi \in C({\mathbb {R}},{\mathbb {R}})$ be such that $0\leqslant \chi \leqslant 1$ on ${\mathbb {R}}$, $\chi =1$ on $(-\infty ,1]$, and $\chi =0$ on $[2,\infty )$. Put $f_n:=\chi (\Vert f\Vert _Z/n)f$, $n\in {\mathbb {N}}$, so that each $f_n$ enjoys all properties of f mentioned in the statement as well, and so that $\Vert f_n\Vert _Z\leqslant 2n$ and $f_n\rightarrow f$, $n\rightarrow \infty$, pointwise on $X\times Y$. Set $h_n(x):=f_n(x,g(x))$, $x\in X$, and $\phi _n(y):=E[f_n(\cdot ,y)]$, $y\in Y$. Then we have the dominations $\Vert f_n(\cdot ,y)\Vert \leqslant \Vert f(\cdot ,y)\Vert$ and $\Vert h_n\Vert _Z\leqslant \Vert h\Vert _Z$ on X, for every $n\in {\mathbb {N}}$. Hence, the dominated convergence theorem for the Bochner-Lebesgue integral implies that $\phi _n(y)\rightarrow \phi (y)$, $n\rightarrow \infty$, for every $y\in Y$, while the dominated convergence theorem for Z-valued conditional expectations implies that ${\mathbb {E}}^{{\mathfrak {C}}}[h_n]\rightarrow {\mathbb {E}}^{{\mathfrak {C}}}[h]$, $n\rightarrow \infty$, P-a.s. Therefore, it only remains to show that ${\mathbb {E}}^{{\mathfrak {C}}}[h_n]=\phi _n(g)$ holds P-a.s., for each fixed $n\in {\mathbb {N}}$. Or, put differently, we may assume without loss of generality that f is bounded, which we shall do in the rest of this proof.

There exists a sequence of ${\mathfrak {C}}$-${\mathfrak {B}}(Y)$-measurable functions $(g_n)_{n\in {\mathbb {N}}}$ such that the image $g_n(X)$ is finite, for every $n\in {\mathbb {N}}$, and such that $g_n\rightarrow g$, $n\rightarrow \infty$, pointwise on X. Let $n\in {\mathbb {N}}$. Then $g_n$ has a standard representation $g_n=\sum _{i=1}^{k_n}1_{A^n_i}y_i^n$ for suitable $k_n\in {\mathbb {N}}$, $y_1^n,\ldots ,y_{k_n}^n\in Y$, and suitable disjoint $A_1^n,\ldots ,A_{k_n}^n\in {\mathfrak {C}}$ such that $A_1^n\cup \dots \cup A_{k_n}^n=X$. Then

$$\begin{aligned} \tilde{h}_n(x):=f(x,g_n(x))=\sum _{i=1}^{k_n}1_{A_i^n}(x)f(x,y_i^n), \quad x\in X. \end{aligned}$$

Since $A_i^n\in {\mathfrak {C}}$ and since $f(\cdot ,y_i^n):X\rightarrow Z$ is ${\mathfrak {C}}$-independent, well-known computation rules for the conditional expectation now imply

$$\begin{aligned} E^{{\mathfrak {C}}}[\tilde{h}_n]=\sum _{i=1}^{k_n}1_{A_i^n}\phi (y_i^n)=\phi (g_n), \quad P\text {-a.s.,} \end{aligned}$$

where we again used that $y_i^n=g_n$ on $A_i^n$ in the second equality. Furthermore, by our present assumptions on f, the functions $\tilde{h}_n$, ${n\in {\mathbb {N}}}$, are uniformly bounded, and thanks to the continuity of $y\mapsto f(x,y)$ for each x, we know that $\tilde{h}_n\rightarrow h$, $n\rightarrow \infty$, pointwise on X. Hence, $E^{{\mathfrak {C}}}[\tilde{h}_n]\rightarrow E^{{\mathfrak {C}}}[h]$, $n\rightarrow \infty$, P-a.s., by the dominated convergence theorem for Z-valued conditional expectations. Finally, we observe that $\phi :Y\rightarrow Z$ is continuous by the boundedness of f and dominated convergence. Thus, $\phi (g_n)\rightarrow \phi (g)$, $n\rightarrow \infty$, pointwise on X. $\square$

Example A.2

Let $(X,{\mathfrak {A}},P)$ and ${\mathfrak {C}}$ be as in Lemma A.1. Let Z be a separable Hilbert space, $A(\varvec{y}):X\rightarrow {\mathcal {B}}(Z)$ be measurable and separably valued, for every $\varvec{y}\in {\mathbb {R}}^\nu$, such that ${\mathbb {R}}^\nu \ni \varvec{y}\mapsto (A(\varvec{y}))(x)$ is strongly continuous for all $x\in X$. Suppose that $A(\varvec{y})$ is ${\mathfrak {C}}$-independent and let $g:=(\varvec{q},\Psi ):X\rightarrow {\mathbb {R}}^\nu \times Z$ be ${\mathfrak {C}}$-measurable with $\int _X\Vert \Psi \Vert _{Z}\mathrm{d}P<\infty$. Finally, assume there exists $C>0$ such that $\Vert A(y)\Vert \leqslant C$, ${\mathbb {P}}$-a.s., for every $y\in {\mathbb {R}}^\nu$. Then we can apply Lemma A.1 to the function f given by $f(x,\varvec{y},\psi ):=A(y)\psi$, $(\varvec{y},\psi )={\mathbb {R}}^\nu \times Z(=:Y)$ with $\phi (\varvec{y},\psi )=E[f(\cdot ,\varvec{y},\psi )]=E[A(\varvec{y})]\psi$. That is,

$$\begin{aligned} E^{{\mathfrak {C}}}[A(\varvec{q})\Psi ]&=E[A(\varvec{y})]\big |_{\varvec{y}=\varvec{q}}\Psi . \end{aligned}$$