text:
\begin{document}

\title{A mathematical theory of the Feynman path integrals for  continuous quantum position measurements}

\author{Wataru Ichinose\thanks{This work was supported by JSPS KAKENHI Grant Number JP18K03361.}}

\date{}

\maketitle

\begin{quote}
{\small Department of Mathematics, Shinshu University,
Matsumoto 390-8621, Japan. \\
 E-mail: ichinose@math.shinshu-u.ac.jp}
\end{quote}\par

	\begin{abstract}
	It has  been stated in the physics literature that
	the probability amplitude  for a quantum system after a continuous position measurement can be expressed by  the restricted or the weighted Feynman path integral  containing information about the result and the resolution of its measurement.  In the present paper  it is proved that the restricted Feynman path integrals can be defined in a rigorous sense.  More specifically,
	we consider a quantum system consisting of some  particles and   simultaneously perform a continuous measurement of the positions of all particles in the time interval $[0,T]$.  Then, it is shown rigorously that the restricted Feynman path integral after the measurement can be defined in the $L^2$ space and satisfies the corresponding non-self-adjoint Schr\"odinger equation. In addition, our results are generalized to a quantum spin system where a continuous position measurement for all spin components of all particles is simultaneously performed, though the results obtained are inadequate for general use.

\end{abstract}

\section{Introduction}

Let $T > 0$ be an arbitrary constant, $0 \leq t \leq T$ and
$x = (x_1,\dotsc,x_d)\in \bR^d$. First, we consider a one-particle system with mass $m > 0$ and charge $q_c \in \bR$ moving in $\bR^d$ with  electric strength $E(t,x) = (E_1,
\dots, E_d) \in \bR^d$ and a  magnetic strength tensor $B(t,x) = (B_{jk}(t,x))_{1\leq j < k \leq d}
\in \bR^{d(d-1)/2}$.
Let   $(V(t,x),A(t,x)) = (V,A_1,
\dots,A_d)  \in \bR^{d+1}$ be an electromagnetic potential, i.e.

\begin{align} \label{1.1}
     & E = -\frac{\partial A}{\partial t} - \frac{\partial V}{\partial
x},\notag \\

          & B_{jk} =  \frac{\partial A_k}{\partial x_j}  -\frac{\partial
A_j}{\partial x_k}
\quad (1 \leq j <  k \leq d),
\end{align}

where  $\partial V/\partial x = (\partial V/\partial x_1,\dots,\partial V/\partial x_d)$.
Then the Lagrangian function and the classical action are given by

\begin{equation} \label{1.2}
      {\cal L}(t,x,\dot{x})=  \frac{m}{2}|\dot{x}|^2 + q_c\dot{x}\cdot A(t,x) - q_cV(t,x),
\   \dot{x}\in \bR^d
\end{equation}
and

\begin{equation} \label{1.3}
   S(t,s;q) = \int_s^t {\cal L}(\theta,q(\theta),\dot{q}(\theta))d\theta,\quad \dot{q}(\theta) = \frac{dq(\theta)}{d\theta}
\end{equation}

for a path $q(\theta) \in \bR^d\ (s \leq \theta \leq t)$, respectively.
The corresponding Schr\"odinger equation is defined by

\begin{align} \label{1.4}
& i\hbar \frac{\partial u}{\partial t}(t)  = H(t)u(t)\notag\\
& := \left[ \frac{1}{2m}\sum_{j=1}^d
      \left(\frac{\hbar}{i}\frac{\partial}{\partial x_j} - q_cA_j(t,x)\right)^2 + q_cV(t,x)\right]u(t),
\end{align}

where $\hbar$ is the Planck constant.
Throughout this paper we always consider solutions to the Schr\"odinger equations in the sense of distribution.
\par

We perform a continuous quantum   measurement of the position of the particle in the time interval $[0,T]$. Let its result be $\{a(t) \in \bR^d;0 \leq t \leq T\}$ and its resolution or its error  $\Delta a > 0$ determined by its precision.  We take $2|x - a(t)|^2/(T\Delta a^2)$ as a weight function $W(t,x)$, and define the effective Lagrangian function under the measurement by

\begin{equation} \label{1.5}
      {\cal L}_w(t,x,\dot{x}) =  {\cal L}(t,x,\dot{x}) + i\hbar W(t,x)
\end{equation}

and  the effective classical action by

\begin{equation} \label{1.6}
   S_w(t,s;q) = \int_s^t {\cal L}_w(\theta,q(\theta),\dot{q}(\theta))d\theta
\end{equation}

for a path $q(\theta) \in \bR^d$ as in \eqref{1.3}. Then, Feynman in the 1st and the 2nd postulates on page 371 of \cite{Feynman 1948} (cf. \S 3.2 of \cite{Feynman-Hibbs}), and Mensky   in \S 4.2 of \cite{Mensky 1993} and \S 5.1.3 in \cite{Mensky 2000} have stated that  the probability amplitude of the particle after the measurement is formally expressed in terms of the path integral by

\begin{equation} \label{1.7}
 K(t,0)f = \int e^{i\hbar^{-1}S_w(t,0;q)}f(q(0)){\cal D}q
\end{equation}

for a given initial amplitude $f$ at $t = 0$ and satisfies
the  non-self-adjoint Schr\"odinger equation corresponding to \eqref{1.5}

\begin{align} \label{1.8}
& i\hbar \frac{\partial u}{\partial t}(t)  = H_w(t)u(t)\notag\\
& := \left[ \frac{1}{2m}\sum_{j=1}^d
      \left(\frac{\hbar}{i}\frac{\partial}{\partial x_j} - q_cA_j(t,x)\right)^2 + q_cV(t,x) -i\hbar W(t,x)\right]u(t).
\end{align}

The path integral expressed by \eqref{1.7}
 is called the restricted or the weighted Feynman path integral (RFPI or WFPI).
\par

Let $L^2 = L^2(\bR^d)$ denote the space of all square integrable functions on
$\bR^d$ with the inner
product $(f,g) := \int f(x)g(x)^*dx$ and the norm $\Vert f\Vert$, where $g(x)^*$ denotes the complex conjugate of $g(x)$.
In the present paper we consider not only $2|x - a(t)|^2/(T\Delta a^2)$  but also  more general functions  $W(t,x)$ as weight functions.
 We will    prove that for any given $f \in L^2$  $K(t,0)f$ expressed by \eqref{1.7} can be determined rigorously
in $C^0_t([0,T]:L^2(\bR^d))$ and satisfies \eqref{1.8} with $u(0) = f$, where $C^j_t([0,T];L^2)$ denotes the space of all $L^2$-valued, $j$-times continuously differentiable functions in $t \in [0,T]$.
\par

Next, we will generalize the above result  to a one-particle spin system with $l$ spin components.
We simultaneously perform a  continuous quantum  measurement of the positions of all spin components of the particle  in the  time interval $[0,T]$.  Then,  the effective Lagrangian function is given by

\begin{equation} \label{1.9}
\mathcal{L}_{sw}(t,x,\dot{x}) = \mathcal{L}_{w}(t,x,\dot{x}) - \hbar H_s(t,x) + i\hbar W_s(t,x),
 \end{equation}

where $\mathcal{L}_{w}(t,x,\dot{x})$ is the Lagrangian function defined by \eqref{1.5}, $H_s(t,x)$ an $l\times l$ Hermitian matrix denoting the spin term and  $W_s(t,x)$ an $l\times l$ Hermitian matrix denoting the weight term acting on the spin components.
The corresponding non-self-adjoint Schr\"odinger equation is written by

\begin{equation}  \label{1.10}
   i\hbar\frac{\partial u}{\partial t}(t) = \bigl[H_w(t)I + \hbar H_{s}(t,x)  - i\hbar W_s(t,x)\bigr]u(t),
\end{equation}

where $H_w(t)$ is the Hamiltonian operator defined by \eqref{1.8}.  We will prove that the RFPI for \eqref{1.9} can be determined  rigorously in $C^0_t([0,T];(L^2(\bR^{d})^l)$ and satisfies the equation \eqref{1.10}.   However, it should be noted that for $W_s(t,x)$ we  assume the tough restrictions

\begin{equation}  \label{1.11}
  |\partial_{x}^{\alpha} w_{sij}(t,x)| \leq C_{\alpha}, \quad i, j = 1,2,\dots,l
\end{equation}

in $\domain$ for all $\alpha$, where $w_{sij}(t,x)$ denotes the $(i,j)$-component of $W_s(t,x)$. Accordingly, the results obtained are inadequate for general use.
\par

	Finally, we  consider a quantum spin system consisting of $N$ particles with $l$ spin components and simultaneously perform a  continuous position measurement for all spin components of all particles in the  time interval $[0,T]$.
	We will prove that the RFPI under this measurement can be determined  rigorously in $C^0_t([0,T];(L^2(\bR^{d})^l\otimes \dots \otimes L^2(\bR^{d})^l)$ and satisfies the  corresponding non-self-adjoint Schr\"odinger equation.
	\par
	We note that if $W(t,x) = 0$ and $W_s(t,x) = 0$, all  results in the present paper give the same results as for the normal Feynman path integrals in \cite{Ichinose 1999, Ichinose 2007, Ichinose 2020}.
\par

   In the present paper we will consider the RFPIs for continuous quantum  position measurements.  On  the other hand, there is another approach to continuous quantum  position measurements as stated below.  We begin by  considering a sequence of $n$ instantaneous position measurements separated by a time $\Delta t$ and then, determine the evolution of the measured system in the continuous limit $n \to \infty$ and $\Delta t \to 0$ (cf. \cite{Caves 1986, Caves 1987-1,Caves 1987-2}, Chapter 3 in \cite{Jacobs}, Chapter 2 in \cite{Mensky 1993} and Chapter 2 in \cite{Mensky 2000}). As for this approach we only note in the present paper that the Feynman path integrals for a sequence of instantaneous position measurements have been proved to be determined rigorously in the $L^2$ space  in \cite{Ichinose 2006}.
   \par

   	If $\mathcal{L}(t,x,\dot{x})$ defined by \eqref{1.2} is a harmonic oscillator and $W(t,x)$  the sum of  $2|x - a(j)|^2/(T\Delta a(j)^2)\ (j= 1,2,\dots,n)$
   with constant vectors $a(j) \in \bR^d$, we can directly calculate the RFPI expressed by \eqref{1.7} by using Gaussian integrals (cf. \S 3.2 in \cite{Feynman-Hibbs}, \S 4.5 and \S 5.4 in \cite{Mensky 1993}).  If $W(t,x) \not= 0$,  we have no mathematical results about  RFPIs except for this example as far as the author knows.
   \par
     In the present paper  the RFPIs are defined by the time-slicing method in terms of piecewise free moving paths or piecewise straight lines.  This  approach  to the Feynman path integrals are  widely used in the physics literature (cf.  \S 2.4 in \cite{Feynman-Hibbs}, \S 3.2 in \cite{Mensky 1993}, Appendix A3 in \cite{Mensky 2000}, \S 9.1 in \cite{Peskin-Schroeder} and \S 5.1 in \cite{Ryder}).
    \par

 	We will prove the results in the present paper, following the proofs in \cite{Ichinose 1997, Ichinose 1999, Ichinose 2007}, where the normal Feynman path integrals, i.e. the RFPIs with $W(t,x) = 0$ and $W_s(t,x) = 0$  were studied.  More specifically, we first introduce the fundamental operator $\mathcal{C}(t,s)$ in \S 3, and then prove its stability and consistency.  Combining these results and the results in \cite{Ichinose 2019} concerning the non-self-adjoint Schr\"odinger equations \eqref{1.8} and \eqref{1.10}, we can complete the proofs of  our results.  In particular, we define the RFPIs for the spin system, following \cite{Ichinose 2007}.  We also note that in the  present paper we will use
  the following delicate result  concerning the $L^2$-boundedness of pseudo-differential operators, which is stated as Theorem 13.13 on p. 322 in \cite{Zworski}.
  \par
 \vspace{0.5cm}
 T{\sc heorem} 1.A.  {\it Suppose $p(x,\xi,x') \in S^0(\bR^{3d})$, i.e.

 \begin{equation} \label{1.12}
\sup_{x,\xi,x'}|\partial_{\xi}^{\alpha}\partial_{x}^{\beta}\partial_{x'}^{\gamma}p(x,\xi,x')| \leq C_{\alpha\beta\gamma} < \infty
 \end{equation}

for all multi-indices $\alpha, \beta$ and $\gamma$, where   $\partial_{\xi}^{\alpha}$ denotes
  $(\partial/\partial \xi_1)^{\alpha_1}
\cdots (\partial/\partial \xi_d)^{\alpha_d}$ for  $\alpha = (\alpha_1,\dots,\alpha_d)$.
  Let $P(X,\hbar D_x,X')$ be the pseudo-differential operator defined by

 \begin{equation*}
\int e^{ix\cdot \xi}\ \dbar\xi \int e^{-ix'\cdot \xi}p(x,\hbar\xi,x')f(x')dx',\quad \dbar\xi = (2\pi)^{-d}d\xi
 \end{equation*}

for $f \in \Sspace(\bR^d)$, where $x\cdot \xi = \sum_{j=1}^ d x_j\xi_j $ and $\Sspace(\bR^d)$ denotes the Schwartz space of all rapidly decreasing functions on $\bR^d$.  Then we have

 \begin{equation} \label{1.13}
\Vert P(X,\hbar D_x,X')\Vert_{L^2\to L^2} = \sup_{x,\xi,x'}|p(x,\xi,x')| + O(\hbar),
 \end{equation}

where $\Vert P\Vert_{L^2\to L^2}$ denotes the operator norm from $L^2$ into $L^2$.}
\vspace{0.5cm}
\par

The plan of the present paper is as follows.  In \S 2 all main results are  stated in Theorems 2.1 - 2.6.  In \S 3 we will prove Theorems 2.1 and 2.2.  In \S 4 Theorems 2.3 - 2.6 will be proved.

  \section{Main theorems}

Hereafter we suppose $\hbar = 1$ and $q_c = 1$ for  simplicity.
We first consider \eqref{1.7}.
Let $t$ in $[0,T]$.  For an arbitrary integer $\nu \geq 1$ we take $\tau_j \in [0,t]\ (j = 1,2,\dots,\nu-1)$ satisfying $0 = \tau_0 < \tau_1 < \dots < \tau_{\nu-1} < \tau_{\nu}= t$,  set $\Delta := \{\tau_j\}_{j=1}^{\nu-1}$ and write $|\Delta|:= \max\{\tau_{j+1}- \tau_j; j = 0,1,\dots,\nu-1\}$. Let $x \in \bR^d$ be fixed.
  We take arbitrary points
$x^{(j)} \in \bR^d\ (j = 0,1,\dotsc,\nu-1)$ and determine the piecewise free moving path or the piecewise straight line $\qdelta(\theta;x^{(0)},\dotsc,x^{(\nu-1)},x) \in \bR^d \ (0 \leq \theta \leq t)$ by joining  $x^{(j)}$ at $\tau_j\ (j = 0,1, \dotsc,\nu,
x^{(\nu)} = x)$  in order.  Let   $S_w(t, s; q)$ be the effective classical action defined by \eqref{1.6}.  Take $\chi \in \Cspace_0(\bR^d)$, i.e. an infinitely differentiable function on $\bR^d$ with compact support,  such that $\chi(0) = 1$ and determine the approximation  of the RFPI \eqref{1.7}  by

 \begin{align} \label{2.1}
\kdelta(t,0)f =  \limepsilon & \prod_{j=0}^{\nu-1}\sqrt{\frac{m}{2\pi i(\tau_{j+1} -
\tau_{j})}}^{\ d}
        \int\cdots\int_{\bR^d} e^{iS_w(t,0;\qdelta)} \notag \\
       & \times f(x^{(0)}) \prod_{j=1}^{\nu-1}\chi(\epsilon x^{(j)})
          dx^{(0)}dx^{(1)}\cdots dx^{(\nu-1)}
 \end{align}

for $f \in \Cspace_0(\bR^d)$. From now on, we always suppose that $\chi$ is a real-valued function belonging to  $\Cspace_0(\bR^d)$ such that $\chi(0) = 1$.
The right-hand side of \eqref{2.1} is called an oscillatory integral and will be denoted by

 \begin{align*}
  & \prod_{j=0}^{\nu-1}\sqrt{\frac{m}{2\pi i(\tau_{j+1} -
\tau_{j})}}^{\ d}
      \text{Os} -  \int\cdots\int_{\bR^d} e^{iS_w(t,0;\qdelta)}  f(x^{(0)})
          dx^{(0)}dx^{(1)}\cdots dx^{(\nu-1)}
 \end{align*}

 (cf. p. 45 of \cite{Kumano-go}).
  \par
   For a multi-index $\alpha = (\alpha_1,\dots,\alpha_d)$ and $x \in \bR^d$ we  write $|\alpha| =
\sum_{j=1}^d
\alpha_j$, $x^{\alpha} =  x_1^{\alpha_1}
\cdots  x_d^{\alpha_d}$ and  $<x> = \sqrt{1+|x|^2}$.
 In the present paper we often use symbols $C, C_{\alpha}, C_{\alpha\beta}$, $C_a$  and $\delta_{\alpha}$ to write down constants, though these values are different in general.
 \par
   \vspace{0.5cm}

Throughout the present paper we assume that $\partial_x^{\alpha}E_j(t,x)\ (j = 1,2,\dots,d), \\ \partial_x^{\alpha}B_{jk}(t,x)\ (1 \leq j < k \leq d)$ and $\partial_x^{\alpha}W(t,x)$ are continuous in $\domain$ for all $\alpha$.  Then, $\partial_x^{\alpha}\partial_tB_{jk}(t,x)\ (1 \leq j < k \leq d)$ are also continuous in $\domain$ for all $\alpha$, because of Faraday's law $\partial_tB_{jk} =  - \partial E_k/\partial x_j + \partial E_j/\partial x_k$, which follows from \eqref{1.1}.
\par

{\bf Assumption 2.A.}  We assume

\begin{equation} \label{2.2}
|\partial_x^{\alpha}E_j(t,x)| \leq C_{\alpha},\  |\alpha| \geq 1,\quad j = 1,2,\dots,d,
\end{equation}

\begin{equation} \label{2.3}
|\partial_x^{\alpha}B_{jk}(t,x)| \leq C_{\alpha}<x>^{-(1+ \delta_{\alpha})},\  |\alpha| \geq 1, \quad 1 \leq j < k \leq d
\end{equation}

in $\domain$ with constants $C_{\alpha} \geq 0$ and $\delta_{\alpha} > 0$.
\par

{\bf Assumption 2.B.}  We assume \eqref{2.2} and

\begin{equation} \label{2.4}
|\partial_x^{\alpha}\partial_tB_{jk}(t,x)| \leq C_{\alpha}<x>^{-(1+ \delta_{\alpha})},\  |\alpha| \geq 1, \quad 1 \leq j < k \leq d
\end{equation}

in $\domain$ with constants $C_{\alpha} \geq 0$ and $\delta_{\alpha} > 0$.
\par

{\bf Assumption 2.C.} We assume that $\partial_x^{\alpha}A_j(t,x)\ (j = 1,2,\dots,d)$ and $\partial_x^{\alpha}V(t,x)$ are continuous in $\domain$ for all $\alpha$ and satisfy

\begin{equation} \label{2.5}
|\partial_x^{\alpha}A_j(t,x)| \leq C_{\alpha},\ |\alpha| \geq 1,\quad j = 1,2,\dots,d,
\end{equation}

\begin{equation} \label{2.6}
|\partial_x^{\alpha}V(t,x)| \leq C_{\alpha}<x>,\quad |\alpha| \geq 1
\end{equation}

in $\domain$ with constants $C_{\alpha} \geq 0$.
\par

{\bf Assumption 2.D.}  We assume

\begin{equation} \label{2.7}
W(t,x) \geq - C(W),
\end{equation}

\begin{equation} \label{2.8}
|\partial_x^{\alpha}W(t,x)|^{p_{\alpha}} \leq C_{\alpha}\bigl\{1 + C(W) + W(t,x)\bigr\},\quad |\alpha| \geq 1,
\end{equation}

\begin{equation} \label{2.9}
|\partial_x^{\alpha}W(t,x)| \leq C_{\alpha}<x>,\quad |\alpha| \geq 1
\end{equation}

in $\domain$ with constants $C(W) \geq 0, C_{\alpha} \geq 0$ and $p_{\alpha} > 1$.
\par

{\it Example 2.1.} The function $2|x - a(t)|^2/(T\Delta a^2)$ with a continuous path $a(t) \in \bR^d$, which was stated in introduction, satisfies Assumption 2.D.
\par

\begin{thm}
Suppose that Assumptions 2.A and 2.D are satisfied.  Then, there exists a constant $\rho^* > 0$ such that  the following statements hold for arbitrary potentials $(V,A)$ with continuous

$V, \partial V/\partial x_j, \partial A_j/\partial t, \partial A_j/\partial x_k\ (j,k = 1,2,\dots,d)$
in $\domain$, all $\Delta$ satisfying $|\Delta| \leq \rho^*$ and all $t \in [0,T]$:
\\
(1) $K_{\Delta}(t,0)f$ defined on $f \in \Czerospace$ by \eqref{2.1} is
determined independently of the choice of $\chi$ and $K_{\Delta}(t,0)f$ can be uniquely extended to a bounded operator on $L^{2}$.
\\

(2) For all $f \in L^{2}$, as $|\Delta| \to 0$, $\kdelta(t,0)f$ converges in $L^{2}$ uniformly in $t \in [0,T]$ to an element $K(t,0)f \in L^{2}$, which we call the  RFPI or the WFPI of $f$.
\\

(3) For all $f \in L^{2}$, $K(t,0)f$  belongs to   $C_{t}^{0}([0,T];L^{2})$. In addition, $K(t,0)f$ is the unique solution  in $C^{0}_{t}([0,T];L^{2})$ to \eqref{1.8} with $u(0) = f$.
\\

(4) Let $\psi(t,x) \in C^1(\domain)$ be a real-valued function such that $\partial_{x_j}\partial_{x_k}\psi(t,x)$ and $\partial_{t}\partial_{x_j}\psi(t,x)$ $(j,k = 1,2,\dotsc,d)$ are continuous in $[0,T] \times \bR^d$ and consider the gauge transformation

\begin{equation}  \label{2.10}
    V' = V -\frac{\partial\psi}{\partial  t}, \quad  A'_j = A_j + \frac{\partial\psi}{\partial  x_j}\quad (j= 1,2,\dots,d).
\end{equation}

We write \eqref{2.1} for this $(V',A')$  as $K'_{\Delta}(t,0)f$.  Then we have the formula

\begin{equation}  \label{2.11}
     K'_{\Delta}(t,0)f  = e^{i\psi (t,\cdot)}K_{\Delta}(t,0)\left(e^{-i\psi (0,\cdot)}f\right)
\end{equation}

for  all $f \in L^2$ as in the case of $W(t,x) = 0$ (cf. (6.16) in \cite{Ichinose 1999}), and we have the analogous relation between the limits $K'(t,0)f$ and $K(t,0)f$.

\end{thm}

Let us introduce the weighted Sobolev spaces

\begin{align} \label{2.12}
B^a(\mathbb{R}^d) & := \{f \in  L^2(\mathbb{R}^d);
 \|f\|_a := \|f\| + \sum_{|\alpha| \leq  2a} \|\partial_x^{\alpha}f\| \notag\\
 &  +
\|<\cdot>^{2a}f\| < \infty\}\ (a = 1,2,\dots)
 \end{align}

as in \cite{Ichinose 1999}. We denote the dual space of $B^a$ by $B^{-a}$ and the $L^2$ space by $B^0$.

\begin{thm}
Suppose that either Assumption 2.A or 2.B is satisfied.  In addition, we suppose Assumptions 2.C and 2.D. Then there exists another constant $\rho^* > 0$  such that the same statements (1) - (4) as in Theorem 2.1 hold for all $\Delta$ satisfying $|\Delta| \leq \rho^*$ and all $t \in [0,T]$.  In addition, for all $f \in B^a(\bR^d)\ (a = 1,2,\dots)$ $K_{\Delta}(t,0)f$  belongs to   $B^a$ and  as $|\Delta| \to 0$, $\kdelta(t,0)f$ converges in $B^a$ uniformly in $t \in [0,T]$ to $K(t,0)f$, which belongs to  $C^0_t([0,T];B^a)$.

\end{thm}

	Next, we consider a one-particle spin system \eqref{1.9} or \eqref{1.10} with $l$ spin components.    Throughout the present paper we assume that $\partial_x^{\alpha}h_{sij}(t,x)$ and $\partial_x^{\alpha}w_{sij}(t,x)$ are continuous in $\domain$ for all $\alpha$ and $i,j = 1,2,\dots,l$, where $h_{sij}$  denotes the
$(i,j)$-component of $H_s$.

For a continuous path $q(\theta) \in \bR^d\ (s \leq \theta \leq t)$ we define an $l \times l$ matrix $\mathcal{F}(\theta,s;q)\ (s \leq \theta \leq t)$ by the solution to

\begin{equation} \label{2.13}
\frac{d}{d\theta}\,\mathcal{U}(\theta) = -\bigl\{iH_s(\theta,q(\theta)) + W_s(\theta,q(\theta))\bigr\}\mathcal{U}(\theta), \quad \mathcal{U}(s) = I
 \end{equation}

with the identity matrix $I$.
\par

	Let $\Delta = \{\tau_j\}_{j=1}^{\nu - 1}$ be a subdivision of $[0,t]$ and $\qdelta = \qdelta(\theta;x^{(0)},\dotsc,x^{(\nu-1)},x) \in \bR^d \ (0 \leq \theta \leq t)$ the piecewise free moving path defined in the early part of this section.  We define the probability amplitude by

\begin{equation} \label{2.14}
\exp *iS_{sw}(t,0;\qdelta) := \bigl(\exp iS_w(t,0;\qdelta)\bigr)\mathcal{F}(t,0;\qdelta)
 \end{equation}

 as in the case of $W_s(t,x) = 0$ (cf. \S 2 in \cite{Ichinose 2007}), where $S_w(t,0;\qdelta)$ is the classical action defined by \eqref{1.6}.  For $f = {}^t(f_1,\dots,f_l) \in \Czerospace^l$ we determine the approximation of the RFPI for this system under the measurement by

 \begin{align} \label{2.15}
\ksdelta(t,0)f =  \limepsilon & \prod_{j=0}^{\nu-1}\sqrt{\frac{m}{2\pi i(\tau_{j+1} -
\tau_{j})}}^{\ d}
        \int\cdots\int_{\bR^d} \bigl(\exp *iS_{sw}(t,0;\qdelta)\bigr) \notag \\
       & \times f(x^{(0)}) \prod_{j=1}^{\nu-1}\chi(\epsilon x^{(j)})
          dx^{(0)}dx^{(1)}\cdots dx^{(\nu-1)}
 \end{align}

 as in \cite{Ichinose 2007}.
The $L^2$-norm  of $f = {}^t(f_1,\dots,f_l) \in (L^2)^l$ is defined by $\Vert f \Vert := \sqrt{\sum_{j=1}^l \Vert f_j\Vert^2}$.

\begin{thm}
Besides Assumptions 2.A and 2.D we suppose that $H_s(t,x)$ and $W_s(t,x)$ satisfy

\begin{equation}  \label{2.16}
  |\partial_{x}^{\alpha} h_{sij}(t,x)| \leq C_{\alpha},\quad i, j = 1,2,\dots,l
\end{equation}

 and \eqref{1.11} for all $\alpha$, respectively.  Let $\rho^* > 0$ be the constant determined in Theorem 2.1.
Then the same statements for $\ksdelta(t,0)f$ as for $\kdelta(t,0)f$ in Theorem 2.1 hold, where $K_s(t,0)f := \lim_{|\Delta|\to 0}\ksdelta(t,0)f \in C^0_t
([0,T]
;(L^2)^l)$ for $f \in (L^2)^l$ is the unique solution in $C^0_t([0,T];(L^2)^l)$ to \eqref{1.10}
with $u(0) = f$.

\end{thm}

{\it Example 2.2.} We perform a continuous quantum measurement of the positions of all  spin components of a particle.   Let $a^{(j)}(t)\in \bR^d\ (j = 1,2,\dots,l)$ be the result for the $j$-th component of spin and $\Delta a$ the resolution of the measurement.
Then  in Theorem 2.3 we take $W(t,x) = 0$ and the diagonal matrix $W_s(t,x)$ with

\begin{equation*}
w_{sjj}(t,x) = \Omega \left(\frac{2|x - a^{(j)}(t)|^2}{T\Delta a^2}\right), \quad j = 1,2,\dots,l,
\end{equation*}

where  $\Omega (\theta) \in \Cspace([0,\infty))$ is an increasing function such that $\Omega (\theta) = \theta$ if $0 \leq \theta \leq 1$ and  $\Omega (x) = L$ if $\theta \geq 2$ with a sufficiently large constant $L > 0$.  These $W(t,x)$ and $W_s(t,x)$ satisfy the assupmtions of Theorem 2.3.
\par
\vspace{0.3cm}

The $B^a$-norm  of $f = {}^t(f_1,\dots,f_l) \in (B^a)^l$ is defined by $\Vert f \Vert_a := \sqrt{\sum_{j=1}^l \Vert f_j\Vert_a^2}$.

\begin{thm}
We suppose that either Assumption 2.A or 2.B is satisfied.  In addition, we suppose  Assumptions 2.C, 2.D, \eqref{1.11} and \eqref{2.16}.  Let $\rho^* > 0$ be the constant determined in Theorem 2.2.  Then the same statements  for $\ksdelta(t,0)f$ as for $\kdelta(t,0)f$ in Theorem 2.2 hold, where we replace $B^a$ with $(B^a)^l$.
\end{thm}

	Finally, we consider a quantum spin system consisting of $N$ particles with $l$ spin components.  We simultaneously perform a continuous quantum measurement of the positions of all spin components of all particles in the time interval $[0,T]$.  Denoting the coordinates of the $j$-th particle by $\bx_j \in \bR^d \ (j = 1 ,2,\dots,N)$
, we write $x = (\bx_1,\bx_2,\dots,\bx_N) \in \bR^{dN}$.
	 Let $W_j(t,\bx_j) \in \bR$ and set

	\begin{align} \label{2.17}
& {\cal L}^{\sharp}_{w}(t,x,\dot{x}) =  \sum_{j=1}^N
 \biggl\{\frac{m}{2}|\dot{\bx}_j|^2 + \dot{\bx}_j\cdot \bA_j(t,\bx_j) -V_j(t,\bx_j) \notag \\
&\qquad + iW_j(t,\bx_j) \biggr\}
   - \sum_{j,k=1,j \not= k}^N V_{jk}(t,\bx_j-\bx_k),
   \end{align}

where $\bA_j(t,\bx_j) \in \bR^d, V_j(t,\bx_j) \in \bR,$ $V_{jk}(t,\bx_j-\bx_k) \in \bR$.   The effective Lagrangian function we consider is

\begin{align} \label{2.18}
& {\cal L}^{\sharp}_{sw}(t,x,\dot{x}) =  {\cal L}^{\sharp}_{w}(t,x,\dot{x}) + \sum_{j=1}^N I_1\otimes \cdots \otimes I_{j-1} \notag \\
 & \qquad \otimes \bigl\{-H_{sj}(t,\bx_j) + iW_{sj}(t,\bx_j) \bigr\}\otimes I_{j+1}\otimes \cdots\otimes I_N,
   \end{align}

 where $H_{sj}(t,\bx_j)$ and $W_{sj}(t,\bx_j)$ are $l \times l$ Hermitian matrices.
 \par

 For a continuous path $\bq_j(\theta) \in \bR^d\ (j = 1,2,\dots,N, s \leq \theta \leq t)$, we define $\mathcal{F}_j(\theta,s;\bq_j)\ (s \leq \theta \leq t)$ by the solution to \eqref{2.13} where $H_s = H_{sj}$ and $W_s = W_{sj}$.
 For the piecewise free moving path $\qdelta = \bigl(\bq_{1\Delta}(\theta;\bx_1^{(0)},\dotsc,\bx_1^{(\nu-1)},\bx_1),\dots, \\
  \bq_{N\Delta}
  (\theta;\bx_N^{(0)},\dotsc,\bx_N^{(\nu-1)},\bx_N)\bigr)\in \bR^{Nd} \ (0 \leq \theta \leq t)$,  we define the probability amplitude by

\begin{align} \label{2.19}
& \exp *iS^{\sharp}_{sw}(t,0;\qdelta) \notag
\\ &:= \bigl(\exp iS^{\sharp}_w(t,0;\qdelta)\bigr)\mathcal{F}_1(t,0;\bq_{1\Delta})\otimes \cdots \otimes \mathcal{F}_N(t,0;\bq_{N\Delta})
 \end{align}

in terms of the tensor product of matrices,
where $S^{\sharp}_w(t,0;\qdelta)$ is the classical action defined from \eqref{2.17}.  Then, we
determine the approximation $K^{\sharp}_{s\Delta}(t,0)f$ of the RFPI by \eqref{2.15}, where $\exp *iS_{sw}(t,0;\qdelta)$ and $f \in \Czerospace^l$ are replaced with  $\exp *iS^{\sharp}_{sw}(t,0;\qdelta)$ and $f = f_1 \otimes \cdots \otimes f_N$ $\bigl(f_j \in \Czerospace^l, j = 1,2,\dots,N\bigr)$, respectively.
\par

	Writing $A(t,x) = \bigl(\bA_1(t,\bx_1),\dots, \bA_N(t,\bx_N)\bigr) \in \bR^{dN}$ and $V(t,x) = \sum_{j=1}^N
	V_j(t,\\
	\bx_j)
		+ \sum_{j,k=1,j \not= k} V_{jk}(t,\bx_j-\bx_k) \in \bR$, we define $E(t,x) \in \bR^{dN}$ and $B_{jk}(t,x) \in \bR\  (1 \leq j < k \leq dN)$ by \eqref{1.1}.  Then we have the following.

	\begin{thm}
	Suppose that Assumptions 2.A is satisfied.  In addition, we assume that each $W_j(t,\bx_j)\ ( j = 1,2,\dots,N)$ satisfies 2.D and that each $W_{sj}(t,\bx_j)$  and $H_{sj}(t,\bx_j)$ satisfies \eqref{1.11} and \eqref{2.16}, respectively.  Let $(L^2)^l\otimes \cdots \otimes(L^2)^l$ denote the tensor product of $N$ copies of $L^2(\bR^d)^l$.
	Then, there exits a constant $\rho'^* > 0$ such that the same statements for $K^{\sharp}_{s\Delta}(t,0)f$ as for $\kdelta(t,0)f$ in Theorem 2.1 hold, where $K^{\sharp}_{s}(t,0)f := \lim_{|\Delta| \to 0} K^{\sharp}_{s\Delta}(t,0)f \in C^0_t([0,T];(L^2)^l\otimes \cdots \otimes(L^2)^l)$ for $f \in (L^2)^l\otimes \cdots \otimes(L^2)^l$ is the unique solution in $C^0_t([0,T];(L^2)^l\otimes \cdots \otimes(L^2)^l)$ to

\begin{align} \label{2.20}
& i \frac{\partial u}{\partial t}(t)  =\Biggl[\sum_{j=1}^N \Bigg\{\frac{1}{2m}
      \left|\frac{1}{i}\frac{\partial}{\partial \bx_j} - \bA_j(t,\bx_j)\right|^2 + V_j(t,\bx_j) -iW_{j}(t,\bx_j)\notag\\
& \quad   + I_1\otimes \cdots \otimes I_{j-1}\otimes \bigl\{H_{sj}(t,\bx_j) -  iW_{sj}(t,\bx_j)\bigl\}\otimes I_{j+1}\otimes \cdots\otimes I_N\Biggr\}  \notag \\
&\qquad  + \sum_{j,k=1,j \not= k}^N V_{jk}(t,\bx_j-\bx_k)
\Biggr]u(t)
\end{align}

with $u(0) = f$.

		\end{thm}

		Noitng that the $N$-fold tensor product $L^2(\bR^d)\otimes \cdots \otimes L^2(\bR^d)$ is equal to $L^2(\bR^{Nd})$ (cf. Theorem II.10 on p.52 in \cite{Reed-Simon I}), we can see

\begin{equation*}
 L^2(\bR^d)^l\otimes \cdots \otimes L^2(\bR^d)^l = \bigl(L^2(\bR^d)\otimes \cdots \otimes L^2(\bR^d)\bigr)^{l^N}
= L^2(\bR^{dN})^{l^N}
\end{equation*}

(cf. 4.2.1 on p.243 in \cite{Horn-Johnson}). In the same way we can define a subspace $B^a(\bR^{dN})^{l^N}\\ (a = 1,2,\dots)$ of $L^2(\bR^d)^l\otimes \cdots \otimes L^2(\bR^d)^l$.  Then we have the following.

\begin{thm}
Suppose that either Assumption 2.A or 2.B is satisfied.  In addition, we suppose  Assumption 2.C for $(V(t,x),A(t,x))$  and that each $W_j(t,\bx_j), W_{sj}(t,\bx_j)$ and $H_{sj}(t,\bx_j)$ satisfies the assumptions stated in Theorem 2.5.  Then, there exits another constant $\rho\,'^* > 0$ such that  the same statements for $K^{\sharp}_{s\Delta}(t,0)f$ as for $\kdelta(t,0)f$ in Theorem 2.2 hold, where we replace $B^a(\bR^d)$ with  $B^a(\bR^{dN})^{l^N}$.
\end{thm}

\begin{rem}
Let us assume in Theorems 2.5 and 2.6 that all $N$ particles are identical.  Accordingly, we suppose $\bA_j = \bA, V_j = V, V_{jk} = \widehat{V}, H_{sj} = H_s, W_j = W$ and $W_{sj} = W_s$ for all $j,k = 1,2,\dots,N$.  We write $L^2(\bR^d)^l\otimes\cdots \otimes L^2(\bR^d)^l$ as $\mathfrak{H}$.  Let $\widehat{P}_{ij}\ (i, j = 1,2,\dots,N)$ be the operator exchanging the $i$-th particle  and the $j$-th one.  That is,  we define

\begin{align*}
& \widehat{P}_{ij}\bigl( f_1(\bx_1)\otimes \cdots \otimes f_i(\bx_i)\otimes \cdots \otimes f_j(\bx_j) \otimes \cdots \otimes f_N(\bx_N)\bigr) \notag \\
& =  f_1(\bx_1)\otimes \cdots \otimes f_j(\bx_i)\otimes \cdots \otimes f_i(\bx_j) \otimes \cdots \otimes f_N(\bx_N)
\end{align*}

for $f_k(\bx_k) \in L^2(\bR^d)^{l}\ (k = 1,2,\dots,N)$ and extend $\widehat{P}_{ij}$ on  $f \in \mathfrak{H}$.  As in the proof of Theorem 4.1 in \cite{Ichinose 2020} (cf. Remark 4.1 in \cite{Ichinose 2020}), we can prove  that if $f  \in \mathfrak{H}$ is symmetric, i.e. $\hat{P}_{ij}f = f$ for all $i$ and $j$, so is $K^{\sharp}_{s}(t,0)f$ and that if $f \in \mathfrak{H}$ is antisymmetric, i.e. $\hat{P}_{ij}f = - f$ for all $i \not= j$, so is $K^{\sharp}_{s}(t,0)f$. Consequently, we see that our RFPIs are expressing bosons and fermions.
\end{rem}

\begin{rem}
We consider polynomially growing potentials

\begin{equation} \label{2.21}
V(t,x) = |x|^{2(l_0+1)} + \sum_{|\alpha| \leq 2l_0 + 1}a_{\alpha}(t)x^{\alpha},
\end{equation}

\begin{equation} \label{2.22}
A_j(t,x) =  \sum_{|\alpha| \leq l_0}b_{j\alpha}(t)x^{\alpha}\quad (j = 1,2,\dots,d)
\end{equation}

with an integer $l_0 \geq 1$ and functions $a_{\alpha}(t) \in \bR, b_{j\alpha}(t) \in \bR$ in $C^1([0,T])$, which do not satisfy either Assumption 2.A, 2.B or 2.C.  We suppose Assumption 2.D, \eqref{1.11} and \eqref{2.16} for $W(t,x), W_s(t,x)$ and $H_s(t,x)$ respectively where we replace \eqref{2.9} with

\begin{equation} \label{2.23}
|\partial_x^{\alpha}W(t,x)| \leq C_{\alpha}<x>^{l_0+1},\quad |\alpha| \geq 1,
\end{equation}

and define $\ksdelta(t,0)f$ by \eqref{2.15}.  Then there exists a constant $\rho^* > 0$ such that we have the same statements as in Theorem 2.3.  In more general, we suppose about potentials $(V,A)$ that Assumptions 2.1 and 2.2 in \cite{Ichinose 2020} are satisfied.  Then, we can prove the same statements as above by using Lemma 5.6, (5.32), (5.37), Theorem 6.5 in \cite{Ichinose 2020} and Theorem 2.1 in \cite{Ichinose 2019},  following the proof of Theorem 2.3 in the present paper.
\end{rem}

  \section{Proofs of Theorems 2.1 and 2.2}

  Let $S(t,s;q)$ and $S_w(t,s;q)$ be the classical actions defined by \eqref{1.3} and \eqref{1.6}, respectively.
  Let $\qts$  be the straight line defined by

\begin{equation}  \label{3.1}
  q^{t,s}_{x,y}(\theta) = y + \frac{\theta-s}{t-s}(x-y), \quad s\leq \theta \leq t
\end{equation}

and write

\begin{equation} \label{3.2}
\gamma^{t,s}_{x,y}: \gamma^{t,s}_{x,y}(\theta) = (\theta,\qts(\theta)) \in \bR^{d+1}, \  s\leq \theta \leq t.
\end{equation}

Then we have

\begin{align}  \label{3.3}
   S(t,s;q^{t,s}_{x,y})
       & = \frac{m|x - y|^2}{2(t - s)} + \int_{\gamma^{t,s}_{x,y}} \bigl(A\cdot dx - Vdt\bigr)
        \notag \\

     & = \frac{m|x - y|^2}{2(t - s)} + (x - y)\cdot\int^1_0 A(s+ \theta\rho,
y+ \theta (x - y))d\theta \notag \\

   & \qquad -  \int^t_s  V\left(\theta, y+ \frac{\theta - s}{t-s} (x - y)\right)d\theta \notag \\

   & = \frac{m|x - y|^2}{2(t - s)} + (x - y)\cdot\int^1_0 A(t - \theta\rho,
x - \theta (x - y))d\theta \notag \\

   & \qquad - \rho \int^1_0  V(t - \theta\rho,x - \theta (x - y))d\theta, \ \rho = t - s,
\end{align}

\begin{align}  \label{3.4}
   & S_w(t,s;q^{t,s}_{x,y})
        = S(t,s;q^{t,s}_{x,y}) + i\int^t_s  W\left(\theta, y+ \frac{\theta - s}{t-s} (x - y)\right)d\theta \notag \\
       &\qquad = S(t,s;q^{t,s}_{x,y}) + i\rho \int^1_0  W(t - \theta\rho,x - \theta (x - y))d\theta.
\end{align}

\par

Let $M \geq 0$ be a constant and suppose that $p(x,w) \in \Cspace(\bR^{2d})$ satisfies

\begin{equation} \label{3.5}
|\partial_{w}^{\alpha}\partial_{x}^{\beta}p(x,w)| \leq C_{\alpha\beta}<x;w>^M, \ (x,w) \in \bR^{2d}
\end{equation}

for all $\alpha$ and $\beta$, where $<x;w> = \sqrt{1 + |x|^2 + |w|^2}$. For $f \in \Czerospace$ we define

\begin{equation}  \label{3.6}
P(t,s)f =
        \begin{cases}
            \begin{split}
              & \sqrt{m/(2\pi i\rho)}^{\ d}
                  \int \bigl(\exp iS_w(t,s; q^{t,s}_{x,y})\bigr) \\
         &\hspace{2cm} \times    p(x,(x-y)/\sqrt{\rho})f(y)dy,\quad  s < t,
                      \end{split}

                  \\
\begin{split}
        & \sqrt{m/(2\pi i)}^{\ d}
    \text{Os}-\int (\exp im|w|^2/2)\\
             &\hspace{2cm}\times p(x,w)dwf(x), \quad s = t.
             \end{split}

        \end{cases}
\end{equation}

 Then, the formal adjoint operator $P(t,s)^{\dag}$ of $P(t,s)$ on $\Czerospace$ is given by

\begin{equation*}
P(t,s)^{\dag}f =
        \begin{cases}
            \begin{split}
              & \sqrt{i m/(2\pi \rho)}^{\ d}
                  \int \bigl(\exp iS_w(t,s; q^{t,s}_{y,x})\bigr)^* \\
         &\hspace{2cm} \times    p(y,(y-x)/\sqrt{\rho})^*f(y)dy,\quad  s < t,
                      \end{split}

                  \\
\begin{split}
        & \sqrt{im/(2\pi )}^{\ d}
    \text{Os}-\int (\exp -im|w|^2/2)\\
             &\hspace{2cm}\times p(x,w)^{*}dwf(x),\quad  s = t.
             \end{split}

        \end{cases}
\end{equation*}

We can prove the following as in the proof of Lemma 2.1 of \cite{Ichinose 1999}.

\begin{lem}
Let $p(x,w)$ be a function satisfying \eqref{3.5}.  We assume \eqref{2.7}, that $\partial_x^{\alpha}V(t,x)$ and  $\partial_x^{\alpha}A_j(t,x)\ (j = 1,2,\dots,d)$  are continuous in $\domain$ for all $\alpha$ and that there exists a constant $M' \geq 0$ satisfying

\begin{equation*}
|\partial_x^{\alpha}V(t,x)| + \sum_{j=1}^d|\partial_x^{\alpha}A_j(t,x)| + |\partial_x^{\alpha}W(t,x)| \leq C_{\alpha}<x>^{M'}
\end{equation*}

in $\domain$ for all $\alpha$.
 Then, for $f \in \Sspace$ $\partial_{x}^{\alpha}(P(t,s)f)$ are continuous in $0 \leq s \leq t \leq T$ and $x \in \bR^{d}$ for all $\alpha$.
\end{lem}

 	In particular, when $p(x,w) = 1$, we write $P(t,s)f$ as $\Cts(t,s)f$.  That is,

\begin{equation}  \label{3.7}
\mathcal{C}(t,s)f =
        \begin{cases}
            \begin{split}
              & \sqrt{m/(2\pi i\rho)}^{\ d}
                  \int \bigl(\exp iS_w(t,s; q^{t,s}_{x,y})\bigr)f(y)dy, \quad s < t, \\
                      \end{split}

                  \\
\begin{split}
        & f,\qquad s = t.
             \end{split}

        \end{cases}
\end{equation}

Then, from \eqref{2.1} we can write

\begin{equation}  \label{3.8}
     \kdelta(t,0)f = \limepsilon{\cal C}(t,\tau_{\nu-1})\chi(\epsilon\cdot){\cal C}(\tau_{\nu-1},\tau_{\nu-2})\chi(\epsilon\cdot)\cdots \chi(\epsilon\cdot)
{\cal C}(\tau_1,0)f
\end{equation}

for $f \in \Czerospace$.

\begin{rem}
In \S 2 we assumed $0 = \tau_0 < \tau_1 < \dots < \tau_{\nu - 1} < \tau_{\nu} = t$.  But, if we define $\kdelta(t,0)f$ by \eqref{3.8} in place of \eqref{2.1}, we can assume $0 = \tau_0 \leq \tau_1 \leq \dots \leq \tau_{\nu - 1} \leq \tau_{\nu} = t$.  Then, we can see that all theorems stated in \S 2 remain valid, following our proofs.
\end{rem}

	For a weight function $W(t,x)$, we set

\begin{equation}  \label{3.9}
    c_w(t,s;x,y) = \exp \left(-\rho\int_0^1W(t - \theta\rho,x - \theta(x - y))d\theta\right), \ \rho = t - s.
    \end{equation}

\begin{lem}
Let $p(x,w)$ be a function satisfying \eqref{3.5}. We assume that $\partial_x^{\alpha}V(t,x), \partial_x^{\alpha}A_j(t,x)$ and  $\partial_x^{\alpha}\partial_tA_j(t,x)$ are continuous in $\domain$ for  $|\alpha| \leq 1$ and $j = 1,2,\dots,d$. Let $f \in \Czerospace$. Then, for any $0 < \epsilon \leq 1$ and $0 \leq s < t \leq T$ we have

\begin{align}  \label{3.10}
    & P(t,s)^{\dag}\chi(\epsilon\cdot)^2P(t,s)f = \left(\frac{m}{2\pi(t-s)}\right)^d\int f(y)dy \int\chi(\epsilon z)^2
    \notag \\
    &\qquad \times  \left(\exp i(x - y)\cdot \frac{m\Phi}{t-s} \right) c_w(t,s;z,x)c_w(t,s;z,y)\notag \\
   &\qquad \times p\left(z,\frac{z - x}{\sqrt{t-s}}\right)^*p\left(z,\frac{z - y}{\sqrt{t-s}}\right)dz,
\end{align}

\begin{equation*}
  \Phi = \Phi(t,s;x,y,z) = (\Phi_1,\dots,\Phi_d),
\end{equation*}

\begin{align}  \label{3.11}
    & \Phi_j =  z_j - \frac{x_j + y_j}{2} + \frac{t - s }{m}
              \int_0^1A_j(s,x + \theta(y - x))d\theta           \notag \\

      & -\frac{(t - s)^2 }{m}\int_0^1\int_0^1
                \sigma_1E_j(\tau(\sigma),\zeta(\sigma))
                       d\sigma_1d\sigma_2 \notag \\

                        & - \frac{t - s }{m}\sum_{k=1}^d (z_k -
x_k)\int_0^1\int_0^1\sigma_1B_{jk}
                     (\tau(\sigma),\zeta(\sigma))d\sigma_1d\sigma_2
\end{align}

or

\begin{align}  \label{3.12}
    & \Phi_j =  z_j - \frac{x_j + y_j}{2} + \frac{t - s }{m}
              \int_0^1A_j(s,x + \theta(y - x))d\theta           \notag \\

      & -\frac{(t - s)^2 }{m}\int_0^1\int_0^1
                \sigma_1E_j(\tau(\sigma),\zeta(\sigma))
                       d\sigma_1d\sigma_2  - \frac{(t - s)^2 }{m}\int_0^1d\theta \sum_{k=1}^d (z_k -
x_k) \notag \\

                        &\qquad \times \int_0^1\int_0^1\sigma_1(1-\sigma_1)\frac{\partial B_{jk}}{\partial t}
                     (s+\theta(1-\sigma_1)\rho,\zeta(\sigma))d\sigma_1d\sigma_2,
\end{align}

where

\begin{equation}  \label{3.13}
    \bigl(\tau(\sigma),\zeta(\sigma)\bigr)
       = \bigl(t - \sigma_1(t - s), z + \sigma_1(x - z)
                  + \sigma_1\sigma_2(y - x)\bigr)\ \in \bR^{d+1}.
\end{equation}

\end{lem}

\begin{proof}
From \eqref{3.4}, \eqref{3.6} and \eqref{3.9}, we can write

\begin{align}  \label{3.14}
    P(t,s)f & = \sqrt{m/(2\pi i\rho)}^{\ d}
                  \int \bigl(\exp iS(t,s; q^{t,s}_{x,y})\bigr)c_w(t,s;x,y) \notag \\
                  & \times p\left(x,\frac{x - y}{\sqrt{t-s}}\right)f(y)dy,\quad \rho = t - s > 0.
\end{align}

Hence, we can easily prove Lemma 3.2 from Lemma 5.2 in \cite{Ichinose 2020}.

\end{proof}

\begin{lem}
We assume that $ \partial_x^{\alpha}A_j(t,x)\ (j = 1,2,\dots,d)$  are continuous for all $\alpha$ and satisfy \eqref{2.5}  in $\domain$.
\\

(1) Suppose that  Assumption 2.A is satisfied.  Let $\Phi_j(t,s;x,y,z)\ (j = 1,2,\dots,d)$ be the functions defined by \eqref{3.11}.  Then, there exist a constant $\rho^* > 0$ such that for all fixed $0 \leq t - s \leq \rho^*$ and $(x,y) \in \bR^{2d}$, the map: $\bR^d \ni z \to \xi = \Phi(\ts;x,y,z) \in \bR^d$ is a homeomorphism, whose inverse will be denoted by the map: $\bR^d \ni \xi \to z = z(\ts;x,\xi,y) \in \bR^d$, and we have

\begin{equation}   \label{3.15}
           \sum_{j=1}^d |\partial_{\xi}^{\alpha}\partial_x^{\beta}\partial_{y}^{\gamma}
            z_j(t,s;x,\xi,y)|  \leq C_{\alpha\beta\gamma},
\  |\alpha + \beta  + \gamma| \geq 1,
           \end{equation}

\begin{equation}   \label{3.16}
           \det \frac{\partial z}{\partial\xi}(t,s;x,\xi,y) = 1 + (t - s)h(t,s;x,\xi,y) > 0,
           \end{equation}

\begin{equation}   \label{3.17}
            |\partial_{\xi}^{\alpha}\partial_x^{\beta}\partial_{y}^{\gamma}
            h(t,s;x,\xi,y)|  \leq C_{\alpha\beta\gamma} < \infty
           \end{equation}

for all $\alpha, \beta$ and $\gamma$ in $0 \leq t - s \leq \rho^*$ and $(x,\xi,y) \in \bR^{3d}$.
\\

(2)  Suppose that Assumption 2.B is satisfied.  Let  $\Phi_j(t,s;x,y,z)\ (j = 1,2,\dots,d)$ be the functions defined by \eqref{3.12}.  Then, we have the same statements as in  (1).

\end{lem}

\begin{proof}
We have already proved (1) in Lemma 3.2, (3.9) and (3.10) of \cite{Ichinose 1999} (cf.  Lemma 3.6, (3.18) and (3.19) of \cite{Ichinose 1997}).
\par
	We will prove (2).  Writing the $5$-th term on the right-hand side of \eqref{3.12} as $-(t-s)^2B'(t,s;x,y,z)/m$, from the assumption \eqref{2.4} we have

\begin{equation}   \label{3.17-2}
            |\partial_{x}^{\alpha}\partial_y^{\beta}\partial_{z}^{\gamma}
            B'_j(t,s;x,\xi,y)|  \leq C_{\alpha\beta\gamma}, \quad |\alpha + \beta + \gamma| \geq 1
  \end{equation}

  in $0 \leq s \leq t \leq T$ and $(x,y,z) \in \bR^{3d}$, where $B' = (B'_1,\cdots,B'_d) \in \bR^d$.   The inequalties \eqref{3.17-2} can be proved as in the proof of \eqref{3.15} of \cite{Ichinose 1997}.  Thereby we can prove (2) as in the proof of (1).

\end{proof}

 	From now on we fix $\rho^* > 0$ determined in Lemma 3.3 throughout the present paper.  The following lemma is crucial in the present paper.

	\begin{lem}
	Assume \eqref{2.7} and \eqref{2.8} where we suppose $C(W) = 0$.  Let $c_w(t,s;x,y)$ be the function defined by \eqref{3.9}.  Then we have

\begin{equation}   \label{3.18}
            0 \leq c_w(t,s;x,y) \leq 1, \
           |\partial_x^{\alpha}\partial_y^{\beta}c_w(t,s;x,y)| \leq C_{\alpha\beta}, \ |\alpha + \beta| \geq 1
 \end{equation}

for $0 \leq s \leq t \leq T$ and $(x,y) \in \bR^{2d}$ with constants $C_{\alpha\beta} \geq 0$.

	\end{lem}

\begin{proof}
It is clear from \eqref{2.7} and \eqref{3.9} that the first inequality of \eqref{3.18} holds.  For $a \geq 0$ we can easily see

\begin{equation}   \label{3.19}
           \sup_{r \geq 0} e^{-r}(T + r)^a = C'_a < \infty
 \end{equation}

with constants $C'_a \geq 0$.  Let $|\alpha| \geq 1$.  Using H\"older's inequality and \eqref{2.8}, we have

\begin{align} \label{3.20}
& \rho \int_0^1|(\partial_x^{\alpha}W)(t - \theta\rho,x - \theta(x-y))|d\theta \notag \\
& \leq \rho \biggl(\int_0^1|(\partial_x^{\alpha}W)(t - \theta\rho,x - \theta(x-y))|^{p_{\alpha}}d\theta \biggr)^{1/p_{\alpha}} \notag \\
& \leq C_{\alpha}^{1/p_{\alpha}}\rho \biggl[\int_0^1\bigl\{1 + W(t - \theta\rho,x - \theta(x-y))\bigr\}d\theta \biggr]^{1/p_{\alpha}} \notag \\
& =  C_{\alpha}^{1/p_{\alpha}}\rho^{1 - 1/p_{\alpha}}\biggl\{\rho + \rho\int_0^1  W(t - \theta\rho,x - \theta(x-y))d\theta \biggr\}^{1/p_{\alpha}} \notag \\
& \leq  C_{\alpha}^{1/p_{\alpha}}\rho^{1 - 1/p_{\alpha}}\biggl\{T + \rho\int_0^1  W(t - \theta\rho,x - \theta(x-y))d\theta \biggr\}^{1/p_{\alpha}}.
\end{align}

Hence, letting $\alpha = (1,0,\dots,0) \in \bR^d$, by \eqref{2.7} and \eqref{3.19} we have

\begin{align} \label{3.21}
& |\partial_{x_1}c_w(t,s;x,y)| \leq \biggl(\exp -\rho\int_0^1W(t - \theta\rho,x - \theta(x-y))d\theta\biggl)
\notag \\
& \times C_{\alpha}^{1/p_{\alpha}}T^{1 - 1/p_{\alpha}}\biggl\{T + \rho\int_0^1  W(t - \theta\rho,x - \theta(x-y))d\theta \biggr\}^{1/p_{\alpha}}
\notag \\
& \leq C_{\alpha}^{1/p_{\alpha}}T^{1 - 1/p_{\alpha}}C'_{1/p_{\alpha}} < \infty.
\end{align}

In the same way we can complete the proof of  the second inequality of \eqref{3.18}, using \eqref{3.19} and \eqref{3.20}.

\end{proof}

\begin{pro}
Suppose that either Assumption 2.A or 2.B is satisfied.  In addition, we suppose Assumption 2.C, \eqref{2.7} and \eqref{2.8}, where $C(W) = 0$ and \eqref{2.6} is replaced with

\begin{equation} \label{3.22}
|\partial_x^{\alpha}V(t,x)| \leq C_{\alpha}<x>^{M_1}, \quad |\alpha| \geq 1
\end{equation}

with a constant $M_1 \geq 1$ independent of $\alpha$.  Let $\rho^* > 0$ be the constant determined in Lemma 3.3 and $\Cts(t,s)$ the operator defined by \eqref{3.7}.  Then, there exists a constant $K_0 \geq 0$ such that

\begin{equation}   \label{3.23}
 \Vert \Cts(\ts)f\Vert \leq e^{K_0(t-s)}\Vert f \Vert, \quad 0 \leq t - s \leq \rho^*
           \end{equation}

           for all $f \in L^2$.

\end{pro}

\begin{proof}
We first suppose that Assumption 2.A is satisfied.  Then, letting $\Phi_j$ be defined by \eqref{3.11}, from \eqref{3.10} we have

\begin{align*}
    & \Cts(t,s)^{\dag}\chi(\epsilon\cdot)^2\Cts(t,s)f = \left(\frac{m}{2\pi(t-s)}\right)^d\int f(y)dy \int\chi(\epsilon z)^2
    \notag \\
    &\qquad \times  \left(\exp i(x - y)\cdot \frac{m\Phi}{t-s} \right) c_w(t,s;z,x)c_w(t,s;z,y)dz
  \end{align*}

  for $f \in \Sspace$.  We will use (1) in Lemma 3.3.  Letting $0 \leq t - s \leq \rho^*$ and making the change of variables: $\bR^d \ni z \to \xi = \Phi(t,s;x,y,z) \in \bR^d$ in the above equation, we have

\begin{align*}
    & \Cts(t,s)^{\dag}\chi(\epsilon\cdot)^2\Cts(t,s)f = \left(\frac{m}{2\pi(t-s)}\right)^d\int f(y)dy \int\chi(\epsilon z)^2
    \notag \\
    & \times  \left(\exp i(x - y)\cdot \frac{m\xi}{t-s} \right) c_w(t,s;z,x)c_w(t,s;z,y)
    \det\frac{\partial z}{\partial \xi}(t,s;x,\xi,y)d\xi
  \end{align*}

  with $z = z(t,s;x,\xi,y)$, which shows

  \begin{align} \label{3.24}
    & \Cts(t,s)^{\dag}\chi(\epsilon\cdot)^2\Cts(t,s)f = \left(\frac{1}{2\pi}\right)^d\int e^{i(x-y)\cdot\eta}d\eta\int\chi(\epsilon z)^2c_w(t,s;z,x)
       \notag \\
    & \ \times c_w(t,s;z,y)\det\frac{\partial z}{\partial \xi}(t,s;x,(t-s)\eta/m,y) f(y)dy,\  0 \leq t - s \leq \rho^*
  \end{align}

  with $z = z(t,s;x,(t-s)\eta/m,y)$.  Hence, noting \eqref{3.15} and \eqref{3.18}, we can easily prove

  \begin{align} \label{3.25}
    &  \limepsilon\Cts(t,s)^{\dag}\chi(\epsilon\cdot)^2\Cts(t,s)f\notag \\
    & = \left(\frac{1}{2\pi}\right)^d\int e^{i(x-y)\cdot\eta}d\eta\int c_w(t,s;z,x)
    c_w(t,s;z,y)
       \notag \\
    & \ \times \det\frac{\partial z}{\partial \xi}(t,s;x,(t-s)\eta/m,y) f(y)dy,\  0 \leq t - s \leq \rho^*
  \end{align}

 with $z = z(t,s;x,(t-s)\eta/m,y)$ in the topology of $\Sspace$, which we write as $\Cts(t,s)^{\dag}\,\Cts(t,s)f$ formally.
 \par
 	Noting \eqref{3.15} - \eqref{3.17} and \eqref{3.18}, let us apply Theorem 1.A to \eqref{3.25}.
Then we have

\begin{align}  \label{3.26}
     \Vert \Cts(\ts)^{\dag}\,\Cts(\ts)f\Vert &\leq \bigl\{1 + 2K_0(t-s)\bigr\}\Vert f\Vert \notag \\
    & \leq e^{2K_0(t-s)} \Vert f\Vert, \quad  0 \leq t - s \leq \rho^*
\end{align}

with a constant $K_0 \geq 0$.  Consequently from \eqref{3.25} we have

\begin{align*}
\Vert\Cts(t,s)f\Vert^2 &\leq \lim_{\epsilon \to 0+}\bigl(\chi(\epsilon\cdot)\Cts(\ts)f,\chi(\epsilon\cdot)\Cts(\ts)f\bigr) \\
    & = \lim_{\epsilon \to 0+}\bigl(\Cts(\ts)^{\dag}\chi(\epsilon\cdot)^2\Cts(\ts)f,f\bigr)  \\
    & = \bigl(\Cts(\ts)^{\dag}\,\Cts(\ts)f,f\bigr) \leq e^{2K_0(t-s)} \Vert f\Vert^2
\end{align*}

for $f \in \Sspace$ by using Fatou's lemma, which shows \eqref{3.23}.
\par
	Next we suppose that Assumption 2.B is satisfied.  Letting $\Phi_j$ be defined by \eqref{3.12}, we can also prove \eqref{3.25} and \eqref{3.26} as in the proof of the first case.  Consequently we can prove \eqref{3.23}.

\end{proof}

\begin{pro}
Let $p(x,w)$ be a function satisfying \eqref{3.5} and $P(t,s)$ the operator defined by \eqref{3.6}.  Then, under the assumptions of Proposition 3.5 we have

\begin{equation} \label{3.27}
\Vert P(t,s)f\Vert_a \leq
     C_a \Vert f\Vert_{M+aM_1}, \quad  0 \leq t - s \leq \rho^*
\end{equation}

for $a = 0,1,2,\dots$ and all $f \in B^{M+aM_1}$ with constants $C_a \geq 0$, where $M_1$ is the constant in \eqref{3.22}.

\end{pro}

\begin{proof}
Setting

\begin{equation} \label{3.28}
p'(t,s;x,w) := p(x,w)c_w(t,s;x,x - \sqrt{\rho}w),
\end{equation}

from \eqref{3.14} we have

\begin{align}  \label{3.29}
    P(t,s)f & = \sqrt{m/(2\pi i\rho)}^{\ d}
                  \int \bigl(\exp iS(t,s; q^{t,s}_{x,y})\bigr)p'\left(t,s;x,\frac{x - y}{\sqrt{t-s}}\right)f(y)dy, \notag \\
                 &\quad \rho = t - s > 0
\end{align}

and also from \eqref{3.18}

\begin{equation} \label{3.30}
|\partial_w^{\alpha}\partial_x^{\beta}p'(t,s;x,w)| \leq C_{\alpha\beta}<x;w>^M
\end{equation}

for all $\alpha$ and $\beta$.
\par

	At first we suppose that Assumption 2.A is satisfied. Then, using \eqref{3.29} and \eqref{3.30}, we can prove \eqref{3.27}  from Theorem 4.4 of \cite{Ichinose 1999}.  We can also prove \eqref{3.27} under Assumption 2.B, noting \eqref{3.17-2} and following the proof of Theorem 4.4 in \cite{Ichinose 1999}.

\end{proof}

\begin{lem}
Let $H_w(t)$ be the operator defined by \eqref{1.8}.  We assume that  for  all $\alpha$ $\partial_x^{\alpha}V(t,x), \partial_x^{\alpha}A_j(t,x)\ (j = 1,2,\dots,d)$ and $ \partial_x^{\alpha}\partial_tA_j(t,x)$  are continuous in $\domain$ and satisfy

\begin{align*}
&|\partial_x^{\alpha}V(t,x)| + \sum_{j=1}^d\bigl(|\partial_x^{\alpha}A_j(t,x)| + |\partial_x^{\alpha}\partial_tA_j(t,x)|  \bigr)+ |\partial_x^{\alpha}W(t,x)|
\notag \\
& \leq C_{\alpha}<x>^{M'}
\end{align*}

with constants $C_{\alpha} \geq 0$ and $M' \geq 0$, where $M'$ is independent of $\alpha$.  Then, there exists a function $r(t,s;x,w)$ satisfying \eqref{3.5} for an $M \geq 0$ such that $\partial_w^{\alpha}\partial_x^{\beta}r(t,s;x,w)$ are continuous in $0 \leq s \leq t \leq T$ and $(x,w) \in \bR^{2d}$ for all $\alpha, \beta$ and we have

\begin{equation}  \label{3.31}
      \left\{i\frac{\partial}{\partial t} - H_w(t)\right\}\Cts(\ts) f
                        = \sqrt{t - s} R(\ts)f
\end{equation}

for $f \in \Czerospace$.

\end{lem}

\begin{proof}
We note \eqref{1.6} and \eqref{3.7}.  Then, replacing $V(t,x)$ with $V(t,x) - iW(t,x)$ in the proof of Lemma 4.1 of \cite{Ichinose 2003}, we can complete the proof of Lemma 3.7.
\end{proof}

\begin{pro}
Besides the assumptions of Proposition 3.5 we assume

\begin{equation*}
 |\partial_x^{\alpha}W(t,x)|
 \leq C_{\alpha}<x>^{M''}
\end{equation*}

in $\domain$ for all $\alpha$ with constants $C_{\alpha} \geq 0$ and $M'' \geq 0$, where $M''$ is independent of $\alpha$.  Then, there exists a function $r(t,s;x,w)$ satisfying the properties stated in Lemma 3.7 and we have

\begin{equation} \label{3.32}
\Vert R(t,s)f\Vert_a \leq
     C_a \Vert f\Vert_{M+aM_1}, \quad  0 \leq t - s \leq \rho^*
\end{equation}

for $a = 0,1,2,\dots$ and all $f \in B^{M+aM_1}$, where $M_1$ is the constant in \eqref{3.22}.

\end{pro}

\begin{proof}
From \eqref{1.1} we have $\partial_tA_j = -E_j - \partial_{x_j}V$.  Hence we see that $\partial_x^{\alpha}\partial_tA_j(t,x)$ are continuous in $\domain$ for all $\alpha$ from the assumptions.  In addition, from \eqref{2.2} and \eqref{3.22} we have

\begin{equation*}
 |\partial_x^{\alpha}\partial_tA_j(t,x)| \leq |\partial_x^{\alpha}E_j(t,x)| + |\partial_x^{\alpha}\partial_{x_j}V(t,x)|
 \leq C_{\alpha}<x>^{M_1}, \ |\alpha| \geq 1.
\end{equation*}

Therefore, Lemma 3.7 holds.  Applying Proposition 3.6 to $R(t,s)f$, we get \eqref{3.32}.

\end{proof}

	We have proved the following in (1) of Theorem 2.1 of \cite{Ichinose 2019}.
	\vspace{0.5cm}
	\par

{\bf Theorem 3.A.}  {\it Suppose Assumption 2.C and \eqref{2.9}.  Then for any $u_0 \in B^a\ (a = 0, \pm1,\pm2,\dots)$
 there exists the unique solution $u(t)$ in $C^0_t([0,T];B^a) \cap C^1_t([0,T];B^{a-1})$  with $u(0) = u_0$ to the equation \eqref{1.8}. This solution $u(t)$ satisfies

 \begin{equation} \label{3.33}
 \Vert u(t)\Vert_a\leq  C_a\Vert u_0\Vert_a,  \ 0 \leq t \leq T.
 \end{equation}

 }
 \par

 	{\bf The proof of Theorem 2.1}.  Let $C(W) \geq 0$ be the constant in Assumption 2.D and $\kdelta(t,0)f$ the operator defined by \eqref{2.1}. We have only to study $e^{-tC(W)}\kdelta(t,0)f$ in place of $\kdelta(t,0)f$ to prove Theorem 2.1.
	Accordingly, since from \eqref{1.5} and \eqref{1.6} we have

\begin{align*}
& iS_w(t,0;\qdelta) - tC(W) \\
& = i\int_0^t\bigl[\mathcal{L}(\theta,\qdelta(\theta),\dot{\qdelta}(\theta)) + i\left\{ W(\theta,\qdelta(\theta)) + C(W)\right\}\bigr]d\theta,
\end{align*}

we can take $W'(t,x) = W(t,x) + C(W) \geq 0$  as a weight function in place of $W(t,x)$.  Hence we may assume $C(W) = 0$ in the proof of Theorem 2.1.
	Let $C(W) = 0$ in the proof  below.
	\par

	Noting Assumption 2.A, from Lemma 6.1 of \cite{Ichinose 1999} we can find a potential $(V,A)$ satisfying Assumption 2.C. We fix this potential $(V,A)$ for a while.  Then, we have \eqref{3.23}, \eqref{3.27} and \eqref{3.31} - \eqref{3.33}.  Hence, if we note \eqref{3.8}, we can prove the statements (1) - (3) of Theorem 2.1, following the proof of Proposition 6.3 of \cite{Ichinose 1999}.
	\par

	Now, let $(V,A)$ be an arbitrary potential for $E$ and $(B_{jk})_{1 \leq j < k \leq d}$ with continuos $V, \partial V/\partial x_j, \partial A_j/\partial t$ and $\partial A_j/\partial x_k$.  Then, using the gauge transformation, as in the proof of Theorem of \cite{Ichinose 1999} we can complete the proof of Theorem 2.1.
	\par

	\vspace{0.3cm}
	Making the change of variables: $\bR^d \ni y \to w = (x-y)/\sqrt{\rho} \in \bR^d$ in \eqref{3.6}, from \eqref{3.3} and \eqref{3.4} we have

\begin{equation}  \label{3.34}
    P(t,s)f  = \sqrt{\frac{m}{2\pi i}}^{\, d}
                  \int e^{i\phi(t,s;x,w)}p(x,w)f(x-\sqrt{\rho}w)dw,\  \rho = t - s > 0
\end{equation}

for $f \in \Czerospace$, where

\begin{align}  \label{3.35}
   & \phi(t,s;x,w)  = \frac{m}{2}|w|^2 + \sqrt{\rho}w\cdot\int_0^1A(t-\theta\rho,x-\theta\sqrt{\rho}w)d\theta
   \notag \\
   & -\rho\int_0^1V(t-\theta\rho,x-\theta\sqrt{\rho}w)d\theta + i\rho\int_0^1W(t-\theta\rho,x-\theta\sqrt{\rho}w)d\theta.
                  \end{align}

\begin{lem}
Suppose Assumption 2.C and \eqref{2.9}. Let $\Cts(t,s)$ be the operator defined by \eqref{3.7}.  Then, for an arbitrary multi-index $\kappa$, both of commutators $[\partial_x^{\kappa},\Cts(t,s)]f$ and $[x^{\kappa},\Cts(t,s)]f$ for $f \in \Czerospace$ are written in the form

\begin{align}  \label{3.36}
   & (t-s)\sum_{|\gamma| < |\kappa|}\widetilde{P}_{\gamma}(t,s)(\partial_x^{\gamma}f) := (t-s) \sum_{|\gamma| < |\kappa|}   \sqrt{\frac{m}{2\pi i}}^{\, d}\notag \\
   & \times \int e^{i\phi(t,s;x,w)}p_{\gamma}(t,s;x,\sqrt{\rho}w)(\partial_x^{\gamma}f)(x-\sqrt{\rho}w)dw,
   \end{align}

   where $p_{\gamma}(t,s;x,\zeta)$ satisfy

\begin{equation}  \label{3.37}
   |\partial_{\zeta}^{\alpha}\partial_x^{\beta}p_{\gamma}(t,s;x,\zeta)| \leq C_{\alpha\beta}<x;\zeta>^{|\kappa| - |\gamma|}
\end{equation}

for all $\alpha$ and $\beta$.

\end{lem}

\begin{proof}
Replacing $V(t,x)$ with $V(t,x) - iW(t,x)$ in the proof of Lemma 3.2 of \cite{Ichinose 2003}, we can prove Lemma 3.9.
\end{proof}

\begin{pro}
Suppose that the assumptions of Theorem 2.2 are satisfied, where $C(W) = 0$.  Then, for $a = 0,1,2,\dots$ there exist constants $K_a \geq 0$ such that

\begin{equation}   \label{3.38}
 \Vert \Cts(\ts)f\Vert_a \leq e^{K_a(t-s)}\Vert f \Vert_a, \quad 0 \leq t - s \leq \rho^*
           \end{equation}

for all $f \in B^a$.
\end{pro}

 \begin{proof}
 As in the proof of Proposition 3.4 of \cite{Ichinose 2003}, we can prove \eqref{3.38} from \eqref{3.23}, \eqref{3.27} with $M_1 = 1$ and Lemma 3.9.
 \end{proof}

\par

{\bf The proof of Theorem 2.2.}  As in the proof of Theorem 2.1, we may assume $C(W) = 0$.  Then, noting \eqref{3.8}, and using \eqref{3.27}, \eqref{3.31} - \eqref{3.33} and \eqref{3.38}, we can complete the proof of Theorem 2.2 as in the proof of Theorem 2.1.

\section{Proofs of Theorems 2.3 - 2.6}

	Let $C(W)$ be the constant in Assumption 2.D and $W_s(t,x)$ the Hermitian matrix in Theorems 2.3 and 2.4. As in the proofs of Theorems 2.1 and 2.2 we may assume

\begin{equation}   \label{4.1}
 C(W) = 0,\quad W_s(t,x) \geq 0
 \end{equation}

 in the proofs of Theorems 2.3 and 2.4.
 \par

 	Using $\mathcal{F}(t,s;\qts)$ defined by the solution to \eqref{2.13} with $\mathcal{U}(s) = I$, we define

\begin{equation}  \label{4.2}
\mathcal{C}_s(t,s)f =
        \begin{cases}
            \begin{split}
              & \sqrt{m/(2\pi i\rho)}^{\ d}
                  \int \bigl(\exp iS_w(t,s; q^{t,s}_{x,y})\bigr) \\
         &\quad \times    \mathcal{F}(t,s;\qts)f(y)dy,\quad  s < t,
                      \end{split}

                  \\

         f, \quad s = t

        \end{cases}
\end{equation}

for $f \in \Czerospace^l$, which is corresponding to $\mathcal{C}(t,s)$ defined by \eqref{3.7}.  Then, we can write $\ksdelta(t,0)f$ defined by \eqref{2.15} as

\begin{equation}  \label{4.3}
     \ksdelta(t,0)f = \limepsilon{\cal C}_s(t,\tau_{\nu-1})\chi(\epsilon\cdot){\cal C}_s(\tau_{\nu-1},\tau_{\nu-2})\chi(\epsilon\cdot)\cdots \chi(\epsilon\cdot)
{\cal C}_s(\tau_1,0)f
\end{equation}

for $f \in \Czerospace^l$ in the same way as we did \eqref{3.8}, using

\begin{align} \label{4.4}
& \mathcal{F}(t,0;\qdelta) \notag \\
& = \mathcal{F}\bigl(t,\tau_{\nu-1};q^{t,\tau_{\nu-1}}_{x,x^{(\nu-1)}}\bigr)\mathcal{F}\bigl(\tau_{\nu-1},\tau_{\nu-2};q^{\tau_{\nu-1},\tau_{\nu-2}}_{x^{(\nu-1)},x^{(\nu-2)}}\bigr) \cdots
  \mathcal{F}\bigl(\tau_{1},0;q^{\tau_{1},0}_{x^{(1)},x^{(0)}}\bigr)
\end{align}

which has been easily proved in Lemma 2.1 of \cite{Ichinose 2007}.

\begin{lem}
(1)  Assume $W_s(t,x) \geq 0$ in $\domain$.  Let $q(\theta) \in \bR^d\ (s \leq \theta \leq t)$ be a continuous path.  Then we have

\begin{equation} \label{4.5}
0 \leq \mathcal{F}(t,s;q)^{\dag}\mathcal{F}(t,s;q) \leq 1,
\end{equation}

\begin{equation} \label{4.6}
\sum_{i=1}^l |\mathcal{F}_{ij}(t,s;q)|^2 \leq 1,\ j = 1,2,\dots,l,
\end{equation}

where $\mathcal{F}_{ij}(t,s;q)$ denotes the $(i,j)$-component of $\mathcal{F}(t,s;q)$. (2) Assume $W_s(t,x) \geq 0$ in $\domain$ and

\begin{equation} \label{4.7}
 |\partial_{x}^{\alpha} h_{sjk}(t,x)| \leq C_{\alpha},\ |\alpha| \geq 1,
\end{equation}

\begin{equation} \label{4.8}
|\partial_{x}^{\alpha} w_{sjk}(t,x)| \leq C_{\alpha},\ |\alpha| \geq 1
\end{equation}

in $\domain$ for  $i,j = 1,2,\dots,l$.  Then we have

\begin{equation} \label{4.9}
|\partial_{x}^{\alpha}\partial_{y}^{\beta} \mathcal{F}(t',s';\qts)| \leq C_{\alpha\beta}(t' - s'),\ |\alpha+ \beta| \geq 1
\end{equation}

for $0 \leq s \leq s' \leq t' \leq t \leq T$ and $(x,y) \in \bR^{2d}$, where $|\Omega|$ denotes the Hilbert-Schmidt norm $\bigl( \sum_{i,j=1}^l|\Omega_{ij}|^2\bigr)^{1/2}$ of  a matrix $\Omega =(\Omega_{ij};i\downarrow j \rightarrow 1,2,\dots,l)$.

\end{lem}

\begin{proof}
(1) We set $\mathcal{U}(t) = \mathcal{F}(t,s;q)$.  From \eqref{2.13} we have

\begin{align} \label{4.10}
& \frac{d}{d\theta}\, \mathcal{U}(\theta)^{\dag}\,\mathcal{U}(\theta) = \mathcal{U}(\theta)^{\dag}\bigl\{iH_s(\theta,q(\theta)) - W_s(\theta,q(\theta))\bigr\}\mathcal{U}(\theta)
\notag \\
&\quad - \mathcal{U}(\theta)^{\dag}\bigl\{iH_s(\theta,q(\theta)) + W_s(\theta,q(\theta))\bigr\}\mathcal{U}(\theta)
\notag \\
& = -2 \mathcal{U}(\theta)^{\dag}W_s(\theta,q(\theta))\mathcal{U}(\theta) \leq 0.
\end{align}

Hence we have \eqref{4.5} because of $\mathcal{U}(s) = I$.  Taking $e_1 = {}^t(1,0,\dots,0) \in \bR^l$, from \eqref{4.5} we have

\begin{equation*}
1 \geq \, <\mathcal{U}(t)^{\dag}\,\mathcal{U}(t)e_1,e_1>\, = \sum_{i=1}^l |\mathcal{F}_{i1}(t,s;q)|^2,
\end{equation*}

where $<\cdot,\cdot>$ denotes the usual inner product of $\bR^l$.  In the same way we can prove \eqref{4.6}.
\par

(2) From \eqref{2.13} we can easily see

\begin{align} \label{4.11}
   & \frac{\partial}{\partial x_j}\mathcal{F} (t',s';\qts)  = -\int_{s'}^{t'} \mathcal{F} (t',\theta;\qts)
   \left[\frac{\partial}{\partial x_j}\bigl\{
iH_s(\theta,\qts(\theta)) +  W_s(\theta,\qts(\theta)) \bigr\}\right]
\notag \\
& \qquad
\times \mathcal{F} (\theta,s';\qts)d\theta
\end{align}

(cf. (3.3) in \cite{Ichinose 2007}).  Then, noting \eqref{3.1}, from \eqref{4.6} - \eqref{4.8} we can prove

\begin{equation*}
|\partial_{x_j} \mathcal{F}(t',s';\qts)| \leq C(t' - s')
\end{equation*}

with a constant $C \geq 0$. In the same way we can prove \eqref{4.9} from \eqref{4.11} by induction.

\end{proof}

\begin{lem}
Assume $W_s(t,x) \geq 0$, \eqref{1.11} and \eqref{2.16}.  Then we have

\begin{equation} \label{4.12}
\left|\partial_x^{\alpha}\partial_y^{\beta}\left\{ \calF
(t,s;q^{t,s}_{x,y})- I \right\}
\right| \leq C_{\alpha,\beta} (t - s)
\end{equation}

in $0 \leq s \leq t \leq T$ and $(x,y) \in \bR^{2d}$ for all $\alpha$ and $\beta$.

\end{lem}

\begin{proof}
From \eqref{2.13} we have

\begin{equation*}
\calF (t',s;q^{t,s}_{x,y})- I =
- \int_{s}^{t'}\bigl\{ iH_s(\theta,\qts (\theta)) + W_s(\theta,\qts (\theta))\bigr\}\calF
(\theta,s;q^{t,s}_{x,y})d\theta.
\end{equation*}

Hence, by \eqref{1.11} , \eqref{2.16} and \eqref{4.6} we see

\begin{equation*}
\left|\calF
(t,s;q^{t,s}_{x,y})- I
\right| \leq C (t - s)
\end{equation*}

in $0 \leq s \leq t \leq T$ and $(x,y) \in \bR^{2d}$ with a constant $C \geq 0$.  The inequalities \eqref{4.12} for $|\alpha + \beta| \geq 1$ follow from \eqref{4.9}.

\end{proof}

\begin{pro}
Besides the assumptions of Proposition 3.5 we suppose $W_s(t,x) \geq 0$,  \eqref{1.11} and \eqref{2.16}.  Let $\rho^* > 0$ be the constant determined in Lemma 3.3 and $\mathcal{C}_s(t,s)$ the operator defined by \eqref{4.2}.  Then, there exists a constant $K'_0 \geq 0$ such that

\begin{equation}   \label{4.13}
 \Vert \Cts_s(\ts)f\Vert \leq e^{K'_0(t-s)}\Vert f \Vert, \quad 0 \leq t - s \leq \rho^*
\end{equation}

           for all $f \in (L^2)^l$.

\end{pro}

\begin{proof}
Using $\Cts(t,s)$ defined by \eqref{3.7}, we write

\begin{align} \label{4.14}
& \Cts_s(t,s) f = \Cts(t,s)f +
\sqrt{\frac{m}{2\pi i\rho}}^{\, d} \int \bigl(\exp iS_w(t,s;
q^{t,s}_{x,y})\bigr)
  \notag \\
& \quad \times
\left\{\calF (t,s;\qts) - I\right\}f(y)dy
\end{align}

for $f \in \Sspace^l$.   Noting \eqref{4.12}, from Proposition 3.6 we see that the $L^2$-norm of the second term on the right-hand side of \eqref{4.14} is bounded by $C(t-s)\Vert f \Vert$ from above with a constant $C \geq 0$. Proposition 3.5 is showing \eqref{3.23}. Hence we have

\begin{align*}
\Vert \Cts_s(t,s) f \Vert &
\leq \mathrm{e}^{K_0(t - s)}\Vert f \Vert + C(t - s)\Vert
f \Vert \\
& \leq \mathrm{e}^{(K_0+C)(t - s)}\Vert f \Vert, \quad 0 \leq t - s \leq \rho^*.
\end{align*}

Therefore, we can prove \eqref{4.13} with a constant $K'_0 \geq 0$.

\end{proof}

\begin{lem}
Besides the assumptions of Lemma 3.7 we assume $W_s(t,x) \geq 0$ and \eqref{4.7} - \eqref{4.8}.  Then, there exist functions $r_{ij}(t,s;x,w)\ (i,j = 1,2,\dots,l)$ satisfying \eqref{3.5} for an $M \geq 0$ such that $\partial_w^{\alpha}\partial_x^{\beta}r_{ij}(t,s;x,w)$ are continuous in $0 \leq s \leq t \leq T$ and $(x,w) \in \bR^{2d}$ for all $\alpha$ and $\beta$, and we have

\begin{align}  \label{4.15}
     & \left\{i\frac{\partial}{\partial t} - H(t) -H_s(t,x) + iW(t,x) + iW_s(t,x)\right\}\Cts_s(\ts) f
     \notag \\
& = \sqrt{t - s} \Bigl(R_{ij}(\ts);i\downarrow j\rightarrow 1,2,\dots,l \Bigr)f \equiv \sqrt{t - s} R(t,s)f
\end{align}

for $f \in \Czerospace^l$, where $R_{ij}(\ts)$ are the operators defined by \eqref{3.6}.
\end{lem}

\begin{proof}
We note that  \eqref{4.6} and \eqref{4.9} hold under our assumptions.  Consequently, replacing $V(t,s)$ and $H_s(t,x)$ with $V(t,x) - iW(t,s)$ and $H_s(t,x) - iW_s(t,x)$, respectively in the proof of Proposition 3.5 of \cite{Ichinose 2007}, we can complete the proof of Lemma 4.4.
\end{proof}

\begin{pro}

Besides the assumptions of Proposition 3.8, we suppose $W_s(t,s) \geq 0$ and \eqref{4.7} - \eqref{4.8}.  Then, there exist functions $r_{ij}(t,s;x,w)\ (i,j = 1,2,\dots,l)$ satisfying the properties stated in Lemma 4.4 and we have

\begin{equation} \label{4.16}
\Vert R(t,s)f\Vert_a \leq C_a\Vert f\Vert_{M+aM_1}, \quad 0 \leq t - s \leq \rho^*
\end{equation}

for $a = 0, 1,2,\dots$ and all $f \in (B^{M+aM_1})^l$, where $M_1$ is the constant in \eqref{3.22}.

\end{pro}

\begin{proof}
As in the proof of Proposition 3.8, we can easily see that the assumptions of Lemma 4.4 hold.  Hence, using Lemma 4.4, from Proposition 3.6 we can prove \eqref{4.16}.
\end{proof}

{\bf Proof of Theorem 2.3.}
We may assume \eqref{4.1}.  If we suppose \eqref{1.11} and \eqref{2.16} besides the assumptions of  Theorem 3.A, then we can easily see that the same statements for the equation \eqref{1.10} as for \eqref{1.8} in Theorem 3.A hold (cf. Theorem 2.1 in \cite{Ichinose 2019}). In addition, we have already proved \eqref{4.3}, \eqref{4.13} and \eqref{4.15} - \eqref{4.16}. Hence, using \eqref{3.27}, we can complete the proof of Theorem 2.3 as in the proof of Theorem 2.1.

\begin{pro}
Besides the assumptions of Theorem 2.2 we suppose \eqref{1.11}, \eqref{2.16} and \eqref{4.1}.  Then, for $a = 0, 1, 2, \dots$ there exist constants $K'_a \geq 0$ such that

\begin{equation} \label{4.17}
\Vert C_s(t,s)f\Vert_a \leq e^{K'_a(t - s)}\Vert f\Vert_{a}, \quad 0 \leq t - s \leq \rho^*
\end{equation}

for all $f \in (B^{a})^l$.
\end{pro}

\begin{proof}
We note that \eqref{4.12} hold.  Then, applying Proposition 3.6 as $M_1 = 1$ to the second term on the right-hand side of \eqref{4.14}, its $B^a$-norm is bounded by $C_a(t - s)\Vert f \Vert_a$ from above  with a constant $C_a \geq 0$ for $0 \leq t - s \leq \rho^*$.  Hence, using \eqref{3.38}, from \eqref{4.14} we can prove

\begin{align*}
\Vert C_s(t,s)f\Vert_a &\leq e^{K_a(t - s)}\Vert f\Vert_{a} + C_a(t - s)\Vert f\Vert_{a}
\\ & \leq e^{(K_a + C_a)(t - s)}\Vert f\Vert_{a},
\quad 0 \leq t - s \leq \rho^*,
\end{align*}

which shows \eqref{4.17}.
\end{proof}

{\bf Proof of Theorem 2.4.}
We may assume \eqref{4.1}.
Noting \eqref{4.3}, from \eqref{3.27} and \eqref{4.15} - \eqref{4.17} we can prove Theorem 2.4 as in the proof of Theorem 2.3.
\par
\vspace{0.5cm}

{\bf Proofs of Theorems 2.5 and 2.6.}
For a continuous path $q(\theta) = \bigl(\bq_1(\theta),\dots,\\
\bq_N(\theta) \bigr) \in \bR^{dN}\ (s \leq \theta \leq t)$ we define $\mathcal{F}^{\sharp}(\theta,s;q)\ (s \leq \theta \leq t)$
by the solution to

\begin{align} \label{4.18}
&\frac{d}{d\theta}\,\mathcal{U}^{\sharp}(\theta) =
- \biggl[\sum_{j=1}^N  I_1\otimes \cdots \otimes I_{j-1}\otimes \bigl\{iH_{sj}(\theta,\bq_j(\theta)) +  W_{sj}(\theta,\bq_j(\theta))\bigl\} \notag\\
& \quad   \otimes I_{j+1}\otimes \cdots\otimes I_N
\biggr]\mathcal{U}^{\sharp}(\theta), \quad \mathcal{U}^{\sharp}(s) = I_1\otimes \cdots \otimes I_N
\end{align}

as in the case of $W_{sj} = 0\ (j = 1,2,\dots,N)$ (cf. \S 4 in \cite{Ichinose 2020}).
Then we have

\begin{equation} \label{4.19}
\mathcal{F}^{\sharp}(\theta,s;q) = \mathcal{F}_1(\theta,s;\bq_{1})\otimes \cdots \otimes \mathcal{F}_N(\theta,s;\bq_{N})
\end{equation}

as shown in (4.4) of \cite{Ichinose 2020}.
In fact, each side of \eqref{4.19} is the solution to \eqref{4.18}, which follows from \eqref{2.13} and
the simple properties of the tensor products of matrices (cf. \S 4.2 in \cite{Horn-Johnson}, \S II.4 and \S VIII.10 in \cite{Reed-Simon I}).
\par

We can rewrite \eqref{2.18} in the form of \eqref{1.9}, setting

\begin{align*}
& H^{\sharp}_s(t,x) := \sum_{j=1}^N I_1\otimes \cdots \otimes I_{j-1}
  \otimes H_{sj}(t,\bx_j)\otimes I_{j+1}\otimes \cdots\otimes I_N, \\

& W^{\sharp}_s(t,x) := \sum_{j=1}^N I_1\otimes \cdots \otimes I_{j-1}
  \otimes W_{sj}(t,\bx_j)\otimes I_{j+1}\otimes \cdots\otimes I_N,
\end{align*}

where we can see from the simple properties of the tensor products that both of $W^{\sharp}_s(t,x)$ and $H^{\sharp}_s(t,x)$ are written as $l^N\times l^N$ Hermitian matrices and satisfy \eqref{1.11} and \eqref{2.16}, respectively and that $W^{\sharp}_s(t,x)$ is bounded below.  In addition, setting $W^{\sharp}(t,x) := \sum_{j=1}^N W_{j}(t,\bx_j)$, we see that \eqref{3.18} for $W^{\sharp}(t,x)$ remains valid.

  Hence, noting \eqref{2.19} and \eqref{4.19},  we see from Theorem 2.3 that Theorem 2.5 holds because of
  $L^2(\bR^{dN})^{l^N} = L^2(\bR^d)^l\otimes \cdots L^2(\bR^d)^l$. In the same way we see that Theorem 2.6 follows from Theorem 2.4.


TITLE: Gravitational time dilation on the surface of a planet
QUESTION [4 upvotes]: Suppose we have a massive non-rotating planet, so that time is dilated on its surface.
There are two cities on the planet, city A and city B, and it takes light 1 ms to travel between them.
But what time will it take from the point of view of an observer in space?
Suppose the city A when sending signal to city B also sends signal to the observer in space. The city B also sends the observer in space a notification when it recieves the signal.
Due to time dilation on the planet's surface, the remote observer should notice that the signal travelled more than 1 ms.
But the speed of light should be the same. Does it mean that the distance between the cities A and B is greater from the point of view of a remote observer? So, the area of the planet is greater as well?

REPLY [0 votes]: A and B are on earth, but must be closer together than 1 ms (300 km) so they can see each other from the tops of reasonable height towers.  An observer on earth using his own meter stick measures a distance $\Delta x$ between A and B.  Using his own clock he also measures the time $\Delta t$ for a light pulse to travel between A and B.  He confirms that $\frac{\Delta x}{\Delta t}$ yields the standard speed of light $c=3 \ 10^8 \ km/sec$.
Then, yes the observer far out in space and using his own clock will see $\Delta t'>\Delta t$.  He will also see (by the angular separation between A and B) that $\Delta x'=\Delta x$ and conclude that the light (which is deep in the earth's gravitational potential compared to the far out observer) traveled between A and B with a speed $c'=\frac{\Delta x'}{\Delta t'}$  where $c'<c$. This is the same effect as seen in the Shapiro Delay where the far out observer sees a radar signal passing near the sun travel at $c'<c$.
What caused the confusion in your question was your statement: “But the speed of light should be the same….”. It isn't.  It is true that the coordinate speed of light is the same $c$ in all frames with a Minkowski metric and that Lorentz transformations leave the Minkowski metric unchanged. However, the Scwarzschild field (and Gravity Waves) do strain transformations (which are not Lorentz transformations) to the coordinates and to the metric indices. For these strains and for Lorentz transformations, the proper length $g’_{\mu\nu} \Delta x’^{\mu}\Delta x’^{\nu}=\eta_{\mu\nu} \Delta x^{\mu}\Delta x^{\nu}$ remains unchanged.   These strain transformations do not leave the Minkowski metric $\eta_{\mu\nu}$ invariant.  Instead the strains turn the Minkowski metric into the familiar Schwarzschild metric.  The new $ g’_{\mu\nu} $ Schwarzschild metric tells you that the far away observer will measure $c'<c$ for the new coordinate speed of light passing at a radius r in the Schwarzschild field.  The proper length along the path of a photon is 0, $\Delta r=0$ between A and B, and $\Delta x’= r \Delta \theta$ in the Schwarzschild coordinates.
$$
0=(1-2GM/r)(c\Delta t’)^2-(\Delta x’)^2
$$
$$
c’=\frac{\Delta x’}{\Delta t’}=(1-2GM/r)^{1/2}c
$$
The summary answer to the question is: The far away observer sees the same distance AB (for any AB direction perpendicular to the radius vector) and the same area of the earth as the observer on earth. The far away observer sees the light take more time $\Delta t'>\Delta t$ to go from A to B, and the light appears to go slower than the usual c such that $\Delta x'=c'\Delta t'=c\Delta t=\Delta x$.


TITLE: How to use the substitution rule for indefinite integrals to obtain the following result?
QUESTION [1 upvotes]: My book claims the following: $$ x = f(t) $$ $$y = g(t)$$
then, by substitution rule $$ \int y \ dx = \int g(t)f'(t) \ dt$$
I cannot find a way to obtain this result. Could someone show all the steps to obtain it?

REPLY [0 votes]: Let's break this down.
We have:
$$ x = f(t) $$ $$y = g(t)$$
And we're looking to express in terms of those functions the following integral:$$ \int y \ dx$$
Let's substitute only $y$ to begin with. So the integral becomes:
$$\int g(t)dx$$
Now let's look at substituting the $dx$ and rearranging.
If $x=f(t)$, then $\frac{dx}{dt} = \frac{df(t)}{dt}.$
$i.e.  \frac{dx}{dt} = f'(t).$
$\therefore  {dx} = f'(t)dt.$
Now to complete the substitution, substitute $dx$ with our result:
$$\int g(t)dx = \int g(t)f'(t)dt$$


TITLE: Positive Ricci curvature on fiber bundles
QUESTION [10 upvotes]: My advisor and I are working on Ricci curvature and an anonymous referee pointed out the following conjecture:

Let $F\hookrightarrow M\stackrel{\pi}{\to}B$ be a fiber bundle from a compact manifold $M$ with fiber $F$, compact structure group $G$ and base $B$. Suppose that:
              i) $B$ has a metric of positive Ricci curvature;
              ii) $F$ has a $G$-invariant metric of positive Ricci curvature.
          Then $M$ carries a metric of positive Ricci curvature.

Both the referee and us are not sure if it is in literature or not. It happens that the conjecture is easily proved using classical arguments.
Therefore, we would like to ask if someone knows a reference, or if the conjecture is well known to be true among specialists.
Partial results we could find are:
1) (Nash) https://projecteuclid.org/euclid.jdg/1214434973
2) W. A. Poor, Some exotic spheres with positive Ricci curvature, Math. Ann. 216 (1975)
245-252.

REPLY [4 votes]: This conjecture is already proved in
Gromoll, Detlef; Walschap, Gerard, Metric foliations and curvature, Progress in Mathematics 268. Basel: Birkhäuser (ISBN 978-3-7643-8714-3/hbk). viii, 174 p. (2009). ZBL1163.53001.
(page 100, Theorem 2.7.3).


TITLE: Find antiderivative to $-2\sin(2x)e^{-4x}$
QUESTION [1 upvotes]: I'm trying to find an antiderivative to $(-\sin(2x)e^{-4x} dx)$. I've started out by using partial integration and come this far (see the picture

but don't know if I'm on the right track. Does anyone have any suggestions?
Thanks!
/Nick

REPLY [0 votes]: It is the imaginary part of the antiderivative of $\, \mathrm e^{2ix}\mathrm e^{4x}=-\mathrm e^{(4+2i) x}$, i.e. it is
$$-\frac 1{2(2+i)}\mathrm e^{(4+2i)x}.$$
There remains to find explicitly this imaginary part.
Note that for the same p price you'll also get $\;\displaystyle\int (-4\cos 2x\mathrm e^{4x})\,\mathrm dx$.


text:
\begin{document}

\newcommand{\eps}{\epsilon}
\newcommand{\ord}{{\cal O}}

\title{One-Dimensional Peg Solitaire}

\author{Cristopher Moore \inst{1,2}
\and
David Eppstein \inst{3}}

\institute{
Computer Science Department, University of New Mexico,
Albuquerque NM 87131 {\tt moore@cs.unm.edu}
\and
Santa Fe Institute, 1399 Hyde Park Road, Santa Fe NM 87501
\and
Dept. of Information and Computer Science, University of California,
Irvine, Irvine CA 92697-3425 {\tt eppstein@ics.uci.edu}}

\maketitle

\begin{abstract}
We solve the problem of one-dimensional peg solitaire.  In particular,
we show that the set of configurations that can be reduced to a single
peg forms a regular language, and that a linear-time algorithm exists
for reducing any configuration to the minimum number of pegs.
\end{abstract}

\section{Introduction}

Peg solitaire is a game for one player.  Each move consists of hopping
a peg over another one, which is removed.  The goal is to reduce the
board to a single peg.  The best-known forms of the game take place on
cross-shaped or triangular boards, and it has been marketed as
``Puzzle Pegs'' and ``Hi-Q.''  Discussions and various solutions can
be found in \cite{kraitchik,gardner,hakmem,bcg,beasley}.

In \cite{guy}, Guy proposes one-dimensional peg solitaire as an open
problem in the field of combinatorial games.  Here we show that the
set of solvable configurations forms a regular language, i.e.\ it can
be recognized by a finite-state automaton.  In fact, this was already
shown in 1991 by Plambeck (\cite{chang}, Introduction and Ch.5) and
appeared as an exercise in a 1974 book of Manna \cite{manna}.  More
generally, B. Ravikumar showed that the set of solvable configurations
on rectangular boards of any finite width is regular \cite{ravikumar},
although finding an explicit grammar seems to be difficult on boards
of width greater than 2.

Thus there is little new about this result.  However, it seems not to
have appeared in print, so here it is.

\begin{theorem}
The set of configurations that can be reduced to a single peg is the
regular language $0^* L 0^*$ where
\begin{eqnarray}
L & = & 1 + 011 + 110 \nonumber \\
& + & 11 (01)^*
\,\Bigl[ 00 + 00(11)^+ + (11)^+00 + (11)^* 1011 + 1101 (11)^*
\Bigr]\, (10)^* 11 \nonumber \\
& + & 11 (01)^* (11)^* 01 + 10 (11)^* (10)^* 11 .
\label{lang}
\end{eqnarray}
Here $1$ and $0$ indicate a peg and a hole respectively, $w^*$ means
`0 or more repetitions of $w$', and $w^+=ww^*$ means `1 or more
repetitions of $w$.'
\label{thm}
\end{theorem}

\pagebreak

\begin{proof}
To prove the theorem, we follow Leibnitz \cite{bcg} in starting with a
single peg, which we denote
\[ 1 \]
and playing the game in reverse.  The first `unhop' produces
\[ 011 \mbox{ or } 110 \]
and the next
\[ 1101 \mbox{ or } 1011. \]
(As it turns out, $11$ is the only configuration that cannot be
reduced to a single peg without using a hole outside the initial set
of pegs.  Therefore, for all larger configurations we can ignore the
$0$'s on each end.)

We take the second of these as our example.  It has two ends,
$10\ldots$ and $\ldots11$.  The latter can propagate itself
indefinitely by unhopping to the right,
\[ 1010101011. \]
When the former unhops, two things happen; it becomes an end of the
form $11\ldots$ and it leaves behind a space of two adjacent holes,
\[ 110010101011. \]
Furthermore, this is the only way to create a $00$.  We can move the
$00$ to the right by unhopping pegs into it,
\[ 111111110011. \]
However, since this leaves a solid block of $1$'s to its left, we
cannot move the $00$ back to the left.  Any attempt to do so reduces it
to a single hole,
\[ 111111101111. \]
Here we are using the fact that if a peg has another peg to its left,
it can never unhop to its left.  We prove this by induction: assume it
is true for pairs of pegs farther left in the configuration.  Since
adding a peg never helps another peg unhop, we can assume that the two
pegs have nothing but holes to their left.  Unhopping the leftmost peg
then produces $1101$, and the original (rightmost) peg is still
blocked, this time by a peg which itself cannot move for the same
reason.

In fact, there can never be more than one $00$, and there is no need
to create one more than once, since after creating the first one the
only way to create another end of the form $10\ldots$ or $\ldots01$ is
to move the $00$ all the way through to the other side
\[ 111111111101 \]
and another $00$ created on the right end now might as well be the
same one.

We can summarize, and say that any configuration with three or more
pegs that can be reduced to a single peg can be obtained in reverse
from a single peg by going through the following stages, or their
mirror reflection:
\begin{enumerate}
\item We start with $1011$.  By unhopping the rightmost peg, we obtain
$10(10)^*11$.  If we like, we then
\item Unhop the leftmost peg one or more times, creating a pair of
holes and obtaining $11(01)^*00(10)^*11$.  We can then
\item Move the $00$ to the right (say), obtaining
$11(01)^*(11)^*00(10)^*11$.  We can stop here, or
\item Move the $00$ all the way to the right, obtaining
$11(01)^*(11)^*01$, or
\item Fill the pair by unhopping from the left, obtaining
$11(01)^*(11)^*1011(10)^*11$.
\end{enumerate}
Equation~\ref{lang} simply states that the set of configurations is
the union of all of these plus $1$, $011$, and $110$, with as many
additional holes on either side as we like.  Then $0^* L 0^*$ is
regular since it can be described by a regular expression
\cite{hopcroft}, i.e.\ a finite expression using the operators $+$ and
$*$.  \qed \end{proof}

Among other things, Theorem~\ref{thm} allows us to calculate the
number of distinct configurations with $n$ pegs, which is
\[ N(n) = \left\{ \begin{array}{lll}
1 & \hspace{5mm} & n = 1 \\
1 & & n = 2 \\
2 & & n = 3 \\
15 - 7n + n^2 & & n \ge 4, \,n \mbox{ even} \\
16 - 7n + n^2 & & n \ge 5, \,n \mbox{ odd}
\end{array} \right. \]
Here we decline to count $011$ and $110$ as separate configurations,
since many configurations have more than one way to reduce them.

We also have the corollary

\begin{corollary}\label{one-peg-strategy}
There is a linear-time strategy for playing peg solitaire in one
dimension.
\end{corollary}

\begin{proof}
Our proof of Theorem~\ref{thm} is constructive in that it tells us how
to unhop from a single peg to any feasible configuration.  We simply
reverse this series of moves to play the game.  \qed
\end{proof}

More generally, a configuration that can be reduced to $k$ pegs
must belong to the regular language $(0^* L 0^*)^k$,
since unhopping cannot interleave the pegs coming from different origins
\cite{chang}.  This leads to the following algorithm:

\begin{theorem}
There is a linear-time strategy for reducing any one-dimensional peg
solitaire configuration to the minimum possible number of pegs.
\end{theorem}

\begin{proof}
Suppose we are given a string $c_0c_1c_2\ldots c_{n-1}$ where each
$c_i\in\{0,1\}$.  Let ${\cal A}$ be a nondeterministic finite
automaton (without $\epsilon$-transitions) for $0^*L0^*$, where $A$ is
the set of states in ${\cal A}$, $s$ is the start state, and $T$ is
the set of accepting states.  We then construct a directed acyclic
graph $G$ as follows: Let the vertices of $G$ consist of all pairs
$(a,i)$ where $a \in A$ and $0 \le i \le n$.  Draw an arc from $(a,i)$
to $(b,i+1)$ in $G$ whenever ${\cal A}$ makes a transition from state
$a$ to state $b$ on symbol $c_i$.  Also, draw an arc from $(t,i)$ to
$(s,i)$ for any $t\in T$ and any $0\le i\le n$.  Since $|{\cal A}| =
\ord(1)$, $|G|=\ord(n)$.

Then any path from $(s,0)$ to $(s,n)$ in $G$ consists of $n$ arcs of
the form $(a,i)$ to $(b,i+1)$, together with some number $k$ of arcs
of the form $(t,i)$ to $(s,i)$.  Breaking the path into subpaths by
removing all but the last arc of this second type corresponds to
partitioning the input string into substrings of the form $0^* L 0^*$,
so the length of the shortest path from $(s,0)$ to $(s,n)$ in $G$ is
$n+k$, where $k$ is the minimum number of pegs to which the initial
configuration can be reduced.  Since $G$ is a directed acyclic graph,
we can find shortest paths from $(s,0)$ by scanning the vertices
$(a,i)$ in order by $i$, resolving ties among vertices with equal $i$
by scanning vertices $(t,i)$ (with $t\in T$) earlier than vertex
$(s,i)$.  When we scan a vertex, we compute its distance to $(s,0)$ as
one plus the minimum distance of any predecessor of the vertex.  If
the vertex is $(s,0)$ itself, the distance is zero, and all other
vertices $(a,0)$ have no predecessors and infinite distance.

Thus we can find the optimal strategy for the initial peg solitaire
configuration by forming $G$, computing its shortest path, using the
location of the edges from $(t,i)$ to $(s,i)$ to partition the
configuration into one-peg subconfigurations, and applying
Corollary~\ref{one-peg-strategy} to each subconfiguration.  Since
$|G|=\ord(n)$, this algorithm runs in linear time.
\end{proof}

In contrast to these results, Uehara and Iwata \cite{uehara} showed
that in two or more dimensions peg solitaire is NP-complete.  However,
the complexity of finding the minimum number of pegs to which a $k
\times n$ configuration can be reduced, for bounded $k > 2$, remains
open.

Finally, we note that Ravikumar has proposed an impartial two-player
game, in which players take turns making peg solitaire moves, and
whoever is left without a move loses.  It is tempting to think that
this two-player game might be PSPACE-complete in two or more
dimensions, and polynomial-time solvable in one.

\paragraph{Acknowledgements.}
We thank Aviezri Fraenkel, Michael Lachmann, Molly Rose, B. Sivakumar,
and Spootie the Cat for helpful conversations, and the organizers of
the 2000 MSRI Workshop on Combinatorial Games.

config: arxiv