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-rw-r--r--fonts/firamath-otf/doc/firamath-otf-doc.tex168
1 files changed, 19 insertions, 149 deletions
diff --git a/fonts/firamath-otf/doc/firamath-otf-doc.tex b/fonts/firamath-otf/doc/firamath-otf-doc.tex
index cc43c139b2..e671662497 100644
--- a/fonts/firamath-otf/doc/firamath-otf-doc.tex
+++ b/fonts/firamath-otf/doc/firamath-otf-doc.tex
@@ -1,12 +1,13 @@
-%% $Id: firamath-otf-doc.tex 1022 2019-04-03 12:00:29Z herbert $
+%% $Id: firamath-otf-doc.tex 673 2023-01-16 19:44:41Z herbert $
\listfiles
\documentclass[english,log-declarations=false]{article}
-\usepackage{amsmath,esvect}
+\usepackage{mathtools,esvect}
\usepackage{FiraSans}
\setmonofont{FiraMono}[ Numbers = {Monospaced},
Scale=MatchUppercase,FakeStretch=0.93]
-\usepackage[fakebold]{firamath-otf}
+\usepackage[fakebold,mathrm=sym]{firamath-otf}
\usepackage{babel}
+\everymath{}\everydisplay{}% FIX for current babel
\usepackage{booktabs}
\usepackage{xltabular}
\usepackage{listings}
@@ -16,7 +17,8 @@
\usepackage{xcolor,url}
\usepackage{varioref,multido}
\newcommand\Macro[1]{\texttt{\textbackslash#1}}
-\usepackage{dtk-extern}
+\usepackage{expl3}
+\usepackage{hvextern}
\newenvironment{demoquote}
{\begingroup
@@ -47,6 +49,7 @@
\def\PackageInfo#1#2{}
\ExplSyntaxOn
+
\cs_new:Npn \__fonttest_close_msg:nn #1#2
{ \msg_redirect_name:nnn {#1} {#2} { none } }
\__fonttest_close_msg:nn { LaTeX / xparse } { not-single-char }
@@ -240,14 +243,12 @@
\ExplSyntaxOff
-
+\def\Lcs#1{\texttt{\textbackslash #1}}
\renewcommand\familydefault{\sfdefault}
\DeclareMathOperator{\Div}{\symup{div}}
\DeclareMathOperator{\Grad}{\symup{grad}}
-
-
\title{OpenType math font Fira}
\author{Herbert Voß}
\usepackage{parskip}
@@ -276,15 +277,18 @@ only the regular version has from todays update all symbols.
Optional arguments are
\begin{description}
+%\item[\texttt{mathrm=sym}] Use characters from firamath for \Lcs{mathrm}
\item[\texttt{fakebold}] Use faked bold symbols
\item[\texttt{usefilenames}] Use filenames for the fonts instead of the symbolic font names
\end{description}
+All other unknown options, e.g. \verb|mathrm=sym| will be passed to the main package \texttt{unicode-math}.
+
The package itself loads by default
\begin{verbatim}
-\RequirePackage{ifxetex,ifluatex,xkeyval,textcomp}
+\RequirePackage{iftex,xkeyval,textcomp}
\RequirePackage{unicode-math}
\end{verbatim}
@@ -317,12 +321,12 @@ The package itself loads by default
\end{align}
\begin{align}
- \Q{}{t}\iiint\<\varrho\diff^3V+\oiint\varrho(\vec{v}\cdot\vec{v}\vec{n})\diff^2A &= 0\\
+ \Q{}{t}\iiint\<\varrho\diff^3V+\oiint\varrho(\vec{v}\cdot\vec{v}\vec{n}) \diff^2 A &= 0\\
\Q{}{t}\iiint\<\varrho\vec{v}\diff^3V+\oiint\varrho\vec{v}(\vec{v}\cdot\vec{n}\,)\diff^2A &=
\iiint\<f_0\diff^3V+\oiint\vec{n}\cdot\TT\diff^2A \\
\Q{}{t}\iiint\<\left(\half v^2+e\right)\varrho\diff^3V+\oiint\left(\half v^2+e\right)
\varrho\left(\vec{v}\cdot\vec{n}\,\right)\diff^2A & =\\
- \multispan2{\hfill${\displaystyle-\oiint\left(\vec{q}\cdot\vec{v}ec{n}\right)\diff^2A+
+ \multispan2{\hfill${\displaystyle-\oiint\left(\vec{q}\cdot\vec{v}\vec{n}\right)\diff^2A+
\iiint\<\left(\vec{v}\cdot\vec{f}_0\right)\diff^3V+\oiint\left(\vec{v}
\cdot\vec{n}~\TT\right)\diff^2A}$}.\nonumber
\end{align}
@@ -336,14 +340,13 @@ package \texttt{xfakebold} which writes some information into the created PDF to
characters. For more informations see the documentation of \texttt{xfakebold}.
\setBold
-%\mathversion{bold}
\begin{align}
- \Q{}{t}\iiint\<\varrho\diff^3V+\oiint\varrho(\vec{v}\cdot\vec{v}\vec{n})\diff^2A &= 0\\
+ \Q{}{t}\iiint\<\varrho\diff^3V+\oiint\varrho(\vec{v}\cdot\vec{v}ec{n})\diff^2A &= 0\\
\Q{}{t}\iiint\<\varrho\vec{v}\diff^3V+\oiint\varrho\vec{v}(\vec{v}\cdot\vec{n}\,)\diff^2A &=
\iiint\<f_0\diff^3V+\oiint\vec{n}\cdot\symup{T}\diff^2A \\
\Q{}{t}\iiint\<\left(\half v^2+e\right)\varrho\diff^3V+\oiint\left(\half v^2+e\right)
\varrho\left(\vec{v}\cdot\vec{n}\,\right)\diff^2A & =\\
- \multispan2{\hfill${\displaystyle-\oiint\left(\vec{q}\cdot\vec{v}\vec{n}\right)\diff^2A+
+ \multispan2{\hfill${\displaystyle-\oiint\left(\vec{q}\cdot\vec{v}ec{n}\right)\diff^2A+
\iiint\<\left(\vec{v}\cdot\vec{f}_0\right)\diff^3V+\oiint\left(\vec{v}
\cdot\vec{n}~\symup{T}\right)\diff^2A}$}.\nonumber
\end{align}
@@ -380,11 +383,11 @@ characters. For more informations see the documentation of \texttt{xfakebold}.
\cdot\vec{n}~\TT\right)\diff^2A}$}.\nonumber
\end{align}
-\subsection{Version bold}
-
+\subsection{Version bold}
\setBold
+
\begin{align}
\Q{}{t}\iiint\<\varrho\diff^3V+\oiint\varrho(\vec{v}\cdot\vec{v}ec{n})\diff^2A &= 0\\
\Q{}{t}\iiint\<\varrho\vec{v}\diff^3V+\oiint\varrho\vec{v}(\vec{v}\cdot\vec{n}\,)\diff^2A &=
@@ -395,6 +398,7 @@ characters. For more informations see the documentation of \texttt{xfakebold}.
\iiint\<\left(\vec{v}\cdot\vec{f}_0\right)\diff^3V+\oiint\left(\vec{v}
\cdot\vec{n}~\symup{T}\right)\diff^2A}$}.\nonumber
\end{align}
+
\unsetBold
@@ -807,138 +811,4 @@ characters. For more informations see the documentation of \texttt{xfakebold}.
-\iffalse
-\subsection{More Samples}
-
-\def\ee{\mathrm{e}}
-\def\ii{\mathrm{i}}
-\def\bm{\symbf}
-\newcommand\innerprod[2]{\left\langle{#1}\middle\vert{#2}\right\rangle}
-\newcommand\brakket[3]{\left\langle{#1}\middle\vert{#2}\middle\vert{#3}\right\rangle}
-% \newcommand{\dd}{\,\mathrm{d}}
-% \newcommand{\norm}[1]{\left\lVert{#1}\right\rVert}
-
-\[ g^{mn} g_{mn} T^{i}_{jk} \]
-
-\[ x \to \infty + \infty - \infty \]
-
-\begin{align*}
- \int_{-\infty}^\infty \ee^{-x^2} \dd{x}
- &= \qty[\int_{-\infty}^\infty \ee^{-x^2} \dd{x} \, \int_{-\infty}^\infty \ee^{-y^2} \dd{y}]^{1/2} \\
- &= \qty[\int_0^{2\pi} \int_0^\infty \ee^{-r^2} r \dd{r}\dd{\theta}]^{1/2} \\
- &= \qty[\pi \int_0^\infty \ee^{-u} \dd{u}]^{1/2} \\
- &= \sqrt{\pi}
-\end{align*}
-
-
-\begin{align*}
-\int_{0}^aJ_0\left[\frac{x_n^{(0)}}{a}r\right]J_0\left[\frac{x_m^{(0)}}{a}r\right]r\dd{r}=\frac{a^2}{2}J_1^2[x_n^{(0)}]\delta_m^n.\\
-\int_{0}^{\infty}\frac{\cos x-\ee^{-x}}{x}\dd{x}=0\\
-\end{align*}
-\[\oint_{\partial\Sigma}\vec E\cdot \dd{\vec{l}}=-\frac{1}{c}\frac{\dd}{\dd t}\iint_{\Sigma}\vec B \cdot \dd{\vec{S}};\]
-\[\partial_{[a}F_{\beta\gamma]}=0;\quad \partial_\alpha F^{\alpha\beta}=\mu_0J^\beta\]
-\[\left(\frac{-\hbar^2}{2m}\nabla^2+V\right)\Psi=i\hbar\dot{\Psi}\]
-\[\begin{split}
-\frac{1}{\mathcal{C}^2}&{}=\frac{\innerprod{g'}{g'}}{\mathcal{C}^2}=\sum_{\bm{k}}\sum_{\bm{k}'}\brakket{g}{c_{\bm{k}',\uparrow}^\dagger c_{\bm{k}',\downarrow} c_{\bm{k},\downarrow}^\dagger c_{\bm{k},\uparrow}}{g}=\sum_{\bm{k}}\brakket{g}{c_{\bm{k},\uparrow}^\dagger c_{\bm{k},\downarrow} c_{\bm{k},\downarrow}^\dagger c_{\bm{k},\uparrow}}{g}\\
-&{}=\sum_{\bm{k}}\brakket{g}{n_{\bm{k},\uparrow}\left(1-n_{\bm{k},\downarrow}\right)}{g}\\
-&{}=\sum_{\norm{\bm{k}}<k_F^\downarrow}\brakket{g}{0}{g}+\sum_{k_F^\downarrow<\norm{\bm{k}}<k_F^\uparrow}\brakket{g}{1}{g}+\sum_{\norm{\bm{k}}>k_F^\uparrow}\brakket{g}{0}{g}\\
-&{}=N_\uparrow-N_\downarrow
-\end{split}\]
-\[\left[ f,g \right]\equiv \sum_{\alpha =1}^{s}{\left( \frac{\partial f}{\partial {{q}_{\alpha }}}\frac{\partial g}{\partial {{p}_{\alpha }}}-\frac{\partial g}{\partial {{q}_{\alpha }}}\frac{\partial f}{\partial {{p}_{\alpha }}} \right)}=\sum\limits_{\alpha =1}^{s}{\begin{vmatrix}
- \partial_{{q}_{\alpha }} f & \partial_{{p}_{\alpha }} f \\
- \partial_{{q}_{\alpha }} g & \partial_{{p}_{\alpha }} g \\
- \end{vmatrix} }=\sum\limits_{\alpha =1}^{s}{\frac{\partial \left( f,g \right)}{\partial \left( {{q}_{\alpha }},{{p}_{\alpha }} \right)}}\]
-\[\begin{split}
-& \frac{{{\text{d}}^{2}}f}{\text{d}{{t}^{2}}}=\frac{\text{d}}{\text{d}t}\left[ f,H \right]=\left[ \left[ f,H \right],H \right]=\hat{H}\hat{H}f={{{\hat{H}}}^{2}}f \\
-& \vdots \\
-& \frac{{{\text{d}}^{n}}f}{\text{d}{{t}^{n}}}=\underbrace{\left[ \left[ \left[ f,H \right],\cdots \right],H \right]}_{n}={{{\hat{H}}}^{n}}f \\
-\end{split}\]
-\[\tilde{U}(r,z)=E_0\dfrac{\omega_0}{\omega(z)}\exp\left[-r^2\left(\dfrac{1}{\omega^2(z)}+\dfrac{\ii k}{2R(z)}\right)-\ii k z+\ii \zeta(z)\right]\]
-\[\omega(z)=\omega_0\sqrt{1+\left(\dfrac{\lambda z}{\pi {\omega_0}^2}\right)^2};\quad R(z)=z\left[1+\left(\dfrac{\pi {\omega_0}^2}{\lambda z}\right)^2\right]\]
-\[\left( \begin{matrix}
-{mg}/{l}\;+k-m\omega _{1}^{2} & -k \\
--k & {mg}/{l}\;+k-m\omega _{1}^{2} \\
-\end{matrix} \right)\left( \begin{matrix}
-{{a}_{11}} \\
-{{a}_{21}} \\
-\end{matrix} \right)=0\]
-\[V=\underbrace{{{V}_{0}}}_{=0}+\underbrace{\sum\limits_{\alpha =1}^{s}{{{\left( \frac{\partial V}{\partial {{q}_{\alpha }}} \right)}_{0}}{{q}_{\alpha }}}}_{=0}+\underbrace{\frac{1}{2}\sum\limits_{\alpha ,\beta =1}^{s}{{{\left( \frac{{{\partial }^{2}}V}{\partial {{q}_{\alpha }}\partial {{q}_{\beta }}} \right)}_{0}}{{q}_{\alpha }}{{q}_{\beta }}}}_{>0}+\cdots \]
-\[T=\frac{1}{2}\sum\limits_{i=1}^{n}{{{m}_{i}}{{{\dot{\bm r}}}_{i}}\cdot {{{\dot{\bm r}}}_{i}}}=\frac{1}{2}\sum\limits_{\alpha ,\beta =1}^{s}{\left[ \sum\limits_{i=1}^{n}{{{m}_{i}}{{\left( \frac{\partial {{\bm r}_{i}}}{\partial {{q}_{\alpha }}} \right)}_{0}}\cdot {{\left( \frac{\partial {{\bm r}_{i}}}{\partial {{q}_{\beta }}} \right)}_{0}}} \right]{{{\dot{q}}}_{\alpha }}{{{\dot{q}}}_{\beta }}}+\cdots \]
-\[\left( \begin{matrix}
-{{u}_{0}} \\
-{{u}_{1}} \\
-\vdots \\
-{{u}_{N-1}} \\
-\end{matrix} \right)=\sum\limits_{k>0}{\left[ \left( \begin{matrix}
- 1 \\
- \cos ka \\
- \vdots \\
- \cos k\left( N-1 \right)a \\
- \end{matrix} \right)\underbrace{{{C}_{k+}}\cos \left( {{\omega }_{k}}t+{{\varphi }_{k+}} \right)}_{\frac{2}{\sqrt{N}}{{q}_{k+}}}+\left( \begin{matrix}
- 0 \\
- \sin ka \\
- \vdots \\
- \sin k\left( N-1 \right)a \\
- \end{matrix} \right)\underbrace{{{C}_{k-}}\cos \left( {{\omega }_{k}}t+{{\varphi }_{k-}} \right)}_{\frac{2}{\sqrt{N}}{{q}_{k-}}} \right]}\]
-\[G(\vec{r},{\vec{r}}',\tau )=\int _{-\infty }^{\infty }\tilde{G}(\vec{r},{\vec{r}}',\omega )e^{-i \tau \omega }d\omega=\int_{-\infty }^{\infty } \frac{e^{-i \tau \omega } e^{i k |\vec{r}-{\vec{r}}'| }}{(2 \pi ) |\vec{r}-{\vec{r}}'| } \, d\omega=\frac{\delta \left(\tau -\frac{R}{c}\right)}{|\vec{r}-{\vec{r}}'| }\]
-\[
-\begin{split}
-\mathcal{F}^{-1}(\ket{j})
-&{}=\frac{1}{\sqrt{2^n}}\sum_{k=0}^{2^n-1}\exp\left(-2\uppi \ii \frac{jk}{2^n}\right)\ket{k}.\\
-&{}=\frac{1}{\sqrt{2^n}}\sum_{k_{n-1}=0}^1\cdots\sum_{k_{0}=0}^1\exp\left(-2\uppi \ii j\sum_{l=0}^{n-1}\frac{2^l k_l}{2^n}\right)\ket{k_{n-1}\cdots k_0}\\
-&{}=\frac{1}{\sqrt{2^n}}\sum_{k_{n-1}=0}^1\cdots\sum_{k_{0}=0}^1\bigotimes_{l=1}^n\left[\exp\left(-2\uppi \ii j\frac{k_{n-l}}{2^l}\right)\ket{k_{n-l}}\right]\\
-&{}=\frac{1}{\sqrt{2^n}}\bigotimes_{l=1}^n\left[\sum_{k_{n-l}=0}^1\exp\left(-2\uppi \ii j\frac{k_{n-l}}{2^l}\right)\ket{k_{n-l}}\right]\\
-&{}=\frac{1}{\sqrt{2^n}}\bigotimes_{l=1}^n\left[\ket{0}_{n-l}+\ee^{-2\uppi \ii j /2^l}\ket{1}_{n-l}\right]\\
-&{}=\frac{1}{\sqrt{2^n}}\bigotimes_{l=1}^n\left[\ket{0}_{n-l}+\ee^{-2\uppi \ii ({0.j_{l-1}\ldots j_0})}\ket{1}_{n-l}\right].
-\end{split}
-\]
-
-\newcommand{\lb}{\left(}
-\newcommand{\rb}{\right)}
-\newcommand{\ec}{\text{,}}
-\newcommand{\ed}{\text{.}}
-\newcommand{\bt}{\lb t\rb}
-\newcommand{\deltaup}{\updelta}
-\newcommand{\piup}{\uppi}
-\newcommand{\ndd}{\,\mathrm{d}}
-\subsubsection*{Problem 1}
-For convenience, first we set $t_i=0$, and in the end, we replace $t_f$ by $t_f-t_i$ and right answer is obtained.
-The classical path is \[x_c\lb t\rb=A \cos\omega t+B\sin \omega t\ec\]where $A$ and $B$ can be determined by plugging $\lb0,x_i\rb$ and $\lb t_f, x_f\rb$ into the equation. The result is
-\[x_c\lb t\rb=x_i \cos\omega t+\frac{x_f-x_i\cos\omega t_f}{\sin\omega t_f}\sin \omega t\ed\]
-We write $x\lb t\rb=x_c\lb t\rb+\deltaup x\bt$. Due to the fact that $\deltaup x$ should vanish at $t=0$ and $t=t_f$, $\deltaup x$ can be expanded as sine series: \[\deltaup x\bt=\sum_{n=1}^\infty a_n\sin\frac{n\piup t}{t_f}\ed\]
-Also, the functional integral can be rewritten as \[\int\mathcal{D}\left[x\bt\right]=c\int\prod_{n=1}^\infty \dd a_n\ed\]
-So, we have
-\[L=\frac{m}{2}\lb\dot{x}_c+\deltaup\dot{x}\rb^2-\frac{m\omega^2}{2}\lb x_c+\deltaup x\rb^2\ec\]
-\[\dot{x}\bt=-\omega x_i \sin\omega t +\omega \frac{x_f-x_i\cos\omega t_f}{\sin\omega t_f}\cos\omega t+\sum_{n=1}^\infty\frac{a_n n \piup}{t_f}\cos\frac{n \piup t}{t_f}\ec\]
-\[S=\int_0^{t_f} L\ndd t\ed\]
-Because $x_c$ is the classical path, $\deltaup S_\text{classical}=0$, so there can't be any the linear term in the expression of $S$, and we also have in mind that the sine and cosine series are orthogonal. So, we can write S as following:
-\[\begin{split}S&{}=\frac{m}{2}\int_0^{t_f}\left[\lb-\omega x_i\sin\omega t+\omega \frac{x_f-x_i\cos\omega t_f}{\sin\omega t_f}\cos\omega t\rb^2+\sum_{n=1}^\infty\lb\frac{a_n n \piup}{t_f}\rb^2\cos^2\frac{n \piup t}{t_f}\right]\ndd t\\%
-&\quad{}-\frac{m\omega^2}{2}\int_0^{t_f}\left[\lb x_i\cos\omega t+ \frac{x_f-x_i\cos\omega t_f}{\sin\omega t_f}\sin\omega t\rb^2+\sum_{n=1}^\infty {a_n}^2\sin^2\frac{n \piup t}{t_f}\right]\ndd t\\%
-&{}=\sum_{n=1}^\infty\int_0^{t_f}\left[\frac{m}{2}\lb\frac{a_n n \piup}{t_f}\rb^2\cos^2\frac{n \piup t}{t_f}-\frac{m\omega^2}{2}{a_n}^2\sin^2\frac{n \piup t}{t_f}\right]\ndd t\\%
-&\quad{}+\frac{m\omega^2}{2}\int_0^{t_f}\left[ {x_i}^2-\lb\frac{x_f-x_i\cos\omega t_f}{\sin\omega t_f}\rb^2\right]\lb\sin^2\omega t-\cos^2\omega t\rb\ndd t\\%
-&\quad{}-\frac{m\omega^2}{2}\int_0^{t_f}4 {x_i}\lb\frac{x_f-x_i\cos\omega t_f}{\sin\omega t_f}\rb\lb\sin\omega t\cos\omega t\rb\ndd t\ed\end{split}\]
-Using
-\[\int_0^{t_f}\lb\sin^2\omega t-\cos^2\omega t\rb\ndd t=-\frac{\sin2\omega t_f}{2\omega}\ec\]
-\[\int_0^{t_f}\sin\omega t\cdot\cos\omega t\ndd t=\frac{\sin^2\omega t_f}{2\omega}\ec\]
-\[\int_0^{t_f}\sin^2\frac{n\piup t}{t_f} \ndd t=\int_0^{t_f}\cos^2\frac{n\piup t}{t_f} \ndd t=\frac{a_n n \piup}{t_f}\ec\]
-we get
-\[S=\sum_{n=1}^\infty\left[\frac{m}{2}\lb\frac{a_n n \piup}{t_f}\rb^2-\frac{m\omega^2}{2}{a_n}^2\right]\frac{t_f}{2}+\frac{m\omega}{2\sin\omega t_f}\left[\lb {x_i}^2+{x_f}^2\rb\cos\omega t_f-2 x_i x_f\right]\ed\]
-\[\begin{split}U={}&\exp\left\{\frac{\ii}{\hbar}\frac{m\omega}{2\sin\omega t_f}\left[\lb {x_i}^2+{x_f}^2\rb\cos\omega t_f-2 x_i x_f\right]\right\}\\%
-&{}\times c\prod_{n=1}^{\infty}\int_{-\infty}^\infty\exp{\frac{\ii}{\hbar}\left[\frac{m}{2}\lb\frac{n \piup}{t_f}\rb^2-\frac{m\omega^2}{2}\right]\frac{t_f {a_n}^2}{2}}\ndd a_n\ed\end{split}\]
-Using the Fresnel integral formula:
-\[\int_{-\infty}^\infty\exp\lb \ii t\rb\ndd t=\sqrt{\piup \ii}\ec\]
-\[\int_{-\infty}^\infty\exp{\frac{\ii}{\hbar}\left[\frac{m}{2}\lb\frac{n \piup}{t_f}\rb^2-\frac{m\omega^2}{2}\right]\frac{t_f {a_n}^2}{2}}\ndd a_n\sim\frac{\sqrt{t_f}}{n}\ec\]
-\[U\lb x_f,t_f;x_i,t_i\rb=c'\lb t_f-t_i\rb\exp\left\{\frac{\ii m\omega}{2\hbar\sin\left[\omega \lb t_f-t_i\rb\right]}\left[\lb {x_i}^2+{x_f}^2\rb\cos\left[\omega\lb t_f-t_i\rb\right]-2 x_i x_f\right]\right\}\ed\]
-Because \[\int\dd x U\lb x_f,t_f;x,t\rb U\lb x,t;x_i,t_i\rb=U\lb x_f,t_f;x_i,t_i\rb\ec\]
-By using the Fresnel integral again:
-\[c'\lb t_f-t\rb c'\lb t-t_i\rb\sqrt{\frac{2 \piup \ii \hbar}{m \omega}\lb\frac{\cos\left[\omega\lb t_f-t\rb\right]}{\sin\left[\omega\lb t_f-t\rb\right]}+\frac{\cos\left[\omega\lb t-t_i\rb\right]}{\sin\left[\omega\lb t-t_i\rb\right]}\rb}=c'\lb t_f-t_i\rb\ec\]
-\[c'\lb t_f-t_i\rb=\sqrt{\frac{m\omega}{2\piup \ii \hbar\sin\left[\omega\lb t_f-t_i\rb\right]}}\ed\]
-Thus
-\[\begin{split}U\lb x_f,t_f;x_i,t_i\rb=&\sqrt{\frac{m\omega}{2\piup \ii \hbar\sin\left[\omega\lb t_f-t_i\rb\right]}}\\&{}\times\exp\left\{\frac{\ii m\omega}{2\hbar\sin\left[\omega \lb t_f-t_i\rb\right]}\left[\lb {x_i}^2+{x_f}^2\rb\cos\left[\omega\lb t_f-t_i\rb\right]-2 x_i x_f\right]\right\}\ed\end{split}\]
-
-
-\fi
-
-
-
\end{document}