<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN">  <!--Converted with LaTeX2HTML 2K.1beta (1.47) original version by:  Nikos Drakos, CBLU, University of Leeds * revised and updated by:  Marcus Hennecke, Ross Moore, Herb Swan * with significant contributions from:   Jens Lippmann, Marek Rouchal, Martin Wilck and others --> <HTML> <HEAD> <TITLE>Mod&#232;les</TITLE> <META NAME="description" CONTENT="Mod&#232;les"> <META NAME="keywords" CONTENT="cours"> <META NAME="resource-type" CONTENT="document"> <META NAME="distribution" CONTENT="global">  <META HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=iso-8859-1"> <META NAME="Generator" CONTENT="LaTeX2HTML v2K.1beta"> <META HTTP-EQUIV="Content-Style-Type" CONTENT="text/css">  <LINK REL="STYLESHEET" HREF="cours.css">  <LINK REL="next" HREF="node5.html"> <LINK REL="previous" HREF="node3.html"> <LINK REL="up" HREF="node3.html"> <LINK REL="next" HREF="node5.html"> </HEAD>  <BODY > <!--Navigation Panel--> <A NAME="tex2html101"   HREF="node5.html"> <IMG WIDTH="37" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="next"  SRC="file:/usr/share/latex2html/icons/next.png"></A>  <A NAME="tex2html97"   HREF="node3.html"> <IMG WIDTH="26" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="up"  SRC="file:/usr/share/latex2html/icons/up.png"></A>  <A NAME="tex2html91"   HREF="node3.html"> <IMG WIDTH="63" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="previous"  SRC="file:/usr/share/latex2html/icons/prev.png"></A>  <A NAME="tex2html99"   HREF="node1.html"> <IMG WIDTH="65" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="contents"  SRC="file:/usr/share/latex2html/icons/contents.png"></A>   <BR> <B> Next:</B> <A NAME="tex2html102"   HREF="node5.html">Approache semi classique</A> <B> Up:</B> <A NAME="tex2html98"   HREF="node3.html">Bases conceptuels</A> <B> Previous:</B> <A NAME="tex2html92"   HREF="node3.html">Bases conceptuels</A>  &nbsp <B>  <A NAME="tex2html100"   HREF="node1.html">Contents</A></B>  <BR> <BR> <!--End of Navigation Panel-->  <H1><A NAME="SECTION00310000000000000000"> Mod&#232;les</A> </H1>  <P> <DIV ALIGN="CENTER"> <B>Formulation Hamiltonienne</B>    </DIV>  <P> Avant de nous lancer dans les myst&#232;res de la m&#233;canique quantique (MQ), il semble appropri&#233; &#224; considerer de point de vue de la m&#233;canique classique (MC) quelques syst&#232;mes physiques  aux quels nous serons int&#233;ress&#233;s.  En g&#233;n&#233;rale dans la MC, l'&#233;nergie totale, <IMG  WIDTH="22" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"  SRC="img9.png"  ALT="$E$">, est represent&#233;e par la summation de son &#233;nergie cin&#233;tique, <IMG  WIDTH="20" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"  SRC="img10.png"  ALT="$T$">, et son &#233;nergie potentielle, <IMG  WIDTH="22" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"  SRC="img11.png"  ALT="$V$">, <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} E = T + V \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.1"></A><IMG  WIDTH="330" HEIGHT="30" BORDER="0"  SRC="img12.png"  ALT="\begin{displaymath} E = T + V   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.1)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> Pour <IMG  WIDTH="24" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"  SRC="img13.png"  ALT="$N$"> particules, l'&#233;nergie cin&#233;tique est donn&#233;e par <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} T = \sum_{i=1}^N \frac{|\vec{p}_i|^2}{2m_i} \, , \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.2"></A><IMG  WIDTH="335" HEIGHT="59" BORDER="0"  SRC="img14.png"  ALT="\begin{displaymath} T = \sum_{i=1}^N \frac{\vert\vec{p}_i\vert^2}{2m_i}   , \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.2)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> o&#249; <IMG  WIDTH="28" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img15.png"  ALT="$m_i$"> est la masse de la particle <IMG  WIDTH="13" HEIGHT="20" ALIGN="BOTTOM" BORDER="0"  SRC="img16.png"  ALT="$i$"> et <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} \vec{p}_i = \left( \begin{array}{c} p^i_x \\p^i_y \\p^i_z \end{array} \right) \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.3"></A><IMG  WIDTH="329" HEIGHT="76" BORDER="0"  SRC="img17.png"  ALT="\begin{displaymath} \vec{p}_i = \left( \begin{array}{c} p^i_x  p^i_y  p^i_z \end{array} \right) \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.3)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> est son impulsion lin&#233;aire.  Dans une seule dimension (<IMG  WIDTH="17" HEIGHT="20" ALIGN="BOTTOM" BORDER="0"  SRC="img18.png"  ALT="$x$">), <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} p = m \frac{d x}{dt} = m v \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.4"></A><IMG  WIDTH="348" HEIGHT="46" BORDER="0"  SRC="img19.png"  ALT="\begin{displaymath} p = m \frac{d x}{dt} = m v   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.4)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> Souvent le potentiel peut &#234;tre deviser dans deux partis, <BR> <DIV ALIGN="CENTER"><A NAME="eq:models.5"></A> <!-- MATH  \begin{eqnarray} V & = & V_1 + V_2 \nonumber \\   V_1 & = & \sum_{i=1}^N v(i) \nonumber \\   V_2 & = & \sum_{i=1}^N \sum_{j=i+1}^N w(i,j) \, , \end{eqnarray}  --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="22" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"  SRC="img20.png"  ALT="$\displaystyle V$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="67" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"  SRC="img22.png"  ALT="$\displaystyle V_1 + V_2$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="25" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"  SRC="img23.png"  ALT="$\displaystyle V_1$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="64" HEIGHT="73" ALIGN="MIDDLE" BORDER="0"  SRC="img24.png"  ALT="$\displaystyle \sum_{i=1}^N v(i)$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="25" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"  SRC="img25.png"  ALT="$\displaystyle V_2$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="135" HEIGHT="73" ALIGN="MIDDLE" BORDER="0"  SRC="img26.png"  ALT="$\displaystyle \sum_{i=1}^N \sum_{j=i+1}^N w(i,j)   ,$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.5)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> o&#249; le potentiel <IMG  WIDTH="37" HEIGHT="40" ALIGN="MIDDLE" BORDER="0"  SRC="img27.png"  ALT="$v(i)$"> agit seulement sur la particule <IMG  WIDTH="13" HEIGHT="20" ALIGN="BOTTOM" BORDER="0"  SRC="img16.png"  ALT="$i$"> et le potentiel <IMG  WIDTH="59" HEIGHT="40" ALIGN="MIDDLE" BORDER="0"  SRC="img28.png"  ALT="$w(i,j)$"> n'agit que sur les particules <IMG  WIDTH="13" HEIGHT="20" ALIGN="BOTTOM" BORDER="0"  SRC="img16.png"  ALT="$i$"> et <IMG  WIDTH="16" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img29.png"  ALT="$j$">.   <P> Pour une mol&#233;cule avec <IMG  WIDTH="24" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"  SRC="img13.png"  ALT="$N$"> &#233;lectrons et <IMG  WIDTH="27" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"  SRC="img30.png"  ALT="$M$"> noyaux, nous pouvons &#233;crire l'&#233;nergie totale &#224; une bonne approximation comme <BR> <DIV ALIGN="CENTER"><A NAME="eq:models.6"></A> <!-- MATH  \begin{eqnarray} E_{totale} & = & \sum_{I=1}^M \frac{|p_I|^2}{2m_I} + \sum_{I=1}^M \sum_{J=I+1}^M \frac{Z_I Z_J e^2}{|\vec{R}_I - \vec{R}_J|}              \nonumber \\              & + & \sum_{i=1}^N \frac{|p_i|^2}{2m} + \sum_{i=1}^N \sum_{j=1}^M \frac{e^2}{|\vec{r}_i-\vec{r}_j|}              \nonumber \\              & - & \sum_{i=1}^N \sum_{I=1}^M \frac{Z_I e^2}{|\vec{R}_I-\vec{r}_i|} \, , \end{eqnarray}  --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="55" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"  SRC="img31.png"  ALT="$\displaystyle E_{totale}$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="253" HEIGHT="73" ALIGN="MIDDLE" BORDER="0"  SRC="img32.png"  ALT="$\displaystyle \sum_{I=1}^M \frac{\vert p_I\vert^2}{2m_I} + \sum_{I=1}^M \sum_{J=I+1}^M \frac{Z_I Z_J e^2}{\vert\vec{R}_I - \vec{R}_J\vert}$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img33.png"  ALT="$\textstyle +$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="212" HEIGHT="73" ALIGN="MIDDLE" BORDER="0"  SRC="img34.png"  ALT="$\displaystyle \sum_{i=1}^N \frac{\vert p_i\vert^2}{2m} + \sum_{i=1}^N \sum_{j=1}^M \frac{e^2}{\vert\vec{r}_i-\vec{r}_j\vert}$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="22" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img35.png"  ALT="$\textstyle -$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="142" HEIGHT="73" ALIGN="MIDDLE" BORDER="0"  SRC="img36.png"  ALT="$\displaystyle \sum_{i=1}^N \sum_{I=1}^M \frac{Z_I e^2}{\vert\vec{R}_I-\vec{r}_i\vert}   ,$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.6)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> o&#249; <IMG  WIDTH="23" HEIGHT="20" ALIGN="BOTTOM" BORDER="0"  SRC="img37.png"  ALT="$m$"> est la masse de l'&#233;lectron et <IMG  WIDTH="60" HEIGHT="40" ALIGN="MIDDLE" BORDER="0"  SRC="img38.png"  ALT="$e=\vert e\vert$"> est (moins) la charge d'un &#233;lectron.  <P> Le cas que nous int&#233;ressons le plus est le cas o&#249; l'&#233;nergie totale, <IMG  WIDTH="22" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"  SRC="img9.png"  ALT="$E$">, est une constante.  <P> <DIV ALIGN="CENTER"> <B>Unit&#233;s</B>  </DIV>  <P> &#199;a vaut la peine &#224; remarquer que nous utilisons les unit&#233;s Gaussien dans l'&#233;q.&nbsp;(<A HREF="node4.html#eq:models.6">1.6</A>). Dans ce syst&#232;me d'unit&#233;s &#233;lectromagn&#233;tique, la force entre deux charges est donn&#233;e par <BR> <DIV ALIGN="CENTER"><A NAME="eq:models.7"></A> <!-- MATH  \begin{eqnarray} \vec{F} & = & \frac{Q_1 Q_2}{r^2} \hat{r} \nonumber \\           & = & \frac{Q_1 Q_2}{r^3} \vec{r}  \, , \end{eqnarray}  --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="21" HEIGHT="48" ALIGN="MIDDLE" BORDER="0"  SRC="img39.png"  ALT="$\displaystyle \vec{F}$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="64" HEIGHT="63" ALIGN="MIDDLE" BORDER="0"  SRC="img40.png"  ALT="$\displaystyle \frac{Q_1 Q_2}{r^2} \hat{r}$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="73" HEIGHT="63" ALIGN="MIDDLE" BORDER="0"  SRC="img41.png"  ALT="$\displaystyle \frac{Q_1 Q_2}{r^3} \vec{r}   ,$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.7)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> o&#249; <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} \vec{r} = \left( \begin{array}{c} x \\y \\z \end{array} \right) \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.7.5"></A><IMG  WIDTH="330" HEIGHT="76" BORDER="0"  SRC="img42.png"  ALT="\begin{displaymath} \vec{r} = \left( \begin{array}{c} x  y  z \end{array} \right)   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.8)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> Le potentiel correspondant &#224; ce force est donn&#233; par <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} V = - \frac{Q_1 Q_2}{r} \, , \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.8"></A><IMG  WIDTH="337" HEIGHT="46" BORDER="0"  SRC="img43.png"  ALT="\begin{displaymath} V = - \frac{Q_1 Q_2}{r}   , \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.9)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> &#224; cause de la d&#233;finition <BR> <DIV ALIGN="CENTER"><A NAME="eq:models.9"></A> <!-- MATH  \begin{eqnarray} \vec{F} & = & - \vec{\nabla} V \nonumber \\     & = & - \left( \begin{array}{c} \partial V/\partial x \\             \partial V/\partial y \\\partial V /\partial z \end{array} \right) \, , \end{eqnarray}  --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="21" HEIGHT="48" ALIGN="MIDDLE" BORDER="0"  SRC="img39.png"  ALT="$\displaystyle \vec{F}$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="53" HEIGHT="48" ALIGN="MIDDLE" BORDER="0"  SRC="img44.png"  ALT="$\displaystyle - \vec{\nabla} V$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="135" HEIGHT="90" ALIGN="MIDDLE" BORDER="0"  SRC="img45.png"  ALT="$\displaystyle - \left( \begin{array}{c} \partial V/\partial x \\ \partial V/\partial y   \partial V /\partial z \end{array} \right)   ,$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.10)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> o&#249;  <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} \vec{\nabla} = \left( \begin{array}{c} \partial /\partial x \\                  \partial /\partial y \\\partial/\partial z \end{array} \right) \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.10"></A><IMG  WIDTH="341" HEIGHT="76" BORDER="0"  SRC="img46.png"  ALT="\begin{displaymath} \vec{\nabla} = \left( \begin{array}{c} \partial /\partial x... ...partial /\partial y  \partial/\partial z \end{array} \right) \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.11)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> est l'op&#233;rateur du gradient.  Dans le syst&#232;me Gaussien, nous utilisons les unit&#233;s centimetre (distance), gramme (poids), seconde (temps), et ``stat coulomb'' ou ``electrostatic unit (esu)'' (charge).  <EM>Le syst&#232;me Gaussien est la basse de la plus part de nos &#233;quations &#233;crites!</EM>  <P> Par contre, un autre syst&#232;me, <EM>le syst&#232;me internationnelle (SI) est pr&#233;ferable pour des calculs pratiques!</EM> Dans le SI les unit&#233;s sont le metre (distance), kilograme (poids), seconde (temps), et coulomb (charge).  Il est  important &#224; comprendre que l'unit&#233; de charge n'a pas la m&#234;me dimensionalit&#233; dans les syst&#232;mes Gaussiens et SI.  Il n'est donc pas surprennant qu'il y a une constante avec unit&#233;s qui entre dans l'expression SI pour la force entre deux charges, <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} \vec{F} = \frac{Q_1' Q_2'}{4\pi \epsilon_0 r^2} \hat{r} \, , \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.11"></A><IMG  WIDTH="337" HEIGHT="50" BORDER="0"  SRC="img47.png"  ALT="\begin{displaymath} \vec{F} = \frac{Q_1' Q_2'}{4\pi \epsilon_0 r^2} \hat{r}   , \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.12)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> o&#249; <IMG  WIDTH="22" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img48.png"  ALT="$\epsilon_0$"> = 8,854 <IMG  WIDTH="22" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img49.png"  ALT="$\times$"> 10<IMG  WIDTH="32" HEIGHT="23" ALIGN="BOTTOM" BORDER="0"  SRC="img50.png"  ALT="$^{-12}$"> C<IMG  WIDTH="14" HEIGHT="23" ALIGN="BOTTOM" BORDER="0"  SRC="img51.png"  ALT="$^2$">/N<IMG  WIDTH="12" HEIGHT="20" ALIGN="BOTTOM" BORDER="0"  SRC="img52.png"  ALT="$\cdot$">m<IMG  WIDTH="14" HEIGHT="23" ALIGN="BOTTOM" BORDER="0"  SRC="img51.png"  ALT="$^2$"> est la permativit&#233; de la vide.  En comparant les &#233;qns.&nbsp;(<A HREF="node4.html#eq:models.7">1.7</A>) et (<A HREF="node4.html#eq:models.11">1.12</A>), on d&#233;duit que  <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} Q' = \sqrt{4\pi \epsilon_0} Q \, , \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.12"></A><IMG  WIDTH="340" HEIGHT="32" BORDER="0"  SRC="img53.png"  ALT="\begin{displaymath} Q' = \sqrt{4\pi \epsilon_0} Q   , \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.13)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> un facteur de conversion entre les deux syst&#232;mes d'unit&#233;s suffaisant pour les fins de ce cours.  Le potentiel correspondant &#224; la force en SI donn&#233;e en &#233;q.&nbsp;(<A HREF="node4.html#eq:models.11">1.12</A>) est <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} V = - \frac{Q_1' Q_2'}{4\pi \epsilon_0 r} \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.13"></A><IMG  WIDTH="336" HEIGHT="50" BORDER="0"  SRC="img54.png"  ALT="\begin{displaymath} V = - \frac{Q_1' Q_2'}{4\pi \epsilon_0 r}   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.14)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P>  <P> <DIV ALIGN="CENTER"> <B>Approximation Born-Oppenheimer</B>  </DIV>  <P> Nous ferons dans la suivante l'approximation de Born et de Oppenheimer.  C'est l'approximation que les &#233;lectrons dans une mol&#233;cule bougent telment vite que l'&#233;nergie des &#233;lectrons &#224; chaque instant ne d&#233;pend que sur les positions des noyaux et pas sur le fait que les noyaux bougent. Dans cette approximation nous pouvons d&#233;finir l'&#233;nergie &#233;lectronique par l'&#233;nergie de la mol&#233;cule sans compter l'&#233;nergie cin&#233;tique des &#233;lectrons.  Donc <BR> <DIV ALIGN="CENTER"><A NAME="eq:models.14"></A> <!-- MATH  \begin{eqnarray} V(\vec{R}_1,\vec{R}_2, ... \vec{R}_M) & = & \sum_{I=1}^M \sum_{J=I+1}^M \frac{Z_I Z_J e^2}{|\vec{R}_I - \vec{R}_J|}              \nonumber \\              & + & \sum_{i=1}^N \frac{|p_i|^2}{2m} + \sum_{i=1}^N \sum_{j=1}^M \frac{e^2}{|\vec{r}_i-\vec{r}_j|}              \nonumber \\              & - & \sum_{i=1}^N \sum_{I=1}^M \frac{Z_I e^2}{|\vec{R}_I-\vec{r}_i|} \, , \end{eqnarray}  --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="142" HEIGHT="48" ALIGN="MIDDLE" BORDER="0"  SRC="img55.png"  ALT="$\displaystyle V(\vec{R}_1,\vec{R}_2, ... \vec{R}_M)$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="164" HEIGHT="73" ALIGN="MIDDLE" BORDER="0"  SRC="img56.png"  ALT="$\displaystyle \sum_{I=1}^M \sum_{J=I+1}^M \frac{Z_I Z_J e^2}{\vert\vec{R}_I - \vec{R}_J\vert}$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img33.png"  ALT="$\textstyle +$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="212" HEIGHT="73" ALIGN="MIDDLE" BORDER="0"  SRC="img34.png"  ALT="$\displaystyle \sum_{i=1}^N \frac{\vert p_i\vert^2}{2m} + \sum_{i=1}^N \sum_{j=1}^M \frac{e^2}{\vert\vec{r}_i-\vec{r}_j\vert}$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="22" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img35.png"  ALT="$\textstyle -$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="142" HEIGHT="73" ALIGN="MIDDLE" BORDER="0"  SRC="img36.png"  ALT="$\displaystyle \sum_{i=1}^N \sum_{I=1}^M \frac{Z_I e^2}{\vert\vec{R}_I-\vec{r}_i\vert}   ,$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.15)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> est &#224; la fois l'&#233;nergie &#233;lectronique et le potentiel dans le quel les noyaux bougent. L'energie des noyaux devient <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} E = \sum_{I=1}^M \frac{|p_I|^2}{2m_I} + V(\vec{R}_1,\vec{R}_2, ... \vec{R}_M) \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.15"></A><IMG  WIDTH="417" HEIGHT="59" BORDER="0"  SRC="img57.png"  ALT="\begin{displaymath} E = \sum_{I=1}^M \frac{\vert p_I\vert^2}{2m_I} + V(\vec{R}_1,\vec{R}_2, ... \vec{R}_M)   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.16)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P>  <P> <DIV ALIGN="CENTER"> <B>Mol&#233;cule diatomique</B>  </DIV>  <P> L'&#233;nergie des noyaux dans une diatomique est donn&#233;e par l'expression, <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} E = T + V(r) \, , \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.16"></A><IMG  WIDTH="342" HEIGHT="33" BORDER="0"  SRC="img58.png"  ALT="\begin{displaymath} E = T + V(r)   , \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.17)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> o&#249; <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} \vec{r} = \vec{R}_1 - \vec{R}_2 \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.17"></A><IMG  WIDTH="331" HEIGHT="31" BORDER="0"  SRC="img59.png"  ALT="\begin{displaymath} \vec{r} = \vec{R}_1 - \vec{R}_2 \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.18)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> est la coordon&#233;e relative et  <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} \vec{R} = \frac{m_1 \vec{R}_1 + m_2 \vec{R}_2}{m_1+m_2} \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.18"></A><IMG  WIDTH="359" HEIGHT="54" BORDER="0"  SRC="img60.png"  ALT="\begin{displaymath} \vec{R} = \frac{m_1 \vec{R}_1 + m_2 \vec{R}_2}{m_1+m_2} \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.19)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> est la coordon&#233;e du centre de masse.  Or, il se passe que l'&#233;nergie cin&#233;tique peut &#234;tre exprim&#233; dans ces deux coordin&#233;es, <BR> <DIV ALIGN="CENTER"><A NAME="eq:models.19"></A> <!-- MATH  \begin{eqnarray} T & = & \frac{|p_1|^2}{2m_1} + \frac{|p_2|^2}{2m_2} \nonumber \\     & = & \frac{1}{2} m_1 |\dot{\vec{R}}_1|^2 + \frac{1}{2} m_2 |\dot{\vec{R}}_2|^2 \nonumber \\     & = & \frac{1}{2} m_1 |\dot{\vec{R}}- \frac{m_2}{m_1+m_2} \dot{\vec{r}}|^2           + \frac{1}{2} m_2 |\dot{\vec{R}}+ \frac{m_1}{m_1+m_2} \dot{\vec{r}}|^2 \nonumber \\     & = & \frac{1}{2} m_1 |\dot{\vec{R}}|^2 - \frac{1}{2} \frac{m_1 m_2}{m_1+m_2} \dot{\vec{R}}\cdot \dot{\vec{r}}           + \frac{1}{2}\frac{m_1m_2^2}{(m_1+m_2)^2} |\dot{\vec{r}}|^2 \nonumber \\     & + & \frac{1}{2} m_2 |\dot{\vec{R}}|^2 + \frac{1}{2} \frac{m_1m_2}{m_1+m_2} \dot{\vec{R}}\cdot\dot{\vec{r}}           + \frac{1}{2}\frac{m_2m_1^2}{(m_1+m_2)^2} |\dot{\vec{r}}|^2 \nonumber \\     & = & \frac{1}{2} (m_1+m_2)|\dot{\vec{R}}|^2            + \frac{1}{2} \left( \frac{m_1 m_2}{m_1+m_2} \right) |\dot{\vec{r}}|^2 \nonumber \\     & = & \frac{1}{2} M |\dot{R}|^2 + \frac{1}{2} \mu |\dot{r}|^2 \, , \end{eqnarray}  --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="20" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"  SRC="img61.png"  ALT="$\displaystyle T$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="108" HEIGHT="67" ALIGN="MIDDLE" BORDER="0"  SRC="img62.png"  ALT="$\displaystyle \frac{\vert p_1\vert^2}{2m_1} + \frac{\vert p_2\vert^2}{2m_2}$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="184" HEIGHT="61" ALIGN="MIDDLE" BORDER="0"  SRC="img63.png"  ALT="$\displaystyle \frac{1}{2} m_1 \vert\dot{\vec{R}}_1\vert^2 + \frac{1}{2} m_2 \vert\dot{\vec{R}}_2\vert^2$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="383" HEIGHT="61" ALIGN="MIDDLE" BORDER="0"  SRC="img64.png"  ALT="$\displaystyle \frac{1}{2} m_1 \vert\dot{\vec{R}}- \frac{m_2}{m_1+m_2} \dot{\vec... ... + \frac{1}{2} m_2 \vert\dot{\vec{R}}+ \frac{m_1}{m_1+m_2} \dot{\vec{r}}\vert^2$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="385" HEIGHT="67" ALIGN="MIDDLE" BORDER="0"  SRC="img65.png"  ALT="$\displaystyle \frac{1}{2} m_1 \vert\dot{\vec{R}}\vert^2 - \frac{1}{2} \frac{m_1... ...ot{\vec{r}} + \frac{1}{2}\frac{m_1m_2^2}{(m_1+m_2)^2} \vert\dot{\vec{r}}\vert^2$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img33.png"  ALT="$\textstyle +$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="384" HEIGHT="67" ALIGN="MIDDLE" BORDER="0"  SRC="img66.png"  ALT="$\displaystyle \frac{1}{2} m_2 \vert\dot{\vec{R}}\vert^2 + \frac{1}{2} \frac{m_1... ...ot{\vec{r}} + \frac{1}{2}\frac{m_2m_1^2}{(m_1+m_2)^2} \vert\dot{\vec{r}}\vert^2$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="306" HEIGHT="61" ALIGN="MIDDLE" BORDER="0"  SRC="img67.png"  ALT="$\displaystyle \frac{1}{2} (m_1+m_2)\vert\dot{\vec{R}}\vert^2 + \frac{1}{2} \left( \frac{m_1 m_2}{m_1+m_2} \right) \vert\dot{\vec{r}}\vert^2$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="156" HEIGHT="61" ALIGN="MIDDLE" BORDER="0"  SRC="img68.png"  ALT="$\displaystyle \frac{1}{2} M \vert\dot{R}\vert^2 + \frac{1}{2} \mu \vert\dot{r}\vert^2   ,$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.20)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> o&#249; <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} M = m_1+m_2 \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.20"></A><IMG  WIDTH="338" HEIGHT="31" BORDER="0"  SRC="img69.png"  ALT="\begin{displaymath} M = m_1+m_2 \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.21)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> est la masse totale et  <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} \mu = \frac{m_1 m_2}{m_1+m_2} \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.21"></A><IMG  WIDTH="335" HEIGHT="44" BORDER="0"  SRC="img70.png"  ALT="\begin{displaymath} \mu = \frac{m_1 m_2}{m_1+m_2} \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.22)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> est la masse reduite. Nous voyons donc que l'&#201;q.&nbsp;(<A HREF="node4.html#eq:models.16">1.17</A>) devient <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} E  = T(R) + T(r) + V(r) \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.22"></A><IMG  WIDTH="387" HEIGHT="33" BORDER="0"  SRC="img71.png"  ALT="\begin{displaymath} E = T(R) + T(r) + V(r)   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.23)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> Comme les variables <IMG  WIDTH="21" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"  SRC="img72.png"  ALT="$R$"> et <IMG  WIDTH="16" HEIGHT="20" ALIGN="BOTTOM" BORDER="0"  SRC="img73.png"  ALT="$r$"> sont ind&#233;pendante et <IMG  WIDTH="22" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"  SRC="img9.png"  ALT="$E$"> est constante, nous avons les trois nouvelles &#233;quations, <BR> <DIV ALIGN="CENTER"><A NAME="eq:models.23"></A> <!-- MATH  \begin{eqnarray} E & = & E_{CM} + E_{rel} \nonumber \\   E_{CM} & = & T(R) \nonumber \\   E_{rel} & = & T(r) + V(r) \, . \end{eqnarray}  --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="22" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"  SRC="img74.png"  ALT="$\displaystyle E$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="101" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"  SRC="img75.png"  ALT="$\displaystyle E_{CM} + E_{rel}$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="46" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"  SRC="img76.png"  ALT="$\displaystyle E_{CM}$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="49" HEIGHT="40" ALIGN="MIDDLE" BORDER="0"  SRC="img77.png"  ALT="$\displaystyle T(R)$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="38" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"  SRC="img78.png"  ALT="$\displaystyle E_{rel}$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="114" HEIGHT="40" ALIGN="MIDDLE" BORDER="0"  SRC="img79.png"  ALT="$\displaystyle T(r) + V(r)   .$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.24)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> Ce qui signifie que le probl&#232;me de la motion des noyaux du diatomique se s&#233;pare dans  deux probl&#232;mes ind&#233;pendantes : le probl&#232;me du motion du centre de masse, <IMG  WIDTH="27" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"  SRC="img30.png"  ALT="$M$">, et le probl&#232;me  du motion d'un pseudo particle de masse <IMG  WIDTH="18" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img80.png"  ALT="$\mu$">.  <P> <DIV ALIGN="CENTER"> <B>La Particule dans une Bo&#238;te</B>  </DIV>  <P> Si la mol&#233;cule est dans un gaz dans une bo&#238;te finie, le potentiel pour le centre de masse de la mol&#233;cule est donn&#233; par <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} V(\vec{R}) = \left\{ \begin{array}{l} 0 \, ; \, \vec{R} \mbox{ est \`a l'int\'erieur de la bo\^{\i}te} \\                      + \infty \, ; \, \vec{R} \mbox{ est \`a l'ext\'erieur de la bo\^{\i}te} \end{array} \right.   \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.24"></A><IMG  WIDTH="481" HEIGHT="58" BORDER="0"  SRC="img81.png"  ALT="\begin{displaymath} V(\vec{R}) = \left\{ \begin{array}{l} 0   ;   \vec{R} \mb... ... \\lq a l'ext\'erieur de la bo\^{\i}te} \end{array} \right.   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.25)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> &#192; l'int&#233;rieur de la bo&#238;te, <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} T(R) = \frac{1}{2} M |\dot{R}|^2 = E_{CM} \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.25"></A><IMG  WIDTH="384" HEIGHT="45" BORDER="0"  SRC="img82.png"  ALT="\begin{displaymath} T(R) = \frac{1}{2} M \vert\dot{R}\vert^2 = E_{CM}   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.26)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> Ces &#233;quations sont d'une importance fondamentale dans la derivation des &#233;quations d'un gaz id&#233;el &#224; partir d'une mod&#232;le mol&#233;culaire.  Mais pour nous, le mod&#232;le d'une particule dans une bo&#238;te va aussi servir comme premi&#232;re approximation pour le potentiel senti par les &#233;lectrons <IMG  WIDTH="18" HEIGHT="20" ALIGN="BOTTOM" BORDER="0"  SRC="img83.png"  ALT="$\pi$"> dans des poly&#232;nes et des  mol&#233;cules aromatiques.  <P> <DIV ALIGN="CENTER"> <B>Rotateur Rigide</B>  </DIV>  <P> Par contre, la motion de la pseudo particule d&#233;crivant la motion interne de la diatomique est gouvern&#233;e par l'&#233;quation, <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} \frac{1}{2} \mu |\dot{r}|^2 + V(r) = E_{rel} \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.26"></A><IMG  WIDTH="369" HEIGHT="45" BORDER="0"  SRC="img84.png"  ALT="\begin{displaymath} \frac{1}{2} \mu \vert\dot{r}\vert^2 + V(r) = E_{rel}   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.27)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> Si nous supposons dans une premi&#232;re approximation que la distance <IMG  WIDTH="16" HEIGHT="20" ALIGN="BOTTOM" BORDER="0"  SRC="img73.png"  ALT="$r$"> est fixe, nous avons le mod&#232;le du rotateur rigide.  L'impulsion angulaire, <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} L = \mu v r \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.27"></A><IMG  WIDTH="315" HEIGHT="32" BORDER="0"  SRC="img85.png"  ALT="\begin{displaymath} L = \mu v r \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.28)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> est conservie (<IMG  WIDTH="61" HEIGHT="40" ALIGN="MIDDLE" BORDER="0"  SRC="img86.png"  ALT="$v=\vert\dot{r}\vert$"> est la vitesse).  Nous avons donc que <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} E_{rot} = \frac{L^2}{2\mu r^2} = E_{rel} - V(r) \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.28"></A><IMG  WIDTH="394" HEIGHT="52" BORDER="0"  SRC="img87.png"  ALT="\begin{displaymath} E_{rot} = \frac{L^2}{2\mu r^2} = E_{rel} - V(r)   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.29)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> est l'&#233;nergie rotationnelle.  Nous pouvons &#233;galement &#233;crire <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} E_{rot} = \frac{L^2}{2 I} \, , \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.28b"></A><IMG  WIDTH="325" HEIGHT="48" BORDER="0"  SRC="img88.png"  ALT="\begin{displaymath} E_{rot} = \frac{L^2}{2 I}   , \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.30)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> o&#249;  <BR> <DIV ALIGN="CENTER"><A NAME="eq:models.29"></A> <!-- MATH  \begin{eqnarray} I & = & m_1 |\vec{R}_1 - \vec{R}|^2 + m_2 |\vec{R}_2 - \vec{R}|^2 \nonumber \\     & = & \mu r^2 \end{eqnarray}  --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="16" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"  SRC="img89.png"  ALT="$\displaystyle I$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="233" HEIGHT="48" ALIGN="MIDDLE" BORDER="0"  SRC="img90.png"  ALT="$\displaystyle m_1 \vert\vec{R}_1 - \vec{R}\vert^2 + m_2 \vert\vec{R}_2 - \vec{R}\vert^2$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="34" HEIGHT="43" ALIGN="MIDDLE" BORDER="0"  SRC="img91.png"  ALT="$\displaystyle \mu r^2$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.31)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> est le moment d'in&#233;rtie.  <P> <DIV ALIGN="CENTER"> <B>Oscillateur Harmonique</B>  </DIV>  <P> Nous pouvons toujours faire une d&#233;velopement de type Taylor autour du longeur de liaison d'&#233;quilibre pour la diatomique, <BR> <DIV ALIGN="CENTER"><A NAME="eq:models.30"></A> <!-- MATH  \begin{eqnarray} V(r) & = & V(r_e) + V'(r_e)(r-r_e) + \frac{1}{2} V''(r_e)(r-r_e)^2 + \frac{1}{6} V'''(r_e)(r-r_e)^3 + \cdots   \nonumber \\   & = & V(r_e) + \frac{1}{2} V''(r_e)(r-r_e)^2 + \frac{1}{6} V'''(r_e)(r-r_e)^3 + \cdots \, , \end{eqnarray}  --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="45" HEIGHT="40" ALIGN="MIDDLE" BORDER="0"  SRC="img92.png"  ALT="$\displaystyle V(r)$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="554" HEIGHT="61" ALIGN="MIDDLE" BORDER="0"  SRC="img93.png"  ALT="$\displaystyle V(r_e) + V'(r_e)(r-r_e) + \frac{1}{2} V''(r_e)(r-r_e)^2 + \frac{1}{6} V'''(r_e)(r-r_e)^3 + \cdots$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="431" HEIGHT="61" ALIGN="MIDDLE" BORDER="0"  SRC="img94.png"  ALT="$\displaystyle V(r_e) + \frac{1}{2} V''(r_e)(r-r_e)^2 + \frac{1}{6} V'''(r_e)(r-r_e)^3 + \cdots   ,$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.32)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> o&#249; la premi&#232;re deriv&#233;e est nulle &#224; cause de la condition que la force, <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} F = - V'(r_e) \, , \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.31"></A><IMG  WIDTH="337" HEIGHT="33" BORDER="0"  SRC="img95.png"  ALT="\begin{displaymath} F = - V'(r_e)   , \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.33)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> soit z&#233;ro &#224; la g&#233;om&#233;trie d'&#233;quilibre.  Pour encore simplifier les choses, nous prenons <IMG  WIDTH="86" HEIGHT="40" ALIGN="MIDDLE" BORDER="0"  SRC="img96.png"  ALT="$V(r_e)=0$"> (il  suffit de rechoisir le z&#233;ro d'&#233;nergie) et nous n&#233;gligeons tous les d&#233;riv&#233;es commen&#231;ant avec  la troisi&#232;me, pour obtenir le mod&#232;le d'oscillateur harmonique, <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} E_{HO} = \frac{p^2}{2\mu} - \frac{k}{2} x^2 \, , \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.32"></A><IMG  WIDTH="356" HEIGHT="52" BORDER="0"  SRC="img97.png"  ALT="\begin{displaymath} E_{HO} = \frac{p^2}{2\mu} - \frac{k}{2} x^2   , \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.34)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> o&#249; <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} x = r-r_e \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.33"></A><IMG  WIDTH="326" HEIGHT="31" BORDER="0"  SRC="img98.png"  ALT="\begin{displaymath} x = r-r_e   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.35)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> Dans l'approximation du rotateur rigide et d'oscillateur harmonique, l'&#233;nergie interne d'un diatomique est donn&#233;e par <BR> <DIV ALIGN="CENTER"><A NAME="eq:models.34"></A> <!-- MATH  \begin{eqnarray} E_{rel} & = & E_{rot} + E_{HO} \nonumber \\           & = & \frac{p^2}{2\mu} + \frac{L^2}{2 \mu r^2 } - \frac{k}{2} (r-r_e)^2  \, . \end{eqnarray}  --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="38" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"  SRC="img78.png"  ALT="$\displaystyle E_{rel}$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="100" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"  SRC="img99.png"  ALT="$\displaystyle E_{rot} + E_{HO}$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="211" HEIGHT="67" ALIGN="MIDDLE" BORDER="0"  SRC="img100.png"  ALT="$\displaystyle \frac{p^2}{2\mu} + \frac{L^2}{2 \mu r^2 } - \frac{k}{2} (r-r_e)^2   .$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.36)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P>  <P> Maintenant cherchons la solution du probl&#232;me de l'oscillateur harmonique en m&#233;canique classique. La loi de Newton est   <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} F = \mu a = \mu \frac{d^2 x}{d t^2}  \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.35"></A><IMG  WIDTH="349" HEIGHT="48" BORDER="0"  SRC="img101.png"  ALT="\begin{displaymath} F = \mu a = \mu \frac{d^2 x}{d t^2}   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.37)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> La force est donn&#233;e par <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} F = - \frac{d V(x)}{dx} = -\frac{d (1/2) kx^2}{dx} = -kx \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.36"></A><IMG  WIDTH="431" HEIGHT="48" BORDER="0"  SRC="img102.png"  ALT="\begin{displaymath} F = - \frac{d V(x)}{dx} = -\frac{d (1/2) kx^2}{dx} = -kx   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.38)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> Nous devons donc resoudre l'&#233;quation <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} -kx(t) = \mu \frac{d^2 x(t)}{dt^2} \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.37"></A><IMG  WIDTH="358" HEIGHT="48" BORDER="0"  SRC="img103.png"  ALT="\begin{displaymath} -kx(t) = \mu \frac{d^2 x(t)}{dt^2}   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.39)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> Nous regarderons en beaucoup de d&#233;tail des techniques diverses pour r&#233;soudre les &#233;quations diff&#233;rentielles.  Il suffit pour l'instant &#224; savoir que la solution g&#233;n&#233;rale de l'&#201;q.&nbsp;(<A HREF="node4.html#eq:models.37">1.39</A>) peut &#234;tre &#233;cris comme, <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} x(t) = A \cos \left( \sqrt{\frac{k}{\mu}} t + \phi \right) \, , \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.38"></A><IMG  WIDTH="386" HEIGHT="57" BORDER="0"  SRC="img104.png"  ALT="\begin{displaymath} x(t) = A \cos \left( \sqrt{\frac{k}{\mu}} t + \phi \right)   , \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.40)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P> o&#249; <IMG  WIDTH="21" HEIGHT="21" ALIGN="BOTTOM" BORDER="0"  SRC="img105.png"  ALT="$A$"> est l'amplitude de la motion est <IMG  WIDTH="18" HEIGHT="37" ALIGN="MIDDLE" BORDER="0"  SRC="img106.png"  ALT="$\phi$"> et la phase de la motion.  La periode de la motion est donn&#233;e par <BR> <DIV ALIGN="CENTER"><A NAME="eq:models.39"></A> <!-- MATH  \begin{eqnarray} \sqrt{\frac{k}{\mu}} \tau & = & 2 \pi \nonumber \\   \tau & = & 2 \pi \sqrt{ \frac{\mu}{k} } \, . \end{eqnarray}  --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="48" HEIGHT="73" ALIGN="MIDDLE" BORDER="0"  SRC="img107.png"  ALT="$\displaystyle \sqrt{\frac{k}{\mu}} \tau$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="27" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img108.png"  ALT="$\displaystyle 2 \pi$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="17" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img109.png"  ALT="$\displaystyle \tau$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="67" HEIGHT="62" ALIGN="MIDDLE" BORDER="0"  SRC="img110.png"  ALT="$\displaystyle 2 \pi \sqrt{ \frac{\mu}{k} }   .$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.41)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> La fr&#233;quence est donn&#233;e par <BR> <DIV ALIGN="CENTER"><A NAME="eq:models.40"></A> <!-- MATH  \begin{eqnarray} \nu & = & \frac{1}{\tau} \nonumber \\        & = & \frac{1}{2\pi} \sqrt{\frac{k}{\mu} } \nonumber \\    \omega & = & 2\pi \nu \nonumber \\           & = & \sqrt{\frac{k}{\mu}} \, . \end{eqnarray}  --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="17" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img111.png"  ALT="$\displaystyle \nu$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="21" HEIGHT="61" ALIGN="MIDDLE" BORDER="0"  SRC="img112.png"  ALT="$\displaystyle \frac{1}{\tau}$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="62" HEIGHT="73" ALIGN="MIDDLE" BORDER="0"  SRC="img113.png"  ALT="$\displaystyle \frac{1}{2\pi} \sqrt{\frac{k}{\mu} }$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG  WIDTH="19" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img114.png"  ALT="$\displaystyle \omega$"></TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="38" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img115.png"  ALT="$\displaystyle 2\pi \nu$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> &nbsp;</TD></TR> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT">&nbsp;</TD> <TD ALIGN="CENTER" NOWRAP><IMG  WIDTH="21" HEIGHT="36" ALIGN="MIDDLE" BORDER="0"  SRC="img21.png"  ALT="$\textstyle =$"></TD> <TD ALIGN="LEFT" NOWRAP><IMG  WIDTH="46" HEIGHT="73" ALIGN="MIDDLE" BORDER="0"  SRC="img116.png"  ALT="$\displaystyle \sqrt{\frac{k}{\mu}}   .$"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.42)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> Donc <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} k = \mu \omega^2 \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.41"></A><IMG  WIDTH="318" HEIGHT="32" BORDER="0"  SRC="img117.png"  ALT="\begin{displaymath} k = \mu \omega^2   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.43)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P>  <P> <DIV ALIGN="CENTER"> <B>Atome Hydrog&#233;no&#239;de</B>  </DIV>  <P> L'expression pour l'&#233;nergie d'un atome hydrog&#233;no&#239;de est donn&#233;e par <BR> <DIV ALIGN="RIGHT">  <!-- MATH  \begin{equation} E = \frac{p^2}{2m} - \frac{Ze^2}{r} \, . \end{equation}  --> <TABLE WIDTH="100%" ALIGN="CENTER"> <TR VALIGN="MIDDLE"><TD ALIGN="CENTER" NOWRAP><A NAME="eq:models.42"></A><IMG  WIDTH="348" HEIGHT="48" BORDER="0"  SRC="img118.png"  ALT="\begin{displaymath} E = \frac{p^2}{2m} - \frac{Ze^2}{r}   . \end{displaymath}"></TD> <TD WIDTH=10 ALIGN="RIGHT"> (1.44)</TD></TR> </TABLE> <BR CLEAR="ALL"></DIV><P></P>  <P> <DIV ALIGN="CENTER"> <B>Atomes et mol&#233;cules &#224; plusieurs &#233;lectrons</B>  </DIV>  <P> Les probl&#232;mes d'un particule dans une bo&#238;te, le rotateur rigide, l'oscillateur harmonique, et les atomes hydrog&#233;no&#239;des ont des solutions QM exactes.  Pour aller plus loin et traiter des atomes &#224; plusieurs &#233;lectrons et m&#234;me des mol&#233;cules(!) nous trouverons que l'&#233;quation de Schr&#246;dinger n'a plus de solution exacte et il faut donc nous familiariser avec des diff&#233;rentes fa&#231;ons pour trouver des solutions approximatives.  <P> <HR> <!--Navigation Panel--> <A NAME="tex2html101"   HREF="node5.html"> <IMG WIDTH="37" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="next"  SRC="file:/usr/share/latex2html/icons/next.png"></A>  <A NAME="tex2html97"   HREF="node3.html"> <IMG WIDTH="26" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="up"  SRC="file:/usr/share/latex2html/icons/up.png"></A>  <A NAME="tex2html91"   HREF="node3.html"> <IMG WIDTH="63" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="previous"  SRC="file:/usr/share/latex2html/icons/prev.png"></A>  <A NAME="tex2html99"   HREF="node1.html"> <IMG WIDTH="65" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="contents"  SRC="file:/usr/share/latex2html/icons/contents.png"></A>   <BR> <B> Next:</B> <A NAME="tex2html102"   HREF="node5.html">Approache semi classique</A> <B> Up:</B> <A NAME="tex2html98"   HREF="node3.html">Bases conceptuels</A> <B> Previous:</B> <A NAME="tex2html92"   HREF="node3.html">Bases conceptuels</A>  &nbsp <B>  <A NAME="tex2html100"   HREF="node1.html">Contents</A></B>  <!--End of Navigation Panel--> <ADDRESS>  2002-11-23 </ADDRESS> </BODY> </HTML> 
