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jeanwsr merged 8 commits into from
Aug 29, 2024
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fix in Chap6 #51

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6b820ba
fix incorrect formula
Usu171 Jul 7, 2024
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update
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Usu171 Jul 7, 2024
6932210
Merge branch 'NominHanggai:master' into master
Usu171 Aug 27, 2024
2c85dcf
fix in Chap6
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54 changes: 25 additions & 29 deletions Chaps/Chap6.tex
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Original file line number Diff line number Diff line change
Expand Up @@ -334,13 +334,13 @@ \subsection{两态系统的图微扰论}

\[
\begin{array}{ccc}
\braket{1|\vs|1}&=V_{11}\longleftrightarrow &\tikz[baseline=(current bounding box.center)]{\path[draw=blue,postaction={on each segment={mid arrow}}] (0,0.33)--node[right]{1}(0,0)--node[right]{1}(0,-.33);}
\braket{1|\vs|1}&=V_{11}\longleftrightarrow &\tikz[baseline=(current bounding box.center)]{\path[draw=blue,postaction={on each segment={mid arrow}}] (0,0.33)--node[right]{1}(0,0)node{\dian}--node[right]{1}(0,-.33);}
\\[5mm]
\braket{1|\vs|2}&=V_{12}\longleftrightarrow &\tikz[scale=.5,baseline=(current bounding box.center)]{\path[draw=blue,postaction={on each segment={mid arrow}}] (-0.77,0.77)--(0,0)--(.77,.77);}
\braket{1|\vs|2}&=V_{12}\longleftrightarrow &\tikz[scale=.5,baseline=(current bounding box.center)]{\path[draw=blue,postaction={on each segment={mid arrow}}] (-0.77,0.77)--node[left]{1}(0,0)node{\dian}--node[right]{2}(.77,.77);}
\\[5mm]
\braket{1|\vs|2}&=V_{12}\longleftrightarrow &
\braket{2|\vs|1}&=V_{21}\longleftrightarrow &\tikz[scale=.5,baseline=(current bounding box.center)]{\path[draw=blue,postaction={on each segment={mid arrow}}] (-0.77,-0.77)--node[left]{2}(0,0)node{\dian}--node[right]{1}(0.77,-0.77);}
\\[5mm]
\braket{1|\vs|1}&=V_{11}\longleftrightarrow &
\braket{2|\vs|2}&=V_{22}\longleftrightarrow &\tikz[baseline=(current bounding box.center)]{\path[draw=blue,postaction={on each segment={mid arrow}}] (0,-0.33)--node[right]{2}(0,0)node{\dian}--node[right]{2}(0,0.33);}
\end{array}
\]
总之,
Expand Down Expand Up @@ -574,11 +574,11 @@ \subsection{两态系统的图微扰论}
(3,3)..controls+(-.75,-.75)and+(-.75,.75)..node[left]{1}(3,0)
(4.5,3)--node[right]{1}(4.5,2)--node[right]{1}(4.5,1)--node[right]{1}(4.5,0) % n=4, 4th
(4.5,0)..controls+(-.75,.75)and+(-.75,-.75)..node[left]{2}(4.5,3)
(6,0)..controls+(.5,.5)and+(.5,-.5)..node[left]{1}(6,2) % n=4, 5th
(6,0)..controls+(.5,.5)and+(.5,-.5)..node[right]{2}(6,2) % n=4, 5th
(6,2)..controls+(.25,-.25)and+(.25,.25)..node[left]{1}(6,1)
(6,1)..controls+(-.5,.5)and+(-.5,-.5)..node[right]{2}(6,3)
(6,3)..controls+(-.75,-.75)and+(-.75,.75)..node[left]{1}(6,0)
(7.5,2)..controls+(.5,-.5)and+(.5,.5)..node[left]{2}(7.5,0) % n=4, 6th
(7.5,2)..controls+(.5,-.5)and+(.5,.5)..node[right]{1}(7.5,0) % n=4, 6th
(7.5,1)..controls+(.25,.25)and+(.25,-.25)..node[left]{2}(7.5,2)
(7.5,3)..controls+(-.5,-.5)and+(-.5,.5)..node[right]{1}(7.5,1)
(7.5,0)..controls+(-.75,.75)and+(-.75,-.75)..node[left]{2}(7.5,3)
Expand All @@ -595,7 +595,7 @@ \subsection{两态系统的图微扰论}
转换规则如下:
\begin{enumerate}
\item 每个点为分子贡献一个因子$\braket{\textit{穿入线的标号}|\vs|\textit{穿出线的标号}}$.
\item 每一对相邻点贡献一个分母上的因子\[ \sum E^{(0)}_\mathrm{空穴}-\sum E^{(0)}_\mathrm{粒子}\]
\item 每一对相邻点贡献一个分母上的因子\[ \sum E^{(0)}_\text{空穴}-\sum E^{(0)}_\text{粒子}\]
其中求和遍及所有与两个点之间的一条水平线相交的空穴线和粒子线.
\item 表达式的正负号为$(-)^{h+l}$, $h$是空穴线的数目, $l$是闭合圈图的数目. 本节的图中$l$总是1, 这里给出更一般的规则以利后用.
\end{enumerate}
Expand All @@ -611,7 +611,7 @@ \subsection{两态系统的图微扰论}
(-2,2.5)node[left]{$A$}--(2,2.5);
\begin{scope}
\path [draw=blue,postaction={on each segment={mid arrow}}]
(0,0)..controls+(.5,.5)and+(.5,-.5)..node[left]{1}(0,2) % n=4, 5th
(0,0)..controls+(.5,.5)and+(.5,-.5)..node[right]{2}(0,2) % n=4, 5th
(0,2)..controls+(.25,-.25)and+(.25,.25)..node[left]{1}(0,1)
(0,1)..controls+(-.5,.5)and+(-.5,-.5)..node[right]{2}(0,3)
(0,3)..controls+(-.75,-.75)and+(-.75,.75)..node[left]{1}(0,0);
Expand All @@ -633,12 +633,12 @@ \subsection{两态系统的图微扰论}
(0,0)circle(1pt) (0,1)circle(1pt) (0,2)circle(1pt) (0,3)circle(1pt);
\begin{scope}
\path [draw=blue,postaction={on each segment={mid arrow}}]
(0,0)..controls+(.5,.5)and+(.5,-.5)..node[left]{1}(0,2) % n=4, 5th
(0,0)..controls+(.5,.5)and+(.5,-.5)..node[right]{2}(0,2) % n=4, 5th
(0,2)..controls+(.25,-.25)and+(.25,.25)..node[left]{1}(0,1)
(0,1)..controls+(-.5,.5)and+(-.5,-.5)..node[right]{2}(0,3)
(0,3)..controls+(-.75,-.75)and+(-.75,.75)..node[left]{1}(0,0);
\end{scope}
\end{tikzpicture} = (-)^3\frac{(V_{12}V_{21})^2}{\left(E_1^{(0)} - E_2^{(0)}\right)\left(2E_1^{(0)} - 2E_2^{(0)}\right)} = \frac{(V_{12}V_{21})^2}{2\left(E_1^{(0)} - E_2^{(0)}\right)^2}
\end{tikzpicture} = (-)^3\frac{(V_{12}V_{21})^2}{\left(E_1^{(0)} - E_2^{(0)}\right)^2\left(2E_1^{(0)} - 2E_2^{(0)}\right)} = -\frac{(V_{12}V_{21})^2}{2\left(E_1^{(0)} - E_2^{(0)}\right)^3}
\end{equation*}

值得注意,
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Expand All @@ -653,11 +653,9 @@ \subsection{两态系统的图微扰论}
(4.5,0)..controls+(-.75,.75)and+(-.75,-.75)..node[left]{2}(4.5,3);
\end{tikzpicture} \hspace{.5cm}
&\equiv \hspace{.5cm}
\begin{tikzpicture}[baseline={(current bounding box.center)},scale=.67]
\filldraw[blue]
(4.50,0)circle(1pt) (4.5,3)circle(1pt);
\path [draw=blue,postaction={on each segment={mid arrow}}] (4.5,3) arc (45:15:2.12cm)node[right,midway]{1} node{\dian} arc (15:-15:2.12cm)node[right,midway]{1}node{\dian} arc (-15:-45:2.12cm)
(4.5,0) arc(225:135:2.12cm)node[left,midway]{2};
\begin{tikzpicture}[baseline={(current bounding box.center)},scale=.67]\diagsize
\path [draw=blue,postaction={on each segment={mid arrow}}] (4.5,3) arc (45:15:2.12cm)node[right,midway]{1}node{\dian} arc (15:-15:2.12cm)node[right,midway]{1}node{\dian} arc (-15:-45:2.12cm)node[right,midway]{1}node{\dian}
(4.5,0) arc(225:135:2.12cm)node[left,midway]{2}node{\dian};
\end{tikzpicture}\vspace{.5cm}\\
\begin{tikzpicture}[baseline={(current bounding box.center)},scale=.67]\diagsize
\filldraw[blue]
Expand All @@ -669,16 +667,14 @@ \subsection{两态系统的图微扰论}
\end{tikzpicture} \hspace{.5cm}
&\equiv \hspace{.5cm}
\begin{tikzpicture}[baseline={(current bounding box.center)},scale=.67]\diagsize
\filldraw[blue]
(4.50,0)circle(1pt) (4.5,3)circle(1pt);
\path [draw=blue,postaction={on each segment={mid arrow}}] (4.5,0) arc (-45:-15:2.12cm)node[right,midway]{2} node{\dian} arc (-15:45:2.12cm)node[right,midway]{2}node{\dian}
(4.5,3) arc(135:165:2.12cm)node[left,midway]{1} arc(165:225:2.12cm)node[midway,left]{1};
(4.5,3) arc(135:165:2.12cm)node[left,midway]{1}node{\dian} arc(165:225:2.12cm)node[midway,left]{1}node{\dian};
\end{tikzpicture}\vspace{.5cm}\\
\begin{tikzpicture}[baseline={(current bounding box.center)},scale=.67]\diagsize
\filldraw[blue]
(6,0)circle(1pt) (6,1)circle(1pt) (6,2)circle(1pt) (6,3)circle(1pt);
\path [draw=blue,postaction={on each segment={mid arrow}}]
(6,0)..controls+(.5,.5)and+(.5,-.5)..node[left]{1}(6,2) % n=4, 5th
(6,0)..controls+(.5,.5)and+(.5,-.5)..node[right]{2}(6,2) % n=4, 5th
(6,2)..controls+(.25,-.25)and+(.25,.25)..node[left]{1}(6,1)
(6,1)..controls+(-.5,.5)and+(-.5,-.5)..node[right]{2}(6,3)
(6,3)..controls+(-.75,-.75)and+(-.75,.75)..node[left]{1}(6,0);
Expand Down Expand Up @@ -919,7 +915,7 @@ \subsection{$N$个态的图微扰论}
arc (135:180:2.12cm)node[left,midway]{1} node{\dian}
arc (180:225:2.12cm)node[left,midway]{1} node{\dian};
\end{tikzpicture}
& = \sum_{n\neq 1}\frac{V_{1n}V_{n1}}{E_1^{(0)} - E_n^{(0)}} = \sum_n'\frac{V_{1n}V_{n1}}{E_1^{(0)} - E_n^{(0)}}
& = \sum_{nm}'\frac{V_{1m}V_{mn}V_{n1}}{(E_1^{(0)} - E_m^{(0)})(E_1^{(0)} - E_n^{(0)})} - \sum_n'\frac{V_{11}V_{1n}V_{n1}}{(E_1^{(0)} - E_n^{(0)})^2}
\end{align}
\end{subequations}
这两个表达式与之前所得到的二阶、三阶能量的表达式(6.
Expand Down Expand Up @@ -1389,7 +1385,7 @@ \section{轨道微扰论: 单粒子微扰}
第零个烯烃单元就与第$n$个单元一样,
第$(n+1)$个实际就是第一个.
在该模型中,
本分子的精确共振能为$2\beta$,
苯分子的精确共振能为$2\beta$,
此外有如下精确共振能的渐近公式：
\begin{align}
\lim\limits_{N\to\infty}E_R = (4/\pi-1)N\beta=0.2732N\beta
Expand Down Expand Up @@ -2303,7 +2299,7 @@ \subsection{Hugenholtz图}
to (0,-5.5)
arc(135:225:3.18198*2/3)
arc(140:220:1.16679)
arc(-40:40:1.16679)
arc(-40:40:1.16679) node{\dian}
to (0,-10)
;
\path[draw=blue]
Expand All @@ -2326,12 +2322,12 @@ \subsection{Hugenholtz图}
arc(-45:45:3.18198) node{\dian}
arc(60:-60:2.59807621135);
\path[draw=blue]
(9,-10)
arc(-45:45:3.18198)
(9,-10) node{\dian}
arc(-45:45:3.18198) node{\dian}
arc(135:225:3.18198*2/3)
arc(-45:45:3.18198*2/3)
arc(150:210:3)
to (9,-7)
arc(150:210:3) node{\dian}
to (9,-7) node{\dian}
arc(135:225:3.18198*2/3)
arc(-45:45:3.18198*2/3)
arc(30:-30:3);
Expand All @@ -2352,9 +2348,9 @@ \subsection{Hugenholtz图}
\begin{center}
\begin{tikzpicture}[baseline={(current bounding box.south)},scale=.6]
\path [draw=blue, mid arrow]
(0,0) arc(-60:60:1.732) node[midway,right]{$s$};
(0,0)node{\dian} arc(-60:60:1.732) node[midway,right]{$s$};
\path [draw=blue, mid arrow]
(0,3) arc(45:-45:2.12) node[left,midway]{$b$};
(0,3)node{\dian} arc(45:-45:2.12) node[left,midway]{$b$};
\path [draw=blue, mid arrow]
(0,3) arc(135:225:2.12) node[right,midway]{$a$};
\path [draw=blue, mid arrow]
Expand Down Expand Up @@ -2473,7 +2469,7 @@ \subsection{Hugenholtz图}
按照如下规``走过"矩阵元:
\begin{align*}
\braket{i\tikzmark{i}j||k\tikzmark{k}l},\quad
\braket{i\tikzmark{j}j||k\tikzmark{l}l}
\braket{ij\tikzmark{j}||kl\tikzmark{l}}
\tikz[overlay,remember picture]
{\draw[->,above arrow,] (i.north) to (k.north);
\draw[->,above arrow,] (j.north) to (l.north);
Expand Down