计算物理(2021/秋季)

主讲老师：龚明，物质科学楼C812,gongm@ustc.edu.cn

助教：饶泽宇，物质科学楼C810，rzy55555@mail.ustc.edu.cn，13615654890

助教：林孝水，物质科学楼C810，lxsphys@mail.ustc.edu.cn，17857091932

上课时间：2(1,2)，4(8,9),2-18周

上课地点：5503

• QQ群号701981243

• 参考书

  1. 《计算物理学》, Hoffmann
  2. 《计算物理学导论》,Tao Pang
  3. 《计算物理学》, 马文淦
  4. 计算物理, 丁泽军
  5. 《数值计算方法与算法》, 张韵华
• project参考论文
• 作业交电子版，发到助教邮箱。内容包括：pdf+源代码。pdf包括题目结果，必要的推导过程、说明等，最好用Latex写。不要只交源代码。

• 9.16 周四

  1. Vandermonde 方程组
     (1) 给定一组\(n\)个数据点，\((x_1,y_1), \cdots,(x_n,y_n)\)，可以求出精确拟合这些点的\(n-1\)次多项式 \[ y_i=\sum_{j=0}^{j=n-1} a_{j} x_{i}^{j} \] 其系数\(a_0,\cdots,a_{n-1}\)组成的矩阵\(a\)满足: \[Va=y,a=V^{-1}y\] \(V\)为Vandermonde矩阵
     
  2. 两点插值
     (1) 过两点\((a,f(a)), (b,f(b))\)的插值函数: \[ \tilde{f}(x) = f(a) + \dfrac{f(b)-f(a)}{b-a}(x-a) \]
     (2) 误差 \( R(x) = \tilde{f}(x) - f(x) = f^{\prime\prime}(\xi) (x-a)(x-b)/2!,\ \xi \in [a,b] \), 局部误差\( |R(x)| = k_2 |x-a| |x-b| \leq \dfrac{k_2}{4} |a-b|^2 < \varepsilon \), 整体误差 \( \int_a^b |R(x)| dx \)
  3. Newton-Cotes插值
     (1) 过\( (x_1, f(x_1)),\ \cdots (x_n,f(x_n))\) \(n\)个点的插值函数： \[ \tilde{f} (x) = b_0 + b_1(x-x_1) + b_2 (x-x_1)(x-x_2) + \cdots + b_{n-1}(x-x_1) \cdots (x-x_{n-1}) \]
     (2) \(b_0 = f[x_1]=f(x_1)\), \(b_1 =f[x_1,x_2]= \dfrac{f(x_i)-f(x_{x_i+1})}{x_i-x_{i+1}} \), \( b_2 = f[x_0,x_1,x_2] =\dfrac{f[x_i,x_{i+1}]-f[x_{i+1},x_{i+2}]}{x_i-x_{i+2}}\)
  4. Lagrange 插值
     (1) \[ \tilde{f}(x) = \sum_{i} w_i l_i(x) = \sum_i f(x_i) \dfrac{\Pi_{j\neq i}(x-x_j)} {\Pi_{j\neq i} (x_i-x_j)} \]
     (2) \(l_i(x) \)性质：
     1. \(l_i(x_j) = \delta_{ij} \)
     2. \( \sum_i l_i(x) = 1 \)
     3. \(l_i(x) \) 是\(n-1\)次多项式
  5. 积分问题
     a. \( \dot{y} = f(x)\Rightarrow y_n = \int_{x_0}^{x_n} f(x) dx \), 离散点插值构成\( f(x)\), Lagrange插值积分(Newton-Cotes 积分)
     b. \(\dot{y} = f(x,y)\)
  6. Newton-Cotes 积分
     (1) 等分点\( x_i = a+ih,\ i=0,1,\cdots,n,\ h=(b-a)/n \), \[ I = \sum_i f(x_i) \int_a^b l_i(x) dx = \sum_i f(x_i) \int_a^b \dfrac{\Pi_{j\neq i}(x-x_j)}{\Pi_{j \neq i}(x_i -x_j)}dx = \sum_i f(x_i) h \int_0^n \dfrac{\Pi_{j\neq i}(t-j)}{\Pi_{j \neq i}(i -j)}dt, \ x=a+th, \ t\in [0,n]\]
     (2) Newton-Cotes系数 (系数表) \[ c^{(n)}_i = \int_0^h \dfrac{\Pi_{j\neq i}(t-j)}{n\Pi_{j \neq i}(i-j)} dt = \dfrac{(-1)^{n-i}}{i!(n-i)! n }\int_0^n \Pi_j(t-j) dt\]
     (3) \(n=1\), 梯形积分, \(I = (f_a+f_b)h/2\); \(n=2\), Simpson 积分，\(I =\dfrac{h}{6} [f(a)+4f(\dfrac{a+b}{2})+f(b) ] \)
  7. Laplace-Runge-Lenz Vector
  8. Runge介绍pdf
  9. Runge介绍docx
  10. Runge生平简介
  11. 从BLAS的一个例子学习编程技巧
  12. 龚老师代码实例pdf
  13. 龚老师代码实例docx
  14. 第一次作业
  15. 课堂笔记1_林
  16. 课堂笔记1_饶

• 9.23 周四

  1. 积分问题
     奇点问题： \[\int_{-1}^{1}\dfrac{1}{\sqrt{|x|}}dx=\int_{\epsilon}^{1}\dfrac{1}{\sqrt{x}}dx+\int_{-1}^{-\epsilon}\dfrac{1}{\sqrt{-x}}dx\]
     变步长积分。
  2. 矩阵
     (1) 解方程：\[Ax=b,x=A^{-1}b\]
     (2) 本征值问题：\[Ax=\lambda x\] 对应两个函数库(BLAS,LAPACK)
  3. 矩阵类型
     (1) 方阵\((n\times n)\):
     (a) 厄米：\(A=A^{+}\),对应于幺正变换 \[U^{+}AU=\lambda\] 实数情况：\(U\in O(N)\)；复数情况：\(U\in U(N)\)。
     (b) 非厄米：\(A\neq A^{+}\),相似变换 \[P^{-1}\lambda P=A\]
     (2) 非方阵\(n\times m,n\neq m\):奇异值分解(SVD) \[A=UDV^{T}\] 分解是唯一的。
  4. 矩阵分解
     (1) 方阵
     (a)LU分解：\(A=LU\);\(L\)为下三角矩阵，\(U\)为上三角矩阵。可以用来求解：\(Ax=b\) \[LUx=b\Rightarrow Ly=b,y=Ux\]
     (b)QR分解：\(A=QR\)；\(Q\)为正交矩阵，\(R\)为上三角矩阵。
     (c)Grand-Schmidt分解
     (2) 非方阵
     SVD分解,对方阵、非方阵都适用。
  5. 特殊矩阵
     (1) 范德蒙特行列式： \[det(A)=\Pi_{i < j}(x_i-x_j) \]
     (2) Pfaffian(\(2n\times 2n\)):\(Pf^{2}(A)=det(A)\);skew matrix:\(A^{T}=-A\)
     (3) 三对角阵，计算复杂度\(O(N^{2})\)，占用内存小，可解析计算。
     (4) Toeplitz matrix:\(A=(t_{i-j})_{N\times N}\);Circular matrix
     (5) 随机矩阵(random matrix):\(A=(\xi_{ij})_{N\times N}\)；\(\xi_{ij}\)独立同分布的随机数。
  6. 矩阵乘法
     计算复杂度：\(O(n^{3})\)
     1969年斯特拉森：\(O(n^{2.81})\)
  7. 作业
     求解google matrix的本征值，讨论其分布（至少\(N>2000\)）。
  8. 作业2pdf
  9. 课堂笔记2_饶
  10. 课堂笔记2_林
  11. 预习一个巧妙的计算，下次课要讲
  12. Google-Matrix简要介绍
  13. 龚老师关于Mathematica的讲解ppt

• 9.28 周二

  1. 一个积分的计算
     (1) Hamiltonian:\( H=-\xi \sigma,\sigma=\pm 1 \)，\(\xi\)为随机数，其概率密度函数\(P(\xi)\)。 \[ Z=e^{-\beta\xi}+e^{\beta\xi}=e^{-\beta F} \] \[ F(\xi)=-\dfrac{1}{\beta}ln(e^{-\beta\xi}+e^{\beta\xi}) \] \[ F=\langle F(\xi) \rangle=\int_{-\infty}^{\infty}-\dfrac{1}{\beta}ln(e^{-\beta\xi}+e^{\beta\xi})P(\xi)d\xi \] 取\(\xi\)为高斯分布： \[ I=\int_{-\infty}^{\infty}ln(e^{-x}+e^{x})e^{-\alpha x^{2}}dx \] 在解析计算之前，可以先进行数值计算。Mathematica函数Plot，NIntegrate,画出数值积分的图像。
     (2) \( \alpha\rightarrow\infty \)，\(x\)局域在\(0\)附近： \[ I=\int_{-\infty}^{\infty}ln(e^{-x}+e^{x})e^{-\alpha x^{2}}dx \] \[ =\int_{-infty}^{\infty}ln(2+x^{2})e^{-\alpha x^{2}}dx \] \[ =\int_{-\infty}^{\infty}(ln2+\dfrac{x^{2}}{2})e^{-\alpha x^{2}}dx \] \[ =\int_{-\infty}^{\infty}ln(2)e^{-\alpha x^2}dx\] \[ =ln2\sqrt{\dfrac{\pi}{\alpha}}\]
     (3) \(\alpha\rightarrow 0\)
     (a) 数值拟合曲线\(I\alpha\sim\alpha\)：\(f(\alpha)=\alpha_{0}+\alpha_{1}\alpha+\alpha_{2}\alpha^{2}\)，猜测\(\alpha_{0}=1,\alpha_{1}=\dfrac{\pi}{4};\dfrac{\pi^{2}}{12}\)
     (b) \[I=\int_{-\infty}^{\infty}[ln(e^{x}+e^{-x})-|x|]e^{-\alpha x^{2}}dx+\int_{-\infty}^{\infty}|x|e^{-\alpha x^{2}}dx\] \[ I=2\int_{0}^{\infty}[ln(e^{x}+e^{-x})-x]e^{-\alpha x^{2}}dx+\dfrac{1}{\alpha} \] \[ I=2\int_{0}^{\infty}ln(1+e^{-2x})dx+\dfrac{1}{\alpha} \] \[ ln(1+x)=\sum_{n}(-1)^{n-1}\dfrac{x^{n}}{n} \] \[ I=2\int_{0}^{\infty}\sum_{n}(-1)^{n-1}\dfrac{e^{-2nx}}{n}dx+\dfrac{1}{\alpha} \] \[ =2\sum_{n}\dfrac{(-1)^{n-1}}{n}\int_{0}^{\infty}e^{-2nx}dx+\dfrac{1}{\alpha} \] \[ =\sum_{n=1}^{\infty}\dfrac{(-1)^{n}}{n^{2}}+\dfrac{1}{\alpha} \] \[ =\dfrac{\pi^{2}}{12}+\dfrac{1}{\alpha}\]
     (c) 还可以对\(ln(1+e^{-2x})\)进行更高阶的展开，得到高阶的结果。
     (4) 总结：数值分析；拟合与猜测；在数值下讨论
  2. 三对角阵
     (1) 三对角阵计算量：\(O(N^{2})\);内存：\( O(N) \)
     (2) Lanczos方法：
     1. \( v \rightarrow v_1 = v/||v||\)
     2. \(A v_1 = \alpha_1 v_1 + \beta_1 v_2\)，要求 \(v_2 \bot v_1, \ v_1^T v_2 = 0\) \(\Rightarrow \alpha_1 = v_1^T A v_1,\ \beta_1=v_2^T A v_1\)
     3. \(3\sim N\) 步： \(A v_n = \gamma_{n-1} v_{n-1} + \alpha_n v_n + \beta_n v_{n+1} \) \(\Rightarrow\) \(\gamma_{n-1} = v_{n-1}^T A v_n,\ \alpha_n=v_n^T A v_n ,\ \beta_n = v_{n+1}^T A v_n\)
     \[AU = U T ,\ U^{-1} A U = T, \ A = UTU^{-1} \Rightarrow {\rm det} (A - \lambda) = {\rm det} (UTU^{-1} -\lambda) = {\rm det }(T-\lambda)\] \[ \begin{eqnarray} T = \left( \begin{array}{cccc} \alpha_1 & \beta_1 & & \\ \gamma_1 & \alpha_2 & \beta_2 & \\ & \gamma_2 & \alpha_3 & \\ & & & \ddots \end{array} \right) \end{eqnarray} , \ U =(v_1,\ v_2,\ v_3\cdots) \] 矩阵\(A\)转化成三对角阵\(T\)，对角化的本征值一样，只需存储系数\(\alpha_n,\ \beta_n,\ \gamma_{n-1}\)
  3. H的本征值;微分方程
     (1) Schrodinger方程： \[ [-\dfrac{\hbar^{2}}{2m}\dfrac{\partial^{2}}{\partial x^{2}}+V(x)]\psi(x)=E\psi(x) \]
     (2) 数值计算需要离散化；无量纲化
  4. 关于Lanczos方法的介绍pdf
  5. 课堂笔记3_饶pdf
  6. 课堂笔记3_林pdf

• 9.30 周四

  1. Schrodinger equation
     (1) 目的：求解1D,2D Schrodinger equation \[ [-\dfrac{\hbar^{2}}{2m}\nabla^{2}+V(\vec{x})]\psi=E\psi \] 有限差分方法：简单直观。
     (2) 1D Schrodinger equation.
     无量纲化： \[ [-\dfrac{1}{2}\dfrac{d^{2}}{dx^{2}}+V(x)]\psi(x)=E\psi(x) \] \[ \psi'(x)=\lim_{h\rightarrow 0}\dfrac{\psi(x+h)-\psi(x)}{h} \] h有限代替\(\psi'(x)\) \[ \psi'(x)=\dfrac{\psi(x+h)-\psi(x)}{h}+O(h^{2}) \] \[ \psi''(x)=\dfrac{\psi(x+h)+\psi(x-h)-2\psi(x)}{h^{2}} \] 带入Schrodinger方程： \[ -\dfrac{1}{2h^{2}}(\psi_{i+1}+\psi_{i-1}-2\psi_{i})+V(x_{i})\psi_{i}=E\psi_{i} \] 其中\( \psi_{i}=\psi(x_{i}) \) \[ H=-\dfrac{1}{2h^{2}} \begin{eqnarray} \left( \begin{array}{cccc} -2 & 1 & & \\ 1 & -2 & 1 & \\ & 1 & -2 & \\ & & & \ddots \end{array} \right) + \left( \begin{array}{cccc} V(x_{1}) & & & \\ & V(x_{2}) & & \\ & & \ddots & \\ & & & V(x_{n}) \end{array} \right) \end{eqnarray} \] 特点(1d):
     (a). 稀疏矩阵，三对角阵，\( O(N^{2}) \)
     (b). \(N\)非常大，\( h=L/N\rightarrow 0 \)
     (3) 无穷深势阱(宽度为\(L\)) \[ -\dfrac{1}{2}\dfrac{d^{2}\psi}{dx^{2}}=E\psi;\quad \psi(0)=\psi(L)=0 \] 可以得到解析解为： \[ \psi\varpropto sin(kx);\quad E=\dfrac{n^{2}\pi^{2}}{2L^{2}} \] 数值方法的结果应该与解析结果一致： \[ -\dfrac{1}{2h^{2}}(\psi_{i+1}+\psi_{i-1}-2\psi_{i})=E\psi_{i} \] 上式数列可以通过特征方程求解 \[ \psi_{i}\varpropto e^{ikhn} \] \[ -\dfrac{1}{2h^{2}}(e^{ikh}+e^{-ikh}-2)=E \] \[E=\dfrac{1}{2h^{2}}(2-2coskh)=\dfrac{1}{h^{2}}(1-coskh)=\dfrac{1}{2}k^{2}\]
     边界条件： \[ -2\psi_{1}+\psi_{2}=E\psi_{1};\quad \psi_{N-1}-2\psi_{N}=E\psi_{N} \] 并将： \[ \psi_{n}=c_{+}e^{ikhn}+c_{-}e^{-ikhn} \] 带入边界条件求解\(k\)。与解析结果相同。
     (4) 2D 二维情况： \[ -\dfrac{1}{2}(\dfrac{d^{2}}{dx^{2}}+\dfrac{d^{2}}{dy^{2}})\psi+V\psi=E\psi \] \[ -\dfrac{1}{2}[\dfrac{1}{h^{2}}(\psi_{i+1j}+\psi_{i-1j}-2\psi_{ij})+\dfrac{1}{h^{2}}(\psi_{ij+1}+\psi_{ij-1}-2\psi_{ij})]+V_{ij}\psi_{ij}=E\psi_{ij} \] \[\Rightarrow -\dfrac{1}{2h^{2}}[\psi_{i+1j}+\psi_{i-1j}+\psi_{ij+1}+\psi_{ij-1}-4\psi_{ij}]+V_{ij}\psi_{ij}=E\psi_{ij}\] 2D特点：
     (a) 稀疏矩阵，非三对角阵。 (b) \( N^{2}\times N^{2};\quad O(N^{6}) \)
     \(\psi\)的矩阵元素的排列次序可以自己拟定。例如： \[ \psi=\begin{eqnarray} \left( \begin{array}{cccc} \psi_{i1}\\ \psi_{i2}\\ \psi_{i3}\\ \vdots \end{array} \right) \end{eqnarray} \]
  2. 求根
     (1) 二分法：\((x_{1},y_{1});(x_{2},y_{2})\)通过判断\(y_{1}y_{2}\)的正负来判断\((x_{1},x_{2})\)之间是否有根。通过判断\((x=\dfrac{x_{1}+x_{2}}{2})\)点左右是否有根来不断二分来确定根的位置。
     求极值可以转换成求导函数的根，但需要判断找到的极值是否是局域的。
     (2) 牛顿法： \[ f(x)=f(x_{0})+f'(x_{0})(x-x_{0}) \] \[ f(x_{0})+f'(x_{0})(x-x_{0})=0 \] \[ x=x_{0}-\dfrac{f(x_{0})}{f'(x_{0})} \] 牛顿的迭代公式： \[ x_{n+1}=x_{n}-\dfrac{f(x_{n})}{f'(x_{n})} \] 非唯一性： \[ x_{n+1}=x_{n}-p\dfrac{f(x_{n})}{f'(x_{n})} \]
  3. Runge-Kutta方法
     (1) 求解： \[ y'=f(x,y) \] \[ y(x_{n+1})=y(x_{n})+f(x_{n}+y_{n})h \] \[ y_{n+1}=y_{n}+f(y_{n})h \] \[ y_{n+1}=y_{n}+\dfrac{1}{2}h[f(x_{n},y_{n})+f(x_{n+1},y_{n+1})] \] \[ y_{n+1}=y_{n}+\dfrac{h}{2}f(x_{n},y_{n})+\dfrac{h}{2}f(x_{n+1},y_{n}+hf(x_{n},y_{n})) \] 更多内容下节课会讲。
  4. 作业
     (1) \(V(x)=\dfrac{1}{2}m\omega^{2}x^{2},m=1\)，对比数值结果与解析结果，体会误差的来源，大小，以及误差与\( L,N,h \)的关系，\(h=L/N\)。取\(h=10^{-3}\backsim 10^{-4}\)保证计算精度。
     (2)\(V(x)=\dfrac{1}{2}m\omega^{2}x^{2}+Acos(kx+\theta)\);(a)数值求解(b)微扰计算，mma，算到二阶。
     (3)2D情况，求两个势的本征态：
     (a)操场(一个长方形加两个半圆)，内部\(V=0\)，外部\(V=\infty\)
     (b)心形线，内部\(V=0\)，外部\(V=\infty\)
     ddl:10月19日
  5. 课堂笔记4_饶pdf
  6. 课堂笔记4_林pdf

• 10.9 周六补周二

  1. Runge-Kutta方法
     (1) 求解： \[ y'=f(t,y) \] \[ y_{n+1}=y_{n}+\int_{t_{n}}^{t_{n}+\Delta t}f(t,y)dt \] \[ y_{n+1}=y_{n}+f(t_{n},y_{n})\Delta t \] 梯形： \[ y_{n+1}=y_{n}+\dfrac{1}{2}[f(t_{n},y_{n})+f(t_{n+1},y_{n+1})]\Delta t \] 采用迭代的方法： \[ y_{n+1}=y_{n}+\dfrac{1}{2}[f(t_{n},y_{n})+f(t_{n}+\Delta t,y_{n}+\Delta tf(t_{n},y_{n}))]+O(\Delta t^{3}) \] 特点：
     (a) 不需要计算导数
     (b) 只要计算两次\(f\)
     (c) 可以进行多次迭代
     (2) Runge-Kutta 23
     2阶算法，精度3阶
     令\(h=\Delta t\) \[ y(t_{n+1})=y(t_{n})+\int_{t_{n}}^{t_{n+1}}f(t,y(t))dt \] \[ y(t_{n+1})=y(t_{n}+h) \] \[ =y(t_{n})+y'h+\dfrac{1}{2}y''h^{2}+O(h^{3}) \] \[ =y(t_{n})+f(t_{n},y_{n})h+\dfrac{1}{2}y''h^{2}+O(h^{3}) \] \[ y''=f_{t}+f_{y}f \] \[ =y+fh+\dfrac{1}{2}(f_{t}+f_{y}f)h^{2}+O(h^{3}) \] 需要计算导数，需要计算3次\(f\)
     RK算法：不算导数 \[ y(t_{n+1})=y(t_{n})+fh+\dfrac{1}{2}(f_{t}+f_{y}f)h^{2} \] \[ =y(t_{n})+h[af(t_{n},y_{n})+bf(t_{n}+h\delta x,y_{n}+h\delta y)]+O(h^{3}) \] \[ =y(t_{n})+h[af(t_{n},y_{n})+bf(t_{n},y_{n})+bf_{t}h\delta x+bf_{y}h\delta y]+O(h^{3}) \] 另有：\( a+b=1 \) \[ hbf_{t}h\delta x=\dfrac{1}{2}f_{t}h^{2} \Rightarrow \delta x=1\] \[ \dfrac{1}{2}f_{y}fh^{2}=bf_{y}h^{2}\delta y \Rightarrow \delta y=f \] \[ y(t_{n+1})=y(t_{n})+h[af(t_{n},y_{n})+bf(t_{n}+h,y_{n}+h)]+O(h^{3}) \] 与之前结果相同
     (3) Runge-Kutta 45 \[ y(t_{n}+h)=y(t_{n})+y'h+\dfrac{1}{2}y''h^{2}+\dfrac{1}{3!}y'''h^{3}+\dfrac{1}{4!}y^{(4)}h^{4}+O(h^{5}) \] \[ y_{n+1}=y_{n}+\dfrac{h}{6}(k_{1}+2k_{2}+2k_{3}+k_{4}) \] 其中： \[ k_{1}=f(t_{n},y_{n}) \] \[ k_{2}=f(t_{n}+\dfrac{1}{3}h,y_{n}+\dfrac{1}{3}hk_{1}) \] \[ k_{3}=f(t_{n}+\dfrac{2}{3}h,y_{n}+\dfrac{1}{3}hk_{1}+hk_{2}) \] \[ k_{4}=f(t_{n}+h,y_{n}+hk_{1}-hk_{2}+hk_{3}) \] 特点：
     (a) 应用最广泛
     (b) 形式不唯一
     (c) 变步长Rk45
     (d) Runge-Kutta方法可以用来计算积分
  2. 扩散方程
     (1) \[ \dfrac{\partial u}{\partial t}=a^{2}\dfrac{\partial^{2}u}{\partial x^{2}} \] \[ \dfrac{\partial^{2}\psi}{\partial x^{2}}=\dfrac{\psi_{n+1}+\psi_{n-1}-2\psi_{n}}{\Delta x^{2}} \] 定义符号：\( u_{i}^{n} \),\(n\)代表时间，\(i\)代表空间。 \[ \dfrac{u_{i}^{n+1}-u_{i}^{n}}{\Delta t}=a^{2}\dfrac{u_{i+1}^{n}+u_{i-1}^{n}-2u_{i}^{n}}{\Delta x^{2}} \] \[ y_{n+1}=y_{n}+hf(t_{n},y_{n})=y_{n}+\dfrac{h}{2}[f(t_{n},y_{n})+f(t_{n+1},y_{n+1})] \] \[ \gamma=\dfrac{\Delta t}{\Delta x^{2}}a^{2} \] \[ u_{i}^{n+1}=u_{i}^{n}+\gamma(u_{i+1}^{n}+u_{i-1}^{n}-2u_{i}^{n})=\gamma(u_{i+1}^{n}+u_{i-1}^{n})+(1-2\gamma)u_{i}^{n} \] 可以写成如下矩阵形式： \[ u^{n+1}=Tu^{n}=T^{2}u^{n-1}=\dots=T^{n}u^{1} \] \(T\)为三对角阵 作分解：\( T=U^{+}\lambda U\Rightarrow T^{n}=U^{+}\lambda^{n}U \)可以较为便捷的求出。
     (2) Crank-Nicolson 方法 \[ \dfrac{u_{i}^{n+1}-u_{i}^{n}}{\Delta t}=\dfrac{a^{2}}{2}[\dfrac{(u_{i+1}^{n+1}+u_{i-1}^{n+1}-2u_{i}^{n+1})}{\Delta x^{2}}+\dfrac{(u_{i+1}^{n}+u_{i-1}^{n}-2u_{i}^{n})}{\Delta x^{2}}] \] 与之前相同，令： \[ \gamma=\dfrac{\Delta t}{\Delta x^{2}}a^{2} \] \[ u_{i}^{n+1}-\dfrac{\gamma}{2}[u_{i+1}^{n+1}+u_{i-1}^{n+1}-2u_{i}^{n+1}]=u_{i}^{n}+\dfrac{\gamma}{2}(u_{i+1}^{n}+u_{i-1}^{n}-2u_{i}^{n}) \] 同样可以写成矩阵形式： \[ T_{L}u^{n+1}=T_{R}u^{n} \] \( T_{L},T_{R} \)为三对角阵 \[ u^{n+1}=T_{L}^{-1}T_{R}u^{n}=Tu^{n} \]
     (3) mathematica：Solve,NSolve;Integrate,NIntegrate;DSolve,NDSolve.
     (4) 分子动力学，flocking behavior,之后会讲
  3. 作业
     (1)一个质量为\(m_{1}\)的小球被轻质硬杆链接在一个固定点上，质量为\(m_{2}\)的小球被轻质硬杆链接在\(m_{1}\)上，杆长分别为\(l_{1},l_{2}\)。链接处可活动，无摩擦。求解2d情况下的运动方程。并尝试探讨不同初态下的时间演化,以及3d的情况。
     (2)求解扩散方程：\( \dfrac{\partial u}{\partial t}=a^{2}\dfrac{\partial^{2}u}{\partial x^{2}} \),数值解与严格解作比对，体会空间时间间隔对精度的影响。
     (3)ddl:10月23日
  4. 课堂笔记5_饶pdf

• 10.12 周二

  1. 回顾
     (1) Runge-Kutta方法
     (2) Crank-Nicolson方法 \[ \dfrac{\partial u}{\partial t}=a^{2}\dfrac{\partial^{2}u}{\partial x^{2}} \] \[ \dfrac{u_{i}^{n+1}-u_{i}^{n}}{\Delta t}=a^{2}\dfrac{u_{i+1}^{n}+u_{i-1}^{n}-2u_{i}^{n}}{\Delta x^{2}} \]
  2. 李雅普诺夫指数 \[ \dfrac{d\vec{x}}{dt}=F(\vec{x})\Rightarrow F(\vec{x}^{*})=0 \] \[ \vec{x}_{n+1}=\vec{x}_{n}+F(\vec{x}_{n})dt \] 有三种可能的结果：
     (a) 随着迭代的进程，远离\(\vec{x}^{*}\)，逐渐发散。
     (b) 临界情况，随着迭代的进程，始终在一个范围内。
     (c) 随着迭代的进程，不断接近\( \vec{x}^{*} \)。
     为了研究什么条件下，会发生这三种情况，设： \[ \vec{x}=\vec{x}^{*}+\vec{y} \] \[ \dfrac{d\vec{y}}{dt}=F(\vec{x}^{*}+\vec{y})=F(\vec{x}^{*})+A\vec{y}=A\vec{y} \] 作相似变换： \[ A=P^{-1}\lambda P \] \[ \Rightarrow \dfrac{d\vec{y}}{dt}=P^{-1}\lambda P\vec{y} \] 令： \[ z=P\vec{y} \] \[ \Rightarrow \dfrac{dz}{dt}=\lambda z \] \[ \Rightarrow \dfrac{d z_{i}}{dt}=\lambda_{i}z\Rightarrow z_{i}=e^{\lambda_{i}t}z_{i}(0) \] 所有的\(\lambda_{i}<0\)，对应稳定情况；\(\lambda_{i}=0\)，极限情况；只要有一个\(\lambda_{i}>0\)，对应不稳定情况。
     两个例子：
     (1) 目的：计算exponent \[ \dfrac{dx}{dt}=x-x^{2}\Rightarrow x^{*}=0,1 \] 令\( x=x^{*}+y=y,(x^{*}=0) \) \[ \Rightarrow \dfrac{dy}{dt}=y-y^{2} \] 因为\( \vert y\vert\ll 1,y\rightarrow 0 \) \[ \dfrac{dy}{dt}=y,\quad y(t)=e^{t}y_{0} \] 再考虑\( x^{*}=1 \),则\(x=1+y\) \[ \dfrac{dy}{dt}=(1+y)-(1+y)^{2}=-y \Rightarrow y=e^{-t}y_{0} \] 以上讨论可以综合写成： \[ y=e^{\lambda t}y_{0} \] 其中\(\lambda=1\)为非稳定情况，\( \lambda=-1 \)为稳定的情况。 这就解释了为什么有时候迭代法很难找到方程的解。
     (2) 目的：了解chaos \[\begin{eqnarray} \left \lbrace \begin{array}{ccc} \dot{x}=\sigma (y-x) \\ \dot{y}=\gamma x -y -xz \\ \dot{z}= xy-bz \end{array} \right. \end{eqnarray} \] \[\begin{eqnarray} \left \lbrace \begin{array}{ccc} \dot{x}=0 \\ \dot{y}=0 \\ \dot{z}=0 \end{array} \right. \end{eqnarray} \] \[ \begin{eqnarray} \left \lbrace \begin{array}{ccc} y-x=0 \\ \gamma x-x-xz=0 \\ -bz+x^{2}=0 \end{array} \right. \end{eqnarray} \] 平庸解：\( (0,0,0)=(x,y,z) \)
     当\(x\neq 0\)，\(z=\gamma-1\),\(x^{2}=b(\gamma-1)\)
     当\( x,y,z\rightarrow 0 \)，\(xy=0;xz=0\) \[\begin{eqnarray} \left \lbrace \begin{array}{ccc} \dot{x}=\sigma (y-x) \\ \dot{y}=\gamma x -y \\ \dot{z}= -bz \end{array} \right. \end{eqnarray} \] \[ \dfrac{d}{dt} \begin{eqnarray} \left( \begin{array}{cc} x \\ y \end{array} \right) = \left( \begin{array}{cc} -\sigma & \sigma \\ \gamma & -1 \end{array} \right) \left( \begin{array}{cc} x \\ y \end{array} \right) \end{eqnarray} \] 例如：取\(\sigma=10,b=\dfrac{8}{3}\) \[ \dfrac{d}{dt} \begin{eqnarray} \left( \begin{array}{ccc} x \\ y \\ z \end{array} \right) = \left( \begin{array}{ccc} -10 & 10 & 0 \\ \gamma & -1 & 0 \\ 0 & 0 & -\dfrac{8}{3} \end{array} \right) \left( \begin{array}{ccc} x \\ y \\ z \end{array} \right) = A\left( \begin{array}{ccc} x \\ y \\ z \end{array} \right) \end{eqnarray} \] 求本征值：\(det(\lambda I-A)=0\) 得到： \[ \begin{eqnarray} \left \lbrace \begin{array}{ccc} \lambda_{1}=-\dfrac{8}{3} \\ \lambda_{2}=\dfrac{-11-\sqrt{81+40\gamma}}{2} \\ \lambda_{3}=\dfrac{-11+\sqrt{81+40\gamma}}{2} \end{array} \right. \end{eqnarray} \] \(\gamma<1\)所有本征值小于零；\(\gamma>1,\lambda_{3}>0\)不稳定。
     另一个点 \[ \vec{x}^{*}=(\sqrt{\dfrac{8}{3}(\gamma-1)},\sqrt{\dfrac{8}{3}(\gamma-1)},\gamma-1) \] \[ \vec{x}=\vec{x}^{*}+(u,v,w) \] \[ \dfrac{d}{dt} \begin{eqnarray} \left( \begin{array}{ccc} u \\ v \\ w \end{array} \right) = \left( \begin{array}{ccc} -10 & 10 & 0 \\ 1 & -1 & -\sqrt{\dfrac{8}{3}(\gamma-1)} \\ \sqrt{\dfrac{8}{3}(\gamma-1)} & \sqrt{\dfrac{8}{3}(\gamma-1)} & -\dfrac{8}{3} \end{array} \right) \left( \begin{array}{ccc} u \\ v \\ w \end{array} \right) \end{eqnarray} \] 求本征值：\(det(\lambda I-A)=0\) \[ \Rightarrow 3\lambda^{3}+41\lambda^{2}+8(\gamma+10)\lambda+160(\gamma-1)=0 \] 如果\(\lambda>1\)
     当\( 1<\gamma<1.3456 \)，3个实根；
     当\(1.3456<\gamma<24.737\)，3个根，\(Re(\lambda_{i})<0\)
     当\( \gamma>24.737 \)，有一个\(Re(\lambda_{i})>0\)，不稳定
  3. 含时间问题
     (1) 单摆
     (2) 弹簧单摆
     (3) 两个小球被弹簧链接，以一定的初速度抛出： \[ T=\dfrac{1}{2}m(\dot{x_{1}}^{2}+\dot{y_{1}}^{2}+\dot{z_{1}}^{2})+\dfrac{1}{2}m(\dot{x_{2}}^{2}+\dot{y_{2}}^{2}+\dot{z_{2}}^{2}) \] \[ U=mgz_{1}+mgz_{2}+\dfrac{1}{2}k(d_{2}-x_{0})^{2} \] \[ d_{2}=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}+(z_{1}-z_{2})^{2}} \] \[ L=T-U;\quad \dfrac{d}{dt}(\dfrac{\partial L}{\partial\dot{q}})=\dfrac{\partial L}{\partial q} \] (4) SIR病毒传播模型
     S:易感染人群；I：感染人；R:康复人 \[ \begin{eqnarray} \left \lbrace \begin{array}{ccc} \dot{S}=-cSI \\ \dot{I}=cSI-gI \\ \dot{R}=gI \end{array} \right. \end{eqnarray} \] 多体：
     (1) 天体： \[ T=\sum_{i}\dfrac{1}{2}m_{i}\dot{\vec{r_{i}}}^{2} \] \[ U=\dfrac{Gm_{i}m_{j}}{\vert \vec{r_{i}}-\vec{r_{j}}\vert} \] \[ m_{i}\ddot{\vec{r_{i}}}=-\sum_{j}\vec{F_{ij}} \] \[ \vec{F_{ij}}=\dfrac{Gm_{i}m_{j}}{\vert \vec{r_{i}}-\vec{r_{j}}\vert^{3}}(\vec{r_{i}}-\vec{r_{j}}) \] \[ \dfrac{\vec{r_{i}}^{n+1}+\vec{r_{i}}^{n-1}-2\vec{r_{i}}^{n}}{dt^{2}}=-\dfrac{1}{2m_{i}}(\sum_{j}\vec{F_{ij}}^{n}+\sum_{j}\vec{F_{ij}}^{n+1}) \] (2) 同步Kuramoto model: \[ \dfrac{d\theta_{i}}{dt}=\omega_{i}-\dfrac{k}{n}\sum_{j}sin(\theta_{i}-\theta_{j}) \] \[ \Rightarrow \theta_{i}^{n+1}=\theta_{i}^{n}+dt[\omega_{i}-\dfrac{k}{n}\sum_{j}sin(\theta_{i}^{n}-\theta_{j}^{n})] \]
  4. 之后要讲的内容
     (1) 分子动力学
     (2) Flocking,鱼群，鸟群群聚运动
     (3) 随机数
     (4) 二次量子化
  5. 作业 画图，Lorenz方程的轨迹，mma有现成的代码，自己调节参数，观察。
     ddl:10月26日
  6. 课堂笔记6_饶pdf
  7. 课堂笔记6_林pdf

• 10.14 周四

  1. 含时问题
     \( y'=f(t,y) \)，采用Rk45，精度\( O(h^{5}) \)
     \( \dfrac{\partial u}{\partial t}=a^{2}\dfrac{\partial^{2}u}{\partial x^{2}} \)，Crank-Nicolson方法，精度\(O(\Delta t^{2})\)，or \(O(\Delta x^{3})\)
     迭代问题: \[ \dfrac{u^{n+1}-u^{n}}{\Delta t}=f(u^{n});\quad u^{n+1}=u^{n}+f(u^{n})\Delta t \] 梯形： \[ u^{n+1}=u^{n}+\dfrac{\Delta t}{2}[f(u^{n+1})+f(u^{n})] \] 核心code：
     Do：\( i=1:N \)
     \(a_{1}^{i}=a_{0}^{i}+f(\vec{a_{0}})\Delta t\)
     \( a_{0}=a_{1}^{i} \)
     End Do
     (1) 天体： \[ \ddot{\vec{r_{i}}}=\dfrac{1}{m_{i}}\sum_{j}\vec{F_{ij}} \] \[ \vec{r_{i}}^{n+1}+\vec{r_{i}}^{n-1}-2\vec{r_{i}}^{n}=\dfrac{\Delta t^{2}}{m_{i}}\sum_{j}\vec{F}(\vec{r_{i}}^{n}-\vec{r_{j}}^{n}) \] 几个因素：
     (a) \( m_{i} \)大，\( \vert \vec{r_{i}}-\vec{r_{j}}\vert \)大
     (b) 时间长
     (c) 归一化
     (2) Kuramoto model \[ \dot{\theta_{i}}=\omega_{i}-\dfrac{k}{N}\sum_{i}(\theta_{i}-\theta_{j}) \] (a) 相变
     (b) 可解 \[ \theta_{i}^{n+1}=\theta_{i}^{n}+\Delta t[\omega_{i}-\dfrac{k}{N}\sum_{j}sin(\theta_{i}^{n}-\theta_{j}^{n})] \]
     (3) 病毒SIR \[ \begin{eqnarray} \left \lbrace \begin{array}{ccc} \dot{S}=-cSI \\ \dot{I}=cSI-gI \\ \dot{R}=gI \end{array} \right. \end{eqnarray} \] \[ S^{n+1}=S^{n}-\Delta tcS^{n}I^{n} \] \[ I^{n+1}=I^{n}+\Delta t(cS^{n}I^{n}-gI^{n}) \] \[ R^{n+1}=R^{n}+gI^{n}\Delta t \] (4) Lorenz eq
     混沌，chaos,\( \lim_{n\rightarrow \infty}x^{n} \)不存在
     (5) Logistic map \[ \dfrac{dx}{dt}=4\gamma x(1-x) \] 韦吕勒，Verhulsr,1845 \[ x_{n+1}=4\gamma x_{n}(1-x_{n}) \] 分叉，Bification,Feigenbaum，费根鲍姆,\( \delta=4.6692 \)
     (6) Bak-Tong-Wiesenfeld(BTW) model
     沙堆模型，Sandpile model
     沙堆可以堆多高？与相变有关？
     \( h_{\vec{r}} \)，\(\vec{r}\)在d维空间中，随机\(\vec{r}\) \[ h_{\vec{r}}\rightarrow h_{\vec{r}}+1 \] \[ h_{\vec{r}}\rightarrow h_{\vec{r}}-h_{c} \] 周围点 \[ h_{\vec{r}+\vec{n}}\rightarrow h_{\vec{r}+\vec{n}}+1 \]
     (7) Flocking behavior,Vicsek model \[ \vec{r_{i}}^{n+1}=\vec{r_{i}}^{n}+\vec{u_{i}}^{n+1}\Delta t \] \[ \vec{u_{i}}^{n+1}=(\sum \vec{v_{j}}^{n})/N+\vec{\xi_{i}} \] \( \vec{\xi_{i}} \)为随机修正 \[ \vec{v}^{n+1}=\dfrac{\vec{r}^{n+1}-\vec{r}^{n}}{\Delta t} \]
     (8) 分子动力学模拟 \[ m_{i}\ddot{\vec{r_{i}}}=\sum_{j}\vec{F_{ij}} \] \[ \vec{F_{ij}}=-\nabla_{i}U \] \[ U(\vec{r})=\sum_{ij}V(\vec{r_{i}}-\vec{r_{j}}) \] 常见相互作用形式：
     (a) Legenalrd-Jones Potential \[ V(r)=4\epsilon[(\dfrac{\sigma}{r})^{12}-(\dfrac{\sigma}{r})^{6}] \] (b) 方势
     (c) Morse Potential \[ V_{0}[1-e^{-a(r-r_{e})}]^{2}-V_{0} \] Verles 算法： \[ \vec{x}(t+dt)+\vec{x}(t-dt)=2\vec{x}(t)+\ddot{\vec{x}}\Delta t+O(\Delta t^{4}) \] 多粒子： \[ \vec{x_{i}}(t+dt)+\vec{x_{i}}(t-dt)=2\vec{x_{i}}(t)+\dfrac{\vec{f_{i}}}{m_{i}}\Delta t^{2}+O(\Delta t^{4}) \] leap frog算法： \[ \vec{v}(t+\dfrac{dt}{2})=\vec{v}(t-\dfrac{dt}{2})+\dfrac{\vec{f}}{m}dt \] \[ \vec{f}=\vec{f}(\vec{x}(t)) \] \[ \vec{x}(t+dt)=\vec{x}(t)+\vec{v}(t+\dfrac{dt}{2})dt \]
     (9) Schrodinger eq: \[ i\dfrac{\partial \psi}{\partial t}=H(t)\psi \] \[ \psi=\sum_{n}c_{n}\vert \varphi_{n}\rangle \] \( \vert \varphi_{n}\rangle \)不含时 \[ \sum_{n}i\dot{c_{n}}\vert\varphi_{n}\rangle=\sum_{m}H\vert \varphi_{m}\rangle c_{m} \] \[ i\dot{c_{n}}=\langle \varphi_{n}\vert H\vert\varphi_{m}\rangle c_{m}=\sum_{m} h_{nm}(t)c_{m} \] \[ c_{n}^{k+1}=c_{n}^{k}-i\sum_{m}h_{nm}(t^{k})c_{m}^{k}dt \]
     (10) kicked rotor model \[ H=\dfrac{p^{2}}{2}+k\cos(x)\sum_{n}\delta(t-n) \] Equation of motion \[ \dot{x}=\dfrac{\partial H}{\partial p}=p \] \[ \dot{p}=-\dfrac{\partial H}{\partial x}=k\sin(x)\sum_{n}\delta(t-n) \] \( \Rightarrow \) \[ p_{n+1}^{-}=p_{n}^{+} \] \[ x_{n+1}^{-}=x_{n}^{+}+p_{n}^{+} \] \[ p_{n}^{+}=p_{n}^{-}+k\sin(x_{n}^{-}) \] \[ x_{n}^{+}=x_{n}^{-} \] \( \Rightarrow \) \[ p_{n+1}=p_{n}+k\sin(x_{n}) \] \[ x_{n+1}=x_{n}+p_{n+1} \] 注意，以上的讨论中只有唯一参数\( k \)
  2. 作业
     (1) Kicked rotor model 讨论
     (2) logistic map 分叉现象，计算
     (3) ddl:10月28日
  3. 课堂笔记7_饶pdf
  4. 课堂笔记7_林pdf
  5. 龚老师讲义pdf

• 10.19 周二

  1. 概率论
     1. 目的：
     (1)\[ \int f(x_{1},x_{2},x_{3},x_{4},x_{5})dx_{1}dx_{2}dx_{3}dx_{4}dx_{5} \]
     (2) Monte Carlo方法
     (3) 求解： \[ \dfrac{\partial f}{\partial t}=f(t)+\xi(t) \] \( \xi(t) \)为随机数 \[ \langle\xi(t)\xi(t')\rangle=D\delta(t-t'),\quad \langle \xi(t)\rangle=0 \]
     (a) 伊藤引理
     (b) Wiennen过程 \[ df=gdt+\sigma d\omega \]
     2. 事件\( x_{i} \)，\( \Omega={x_{i}} \)
     概率\( p_{i} \)，\( p_{i}\geq 0 \)且\( \sum_{i}p_{i}=1 \)
     两个重要的定理：中心极限定理；大数定理。
     中心极限定理： \[ \forall \epsilon>0,\quad \lim_{N\rightarrow\infty}P(\vert \bar{x}-\mu\vert<\epsilon)=1 \] 其中满足正态分布： \[ P(\bar{x})\sim N(\mu,\dfrac{\sigma^{2}}{N}) \] 注意：当测量的精度远大于测量值时，无法进行有效测量，例如用一把直尺（最小刻度为\(1mm\)）直接测量一张纸的厚度。
     补充：大偏差理论
     (a) Hugo Touchette. The large deviation approach to statistical mechanics
     (b) 朱金杰. 基于大偏差的随机动力学研究
     Gauss分布\( N(\mu,\dfrac{\sigma^{2}}{N}) \)，在偏差过大时，真实分布会偏离高斯分布。
     例子：随机比特\(0/1\)，\(n\)次,\(m\)次为\(0\)，\(m=rn\). \[ P(r)=P(m)=\dfrac{1}{2^{n}}C_{n}^{m}=\dfrac{1}{2^{n}}C_{n}^{rn}=\dfrac{1}{2^{n}}\dfrac{n!}{(rn)!(n-rn)!} \] Stirling公式： \[I=n!\] \[ lnI=\sum_{i=1}^{n}ln(i)\approx \int_{1}^{n}\ln xdx=n(\ln n-1)+1\approx n(\ln n-1) \] \[ \Rightarrow \ln n!\approx n(\ln n-1) \] 概率密度： \[ P(r)=e^{-nI(r)} \] \[ I(r)=\ln I+r\ln r+(1-r)\ln(1-r) \] 性质：
     (a) \( I(r)\geq 0 \)
     (b) \(I(r)\), rate function, entropy
     恢复中心极限定理： \[ n\rightarrow \infty \] \[ P(r)=e^{-nI(r)}\approx e^{-nI_{min}(r)} \] \[ \dfrac{\partial I}{\partial r}=0\Rightarrow \ln r-\ln(1-r)=0 \] \[ \Rightarrow r=\dfrac{1}{2} \] 令\( r=\dfrac{1}{2}+x \)，保留到二阶： \[ I(x)=2x^{2} \] \[ \Rightarrow P(x)=e^{-2nx^{2}} \] 平均值\( \bar{x}=0 \)
     方差：\( \sigma^{2}=\dfrac{1}{2n} \)
     3.高斯分布：
     \[ s_{n}=\dfrac{1}{n}\sum_{i=1}^{n}x_{i} \] \( x_{i} \)是独立同分布的高斯分布 \[ P(s_{n}=s)=\int \delta[\dfrac{1}{n}(x_{1}+\dots+x_{n})-s]P(x_{1})\dots P(x_{n})dx_{1}\dots dx_{n} \] \[ =\dfrac{1}{2\pi}\int dk e^{ik(\dfrac{1}{n}\sum_{i}x_{i}-s)}P(x_{1})\dots P(x_{n})dx_{1}\dots dx_{n} \] \[ =\dfrac{1}{2\pi}\int dk e^{-iks}(\int e^{ikx/n}P(x)dx)^{n} \] \[ =\sqrt{\dfrac{n}{2\pi \sigma^{2}}}e^{-nI(s)+\dfrac{1}{2}\ln n-\dfrac{1}{2}\ln (2\pi\sigma^{2})} \] \( \dfrac{1}{2}\ln n-\dfrac{1}{2}\ln (2\pi\sigma^{2}) \)为非主要部分 \[ I(s)=\dfrac{(s-\mu)^{2}}{2\sigma^{2}} \] 4.指数分布： \[ P(x)=\dfrac{1}{\mu}e^{-x/\mu},\quad x\geq 0 \] \[ P(s_{n}=s)\approx e^{-nI(s)} \] \[ I(s)=\dfrac{s}{\mu}-1-\ln(\dfrac{s}{\mu}) \] 在\( \dfrac{s}{\mu}\approx 1 \) \[ P(s)\approx e^{-\dfrac{n}{2}(\dfrac{s}{\mu}-1)^{2}} \] 属于高斯分布
     5.moment 矩；cumulant 累积量； moment expansion; cumulant expansion \[ \langle e^{x}\rangle=\sum_{n=0}^{+\infty}\dfrac{\langle x^{n}\rangle}{n!}=\sum_{n=0}^{+\infty}\dfrac{m_{n}}{n!}=e^{\Omega} \] 矩:\( m_{n}=\langle x^{n}\rangle \) \[ \langle x^{2}\rangle=\langle x\rangle^{2}+\sigma^{2} \] \[ \langle x^{3}\rangle=\langle x\rangle^{3}+3\langle x\rangle\sigma^{2}+\sigma_{3} \] 各阶矩之间并不独立，并且高阶矩会包含一些低阶矩的成分，这通常会导致一些问题。 \( \langle e^{x}\rangle=e^{\Omega} \),\( \Omega \)只是累积量的函数
     6.布朗运动： \[ D=\dfrac{RT}{N_{A}\sigma\pi\eta a} \]
     7. G. Parisi(Kardar-Parisi-Zhang eq) \[ \dfrac{\partial h}{\partial t}=\nu^{2}\partial^{2}h+\xi+\dfrac{\lambda}{2}(\partial h)^{2} \] 8. 作业：
     \( x_{1},\dots,x_{n} \)，\(n\)个随机数，从小到大排列，计算： \[ \Delta=\max{x_{i+1}-x_{i}} \] 证明\(\Delta\)分布是Trary-Widom分布，且和\(x_{i}\)的分布无关。
     ddl:11月7日
  2. 课堂笔记8_饶pdf
  3. 非线性系统的不动点pdf
  4. 诺奖得主Parisi在中科院理论物理所的一次合作

• 10.21 周四

  1. 随机过程
     (1) Brown motion
     (a) 历史；1905 Einstein，工作；朗之万方程
     (b) 意义：微观变化与宏观对应
     扩散方程： \[ \dfrac{\partial \phi}{\partial t}=D\dfrac{\partial^{2}\phi}{\partial x^{2}} \] \[ m\ddot{x}=-\alpha \dot{x}+f+\xi \] 等式右边第一项：阻力；第二项：势场力；第三项：随机力 \[ \langle \xi(t)\rangle=0;\quad \langle\xi(t)\xi(t')\rangle=D\delta(t-t') \] (a) \( f=0 \)的情况 \[ m\ddot{x}=-\alpha\dot{x}+\xi \] \[ \dot{x}=v \] \[ m\dot{v}=-\alpha v+\xi \] 格林函数方法： \[ L=(m\dfrac{d}{dt}+\alpha),\quad Lu=f \] \[ Lg(x-x')=\delta(x-x') \] \[ u=u_{0}+\int g(x-x')f(x')dx'\quad Lu_{0}=0 \] 或者： \[ m\dot{v}=-\alpha v+\xi \] \[ v\sim e^{-(\alpha/m)t} \] 设 \[v(0)=u(0)\] \[ \Rightarrow m e^{-\dfrac{\alpha}{m}t}\dot{u}-\alpha e^{-\dfrac{\alpha}{m}t}u=-\alpha e^{-\dfrac{\alpha}{m}t}u+\xi \] \[ \dot{u}=\dfrac{1}{m}e^{\dfrac{\alpha}{m}t}\xi \] \[ u=u_{0}+\int_{0}^{t}\dfrac{1}{m}e^{\dfrac{\alpha}{m}t'}\xi(t')dt' \] 完整解： \[ v=v_{0}e^{-\dfrac{\alpha}{m}t}+\dfrac{1}{m}\int_{0}^{t}e^{-\dfrac{\alpha}{m}(t-t')}\xi(t')dt' \] (a) 有解，但是解是随机的（非确定的）
     (b) 观测量只在平均上是有意义的
     平均值：
     (a) \[ \bar{v}(t)=\langle v_{0}e^{-\dfrac{\alpha}{m}t}\rangle+\int_{0}^{t}\dfrac{1}{m}e^{-\dfrac{\alpha}{m}(t-t')}\langle \xi(t')\rangle dt' \] 由于\( \langle \xi(t)\rangle=0 \) \[ \bar{v}(t)=v_{0}e^{-\dfrac{\alpha}{m}t} \] (b) \[ \bar{v^{2}}=v_{0}^{2}e^{-\dfrac{2\alpha}{m}t}+\langle(\int_{0}^{t}e^{-\dfrac{\alpha}{m}(t-t')}\xi(t')dt')^{2}\rangle\dfrac{1}{m^{2}} \] \[ =v_{0}^{2}e^{-\dfrac{2\alpha}{m}t}+\int_{0}^{t}dt_{1}'dt_{2}'e^{-\dfrac{\alpha}{m}(t-t_{1}')-\dfrac{\alpha}{m}(t-t_{2}')}\dfrac{1}{m^{2}}\langle \xi(t_{1}')\xi(t_{2}')\rangle \] 由于\( \langle \xi(t_{1}')\xi(t_{2}')\rangle=D\delta(t_{1}'-t_{2}') \) \[ =v_{0}^{2}e^{-2\dfrac{\alpha}{m}t}+\dfrac{D}{m^{2}}\int_{0}^{t}dt_{1}e^{-\dfrac{2\alpha}{m}(t-t_{1})} \] \[ \int_{0}^{t}e^{-\alpha t'}dt'=\dfrac{1}{\alpha}(1-e^{-\alpha t}) \] 若\( \alpha=0 \),则\( \bar{v}(t)=v_{0} \),\( \bar{v^{2}}=v_{0}^{2}+\dfrac{D}{m^{2}}t \) 平均速度不变，方差正比于\(t\)
     另外： \[ \bar{x}=\dfrac{1}{n}(\xi_{1}+\dots+\xi_{n})\sim N(\mu,\dfrac{\sigma^{2}}{n}) \] \[ x=(\xi_{1}+\dots+\xi_{n})\sim N(n\mu,n\sigma^{2}) \] \[ \sigma_{x}^{2}=n\sigma^{2}\propto n \] \[ Y=(\xi_{1}+\dots+\xi_{n})dt=\int_{0}^{t}\xi(t')dt'\sim N(n\mu dt,n\sigma^{2}dt) \] 由于\( ndt=t \) \[ \Rightarrow (\int_{0}^{t}\xi(t')dt')^{2}\propto t \] (2) 在实验上测量的物理量\(C_{V}\)到底是什么？方差？平均值？ \[ C_{V}=(\dfrac{\partial E}{\partial T})_{V}=-\dfrac{\partial \beta}{\partial T}(\dfrac{\partial E}{\partial \beta})_{V}=-\dfrac{1}{T^{2}}(\dfrac{\partial E}{\partial \beta})_{V} \] \[ E=\dfrac{Tr(e^{-\beta H}H)}{Tr(e^{-\beta H})} \] \[ (\dfrac{\partial E}{\partial \beta})_{V}=-\dfrac{Tr(H^{2}e^{-\beta H})}{Tr(e^{-\beta H})}+[\dfrac{Tr(H e^{-\beta H})}{Tr(e^{-\beta H})}]^{2} \] \[ (\dfrac{\partial E}{\partial \beta})_{V}=-(\langle E^{2}\rangle-\langle E\rangle^{2}) \] \[ C_{V}=\dfrac{1}{T^{2}}(\langle E^{2}\rangle-\langle E\rangle^{2}) \] (3) 离散化： 令\( m=1,\alpha=0 \) \[ \dot{v}=\xi\Rightarrow v=v_{0}+\int_{0}^{t}\xi(t')dt' \] 观测\[ \bar{v}=v_{0};\quad \bar{v^{2}}=v_{0}^{2}+Dt \] \[ \langle \xi(t)\rangle=0,\quad \langle\xi(t)\xi(t')\rangle=D\delta(t-t') \] 于是有： \[ v_{n+1}=v_{n}+\xi_{n}\Delta t，\quad v_{n}=v_{n-1}+\xi_{n-1}\Delta t,\quad\dots,\quad v_{1}=v_{0}+\xi_{0}\Delta t \] \[ \langle\xi_{n}\rangle=0,\quad \langle\xi_{n}\xi_{m}\rangle=D\delta_{nm} \] \[ v_{n}=(\sum_{i=0}^{n-1}\xi_{i})\Delta t+v_{0},\quad \Rightarrow \bar{v_{n}}=v_{0} \] \[ \bar{v_{n}^{2}}=v_{0}^{2}+\sum_{i=0}^{n-1}\sum_{j=0}^{n-1}\langle \xi_{i}\xi_{j}\rangle(\Delta t)^{2}=v_{0}^{2}+Dt \] 另一种写法： \[ v_{n}=v_{n-1}+\sqrt{\dfrac{D}{dt}}dt\xi_{n-1} \] 其中\(\xi\sim N(0,1)\)
     (4) 讨论 \[ m\ddot{x}=-\alpha \dot{x}+f(x)+\xi(t) \] \[ \Rightarrow \dot{x}=v;\quad \dot{v}=-\dfrac{\alpha}{m}v+f(x)+\xi(t) \] \[ x_{n+1}=x_{n}+v_{n}dt \] \[ v_{n+1}=v_{n}-\dfrac{\alpha}{m}v_{n}dt+f(x_{n})dt+\sqrt{\dfrac{D}{dt}}dt\xi_{n} \] \( \xi_{n}\sim N(0,1) \) (5)扩散方程与随机过程 \[ \dfrac{\partial \phi}{\partial t}=D\dfrac{\partial^{2}\phi}{\partial x^{2}} \] \[ \phi=A(t)e^{-ax^{2}/t}=\sqrt{\dfrac{\pi t}{a}}e^{-ax^{2}/t};\quad a=\dfrac{1}{4D} \] 归一化： \[ \phi=\dfrac{1}{\sqrt{4D\pi t}}e^{-\dfrac{x^{2}}{4Dt}} \] \[ \langle x^{2}\rangle=2Dt \] Fick‘s law & 流守恒，即： \[ j=-D\dfrac{\partial \rho}{\partial x};\quad \dfrac{\partial \rho}{\partial t}+\dfrac{\partial j}{\partial x}=0 \] \(\phi(x,t),\phi(x,t+dt)\)，\(t+dt\)时刻，\(x\)位置的粒子来源于\(t\)时刻\(x-\Delta\)与\(x+\Delta\)位置 \[ \phi(x,t+dt)=\int P(\Delta)\phi(x+\Delta,t)d\Delta \] \[ =\int P(\Delta)[\phi(x,t)+\dfrac{\partial \phi}{\partial x}\Delta+\dfrac{1}{2}\dfrac{\partial^{2}\phi}{\partial x^{2}}\Delta^{2}]d\Delta \] \[ =\phi(x,t)+\dfrac{1}{2}\dfrac{\partial^{2}\phi}{\partial x^{2}}2Dt \] \[ \dfrac{\partial \phi}{\partial t}=D\dfrac{\partial^{2}\phi}{\partial x^{2}} \]
  2. 作业 \( U=ax^{2}+bx^{4} \),\(a<0,m=1\),求解 \[ \ddot{x}=-\alpha \dot{x}-\nabla U+\xi(t) \] 讨论double well中的动力学过程，参数自己调节
     ddl:11月7日
  3. 课堂笔记9_饶pdf

• 10.26 周二

  1. Ito lemma
     (1) \( \dfrac{ds}{dt}=(r+\xi)s\Rightarrow d\ln s=r+\xi\Rightarrow s=s_{0}e^{rt+\int_{0}^{t}\xi(t')dt'} \)
     ill-defined:\( \dfrac{dx}{dt}=a(x,t)+b(x,t)\xi \) 改写成增量形式：\( dx=adt+b(x,t)dw;\quad dw=\xi dt \)
     (2) Wiener过程,\( dw=\xi dt \) \[ w(t)=\int_{0}^{t}\xi(t')dt' \] (a) \( \langle w(t)\rangle=0 \); (b) \( \langle w^{2}(t)\rangle=\int_{0}^{t}\langle \xi(t1)\xi(t2)\rangle dt_{1}dt_{2}=\sigma^{2}t \)
     (c) \(\langle w(t)w(s)\rangle=\sigma^{2}min(t,s)\); (d)\(\langle w^{2}(t)\rangle-\langle W^{2}(s)\rangle=\sigma^{2}(t-s)\)
     (e) \(\langle dw^{2}\rangle=\sigma^{2}dt\)
     \( dx=adt+bdw \) \[ df=\dfrac{\partial f}{\partial x}dx+\dfrac{\partial f}{\partial t}dt+\dfrac{1}{2}\dfrac{\partial^{2}f}{\partial x^{2}}(dx)^{2}+\dfrac{1}{2}\dfrac{\partial^{2}f}{\partial t^{2}}(dt)^{2}+\dfrac{\partial^{2}f}{\partial x\partial t}dxdt \] 其中：\((dx)^{2}=a^{2}dt^{2}+b^{2}dw^{2}+2abdtdw\) \[ df=\dfrac{\partial f}{\partial x}dx+{\dfrac{\partial f}{\partial t}+\dfrac{b^{2}}{2}\dfrac{\partial^{2}f}{\partial x^{2}}}dt \] 考虑：\( ds=rsdt+\sigma s dw \)，取\( f=\ln s \) \[ df=\dfrac{\partial f}{\partial s}ds+\dfrac{1}{2}\dfrac{\partial^{2}f}{\partial s^{2}}ds^{2}=\dfrac{1}{s}ds-\dfrac{1}{2}\dfrac{1}{s^{2}}(ds)^{2}=(rsdt+\sigma sdw)/s-\dfrac{1}{2s^{2}}(rsdt+\sigma s dw)^{2}=(r-\dfrac{\sigma^{2}}{2})dt+\sigma dw \] \[ \Rightarrow s=s_{0}e^{(r-\sigma^{2})t+\sigma w(t)} \] (3) \[ \int_{0}^{T}wdw=\sum_{n}w_{n}(w_{n}-w_{n-1})\quad or\quad =\sum_{n}w_{n}(w_{n+1}-w_{n}) \] \[ \sum_{n}\langle w_{n}(w_{n}-w_{n-1})\rangle=\sum_{n}(\langle w_{n}^{2}\rangle-\langle w_{n}w_{n-1}\rangle)=\sum_{n}[ndt-(n-1)dt]=T \] \[ \sum_{n}\langle w_{n}(w_{n+1}-w_{n})\rangle=\sum_{n}(\langle w_{n}w_{n+1}\rangle-\langle w_{n}^{2}\rangle)=\sum_{n}(ndt-ndt)=0 \] Ito lemma:\( f=\dfrac{1}{2}w^{2} \) \[ df=\dfrac{\partial f}{\partial w}dw+\dfrac{1}{2}\dfrac{\partial^{2}f}{\partial w^{2}}(dw)^{2}=wdw+\dfrac{1}{2}dt \] \[ \int_{0}^{T}df=\int_{0}^{T}wdw+\dfrac{T}{2}\Rightarrow \int_{0}^{T}wdw=\dfrac{w_{T}^{2}}{2}-\dfrac{T}{2}\Rightarrow \dfrac{T}{2}-\dfrac{T}{2}=0 \] 参考文献：
     (a) Timothy Sauer. Computational solution of stochastic differential equations
     (b) 刘玲. 随机微分方程的数值解（导师王冉）
     (c) Xin Bian. 111 years of Brownian motion
     介绍：
     (a) Kramers. Brownian motion in a field theory of force and the diffusion model of chemical reactions, 1940.
     (b) Schrodinger. WHAT IS LIFE?
     (c) Ming chen wang(王明贞), Uhlenbeck. On the theory of Brownian motion, 1945.
     (4) 刘维尔方程,\( \rho(x,v;t) \) \[ \int\rho(x,v;t)dxdv=1;\quad \dfrac{\partial \rho}{\partial t}+\nabla\cdot \vec{J}=0 \] \[ \dfrac{d\rho}{dt}=\dfrac{\partial \rho}{\partial t}+\dfrac{\partial \rho}{\partial x}\dot{x}+\dfrac{\partial \rho}{\partial v}\dot{v}=0 \] \[ =\dfrac{\partial \rho}{\partial t}+\dfrac{\partial \rho}{\partial x}(\dot{x}\rho)+\dfrac{\partial \rho}{\partial v}(\dot{v}\rho)=0 \] \( \nabla\cdot \vec{J}\Rightarrow \nabla=(\dfrac{\partial}{\partial x},\dfrac{\partial}{\partial v}),\quad \vec{J}=(\dot{x}\rho,\dot{v}\rho) \)
     (5) 求Fokker-Planck eq \[ P(x,t+\tau)=\int P(x+\Delta,t)P(\Delta)d\Delta=\int[P(x,t)+\dfrac{\partial P}{\partial x}\Delta+\dfrac{1}{2}\dfrac{\partial^{2}P}{\partial x^{2}}\Delta^{2}]P(\Delta)d\Delta=P(x,t)+\dfrac{1}{2}\langle \Delta^{2}\rangle\dfrac{\partial^{2}P}{\partial x^{2}}=P(x,t)+\dfrac{\partial P}{\partial t}\tau \] \( \langle \Delta^{2}\rangle=2D\tau \)得到扩散方程。 \[ P(x,t+\tau)=\int P(x+\Delta,t)\rho(\Delta,x,t,\tau)d\Delta \] \[ =P(x,t)+\sum_{n=1}^{\infty}(\dfrac{\partial}{\partial x})^{n}P(x,t)\int \Delta^{2}\rho(\Delta,x,t,\tau)d\Delta=P(x,t)+\sum_{n=1}^{\infty}(\dfrac{\partial}{\partial x})^{n}P(x,t)M_{n}(x,t,\tau) \] \[ \dfrac{\partial P}{\partial t}\tau=\sum_{n=1}^{\infty}(\dfrac{\partial}{\partial x})^{n}P(x,t)M_{n}(x,t,\tau)\Rightarrow \dfrac{\partial P}{\partial t}=\sum_{n=1}^{\infty}(\dfrac{\partial}{\partial x})^{n}P(x,t)D_{n}^{(1)}(x,t) \] 贝叶斯定理： \[ P(A)=\sum_{B}P(A|B)P(B) \] \[ P(x,t+\tau)=\int P(y,t)P(x,t+\tau|y,t)dy \] \( \rho(\Delta,x,t,\tau)=P(x+\Delta,t+\tau) \)，条件概率，传播子。
  2. Computional Solution of Stochastic differential equations pdf
  3. 刘玲. 随机微分方程的数值解
  4. Xin Bian. 111 years of Brownian motion pdf
  5. Kramers. Brownian motion in a field of force and the diffusion model of chemical reactions
  6. On the theory of Brownian motion pdf
  7. Direct measurement of Kramers turnover with a levitated nanoparticle pdf
  8. Einstein, Perrin, and the reality of atoms: 1905 revisited pdf
  9. 课堂笔记10_饶pdf

• 10.28 周四

  1. 随机过程
     (1) MC Simulation, 随机数作计算；量子耗散（量子信息）
     (2) 具体计算：
     (a) \( dx=fdt+\sigma dw;\quad dx=vdt;\quad dv=(-\alpha v+f)dt+\sigma dw \)
     (b) \( \int f(x_{1},\cdots,x_{n})dx_{1}\cdots dx_{n},\quad 3\leq n\leq 10 \)
     (c) \( Z=Tr(e^{-\beta H})=\int dx_{1}\cdots dx_{n}e^{-\beta H(x_{1},\cdots,x_{n})} \);简单的问题如Ising Model的模拟
     (3) Fokker-Planck方程
     (a) 目的：宏观——微观
     (b) \( \rho(x,p,t)\Rightarrow d\rho=0\Rightarrow \)动态平衡
     \( \dfrac{\partial \rho}{\partial t}+\nabla\cdot\vec{J}=0\Rightarrow \vec{J}=(\vec{v}\rho,\vec{a}\rho) \)，只有一阶导数
     \( \dfrac{\partial \phi}{\partial t}=D\dfrac{\partial^{2}\phi}{\partial x^{2}} \)二阶导数
     (4) coupled Langevin eq: \[ \begin{eqnarray} \left \lbrace \begin{array}{cc} \dfrac{dx}{dt}=F(x,y)+\xi_{1} \\ \dfrac{dy}{dt}=G(x,y)+\xi_{2} \end{array} \right. \end{eqnarray} \] \( \dfrac{\partial}{\partial t}P(x,y,t)= \)线性(drift)\( + \)扩散项(diffusion)
     \( -\dfrac{\partial}{\partial x}(FP)\quad -\dfrac{\partial}{\partial y}(GP) \Rightarrow\)线性；\( D_{1}\dfrac{\partial^{2}P}{\partial x^{2}}+D_{2}\dfrac{\partial^{2}P}{\partial y^{2}} \Rightarrow\)可能的扩散项
     两个极限：
     (a) \( F=0,\quad G=0 \)，纯扩散
     (b) \( D_{1}=0,\quad D_{2}=0 \)，刘维尔方程
     \( \dfrac{d}{dt}P(x,y,t)=0\Rightarrow \dfrac{\partial P}{\partial t}+\dfrac{\partial}{\partial x}(FP)+\dfrac{\partial}{\partial y}(GP)=0 \)
     (5) 量子Langevin eq
     Caldeira-Leggett model(CL model)，一个振子与环境相互作用；可解 参考文献： (a) Caldeira-Leggett model Wiki
     (b) Caldeira, Leggett, PRL, 1981 pdf
     (c) Benguria, Kac, PRL, 1981 pdf
     (d) Ford, Kac, On the Quantum Langevin Equation, 1987 pdf \[ m\ddot{x}=-\alpha \dot{x}+f+\xi \] 以上方程唯象导出，可否理论推导；量子化
     量子化： \[ H=\dfrac{P^{2}}{2m}+\dfrac{1}{2}m\omega^{2}x^{2}=\hbar\omega(a^{+}a+\dfrac{1}{2}) \] \[ c^{2}+d^{2}=(c+id)(c-id)+i[c,d]=a^{+}a+\dfrac{1}{2} \] \[ \Rightarrow c=\sqrt{\dfrac{m\omega}{2\hbar}}x;\quad d=-\dfrac{p}{\sqrt{2m\hbar \omega}}\Rightarrow a=\sqrt{\dfrac{m\omega}{2\hbar}}x-\dfrac{ip}{\sqrt{2m\hbar \omega}};\quad a^{+}=\sqrt{\dfrac{m\omega}{2\hbar}}x+\dfrac{ip}{\sqrt{2m\hbar \omega}} \] \[ H=\dfrac{p^{2}}{2m}+V(x)+\sum_{j=1}^{n}[\dfrac{p_{j}^{2}}{2m_{j}}+\dfrac{1}{2}k_{j}(q_{j}-x)^{2}] \] 以上方程是确定的，可数值计算
     作业：数值模拟以上方程，\(N=100\)，\( m_{i},k_{i} \)的取值自己决定。ddl:11月14日。 \[ \begin{eqnarray} \left \lbrace \begin{array}{cc} \dot{p_{i}}=-\dfrac{\partial H}{\partial q_{i}} \\ \dot{q_{i}}=\dfrac{\partial H}{\partial p_{i}} \end{array} \right. \end{eqnarray} \] \[ \begin{eqnarray} \left \lbrace \begin{array}{cccc} \dot{p_{i}}=-\dfrac{\partial H}{\partial q_{i}}=-k_{i}(q_{i}-x) \\ \dot{q_{i}}=\dfrac{\partial H}{\partial p_{i}}=\dfrac{p_{i}}{m_{i}} \\ \dot{x}=\dfrac{\partial H}{\partial p}=\dfrac{p}{m} \\ \dot{p}=-\dfrac{\partial H}{\partial x}=-V'(x)-\sum_{j}k_{j}(x-q_{j}) \\ \end{array} \right. \end{eqnarray} \] \[ m_{i}\ddot{q_{i}}+k_{i}q_{i}=k_{i}x \] \(\Rightarrow q_{i}=\)通解\( + \)特解；\( q_{i}^{0}=A\cos\omega t+B\sin\omega t \) \[ \Rightarrow q_{j}(t)=q_{j}(0)\cos(\omega_{j}t)+P_{j}(0)\sin(\omega_{j}t)/m_{j}\omega_{j}+x(t)-x(0)\cos(\omega_{j}t)-\int_{0}^{t}\cos(\omega_{j}(t-t'))\dot{x}(t')dt' \] \[ \dot{x}=\dfrac{p}{m};\quad \dot{P}=-V'(x)+\sum_{j}k_{j}q_{j}^{0}-[\sum_{j}k_{j}\cos(\omega_{j}t)]x(0)-\int_{0}^{t}\sum_{j}k_{j}\cos(\omega_{j}(t-t'))\dot{x}(t')dt' \] \[ \Rightarrow \dot{P}=-V'(x)+\xi-B(t)x(0)-\int_{0}^{t}B(t-t')\dot{x}(t')dt' \] \[ \xi=\sum_{j}k_{j}q_{j}^{0};\quad B(t)=\sum_{j}k_{j}\cos(\omega_{j}t) \] 若\( B(t-t')=\alpha\delta(t-t') \)，则重新得到\( \dot{P}=-V'+\xi-\alpha\dot{x} \) \[ F=m\ddot{x}+\int_{0}^{t}B(t-t')\dot{x}(t')dt'+V'(x)+B(t)x(0)=F \] 随机力：\( F=\sum_{j}k_{j}q_{j}^{0}=\sum_{j}k_{j}[q_{j}(0)\cos(\omega_{j}t)+\dfrac{P_{j}(0)}{m_{j}\omega_{j}}\sin(\omega_{j}t)] \),其中\( [q_{j},P_{j}]=i\hbar \delta_{ij} \) \[ q_{j}\sim (b_{j}+b_{j}^{+}),P_{j}\sim (b_{j}-b_{j}^{+})\Rightarrow F=\sum_{j}f_{j}(b_{j}e^{-i\omega_{j}t})+b_{j}^{+}e^{i\omega_{j}t} \]
  2. 混沌简介pdf
  3. 课堂笔记11_饶pdf

• 11.2 周二

  1. 量子Langevin方程
     (1) 推广Langevin方程
     (2) 更普遍的形式，Kernel,\( B(t) \)函数
     (3) 量子化(力)
     (4) \( N\rightarrow \infty \)什么影响物理 \[ m\ddot{x}+\int_{0}^{t}B(t-t')\dot{x}(t')dt'+B(t)x(0)+\nabla U=\xi;\quad \xi(t)=\sum_{l}\xi_{l}(b_{l}e^{-i\omega_{l}t}+b_{l}^{+}e^{i\omega_{l}t}) \] \( \langle\xi(t)\rangle=\sum_{l}\xi_{l}\langle b_{l}e^{-i\omega_{l}t}+b_{l}^{+}e^{i\omega_{l}t}\rangle; \quad \langle O\rangle=Tr(O\rho)=Tr(Oe^{-\beta H})/Tr(e^{-\beta H}) \) \[ b_{l}=b_{l}^{+}=0 \] \[ \langle \xi(t)\xi(t')\rangle=\sum_{ll'}\xi_{l}\xi_{l'}\langle(b_{l}e^{-i\omega_{l}t}+b_{l}^{+}e^{i\omega_{l}t})(b_{l'}e^{-i\omega_{l'}t'}+b_{l'}^{+}e^{i\omega_{l'}t'})\rangle \] 由于\( l\neq l';\quad \langle b_{l}b_{l'}^{+}\rangle=\langle b_{l}\rangle\langle b_{l'}^{+}\rangle=0 \) \[ \Rightarrow \sum_{l}\xi_{l}^{2}\langle (b_{l}e^{-i\omega_{l}t}+b_{l}^{+}e^{i\omega_{l}t})(b_{l}e^{-i\omega_{l}t'}+b_{l}^{+}e^{i\omega_{l}t'})\rangle \] \( \langle b_{l}^{+}b_{l}\rangle=n_{l};\quad \langle b_{l}b_{l}^{+}\rangle=n_{l}+1 \) \[ n_{l}=\dfrac{Tr(b_{l}b_{l}^{+}e^{-\beta \omega_{l}b_{l}^{+}b_{l}})}{Tr(e^{-\beta \omega_{l}b_{l}^{+}b_{l}})}=\dfrac{\sum_{n=0}^{+\infty}e^{-\beta \omega_{l}n}n}{\sum_{n=0}^{+\infty}e^{-\beta\omega_{l}n}}=\dfrac{1}{e^{\beta\omega_{l}}-1} \] \[ \Rightarrow \langle \xi(t)\xi(t')\rangle=\sum_{l}\xi_{l}^{2}[(n_{l}+1)e^{-i\omega_{l}(t-t’)}+n_{l}e^{i\omega_{l}(t-t’)}] \] 对于高温情形，\( \beta\rightarrow 0,T\rightarrow \infty;\quad n_{l}\approx n_{l}+1\approx \dfrac{k_{B}T}{\omega_{l}} \) \[ \langle \xi(t)\xi(t')\rangle\approx \sum_{l}\xi_{l}^{2}\dfrac{2k_{B}T}{\omega_{l}}cos[\omega_{l}(t-t’)] \] 无法严格证明上式\( \propto\delta(t-t') \)，但上式是一个比较好的近似，可以证明长时平均\(=0\),记忆时间\( T_{l}\approx (\dfrac{2\pi}{\omega_{l}}) \)，其中\( \omega_{l} \)取最小值。
  2. Spin-Boson Model \[ H=\dfrac{\Delta}{2}\sigma_{x}+\dfrac{\epsilon}{2}\sigma_{z}+\sum_{l}\omega_{l}b_{l}^{+}b_{l}+\sum_{l}g_{l}\sigma_{z}(b_{l}+b_{l}^{+}) \] (1) 每一项的物理意义；(2) 找特殊解，\( g_{l}=0;\quad \Delta=0 \)
     \(\Delta=0\)时，\( H=H_{\uparrow}+H_{\downarrow} \) \[ H_{\uparrow}=\dfrac{\epsilon}{2}+\sum_{l}\omega_{l}b_{l}^{+}b_{l}+\sum_{l}g_{l}(b_{l}^{+}b_{l});\quad H_{\downarrow}=-\dfrac{\epsilon}{2}+\sum_{l}\omega_{l}b_{l}^{+}b_{l}-\sum_{l}g_{l}(b_{l}^{+}b_{l}) \] (3)找极限
     简化：\[ H=\omega b^{+}b+g(b^{+}+b) \] 令：\( a^{+}=b^{+}+\dfrac{g}{\omega},\quad a=b+\dfrac{g}{\omega} \),\( \omega a^{+}a=\omega b^{+}b+g(b^{+}+b)+\dfrac{g^{2}}{\omega^{2}} \) \[ H=\omega a^{+}a-\dfrac{g^{2}}{\omega^{2}} \] 平移算子： \[ e^{a\dfrac{\partial}{\partial x}}f(x)=(1+a\dfrac{\partial}{\partial x}+\dfrac{a^{2}}{2}\dfrac{\partial^{2}}{\partial x^{2}}+\cdots)f(x)=f(x+a) \] \( U=e^{-i\alpha p},\quad U^{+}xU=x+\alpha \)
     \( U^{+}HU=\omega \tilde{b}^{+}\tilde{b}+const \),\(H\)本征态\(U|n\rangle=e^{-\alpha_{l}(b_{l}-b_{l}^{+})}|n\rangle,\quad \alpha_{l}\in R\)
     数值方法，变分法
     \[ \vert \psi\rangle=\vert\uparrow\rangle\varphi_{\uparrow}+\vert\downarrow\rangle\varphi_{\downarrow} \] \[ \varphi_{\uparrow}=\sum_{n}A_{n}e^{\sum_{l}f_{nl}b_{l}^{+}-f_{nl}^{*}b_{l}}\vert 0\rangle_{Bath};\quad \varphi_{\downarrow}=\sum_{n}B_{n}e^{\sum_{l}g_{nl}b_{l}^{+}-g_{nl}^{*}b_{l}}\vert 0\rangle_{Bath} \] 定义：\( L=\dfrac{i}{2}\langle\psi\vert\overset{\leftrightarrow}{\partial}_{t}\vert\psi\rangle-\langle\psi\vert H\vert\psi\rangle=L(U,U^{*},t);\quad U=A_{n},B_{n},f,g \) \[ \dfrac{d}{dt}(\dfrac{\partial L}{\partial U_{n}^{*}})-\dfrac{\partial L}{\partial U_{n}}=0 \] \[ i\dfrac{\partial}{\partial t}\vert \psi\rangle=H\vert\psi\rangle,\quad \vert\psi\rangle=\sum_{n}c_{n}\vert n \rangle\Rightarrow i\dot{c}_{n}=\sum_{m}t_{nm}c_{m},\quad t_{nm}=\langle m\vert H\vert n\rangle \] \[ E=\langle \psi\vert H\vert\psi\rangle=\sum_{nm}t_{nm}c_{n}^{*}c_{m},\quad i\dot{c}_{n}=\dfrac{\partial E}{\partial c_{n}^{*}},i\dot{c}_{n}^{*}=-\dfrac{\partial E}{\partial c_{n}} \] \[ \langle\psi\vert\overset{\leftrightarrow}{\partial}_{t}\vert\psi\rangle=\langle\psi\vert\partial_{t}\vert\psi\rangle-\langle\partial_{t}\psi\vert\psi\rangle=\sum_{n}c_{n}^{*}\dot{c}_{n}-\dot{c}_{n}^{*}c_{n} \] \[ L=\dfrac{i}{2}(\sum_{n}c_{n}^{*}\dot{c}_{n}-\dot{c}_{n}^{*}c_{n})-t_{nm}c_{n}^{*}c_{m}=-i\sum_{n}\dot{c}_{n}^{*}c_{n}-t_{nm}c_{n}^{*}c_{m} \] \[ \dfrac{d}{dt}(\dfrac{\partial L}{\partial \dot{c}_{m}^{*}})=-i\dot{c}_{n},\quad \dfrac{\partial L}{\partial c_{n}^{*}}=-\sum_{m}t_{nm}c_{m}\Rightarrow i\dot{c}_{n}=-\sum_{m}t_{nm}c_{m}\] 投针法计算\( \pi \)，线间距为\(d\),针长度为\( l \)，\(N\)是总的个数，\(N_{c}\)是有交叉的个数，\( \pi=\dfrac{2l}{d}\dfrac{N}{N_{c}} \)
  3. 课堂笔记12_饶pdf
  4. 课堂笔记12_林pdf

• 11.4 周四

  1. 计算\(\pi\)的方法 除了上一次课讲到的蒲丰投针以外，还有一种更加直观的办法。
     在平面上随机产生点，正方形的面积为\(4r^{2}\)，其内切圆的面积为\(\pi r^{2}\)，\(N_{C}\)为落在圆内的点的数目，\(N\)为落在正方形内的点的总数目 \[ \dfrac{N_{C}}{N}=\dfrac{\pi}{4} \]
  2. Monte Carlo方法
     (1) 随机数作具体计算时，主要是积分，以及求配分函数。
     (2) 使用Monte Carlo方法的原因
     传统方法，维度为\(d\)，每个维度范围\(L\)，间隔\(h\)，其误差一般大致为\(O(h^{4})\),计算次\((\dfrac{L}{h})^{d}\)。给定\(N\)，\(h\propto N^{-1/d}\)，误差\(\propto\dfrac{1}{\sqrt{N}}\)。
     可以发现对于高维的情况，计算时间非常长，而且误差收敛很慢。而随机数的误差\(\propto \dfrac{1}{\sqrt{N}}\)，与维度无关，适合高维计算。
     (3) 目标：(a) 产生任意分布\(P(x)\); (b) 高维积分\(d\geq 5\); (c) Ising Model： \[ d=1\Rightarrow H = -J\sum_{i } \sigma_i \sigma_{i+1};\quad d=2\Rightarrow H = -J\sum_{\langle i,j \rangle} \sigma_i \sigma_j - h \sum_i \sigma_i \] 其中一维无相变，二维有相变，感兴趣的同学可以进行验证。
     (4) 积分 \[ \int_{a}^{b}f(x)p(x)dx=\dfrac{b-a}{N}\sum_{i}f_{i}g_{i} \] 其中\(p(x)\)与态密度（Density of States）相关， \[ \rho(\epsilon)=\dfrac{1}{N}\sum_{i}\delta(\epsilon-\epsilon_{i}) \] 另一种方法是 \[ I=\dfrac{1}{N}\sum_{i}f(x_{i}) \] 其中\(x_{i}\sim p(x)\)，属于非等间距的情况 \[ I=\int_{a}^{b}f(x)dx=\int_{a}^{b}\dfrac{f(x)}{p(x)}p(x)dx=\dfrac{1}{N}\sum_{i}\dfrac{f(x_{i})}{p(x_{i})} \] (5) 统计物理： \[ Z=\int \dfrac{dxdp}{2\pi\hbar}e^{-\beta H(\vec{x},\vec{p})} \] \[ Z=Tr(e^{-\beta H}) \] \[ \bar{O}=\int O(x,p)e^{-\beta H(x,p)}dxdp \]
  3. Metropolis (1) \(U(0,1)\)均匀分布于\([0,1)\)的随机数,产生均匀分布于\([-W,W)\)的随机数： \[ p(x)=\begin{eqnarray} \left \lbrace \begin{array}{cc} \dfrac{1}{2W},\quad \vert x\vert\leq W \\ 0,\quad \vert x\vert>W \end{array} \right. \end{eqnarray} \] \( 2W(U(0,1)-0.5) \)
     (2) Guass分布，Box-Muller变换
     (3) 伪随机数产生\(U(0,1)\) \[ x_{n+1}=\dfrac{(a x_{n}+c)\ mod\ m}{m} \rightarrow U(0,1) \] (4) Metropolis \[ x_{1}\rightarrow x_{2}\rightarrow x_{3}\rightarrow \cdots \rightarrow x_{n}\rightarrow x_{n+1}\] 当\(\xi\)小于\(r\),\( x_{n+1}=x^{*}\),当\(\xi\)大于\(r\),\( x_{n+1}=x_{n}\).其中\( \xi\in U(0,1) \)，\(r=min[1,\dfrac{p(x^{*})}{p(x_{n})}]\)，\(n>N_{C}\)时\(x_{n}\)的分布符合预期
     (5) 细致平衡：\[ \dfrac{dp_{n}}{dt}=\sum_{m}(p_{m}T_{m\rightarrow n}-p_{n}T_{n\rightarrow m});\quad p_{m}T_{m\rightarrow n}=p_{n}T_{n\rightarrow m} \] (6) 双值随机模型，详见老师的PPT
  4. 课堂笔记13_饶pdf
  5. 蒙特卡洛（Monte Carlo, MCMC）方法的原理和应用
  6. 关于mc的PPT

• 11.9 周二

  1. MC Integral
     \[I=\int f(x)p(x)dx=\dfrac{1}{N}\sum_{i}f(x_{i})\] (a)产生一组数;(b)\(\lbrace x_{i}\rbrace\rightarrow P(x)\);(c)平稳分布\(\sim\)收敛
     考虑\(f(\vec{x})\)是一个明确的物理对象,有明确的物理意义：
     (a)\( f(\vec{x})=e^{-\beta H(\vec{x})} \)，积分\(I=Tr(e^{-\beta H})=\int d\vec{x}e^{-\beta H(\vec{x})}\);(b)Path Integral MC
     (1)Ising Model: Ising证明，1d情况\[H=-J\sum_{i}\sigma_{i}\sigma_{i+1}\]无相变，其中\(\sigma_{i}=\pm 1\)，转移矩阵方法.
     (2) Heisenberg Model: \[ H=-J\sum_{\langle ij\rangle}\vec{\sigma}_{i}\vec{\sigma}_{j} \]
     (3) Metropolis–Hastings algorithm: \[ x_{n+1}=x^{new}, \xi < r=min\lbrace 1,\dfrac{P(x^{new})}{x_{n}}\rbrace;\quad x_{n+1}=x_{n},\xi > r \] \[ m=\sum m(\sigma)e^{-\beta H(\sigma)}/Z;\quad H(\sigma)=-J\sum_{\langle ij\rangle}\sigma_{i}\sigma_{j} \]
     算法：(a) 初始化\(\sigma=(\sigma^{1},\sigma^{2},\sigma^{3},\cdots,\sigma^{N})\)
     (b) 随机找一个\(i\)自旋翻转：\(\sigma_{n}^{i}\rightarrow -\sigma_{n}^{i}\)
     (c) \[ r=min\lbrace 1,\dfrac{P(x^{new})}{P(x_{n})}\rbrace=min\lbrace 1,e^{-\beta \Delta E}\rbrace \] (d) \[ \vec{\sigma}_{n+1}=\vec{\sigma}',\xi < r;\quad \vec{\sigma}_{n+1}=\vec{\sigma}_{n},\xi > r\]
     (4) Path Integral MC \[ Z=\int dq_{1}dq_{2}\cdots dq_{n}e^{-S};\quad S=\Delta\tau\sum_{i}[\dfrac{m}{2}\dfrac{(q_{i+1}-q_{i})^{2}}{\Delta\tau^{2}}+V(q_{i})];\quad \vec{q}=( \vec{q}_{1},\vec{q}_{2},\cdots,\vec{q}_{n}),q_{i}\in R \] \[ Z=Tr(e^{-\beta H});\quad H=\dfrac{p^{2}}{2m}+V(x) \] \[ Z=\int dx_{0}\langle x_{0}\vert e^{-\beta H}\vert x_{0}\rangle=\int dx_{0}\langle x_{0}\vert e^{-\Delta\tau H}e^{-\Delta\tau H}\cdots e^{-\Delta\tau H}\vert x_{0}\rangle \] 在每个\(e^{-\Delta\tau H}\)之间插入一组完备基可以得到： \[ Z=\Pi_{n=0}^{N-1}\int dx_{n}\langle x_{n+1}\vert e^{-\Delta\tau H}\vert x_{n}\rangle \] \[ e^{-\Delta\tau H}=e^{-\Delta\tau T}e^{-\Delta\tau V}e^{O(\Delta\tau^{2})}\Rightarrow Z=\Pi_{n=0}^{N-1}\int dx_{n}\langle x_{n+1}\vert e^{-\Delta\tau T}\vert x_{n}\rangle e^{-\Delta\tau V(x_{n})}\] \[ \langle x_{n+1}\vert e^{-\Delta\tau T}\vert x_{n}\rangle=\int dp_{n}e^{-\Delta\tau\dfrac{p_{n}^{2}}{2m}}\langle x_{n+1}\vert p_{n}\rangle\langle p_{n}\vert x_{n}\rangle=\int\dfrac{dp_{n}}{2\pi}\exp(-\Delta\tau\dfrac{p_{n}^{2}}{2m})\exp(ip_{n}(x_{n}-x_{n+1}))\sim \exp(-\dfrac{m(x_{n}-x_{n+1})^{2}}{2\Delta\tau}) \] \[ \Rightarrow Z=\Pi_{n=0}^{N-1}\int dx_{n}\exp(-\Delta\tau[\dfrac{m}{2}(\dfrac{x_{n+1}-x_{n}}{\Delta\tau})^{2}+V(x_{n})])=\int Dx e^{-\Delta\tau S[x]} \] (5)总结： \[ I=\int f(\vec{x})p(\vec{x})d\vec{x}=\dfrac{1}{N}\sum_{i=1}^{N}f(\vec{x}_{i}) \] 应用：(1)积分；(2)\( f\sim e^{-\beta H} \)；(a)Ising Model;(b)Path Integral;(3)求\(V(\vec{x})\)的极值(Kinetic MC)
     (1)涨落耗散定理： \[ \langle x^{2}\rangle=2\sigma t \] (2)Ito lemma;(3)MC simulation
  2. 作业 (1) 求高维积分
     (2) 产生任意给定的\(P(x)\)
     (3) 求\(2d,3d\)，Ising Model的相变
     ddl:11月28日
  3. 课堂笔记14_饶pdf
  4. 课堂笔记14_林pdf

• 11.11 周四

  1. Review
     (1) 积分：\( I=\int f(\vec{x})p(\vec{x})d\vec{x}=\dfrac{1}{N}\sum_{i}f(\vec{x}_{i})\Rightarrow \vec{x}_{n}\rightarrow \vec{x}_{n+1} \)
     在具体情况中，\(p(\vec{x})\)有具体的含义，例如\(p(\vec{x})=e^{-\beta H(\vec{x})}/Z\)
     又比如：\(M=\int(\sum_{i}\sigma_{i})e^{-\beta H(\vec{\sigma})}/Z=\dfrac{1}{N}\sum_{\sigma_{i}}Sum(\sigma_{i})\)
     \(\vec{x}\)的可能形式，物理意义:
     (1) 积分,\(\vec{x}\in R\)
     (2) Ising Model,\(\vec{x}=(\sigma_{1},\sigma_{2},\cdots,\sigma_{N}),\quad \sigma_{i}=\pm 1\)
     (3) Heisenberg Model
  2. 相变，相图
     在临界点会出现发散现象 \[ T < T_{c}\Rightarrow A\propto \dfrac{1}{\vert T-T_{c}\vert^{\alpha_{-}}};\quad T > T_{c}\Rightarrow A\propto\dfrac{1}{\vert T-T_{c}\vert^{\alpha_{+}}} \] \( \alpha_{+}=\alpha_{-},\quad or\quad \alpha_{+}\neq\alpha_{-} \)，广泛的来看\(\alpha\)不是普适常数
     (1) Landau相变理论：引入序参量\(m\) \[ H=a(T-T_{c})^{2}m^{2}+bm^{4},\quad a > 0,\quad b > 0 \] \( T > T_{c} \)时，只有\(m=0\)这个点是稳定平衡点，\(T < T_{c}\)时，有正负两个平衡点. 应用：(1) 二级相变;(2)化学反应，Kramers theory, Fokker-Planck equation;(3)一级相变，液晶： \[ H=am^{2}+bm^{3}+cm^{4};\quad \dfrac{\partial H}{\partial m}=0\Rightarrow m[2a+3bm+4cm^{2}]=0 \] 回到\( H=a(T-T_{c})^{2}m^{2}+bm^{4} \) \[ \Rightarrow \dfrac{\partial H}{\partial m}=0\Rightarrow 2a(T-T_{c})m+4bm^{3}=0\Rightarrow m=0\quad or\quad m=\pm\sqrt{\dfrac{2a(T_{c}-T)}{4b}} \] \[ \Rightarrow H=\dfrac{a^{2}}{4b}(T-T_{c})^{2}\Rightarrow m\propto \sqrt{T_{c}-T}\Rightarrow \dfrac{\partial m}{\partial T}\propto \dfrac{1}{\sqrt{T_{c}-T}}\quad When \quad T < T_{c};\quad Otherwise\quad =0 \] 可以得到一个普世常数\( \alpha=\dfrac{1}{2} \);Onsager, Yang，2d Ising Model \( \mu=\dfrac{1}{8} \)
     (2) 发散的本质：相干长度的发散
     \( C_{V}\propto\dfrac{1}{\vert T-T_{c}\vert^{\mu}} \) \[ C_{V}=\dfrac{\partial E}{\partial T}\propto \langle E^{2}\rangle-\langle E\rangle^{2};\quad \bar{E}=\dfrac{Tr(He^{-\beta H})}{Tr(e^{-\beta H})} \] \[ E=\int \epsilon(\vec{r})d\vec{r}=\int d\vec{r}d\vec{r}'\lbrace \langle \epsilon(\vec{r})\epsilon(\vec{r'})\rangle-\langle \epsilon(\vec{r})\rangle\langle\epsilon(\vec{r}')\rangle\rbrace=V\int d\vec{r}\lbrace\langle \epsilon(\vec{r})\epsilon(0)\rangle-\langle\epsilon(\vec{r})\rangle\langle\epsilon(0)\rangle\rbrace =V\int g(\vec{r})d\vec{r}\propto \dfrac{V}{\vert T-T_{c}\vert^{\mu}} \] \[ g(\vec{r})=g_{0}e^{-\vert\vec{r}\vert/\xi};\quad \int g(\vec{r})d\vec{r}\propto \int e^{-\vert\vec{r}\vert/\xi}d\vec{r}\propto\xi^{d}\int e^{-\vert \vec{x}\vert}d\vec{x}\Rightarrow \xi^{d}\propto\dfrac{1}{\vert T-T_{c}\vert^{\mu}} \] 参考文献：Ornstein,Zernike. The influence of accidental deviations of density on the equation of state \[ (P+\dfrac{a}{V^{2}})(V-b)=rT\Rightarrow P+\dfrac{a}{V^{2}}=\dfrac{rT}{V-b}=\dfrac{rT}{V}(1+\dfrac{b}{V}) \] \[ P+\dfrac{a}{V^{2}}=\dfrac{rT}{V}(1+\dfrac{b}{V})+\dfrac{1}{6V}\int mg(\vec{r}_{nk})\geq \vec{r}_{nk}\geq\vec{\bar{r}}(\vec{r}_{nk})d\vec{V}_{k} \] (3) 标度假设：
     标度不变性：坐标标度变换，结果参数变形式不变，以扩散方程为例： \[ \dfrac{\partial }{\partial t}\psi=D\dfrac{\partial^{2}\psi}{\partial x^{2}} \] \[ x\rightarrow\lambda x;\quad t\rightarrow \lambda^{\nu}t;\quad \psi\rightarrow\lambda^{\alpha}\psi;Rightarrow x'=\lambda x,t'=\lambda^{\nu}t,\psi'(x',t')=\psi'(\lambda x,\lambda^{\nu}t)=\lambda^{\alpha}\psi \] \[ \Rightarrow \dfrac{\partial \psi}{\partial t}=\lambda^{\nu-2}D(\lambda)\dfrac{\partial^{2} \psi}{\partial x^{2}}\Rightarrow D\lambda^{2-\nu}=D(\lambda) \] 标度性：\( \psi(\lambda x,\lambda^{\alpha}y)=\lambda^{\beta}\psi(x,y) \) . 取 \(\lambda x=1\),\( \psi(1,\dfrac{y}{x^{\alpha}})=x^{-\beta}\psi(x,y) \),\( x^{\beta}\psi(x,y)=f(\dfrac{y}{x^{\alpha}}) \).
     \( f=\dfrac{t}{2}m^{2}+um^{4}-hm;t=a(T-T_{c});\dfrac{\partial f}{\partial m}=0\Rightarrow m(tm+3um^{3}-h)=0 \)
  3. 作业
     相互作用气体方程：\( (P+\dfrac{a}{V^{2}})(V-b)=k_{B}T \),\(k=-\dfrac{1}{V}(\dfrac{\partial V}{\partial P})_{T}=\dfrac{1}{\vert T-T_{c}\vert^{\nu=1}}\)
     实验：\(\nu=1.2\).
     \( \Delta V\propto \vert T-T_{c}\vert^{1/2} \),实验：\( \Delta V\propto \vert T-T_{c}\vert^{0.32} \)
     求相变点附近的\( \Delta V,k \)等的发散行为：
     ddl:11月28日
  4. 课堂笔记15_饶pdf
  5. 作业pdf

• 11.16 周二

  1. 标度假设
     (1) 相变（关联长度发散）：
     (a) 平均场，Landau理论
     (b) 数值方法：MC,DMRG,ED等
     (c) 标度假设
     (2) 标度假设：自由能是\(t,h\)的广义齐次函数，\(t=\dfrac{T-T_{c}}{T_{c}}\) \[ C_{v}\sim \vert t\vert^{-\alpha};\quad m\sim (-t)^{\beta};\quad \chi\sim\vert t\vert^{-\gamma};\quad m\sim h^{1/\delta}\vert_{t=0};\quad \xi\sim \vert t\vert^{-\nu d} \] 讨论\( \lbrace \alpha,\beta,\nu,\delta,\gamma \)之间的联系.
     参考： Kardar书The scaling hypothesis一章 \[ f=min(\dfrac{t}{2}m^{2}+um^{4}-hm),\quad e^{-\beta F}=Z \] \[ f(\lambda t,\lambda^{3/2}h)=\lambda^{2}f(t,h) \] \[ f(\lambda t,\lambda^{3/2}h)=min(\dfrac{t}{2}\lambda m^{2}+um^{4}-\lambda^{3/2}hm)=\lambda^{2}min(\dfrac{t}{2}+um^{4}-hm) \] 取\(\lambda t=1\),\( f(t,h)=t^{2}f(\dfrac{h}{\vert t \vert^{3/2}}) \)
     (a) \( h=0,t\neq 0 \Rightarrow f\sim t^{2-\alpha}f(0) \);(b)\( t=0,h\neq 0\Rightarrow t^{2-\alpha}f(\infty) \) \[ h=0,f=\dfrac{t}{2}m^{2}+um^{4},\dfrac{\partial f}{\partial m}=0\Rightarrow f=\dfrac{t}{2}(-\dfrac{t}{4u})+u(-\dfrac{t}{4u})^{2}=-\dfrac{t^{2}}{16u}\Rightarrow f(0)=-\dfrac{1}{16u}\] \[ t=0,h\neq 0;\quad f=um^{4}-hm;\quad \dfrac{\partial f}{\partial m}=0\Rightarrow m=(\dfrac{h}{4u})^{1/3}\Rightarrow f=u(\dfrac{h}{4u})^{4/3}-h(\dfrac{h}{4u})^{1/3} \] \(f=\vert t\vert^{2-\alpha}g(\dfrac{h}{t^{\Delta}})\) \[ \Rightarrow C_{v}\sim\dfrac{\partial^{2}f}{\partial t^{2}}\vert_{h=0}\propto \vert t\vert^{-\alpha} \] \[ m\sim\dfrac{\partial f}{\partial h}\vert_{h=0}\sim t^{2-\alpha-\Delta}\dfrac{\partial g(h/t^{\Delta})}{\partial h}\sim t^{2-\alpha-\Delta}\Rightarrow \beta=2-\alpha-\Delta \] \[ \chi\sim\dfrac{\partial m}{\partial h}\sim t^{2-\alpha-2\Delta}\Rightarrow \gamma=2\Delta+\alpha-2 \] \[ \alpha+2\beta+\gamma=2 \] Widom关系： \[ m\propto h^{1/\delta}=h^{\dfrac{2-\alpha-\Delta}{\Delta}}\Rightarrow \delta=\dfrac{\Delta}{2-\alpha-\Delta}\Rightarrow \delta-1=\gamma/\beta \]

     \	\( \alpha \)	\( \beta \)	\( \gamma \)	\( \delta \)	\(\nu\)
     d=2,Ising	\(0\)	\(1/8\)	\(7/4\)	\(5\)	\(1\)
     d=3,Ising	\(0.12\)	\(0.31\)	\(1.25\)	\(5\)	\(0.64\)
     d=3,XY	\( 0.00 \)	\( 0.33 \)	\( 1.33 \)	\( 5 \)	\( 0.66 \)
     d=3,Heisenberg	\( -0.14 \)	\( 0.35 \)	\( 1.4 \)	\( 5 \)	\( 0.7 \)

     Josephson relation:\( 2-\alpha=d\nu \) \[ -\beta F=\ln Z=(\dfrac{L}{\xi})^{d}g_{s}+(\dfrac{L}{a})^{d}g_{a}\] \( \xi\propto t^{-\nu} \quad f\sim\xi^{-d}=t^{\nu d}f(0) \) \[ G(x)=\langle m(x)m(0)\rangle-\langle m(x)\rangle\langle m(0)\rangle\propto \dfrac{1}{\vert x\vert^{d-2+\eta}} \]
  2. 实验对参数的测量pdf
  3. The influence of accidental deviations of density on the equation of state pdf
  4. Phase Transitions pdf
  5. First Order Phase Transitions pdf
  6. Mean Field Approximation (MFA) in the theory of magnetism pdf
  7. Phase Transitions and Collective Phenomena pdf
  8. First Order Phase Transitions pdf
  9. 课堂笔记16_饶 pdf
  10. 课堂笔记16_林 pdf
  11. Scaling Laws for Ising Models Near \( T_{c} \) pdf

• 11.18 周四

  1. 随机矩阵
     矩阵的矩阵元是随机数,\( A=(x_{ij})_{N\times N} \). 1964年,Wigner做了重要工作.
     考虑随机矩阵\(A\)满足一些基本条件，(a) \( A=A^{+} \); (b) 时间反演对称性，或者其他对称性; (c) \( x_{ij} \)独立同分布, \( \sim N(0,\sigma) \).
     Dyson,将\( P(\epsilon_{1},\cdots,\epsilon_{N}) \)对应Fokker-Planck方程.
     (1) 核心是求出随机矩阵\( H \)对应的联合概率密度分布:\( P(\lambda_{1},\lambda_{2},\cdots,\lambda_{N}) \)
     可以先作猜测：对于我们考虑的随机矩阵情况，简并很难发生，且\( \lambda_{i} \)较大的概率应较小，所以可以猜测如下形式： \[ P(\lambda)=P(\lambda_{1},\lambda_{2},\cdots,\lambda_{N})\propto (\Pi_{i < j}\vert \lambda_{i}-\lambda_{j}\vert)^{\beta}\exp[-A\sum_{i}\lambda_{i}^{2}] \] (2) 实厄米的情况： \[ H= \begin{eqnarray} \left( \begin{array}{cc} x_{1} & x_{2} \\ x_{2} & x_{3} \end{array} \right) \end{eqnarray} \] 考虑矩阵元为独立同分布的高斯分布的随机数，求本征值的联合概率分布：\(P(\lambda)\).
     引入正交阵：\( O \),\( O^{T}HO=\lambda, OO^{T}=1 \),\( \tilde{H}=O^{T}HO \). 本征值：\( \lambda(\tilde{H})=\lambda(H) \) \[ P(O^{T}HO)\sim \exp[-V(O^{T}HO)]=P(H) \] \[ P(H)\propto \exp[-A Tr(H^{2})],\quad Tr(H^{2})=(x_{1}^{2}+2x_{2}^{2}+x_{3}^{2})\Rightarrow P(H)\propto \exp[-A(x_{1}^{2}+2x_{2}^{2}+x_{3}^{2})]\propto\exp[-A(\lambda_{1}^{2}+\lambda_{2}^{2})] \] 用\( \epsilon_{1},\epsilon_{2} \)来表示两个本征值： \[ \epsilon_{1}=\dfrac{1}{2}(x_{1}+x_{3})-\sqrt{x_{2}^{2}+\dfrac{1}{4}(x_{1}-x_{3})^{2}};\quad \epsilon_{2}=\dfrac{1}{2}(x_{1}+x_{3})+\sqrt{x_{2}^{2}+\dfrac{1}{4}(x_{1}-x_{3})^{2}} \] \[ P(\lambda)=\int P(H)dH[ \delta(\lambda_{1}-\epsilon_{1})\delta(\lambda_{2}-\epsilon_{2})+\delta(\lambda_{1}-\epsilon_{2})\delta(\lambda_{2}-\epsilon_{1}) ] \] 这样积分会很困难，需要换一种方法： \[ O= \begin{eqnarray} \left( \begin{array}{cc} \cos{\theta} & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{array} \right) \end{eqnarray} \] \[ \Rightarrow H= \begin{eqnarray} \left( \begin{array}{cc} \epsilon_{1}\cos^{2}(\theta)+\epsilon_{2}\sin^{2}(\theta) & (\epsilon_{2}-\epsilon_{1})\sin(\theta)\cos(\theta) \\ (\epsilon_{2}-\epsilon_{1})\sin(\theta)\cos(\theta) & \epsilon_{1}\sin^{2}(\theta)+\epsilon_{2}\cos^{2}(\theta) \end{array} \right) \end{eqnarray} \] \[ P(H)dH=\exp[-A(\epsilon_{1}^{2}+\epsilon_{2}^{2})]dx_{1}dx_{2}dx_{3}=\exp[-A(\epsilon_{1}^{2}+\epsilon_{2}^{2})]Jd\epsilon_{1}d\epsilon_{2}d\epsilon_{3} \] \[ J=\vert\epsilon_{1}-\epsilon_{2}\vert \] \[ P(\lambda_{1},\lambda_{2})=C\vert\lambda_{1}-\lambda_{2}\vert \exp[-A(\lambda_{1}^{2}+\lambda_{2}^{2})] \] 推广到高维： \[ P(\lambda)\propto \Pi_{i < j}\vert x_{i}-x_{j}\vert\exp(-A\sum_{i}\lambda_{i}^{2}) \] (3) 复厄米的情况： \[ H= \begin{eqnarray} \left( \begin{array}{cc} x_{1} & x_{2}+ix_{3} \\ x_{2}-ix_{3} & x_{4} \end{array} \right)=U^{+}\left( \begin{array}{cc} \epsilon_{1} & 0 \\ 0 & \epsilon_{2} \end{array} \right)U(\theta,\phi) \end{eqnarray} \] \[ P(H)dH=e^{-A(\epsilon_{1}^{2}+\epsilon_{2}^{2})}dx_{1}dx_{2}dx_{3}dx_{4}=f(\epsilon_{1},\epsilon_{2},\theta,\phi)d\epsilon_{1}d\epsilon_{2}d\theta d\phi \] \[ U(\theta,\phi)= \begin{eqnarray} \left( \begin{array}{cc} \cos{\theta}\exp{(i\phi/2)} & -\sin{\theta}\exp{(-i\phi/2)} \\ \sin{\theta}\exp{(i\phi/2)} & \cos{\theta}\exp{(-i\phi/2)} \end{array} \right) \end{eqnarray} \] \[ J=(\epsilon_{1}-\epsilon_{2})^{2}\sin(\theta)\cos(\theta)\Rightarrow P\propto \vert\lambda_{1}-\lambda_{2}\vert^{2}\exp[-A(\lambda_{1}^{2}+\lambda_{2}^{2})] \] 一般情况： \[ P\propto \Pi_{i < j}\vert \lambda_{i}-\lambda_{j}\vert^{\beta}\exp[-A\sum_{i}\lambda_{i}^{2}+B\sum_{i}\lambda_{i}] \] \( N\times N \)情况的讨论之后讲.
  2. Statistical Theory of the Energy Levels of Complex Systems. I pdf
  3. A Brownian-Motion Model for the Eigenvalues of a Random Matrix pdf
  4. Random Matrices and the Statistical Theory of Energy Levels pdf
  5. Numerical Solution of Dyson Brownian Motion and a Sampling Scheme for Invariant Matrix Ensembles pdf
  6. Random Matrix Theory: Wigner-Dyson statistics and beyond. Lecture notes given at SISSA (Trieste, Italy)
  7. Random matrices Lecture notes
  8. 课堂笔记17_饶
  9. 关于同步效应的文章

• 11.23 周二

  1. 随机矩阵的本征值的联合概率分布
     (1) 目标：
     (a) 设\(\theta_{i}\)为\(H\)的本征值: \[ P(\theta_{1},\cdots,\theta_{N})\propto \Pi_{i < j}\vert \theta_{i}-\theta_{j}\vert\exp(-A\sum_{i}\theta_{i}^{2}) \]
     (b) Dyson：与Brownian Motion 对应.
     (2) Jacobian Matrix: \[ dx_{1}dx_{2}\cdots dx_{N}=Jdy_{1}dy_{2}\cdots dy_{N};\quad J=det(\dfrac{\partial x_{i}}{\partial y_{i}}) \] \( x,y \)为向量,\( O \)为正交阵，\( O^{T}=O \)，\( dx_{i}=d(Oy)_{i}=O_{ij}dy_{i}\Rightarrow \dfrac{\partial x_{i}}{\partial y_{i}}=O_{ij} \),\( dX=JdY,\quad J=det O=1 \)
     \( X,Y \)推广为矩阵：\( X=OY,\quad dX=f(O)dY \),\( x_{ij}=(OY)_{ij}=O_{k}y_{ki},\quad \dfrac{\partial x_{ij}}{\partial y_{kj}}=O_{ik} \)
     用二维的例子来说明问题： \[ X= \begin{eqnarray} \left( \begin{array}{cc} O_{11} & O_{12} \\ O_{21} & O_{22} \end{array} \right) \left( \begin{array}{cc} y_{11} & y_{12} \\ y_{21} & y_{22} \end{array} \right)= \left( \begin{array}{cc} x_{1} & x_{2} \\ x_{3} & x_{4} \end{array} \right) \end{eqnarray} \] \[ \begin{eqnarray} \left( \begin{array}{cccc} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right) = \left( \begin{array}{cccc} O_{11} & 0 & O_{12} & 0 \\ 0 & O_{11} & 0 & O_{12} \\ O_{21} & 0 & O_{22} & 0 \\ 0 & O_{21} & 0 & O_{22} \\ \end{array} \right) \left( \begin{array}{cccc} y_{1} \\ y_{2} \\ y_{3} \\ y_{4} \\ \end{array} \right)=O\otimes I \left( \begin{array}{cccc} y_{1} \\ y_{2} \\ y_{3} \\ y_{4} \\ \end{array} \right) \end{eqnarray} \] \[ \Rightarrow dX=det(O\otimes I)dY=(det(O))^{2}dY=dY \] 另一种思路： \[ X=OY,dX=f(O)dY;\quad Y=O^{T}X,dY=f(O^{T})dX\Rightarrow f(O)f(O^{T})=1\Rightarrow f(O)=(det(O))^{k} \] 另一种可能； \[ X=O^{T}YO,H'=O^{T}HO;\quad Z=OX=YO;\quad dZ=f(O)dX,dZ=f(O)dY;\quad dX=f(O)f(O^{T})dY=dY \]
     (3) 引入\( H=O^{T}\theta O=(h_{ij})_{N\times N} \)，且\( \theta=diag(\theta_{1},\cdots,\theta_{N}) \)
     \( H \)中的独立元素个数:\( \dfrac{N(N+1)}{2} \)；\( O \)中的独立元素个数:\( \dfrac{N(N-1)}{2} \)
     这些元素为：\( \theta_{1},\cdots,\theta_{N};p_{1},\cdots,p_{l},l=\dfrac{N(N-1)}{2} \),其中：\( p_{l}\in O \) \[ J(\theta,p)=det(\dfrac{\partial H}{\partial \theta},\dfrac{\partial H}{\partial p}) \] \[ \dfrac{\partial H}{\partial p_{\mu}}=\dfrac{\partial O^{T}}{\partial P_{\mu}}\theta O+O^{T}\theta \dfrac{\partial O}{\partial P_{\mu}}=O^{T}(O\dfrac{\partial O^{T}}{\partial p_{\mu}}\theta)O+O^{T}\theta\dfrac{\partial O}{\partial p_{\mu}}O^{T}O \] \[ O\dfrac{\partial O^{T}}{\partial p_{\mu}}=S^{\mu};\quad \dfrac{\partial O}{\partial p_{\mu}}O^{T}=-S^{\mu};\quad O\dfrac{\partial O^{T}}{\partial p_{\mu}}+\dfrac{\partial O}{\partial p_{\mu}}O^{T}=\dfrac{\partial OO^{T}}{\partial p_{\mu}}=0 \] \[ \Rightarrow O(\dfrac{\partial H}{\partial p_{\mu}})O^{T}=S^{\mu}\theta-\theta S^{\mu};\quad O(\dfrac{\partial H}{\partial \theta_{\nu}})O^{T}=\dfrac{\partial \theta}{\partial \theta_{\nu}} \] \[ (O\dfrac{\partial H}{\partial p_{\mu}}O^{T})_{\alpha\beta}=S_{\alpha\beta}^{\mu}(\theta_{\beta}-\theta_{\alpha});\quad (O\dfrac{\partial H}{\partial \theta_{\nu}}O^{T})_{\alpha\beta}=\delta_{\alpha\beta}\delta_{\alpha\nu} \] \[ \Rightarrow J=det \begin{eqnarray} \left( \begin{array}{cc} \delta_{\alpha\beta}\delta_{\alpha\nu} \\ S_{\alpha\beta}^{\mu}(\theta_{\beta}-\theta_{\alpha}) \end{array} \right)\propto \Pi_{\alpha < \beta}(\theta_{\alpha}-\theta_{\beta}) det \left( \begin{array}{cc} \delta_{\alpha\beta}\delta_{\alpha\nu} \\ S_{\alpha\beta}^{\mu} \end{array} \right) \end{eqnarray} \] \[ \Rightarrow P(\lambda)=\int \delta(\lambda-\theta)\exp(-A\sum_{j}\theta_{j}^{2})\Pi_{\alpha < \beta}\vert\theta_{\alpha}-\theta_{\beta}\vert f(p)d\theta dp\propto\int \delta(\lambda-\theta)\exp(-A\sum_{j}\theta_{j}^{2})\Pi_{\alpha < \beta}\vert\theta_{\alpha}-\theta_{\beta}\vert d\theta \] (4) 与Fokker-Planck方程，与Brownain motion对应，本征值\( \Rightarrow \)坐标. \[ P\propto \Pi_{i < j}\vert x_{i}-x_{j}\vert\exp(-\dfrac{\beta}{2}\sum_{i}x_{i}^{2})\propto e^{-\beta W} \] 其中\( \beta \)与温度相关，\( W \)：Potential. \[ E_{i}=-(\dfrac{\partial W}{\partial x_{i}})=-x_{i}+\dfrac{1}{\beta}\sum_{j\neq i}(\dfrac{1}{x_{j}-x_{i}}) \] \( p \)是Fokker-Planck方程的解： \[ f\dfrac{\partial p}{\partial t}=\sum_{j}\beta^{-1}\dfrac{\partial^{2}p}{\partial x_{j}^{2}}-\dfrac{\partial }{\partial x_{j}}(E_{j}p) \] \( \dfrac{d^{2}x_{j}}{dt^{2}}=-f\dfrac{d x_{j}}{dt}+E(x_{j})+\xi(t),\quad E_{j}=\dfrac{\partial W}{\partial x_{j}} \);\( \langle \xi(t)\xi(t')\rangle=\dfrac{2k_{B}T}{f}\delta(t-t') \)
     (a) \( f\langle \delta x_{j}\rangle=E(x_{j})\delta t;\quad f\langle \delta x_{j}^{2}\rangle=2k_{B}T\delta t \)
  2. 作业
     (1) 验证二维，三维实厄米随机矩阵的分布满足：\( P(\lambda_{1},\lambda_{2})\sim\vert \lambda_{1}-\lambda_{2}\vert\exp(-A(\lambda_{1}^{2}+\lambda_{2}^{2})) \)
     (2) 验证二维，三维复厄米随机矩阵的分布满足：\( P\sim\Pi_{i < j}\vert \lambda_{i}-\lambda_{j}\vert^{2}\exp(-A\sum_{i}\lambda_{i}^{2}) \)
     (3) \(1000\times 1000\)的随机矩阵：\( \lambda_{1} < \lambda_{2} < \cdots <\lambda_{N} \)
     (a) \( \Delta=\lambda_{i+1}-\lambda_{i} \)计算分布，\( P(\Delta)\sim\vert \Delta\vert^{\alpha}e^{-B\Delta^{2}} \)
     (b) \( \Delta=max{ \lambda_{i+1}-\lambda_{i} } \)，求\( P(\Delta) \)的分布
     ddl:12月13日
  3. 课堂笔记18_饶 pdf

• 11.25 周四

  1. 二次量子化
     (1) 目标：解释二次量子化；核心：找共轭量
     (2) Hamiltionian equation: \[ \dot{x}=\dfrac{\partial H}{\partial p}=\lbrace x,H\rbrace;\quad \dot{p}=-\dfrac{\partial H}{\partial x}=\lbrace p,H\rbrace \] 柯西-黎曼定理： \[ \dfrac{\partial f_{1}}{\partial x}=\dfrac{\partial f_{2}}{\partial y};\quad \dfrac{\partial f_{1}}{\partial y}=-\dfrac{\partial f_{2}}{\partial x} \] (3) 目的：
     (a) 对给定模型的量子化； (b) 对角化，求Bloch能带
     (c) 理解无序，Anderson Localization (d) 常见多体模型对角化
     (e) Bethe-ansatz数学化 (f) 平均场理论与相变
     (4) 标志性实验：(a)黑体辐射；(b)光电效应
     (5) \[ H=\dfrac{p^{2}}{2m}+V(x); \quad \dot{q}_{i}=\dfrac{\partial H}{\partial p_{i}}=\lbrace q_{i},H\rbrace;\quad\dot{p}_{i}=-\dfrac{\partial H}{\partial q_{i}}=\lbrace p_{i},H\rbrace \] \[ \lbrace A,B\rbrace=\sum_{i}\dfrac{\partial A}{\partial q_{i}}\dfrac{\partial B}{\partial p_{i}}-\dfrac{\partial A}{\partial p_{i}}\dfrac{\partial B}{\partial q_{i}} \] \[ \lbrace q_{i},q_{j}\rbrace=0;\quad \lbrace p_{i},p_{j}\rbrace=0;\quad \lbrace q_{i},p_{j}\rbrace=\delta_{ij} \] 对于谐振子： \[ H=\dfrac{p^{2}}{2m}+\dfrac{1}{2}m\omega^{2}x^{2}=\hbar\omega(a^{+}a+\dfrac{1}{2}) \] \[ a^{+}=\sqrt{\dfrac{m\omega}{2\hbar}} x -\dfrac{ip}{\sqrt{2\hbar m\omega}};\quad a=\sqrt{\dfrac{m\omega}{2\hbar}} x+\dfrac{ip}{\sqrt{2\hbar m\omega}} \] \[ [a_{i},a_{j}]=0;\quad [a_{i}^{+},a_{j}^{+}]=0;\quad [a_{i},a_{j}^{+}]=\delta_{ij} \] (6) 对于非线性系统： \[ H=H=\dfrac{p^{2}}{2m}+\dfrac{1}{2}m\omega^{2}x^{2}+\lambda x^{4}=\hbar\omega(a^{+}a+\dfrac{1}{2})+\lambda(\dfrac{\hbar}{2m\omega})^{2}(a+a^{+})^{4} \] \[ a\vert n\rangle=\sqrt{n}\vert n-1\rangle ;\quad a^{+}\vert n\rangle =\sqrt{n+1}\vert n+1\rangle \] 若使用\( \phi_{n}(x) \)则计算过于复杂： \[ \phi_{n}(x)\sim H_{n}(x)e^{-Ax^{2}};\quad \langle\phi_{n}\vert x^{4}\vert \phi_{n}\rangle\sim\int H_{n}^{2}(x)e^{-2Ax^{2}}dx \] 采用算子： \[ \lambda(\dfrac{\hbar}{2m\omega})^{2}\langle n\vert (a+a^{+})^{4}\vert n\rangle\propto\lambda(\dfrac{\hbar}{2m\omega})^{2}[n(n-1)+(2n+1)^{2}+(n+1)(n+2)] \] (7) 多体： \[ \phi_{n}(x)\phi_{m}(y)+\phi_{n}(y)\phi_{m}(x)\Rightarrow \vert nm \rangle \] \[ \sum \phi_{n1}(x_{p1})\phi_{n2}(x_{p2})\cdots\phi_{nl}(x_{pl})\Rightarrow\vert n_{1}n_{2}\cdots n_{l}\rangle;\quad a_{i}\vert n_{1}n_{2}\cdots n_{l}\rangle=\sqrt{n_{i}}\vert n_{1},\cdots,n_{i}-1,\cdots,n_{l}\rangle \] \[ H=\dfrac{p_{1}^{2}}{2m}+\dfrac{p_{2}^{2}}{2m}+\dfrac{1}{2}m\omega_{1}^{2}x_{1}^{2}+\dfrac{1}{2}m\omega_{2}^{2}x_{2}^{2}+\lambda x_{1}^{2}x_{2}^{2}=\hbar\omega_{1}(a_{1}^{+}a_{1}+\dfrac{1}{2})+\hbar\omega_{2}(a_{2}^{+}a_{2}+\dfrac{1}{2})+\lambda\dfrac{\hbar}{2m\omega_{1}}\dfrac{\hbar}{2m\omega_{2}}(a_{1}+a_{1}^{+})^{2}(a_{2}+a_{2}^{+})^{2} \] \[ \epsilon=\hbar\omega_{1}(n_{1}+\dfrac{1}{2})+\hbar\omega_{2}(n_{2}+\dfrac{1}{2})+\lambda\dfrac{\hbar}{2m\omega_{1}}\dfrac{\hbar}{2m\omega_{2}}\langle n_{1}n_{2}\vert (a_{1}+a_{1}^{+})^{2}(a_{2}+a_{2}^{+})^{2}\vert n_{1}n_{2}\rangle \]
• 课堂笔记19_饶 pdf
• The Quantum Theory of the Emission and Absorption of Radiation pdf

• 11.30 周二

  1. 二次量子化
     \( x=\sqrt{\dfrac{\hbar}{2m\omega}}(a+a^{+});\quad p=i\sqrt{\dfrac{m\hbar \omega}{2}}(a^{+}-a) \)
     可以避免复杂的波函数表示，方便处理多体系统，同时在物理上，直观体现量子化.
     (1) 模型1: \[ H=\hbar \omega(a^{+}a+\dfrac{1}{2})+g_{eff}(a+a^{+})^{4} \] \[ (a+a^{+})^{2}=a^{2}+2a^{+}a+a^{+}a^{+}+1 \]

     \	\( a^{2} \)	\( 2a^{+}a \)	\( (a^{+})^{2} \)	\( 1 \)
     \(a^{2}\)	\(a^{4}\)	\( 2a^{2}a^{+}a \)	\( a^{2}(a^{+})^{2} \)	\( a^{2} \)
     \( 2a^{+}a \)	\( 2a^{+}a^{3} \)	\( 4(a^{+}a)^{2} \)	\( 2a^{+}a(a^{+})^{2} \)	\( 2a^{+}a \)
     \( (a^{+})^{2} \)	\( (a^{+})^{2}a^{2} \)	\( 2(a^{+})^{3}a \)	\( (a^{+})^{4} \)	\( (a^{+})^{2} \)
     \( 1 \)	\( a^{2} \)	\( 2a^{+}a \)	\( (a^{+})^{2} \)	\( 1 \)

     对于\( H=\hbar \omega(a^{+}a+1);\quad i\dot{a}=[a,H]=\hbar \omega a;\quad a(t)=e^{-i\hbar \omega t}a(0);\quad [a,a^{+}]=1 \)
     \(i\dot{a}=\hbar \omega a+[a,(a^{+})^{4}]+[a,2a^{+}aa^{+}]+\cdots\),每一项会贡献不同频率的振动，例如\( e^{-i3\hbar\omega t} \)
     Rotating Wave近似(旋波),\( \dot{x}=-ax+f(t)x;\quad x=e^{-at}+\int_{0}^{t}f(\tau)d\tau x_{0} \),若\( f(t) \)为快速变化，则可略. \[ H=\hbar \omega(a^{+}a+\dfrac{1}{2})+4g_{eff}a^{+}a+4g_{eff}(a^{+}a)^{2}+g_{eff}((a^{+})^{2}a^{2}+a^{2}(a^{+})^{2}) \] \[ a^{2}(a^{+})^{2}\vert n \rangle=(n+2)(n+1)\vert n\rangle;\quad (a^{+})^{2}a^{2}\vert n\rangle =n( n-1 )\vert n\rangle;\Rightarrow (a^{2}(a^{+})^{2}+(a^{+})^{2}a^{2})\vert n\rangle=(n^{2}+2n+2)\vert n\rangle;\quad (a^{2}(a^{+})^{2}+(a^{+})^{2}a^{2})=(a^{+}a)^{2}+2a^{+}a+2 \]
     (2) \[ H=\dfrac{p^{2}}{2m}+\dfrac{1}{2}m\omega^{2}x^{2}+(B_{0}+B_{1}x)\sigma_{x}+\Omega\sigma_{z}=\hbar\omega (a^{+}a+\dfrac{1}{2})+B_{0}\sigma_{x}+\Omega\sigma_{z}+B_{1}x\sigma_{x} \] \( B_{1}x\sigma_{x}=B_{1}\sqrt{\dfrac{\hbar}{2m\omega}}(a^{+}\sigma^{-}+a\sigma^{+})+ \)反旋波；\( \sigma_{x}=\sigma_{+}+\sigma_{-};\quad \sigma_{+}=\dfrac{1}{2}(\sigma_{x}+i\sigma_{y});\quad \sigma_{-}=\dfrac{1}{2}(\sigma_{x}-i\sigma_{y}) \)
     数值计算：\( \vert \phi \rangle =\sum_{n\sigma}C_{n\sigma}\vert \varphi_{n}(x)\rangle \chi_{\sigma} \), 其中:\( \chi_{\downarrow}=\begin{eqnarray} \left( \begin{array}{cc} 0 \\ 1 \end{array} \right) ;\quad \chi_{\uparrow}=\left( \begin{array}{cc} 1 \\ 0 \end{array} \right) \end{eqnarray} \) \[ H=\vert \varphi_{n}(x)\rangle = \hbar \omega(n+1\dfrac{1}{2})\vert\varphi_{n}(x)\rangle \] 重新编号：\( \vert\varphi_{0\downarrow}\rangle\Rightarrow\vert \varphi_{1}\rangle;\quad \vert\varphi_{0\uparrow}\rangle\Rightarrow\vert \varphi_{2}\rangle;\quad \vert\varphi_{1\downarrow}\rangle\Rightarrow\vert \varphi_{3}\rangle;\cdots \) \( H\phi=E\phi;\quad \phi=\sum C_{n}\vert\varphi_{n}\rangle; \langle \varphi_{n}\vert \varphi_{m}\rangle=\delta_{nm}\rightarrow \sum_{n}C_{n}\langle \varphi_{m}\vert H\vert \varphi_{n}\rangle=EC_{m};\quad [H]_{mn}=\langle \varphi_{m}\vert H\vert \varphi_{n}\rangle [C],\quad [H][C]=E[C] \)
     二次量子化： \[ \vert\phi\rangle=\sum_{n\sigma}C_{n\sigma}\vert n\sigma\rangle;\quad a^{+}\vert n\sigma\rangle=\sqrt{n+1}\vert n+1,\sigma\rangle;\quad \sigma_{z}\vert n\sigma\rangle=\sigma\vert n\sigma\rangle;\quad \sigma^{+}\vert n\downarrow\rangle=\vert n\uparrow\rangle;\quad \sigma^{+}\vert n\uparrow\rangle=0 \] (3) \( H=\sum_{i}\dfrac{p_{i}^{2}}{2m}+\dfrac{k}{2}(x_{i+1}-x_{i})^{2}\Rightarrow H=\sum_{i}(\dfrac{p_{i}^{2}}{2m}+kx_{i}^{2})-\sum_{i}kx_{i}x_{i+1} \), 代入\( k=\dfrac{1}{2}m\omega^{2} \), \( H=\sum_{i}\hbar \omega(a_{i}^{+}a_{i}+\dfrac{1}{2})-\dfrac{1}{4}\hbar \omega\sum_{i}(a_{i}a_{i+1}+a_{i}a_{i+1}^{+}+a_{i}^{+}a_{i+1}+a_{i}^{+}a_{i+1}^{+}) \)
     可以得到：\( \omega_{k} = \hbar \omega(1-\cos k)\Rightarrow \hbar \omega \dfrac{k^{2}}{2} \)
     (4) LC电路： \[ H=\dfrac{L}{2}\dot{q}^{2}+\dfrac{q^{2}}{2C}\Leftrightarrow H=\dfrac{m}{2}\dot{x}^{2}+\dfrac{1}{2}m\omega^{2}x^{2}\] \[ L\sim m;\quad x\sim q;\quad RI\sim \eta \dot{x} \]
  2. 课堂笔记20_饶 pdf

• 12.2 周四

  1. 二次量子化 \[ x=\sqrt{\dfrac{\hbar}{2m\omega}}(a+a^{+});\quad p=i\sqrt{\dfrac{\hbar m\omega}{2}}(a^{+}-a) \] (a) 振子\(+\)非线性相互作用项； (b) Spin-harmonic oscillator; (c) 一维小球与弹簧组成的链； (d) LC电路
     旋波近似：忽略\( a^{+}\sigma^{+},a_{i}^{+}a_{j}^{+} \)，但注意在有些情况下不可忽略.
     本次课的参考文献：
     The Quantum Theory of the Emission and Absorption of Radiation pdf
     J. Schmalian (2016) Second Quantization
     (1) \( H(p_{i},q_{i}) \), Hamiltonian equation: \[ \dot{q}_{i}=\dfrac{\partial H}{\partial p_{i}}=\lbrace q_{i},H\rbrace;\quad \dot{p}_{i}=-\dfrac{\partial H}{\partial q_{i}}=\lbrace p_{i},H\rbrace \] \( i\hbar \dfrac{\partial }{\partial t}\psi=H\psi \), \(\psi\)用一组正交基展开：\(\psi=\sum_{n}c_{n}\vert \varphi_{n}\rangle\) \[ E=\langle \psi\vert H\vert \psi\rangle =\sum_{nm}c_{n}^{*}c_{m}\langle \varphi_{n}\vert H\vert \varphi_{m}\rangle=\sum_{nm}c_{n}^{*}c_{m}H_{nm} \] \[ i\hbar \sum_{n}\dot{c}_{n}\vert \varphi_{n}\rangle=\sum_{m}c_{m}H\vert \varphi_{m}\rangle\Rightarrow i\hbar \dot{c}_{n}=\sum_{m}c_{m}\langle \varphi_{n}\vert H\vert \varphi_{m}\rangle =\sum_{m}c_{m}H_{nm} \] 我们可以将\( c_{n}, c_{n}^{*} \)类比到\( q_{n}, p_{n} \) \[ i\hbar \dot{c}_{n}=\dfrac{\partial E}{\partial c_{n}^{*}},\quad i\hbar \dot{c}_{n}^{*}=-\dfrac{\partial E}{\partial c_{n}}\Rightarrow q_{n}\rightarrow c_{n},\quad p_{n}\rightarrow i\hbar c_{n}^{*}\Rightarrow \dot{c}_{n}=\dfrac{\partial E}{\partial (i\hbar c_{n}^{*})},\quad (i\hbar \dot{c}_{n}^{*})=-\dfrac{\partial E}{\partial c_{n}} \] 由\( [ q , p ]=i\hbar;\Rightarrow [c_{n},i\hbar c_{n}^{*}]=i\hbar\Rightarrow [c_{n},c_{n}^{*}]=1\)
     (2) \( \psi(x)=\sum_{n}c_{n}\varphi(x),\quad \psi^{+}(y)=\sum_{n}c_{n}^{*}\varphi_{n}^{*}(y) \) \[ [\psi (x),\psi^{+}(y)]=\sum_{nm}\varphi_{n}(x)\varphi_{m}^{+}(y)[c_{n},c_{m}^{*}]=\sum_{n}\langle x\vert\varphi_{n}\rangle\langle\varphi_{n}\vert y\rangle=\langle x\vert \sum_{n}\vert \varphi_{n}\rangle\langle \varphi_{n}\vert y\rangle=\langle x\vert y\rangle=\delta(x-y) \] \( [\psi(x),\psi(y)]=[\psi^{+}(x),\psi^{+}(y)]=0 \)
     注意：数学上以及早期物理上常用\( c_{n}^{*} \), 而如今物理上常用\( c_{n}^{+} \). (3) 粒子数表象：
     对于\( i\hbar \dot{c}_{n}=\dfrac{\partial E}{\partial c_{n}^{*}},\quad i\hbar \dot{c}_{n}^{*}=-\dfrac{\partial E}{\partial c_{n}} \), 令\( c=\sqrt{N}e^{i\theta},\quad c^{*}=\sqrt{N}e^{-i\theta} \)
     得到：\( \dot{N}=-\dfrac{\partial E}{\partial \theta},\quad \dot{\theta}=\dfrac{\partial E}{\partial N},\quad [\theta,N]=i\hbar \)
     不同的文献有不同的定义： \[ c=\sqrt{N\pm 1}e^{i\theta},\quad c^{+}=e^{-i\theta}\sqrt{N\pm 1} \] \[ c=\sqrt{N\pm 1}e^{-i\theta},\quad c^{+}=e^{i\theta}\sqrt{N\pm 1} \] \[ c=e^{i\theta}\sqrt{N},\quad c^{+}=\sqrt{N}e^{-i\theta} \] \[ c=e^{-i\theta}\sqrt{N},\quad c^{+}=\sqrt{N}e^{i\theta} \] \( c^{+}c=N,\quad [c,c^{+}]=1 \) \[ c\vert k\rangle =f(k)\vert k-1\rangle \Rightarrow (cc^{+}-c^{+}c)\vert k\rangle=\vert k\rangle\Rightarrow cc^{+}\vert k\rangle-c^{+}f(k)\vert k-1\rangle=\vert k\rangle\Rightarrow c^{+}\vert k\rangle \propto\vert k+1\rangle \]
     (4) 对于多体系统(F-D统计orB-E统计)：
     \( \vert n_{1},n_{2},\cdots,n_{L}\rangle \). eg:\( \vert n_{1},n_{2}\rangle=\vert n_{1}\rangle\otimes \vert n_{2}\rangle \) \[ Bose: \langle x,y\vert n_{1},n_{2}\rangle\propto \phi_{n_{1}}(x)\phi_{n_{2}}(y)+\phi_{n_{2}}(x)\phi_{n_{1}}(y);\quad Fermi: \phi_{n_{1}}(x)\phi_{n_{2}}(y)-\phi_{n_{2}}(x)\phi_{n_{1}}(y) \] (a) For Boson: \( a\vert n\rangle =\sqrt{n}\vert n-1 \rangle,\quad a^{+}\vert n\rangle =\sqrt{n+1}\vert n+1\rangle\) \[ a_{p}\vert \lbrace n\rbrace\rangle =\sqrt{n_{p}}\vert n_{1}, n_{2},\cdots,n_{p}-1,\cdots\rangle;\quad a_{p}^{+}\vert \lbrace n\rbrace\rangle =\sqrt{n_{p}+1}\vert n_{1},n_{2},\cdots,n_{p}+1,\cdots\rangle;\quad a_{q}^{+}a_{p}\vert \lbrace n\rbrace\rangle=\sqrt{n_{p}}\sqrt{n_{q}+1}\vert n_{1},\cdots,(n_{p}-1),\cdots,(n_{q}+1),\cdots\rangle;\quad a_{p}a_{q}^{+}\vert \lbrace n\rbrace\rangle=\sqrt{n_{p}}\sqrt{n_{q}+1}\vert n_{1},\cdots,(n_{p}-1),\cdots,(n_{q}+1),\cdots\rangle\] (b) For Fermion: \( a_{p}^{+}a_{q}+a_{q}a_{p}^{+}=0 \), if\( q\neq p \). \[ a\vert n\rangle=n\vert n-1\rangle,\quad a^{+}\vert n\rangle=(1-n)\vert n+1 \rangle;\quad a_{p}\vert \lbrace n\rbrace\rangle =(-1)^{\nu_{p}}n_{p}\vert n_{1},\cdots,(n_{p}-1),\cdots\rangle,\quad a_{p}^{+}\vert \lbrace n\rbrace\rangle=(-1)^{\nu_{p}}(1-n_{p})\vert n_{1},\cdots,(n_{p}+1),\cdots\rangle,\quad \nu_{p}=\sum_{i=1}^{p-1}n_{p} \]
  2. 课堂笔记21_饶 pdf

• 12.7 周二

  1. 总结与回顾
     (1) Born:\( [ q , p ] =i\hbar \); Dirac: 任意共轭量都有类似关系.
     (2) \(c, c^{+}\)算符, \( \vert n\rangle \)粒子数. 这样做的好处：(a) 简单的物理语言，无复杂语言，量子化. (b) 天然是多体物理的描述语言.
     (3) 描述： \[ Fermion: (c_{1}^{\dagger})^{n_{1}}(c_{2}^{\dagger})^{n_{2}}\cdots (c_{L}^{\dagger})^{n_{L}}\cdots \vert 0\rangle\triangleq \vert n_{1}n_{2}\cdots n_{L}\cdots\rangle,\quad n_{i}=0,1; \quad Boson:(c_{1}^{\dagger})^{n_{1}}(c_{2}^{\dagger})^{n_{2}}\cdots (c_{L}^{\dagger})^{n_{L}}\cdots \vert 0\rangle\triangleq \vert n_{1}n_{2}\cdots n_{L}\cdots\rangle,\quad n_{i}=0,1,2,3,\cdots; \] Fermi Sphere:\( \vert G\rangle =\Pi_{\vert k\vert\leqslant k_{F}}c_{k\sigma}^{\dagger}\vert 0\rangle \)
     (4) 操作：Boson: \( a_{p}^{\dagger}a_{q}\vert \phi\rangle=a_{q}a_{p}^{\dagger}\vert \phi\rangle \); Fermion: \( a_{p}^{+}a_{q}\vert \phi\rangle=-a_{q}a_{p}^{\dagger}\vert \phi\rangle \)
     \[ a_{p}\vert n_{1},n_{2},\cdots,n_{L},\cdots\rangle=\sqrt{n_{p}}\vert n_{1},n_{2},\cdots,n_{p}-1,\cdots\rangle;\quad Fermion: a_{p}\vert n_{1},n_{2},\cdots,n_{p},\cdots\rangle=n_{p}(-1)^{\sum_{i=1}^{n-1}n_{i}}\vert n_{1},n_{2},\cdots,n_{p}-1,\cdots\rangle \] \[ \vert \phi\rangle=a_{p}^{\dagger}a_{q}\vert G\rangle,\quad \vert \phi '\rangle=a_{s}^{\dagger}a_{t}\vert G\rangle\Rightarrow \langle \phi\vert \phi '\rangle=\langle G\vert a_{q}^{\dagger}a_{p}a_{s}^{\dagger}a_{t}\vert G\rangle=\delta_{ps}\delta_{qt} \]
  2. 对任意\( H \)的量子化
     (1) \[ F=\langle \psi \vert H\vert \psi\rangle =c_{i}^{\dagger}c_{j}\langle i\vert H\vert j\rangle =c_{i}^{*}c_{j}H_{ij};\quad \vert \psi\rangle =\sum_{i}c_{i}\vert i\rangle ;\quad \dot{c}_{i}=\dfrac{\partial F}{\partial (i\hbar c_{i}^{*})},\quad i\hbar \dot{c}_{i}^{*}=-\dfrac{\partial F}{\partial c_{i}}\Rightarrow q=c_{i},p=i\hbar c_{i}^{*};\quad [q,p]=i\hbar,[c_{i},c_{i}^{*}]=1 \] 其中：\( [\psi(x),\psi^{\dagger}(y)]=\delta(x-y) \)与\( [c_{i},c_{i}^{*}]=1 \)等价. (2) 量子化单粒子： \[ H\vert \varphi_{n}\rangle =\epsilon_{n}\vert \varphi_{n}\rangle\Rightarrow F=\int \psi^{\dagger}H\psi dx =\langle \psi\vert H\vert \psi\rangle =\sum_{n}c_{n}^{+}c_{n}\langle \varphi_{n}\vert H\vert \varphi_{n}\rangle =\sum_{n}\epsilon_{n}c_{n}^{\dagger}c_{n}; \] 其中\( \psi=\sum_{n}c_{n}\vert \varphi_{n}\rangle \).
     (3) 周期性模型：\( H=\dfrac{p^{2}}{2m}+V\cos(kx) \), \( \psi=\sum_{n}c_{n}\vert \varphi_{n}\rangle \)
     \( \vert \varphi_{n}\rangle \)是第\( n \)个well里的基态，wannier函数比对应的基态在尾部衰减的更慢，且有振荡.
     紧束缚模型: \[ F=\sum_{n,m}c_{n}^{*}c_{m}\langle \varphi_{n}\vert H\vert \varphi_{n}\rangle=\sum_{n}-\mu c_{n}^{\dagger}c_{n}-t(c_{n}^{\dagger}c_{n+1}+h.c ) \] (4) \( H=-t\sum_{n}(c_{n}^{\dagger}c_{n+1}+h.c)-\mu c_{n}^{\dagger}c_{n} \)
     Fourier transformation: \[ c_{n}=\dfrac{1}{\sqrt{N}}\sum_{k}e^{ikn}c_{k} \] 物理意义：实空间\( \leftrightarrow \)k空间; 扩展\( \leftrightarrow \)局域. \[ \Rightarrow H=[ -t \sum_{k,q}c_{k}^{\dagger}c_{q}\sum_{n}e^{-ikn+iq(n+1)}+h.c-\mu\sum_{k,q}c_{k}^{\dagger}c_{q}\sum_{n}e^{-ikn+iqn} ]\dfrac{1}{N} \] \[ \Rightarrow H= -t\sum_{k}c_{k}^{\dagger}c_{k}e^{ik}+h.c - \mu\sum_{k}c_{k}^{\dagger}c_{k}=\sum_{k}(-2t\cos{k}-\mu)c_{k}^{\dagger}c_{k} \]
  3. 作业：Anderson localization \[ H=-t\sum_{n}(c_{n}^{\dagger}c_{n+1}+c_{n+1}^{\dagger}c_{n})+\sum_{n}\epsilon_{n}c_{n}^{\dagger}c_{n} \] \( \epsilon_{n}= \)(1) White noise, \( \epsilon_{n}\in U[ -W , W ] \)
     (2) \( \epsilon_{n}=V\cos(\alpha n \pi),\quad \alpha=\dfrac{\sqrt{5}-1}{2} \)
     (3) \( \epsilon_{n}=V\cos(\alpha \vert n\vert ^{\nu}),\quad \nu=0.1 \sim 0.5 \), \( \alpha \)自选.
     ddl: 12月26日.
  4. 多体相互作用 \[ \vert \vec{r},\vec{r}'\rangle =\psi^{\dagger}(\vec{r})\psi(\vec{r}')\vert 0\rangle;\quad [\psi(x),\psi^{\dagger}(y)]=\delta(x-y) ;\quad [c_{i},c_{j}^{+}]=\delta_{ij}\] \[ \vert\psi_{\alpha \alpha'}\rangle =a_{\alpha}^{\dagger}a_{\alpha'}^{\dagger}\vert 0\rangle;\quad \langle r,r'\vert \psi_{\alpha,\alpha'}\rangle=\dfrac{1}{\sqrt{2}}(\varphi_{\alpha}(r)\varphi_{\alpha'}(r')+\varphi_{\alpha}(r')\varphi_{\alpha'}(r)) \] \[ \langle \psi(x)\vert H\vert \psi(x)\rangle=\int \psi^{\dagger}(x)H\psi(x)dx \] \[ \int \psi_{\alpha\alpha'}^{\dagger}(x,y)V(x-y)\psi_{\alpha\alpha'}(x,y)dxdy \] 三体： \[ \int \psi_{\alpha\alpha'\alpha''}^{\dagger}(x,y,z)V(x,y,z)\psi_{\alpha\alpha'\alpha''}(x,y,z)dxdydz \]
  5. 课堂笔记22_饶 pdf

• 12.9 周四

  1. 二次量子化
     (1) Dirac:任意对偶量(正则动量与正则坐标)都可以量子化.
     (2) 单粒子：\( H=\langle \psi(x)\vert H\vert \psi(x)\rangle=\sum_{n,m}c_{n}^{*}c_{m}\langle \varphi_{n}\vert H\vert \varphi_{m}\rangle \); \( H=\int \psi^{\dagger}(x)H\psi(x)dx\Rightarrow \sum_{n,m}c_{n}^{\dagger}c_{m}\langle \varphi_{n}\vert H \vert \varphi_{m}\rangle \)
     (3) \( \psi^{\dagger}(x) \)在\( x \)点产生一个粒子.
     (4) 对于\( H \)，单粒子模型: \[ H\vert \varphi_{\alpha}\rangle=\epsilon_{\alpha}\vert \varphi_{\alpha}\rangle\Rightarrow \sum_{\alpha}H\vert \varphi_{\alpha}\rangle\langle\varphi_{\alpha}\vert=\sum_{\alpha}\epsilon_{\alpha}\vert \varphi_{\alpha}\rangle \langle\varphi_{\alpha}\vert;\quad \sum_{\alpha}\vert \varphi_{\alpha}\rangle\langle\varphi_{\alpha}\vert =1\Rightarrow H=\sum_{\alpha}\epsilon_{\alpha}\vert \varphi_{\alpha}\rangle\langle\varphi_{\alpha}\vert \] 令\( P_{\alpha\beta}=\vert \alpha\rangle \langle\beta\vert \), 则\( H=\sum_{\alpha}\epsilon_{\alpha}P_{\alpha\alpha} \). \( \vert \alpha\rangle\langle \beta\vert \beta\rangle=\vert \alpha\rangle\Rightarrow a_{\alpha}^{\dagger}a_{\beta}\vert \beta\rangle=a_{\alpha}^{\dagger}a_{\beta}\vert 0_{\alpha}1_{\beta}\rangle=a_{\alpha}^{\dagger}\vert 0_{\alpha}0_{\beta}\rangle=\vert \alpha\rangle \)
     采用一般基矢：\( \vert n\rangle\langle n\vert \) \[ H=\sum_{n,m}\vert n\rangle\langle n\vert H\vert m\rangle\langle m\vert=\sum_{n,m}\langle n\vert H\vert m\rangle\vert n\rangle\langle m\vert =\sum_{n.m}\langle n\vert H\vert m\rangle c_{n}^{\dagger}c_{m} \] 对于单粒子情况：(a) 完备基; (b) 对应关系： \( c_{n}^{\dagger}c_{m}\Rightarrow \vert n\rangle\langle m\vert;\quad c_{n}^{\dagger}\vert 0\rangle=\vert n\rangle \)
     (5) 有相互作用，两体：
     空间：\( H_{1}\otimes H_{2} \); 基：\( \sum_{n,m}\vert n,m\rangle\langle n,m\vert =1 \); 考虑相互作用\( V( x - y ) \): \[ V=IV( x - y )I=\sum \vert \alpha,\beta\rangle\langle \alpha,\beta\vert V\vert \gamma,\delta\rangle\langle\gamma,\delta\vert=\sum \langle \alpha,\beta\vert V\vert \gamma,\delta\rangle \vert \alpha,\beta\rangle\langle\gamma,\delta\vert=\sum V_{\alpha\beta,\gamma\delta}\vert \alpha,\beta\rangle\langle\gamma,\delta\vert=\sum V_{\alpha\beta,\gamma\delta}a_{\alpha}^{\dagger}a_{\beta}^{\dagger}a_{\gamma}a_{\delta} \] \[ H=\int \psi^{\dagger}(x)H\psi(x)dx, V=\int \psi^{\dagger}(x)\psi^{\dagger}(y)V( x - y )\psi(y)\psi(x)dxdy;\quad \psi(x)=\sum_{\alpha}a_{\alpha}\varphi_{\alpha}(x)\Rightarrow V=\int a_{\alpha}^{\dagger}a_{\beta}^{\dagger}a_{\gamma}a_{\delta}\langle \varphi_{\alpha}(x)\varphi_{\beta}(y)\vert V( x - y )\vert \varphi_{\gamma}(y)\varphi_{\delta}(x)\rangle=\sum a_{\alpha}^{\dagger}a_{\beta}^{\dagger}a_{\gamma}a_{\delta}\langle \alpha\beta\vert V\vert \gamma\delta\rangle \] (6) \( \alpha=\sum_{n}\vert n\rangle \langle n\vert \alpha\rangle=\sum_{n}\langle n\vert \alpha\rangle \vert n\rangle \), 令\( \alpha=x \), \( c_{x}^{\dagger}=\sum_{n}\langle x\vert x\rangle c_{n}^{\dagger} \), 则\( \psi^{\dagger}(x)\vert 0\rangle=\sum_{n}\langle n\vert x\rangle c_{n}^{\dagger}\vert 0\rangle=\sum_{n}\langle n \vert x\rangle\vert n\rangle=\vert x\rangle \)
     (7) 二次量子化的标准方法：
     \( H=h(x)+V( x - y) \): \[ H=\int \psi^{\dagger}(x)h(x)\psi(x)+\int \psi^{\dagger}(x)\psi^{\dagger}(y)V( x -y )\psi(y)\psi(x)dxdy;\quad \psi(x)=\sum_{n}c_{n}\varphi_{n}(x);\quad \psi^{\dagger}(x)\psi^{\dagger}(y)=\sum_{n,m}\varphi_{n}(x)\varphi_{m}(y)c_{n}^{\dagger}c_{m}^{\dagger}\vert 0\rangle \] \[ H=\sum_{n,m}c_{n}^{\dagger}c_{m}\langle n\vert H\vert m\rangle+\sum c_{n}^{\dagger}c_{m}^{\dagger}c_{k}c_{l}\langle nm\vert V\vert kl\rangle \] 取\( h \)的本征态：\( \Rightarrow \sum\epsilon_{n}c_{n}^{\dagger}c_{n}+Interaction \)
  2. 平均场理论: Ising Model
     (1) \( H=h(\sigma_{1}+\sigma_{2})+J\sigma_{1}\sigma_{2} \), \( \sigma_{1},\sigma_{2}=\pm 1 \)
     \( x=\delta x+\overline{x};y=\delta y+\overline{y} \), \( xy=\overline{x}\overline{y}+\overline{x}\delta y+\overline{y}\delta x+\delta x\delta y \)
     忽略小量：\( \Rightarrow xy=\overline{x}y+\overline{y}x-\overline{x}\overline{y}\)
     \[ H=h(\sigma_{1}+\sigma_{2})+Jm(\sigma_{1}+\sigma_{2})-m^{2}J;\quad m=\langle \sigma_{1}\rangle=\langle\sigma_{2}\rangle \] \[ H=h(\sigma_{1}+\sigma_{2}+\sigma_{3})+J(\sigma_{1}\sigma_{2}+\sigma_{2}\sigma_{3}+\sigma_{3}\sigma_{1})+k\sigma_{1}\sigma_{2}\sigma_{3}=(h+2mJ+km^{2})(\sigma_{1}+\sigma_{2}+\sigma_{3})-3m^{2}J-2km^{3} \] (2) 单spin: \[ H=A\sigma;\quad Z=e^{-A\beta}+e^{A\beta}=e^{-\beta F}\Rightarrow F=-\dfrac{1}{\beta}\ln(e^{A\beta}+e^{-A\beta})\Rightarrow F=-m^{2}J-\dfrac{2}{\beta}\ln[e^{\beta(h+mJ)}+e^{-\beta(h+mJ)}];\quad \dfrac{\partial F}{\partial m}=0 \] \[ P_{\uparrow}=e^{-\beta A}/Z,P_{\downarrow}=e^{\beta A}/Z;\quad m=P_{\uparrow}(+1)+P_{\downarrow}(-1)=\dfrac{e^{-\beta(h+mJ)}-e^{\beta(h+mJ)}}{e^{-\beta(h+mJ)}+e^{\beta(h+mJ)}} \]
  3. 作业：2d Ising Model
     (1) 正方形格子：\( H=-J\sum\sigma_{i}\sigma_{j}+h\sum_{i}\sigma_{i}+k\sum\sigma_{i}\sigma_{j}\sigma_{k}\sigma_{l} \)
     (2) 六边形格子：\( H=-\sum J\sigma_{i}\sigma_{j}+h\sum_{i}\sigma_{i} \)
     平均场理论，求(a)\( \dfrac{\partial F}{\partial m}=0 \);(b)相图.
     详细说明见课堂笔记，ddl:12月26日.
  4. 课堂笔记23_饶 pdf

• 12.14 周二

  1. 平均场方法
     量子力学，多体，\( d^{L} \),指数级，计算量过大.
     (a) 平均场：\( d^{L}\rightarrow d \), 多体\( \rightarrow \)单体.
     (b) 精确对角化， Lanczos 方法, MC计算; 有限计算\( + \)外推.
     (1) 上节课讲过的平均场方法：
     (a) \( x^{3}=2x^{2}+5,\quad x=m,\quad m^{2}x=2x^{2}+5,\quad x=m=f(m) \).
     (b) 2 spin: \( H=h(\sigma_{1}+\sigma_{2})+J\sigma_{1}\sigma_{2}=h(\sigma_{1}+\sigma_{2})+Jm(\sigma_{1}+\sigma_{2})-Jm^{2} \).
     (c) \( H=h(\sigma_{1}+\sigma_{2}+\sigma_{3})+J(\sigma_{1}\sigma_{2}+\sigma_{2}\sigma_{3}+\sigma_{3}\sigma_{1})=h(\sigma_{1}+\sigma_{2}+\sigma_{3})+2Jm(\sigma_{1}+\sigma_{2}+\sigma_{3})-3Jm^{2};\quad k\sigma_{1}\sigma_{2}\sigma_{3}=km^{2}(\sigma_{1}+\sigma_{2}+\sigma_{3})-2km^{3} \).
     (d) \( H=h(\sigma_{1}+\sigma_{2}+\sigma_{3}+\sigma_{4})+J(\sigma_{1}\sigma_{2}+\sigma_{2}\sigma_{3}+\sigma_{3}\sigma_{4}+\sigma_{4}\sigma_{1})+k\sigma_{1}\sigma_{2}\sigma_{3}\sigma_{4} \).
  2. Bose-Hubbard Model 平均场
     (1) \( H=\dfrac{p^{2}}{2m}+U(\vec{x})+g\delta(\vec{x}-\vec{y}) \), \( \psi=\sum_{i}c_{i}\varphi_{i} \), \( \varphi_{i}(x)=W(\vec{x}-\vec{R}_{i}) \). \[ H_{0}=\int \psi^{\dagger}H_{0}\psi dx=\sum_{ij}c_{i}^{\dagger}c_{j}\int \varphi_{i}^{*}[\dfrac{p^{2}}{2m}+U(x)]\varphi_{i}dx \] (a) \( i=j\Rightarrow -\mu c_{i}^{\dagger}c_{i} \); (b) \( i,j \)近邻, \( -tc_{i}^{\dagger}c_{j} \). \[ \int \psi^{\dagger}(x)\psi^{\dagger}(y)g\delta(x-y)\psi(x)\psi(y)dxdy=c_{i}^{\dagger}c_{j}^{\dagger}c_{k}c_{l}g\int \varphi_{i}^{*}(x) \varphi_{j}^{*}(y) \delta(x-y) \varphi_{k}(y)\varphi_{l}(x)dxdy,\quad i=j=k=l,\quad c_{i}^{\dagger}c_{i}^{\dagger}c_{i}c_{i}=n_{i}(n_{i}-1)\Rightarrow H=-t\sum_{\langle ij\rangle}c_{i}^{\dagger}c_{j}-\mu\sum_{i}n_{i}+U\sum_{i}n_{i}(n_{i}-1) \] (2) \( t=0 \)时, 本征态\(\vert n_{1},n_{2},\cdots,n_{L}\rangle\), \( E=-\mu\sum_{i}n_{i}+U\sum_{i}n_{i}(n_{i}-1) \), \( E/L=-\mu n+Un(n-1) \). \[ -\mu n +Un(n-1)\leq -\mu (n+1)+U(n+1)n,\quad -\mu n+Un(n-1)\leq -\mu(n-1)+U(n-1)(n-2),\quad n\in [ \dfrac{\mu}{2U},\dfrac{\mu}{2U}+1 ] \] (3) \( t\rightarrow \infty, \quad H=-t\sum_{\langle ij\rangle}c_{i}^{\dagger}c_{j} \).
     (4) 平均场：\(-tc_{i}^{\dagger}c_{j}\Rightarrow -t(\Delta^{*}c_{i}+c_{j}^{\dagger}\Delta-\vert \Delta\vert^{2})\).
     \( Z \)为配位数： \[ H_{i}=-\mu n_{i}+Un_{i}(n_{i}-1)-t[ Z(\Delta^{*}c_{i}+\Delta c_{i}^{\dagger})-Z\vert \Delta \vert^{2} ];\quad H=-\mu n+Un(n-1)-tZ(\Delta^{*}c+\Delta c^{\dagger})+tZ\vert \Delta\vert^{2} \] \[ E_{g}=E_{g}^{(0)}+E_{g}^{(1)}+E_{g}^{(2)} \] \[ E_{g}^{(0)}=-\mu n+Un(n-1)+tZ\vert \Delta\vert^{2},(t>0);\quad V'=-tZ(\Delta^{*}c+\Delta c^{\dagger});\quad E_{g}^{(1)}\Rightarrow\langle n\vert V'\vert n\rangle =0\] \[ E_{g}^{(2)}\sim -t^{2}Z^{2}\vert \Delta\vert^{2}f(n,\mu,U) \Rightarrow E_{g}\sim (tZ-t^{2}Z^{2}f)\vert \Delta\vert^{2}=c\vert \Delta\vert^{2};\quad E_{g}^{(4)}=A^{(4)}\vert\Delta\vert^{4} \] \( c>0,\vert \Delta\vert =0;\quad c<0,\vert \Delta\vert \neq 0 \). \( E_{g}=E_{g}^{(0)}+a\vert \Delta\vert^{2}+b\vert \Delta\vert^{4} \), Landau二级相变理论.
     相边界： \( 1-tZf(\mu,n,U)=0 \)
  3. 课堂笔记24_饶 pdf

• 12.16 周四

  1. 平均场
     (1) spin model: 在平均场中，不同维度只会导致配位数\( Z \)的差别. 1d Ising Model 无相变，但使用平均场会导致有.
     (2) Bose-Hubbard Model: 有限温平均场不能得到完整的相图.
     平均场的优点：(a) 简单； (b) 可解性，图像简单； (c) 可以做许多种不同的平均场.
  2. 精确对角化
     不同方法的差别：Hilbert space 不同. \(\psi=\sum_{i}c_{i}\vert \varphi_{i}\rangle\). 优点：wave function; 缺点：只能算较小的系统.
     (1) \(N=1\):
     (a) Spin Model (N=1):\( H=\vec{B}\cdot\vec{\sigma} \)
     (b) Bose-Hubbard Model: \( H=t(c+c^{\dagger})+\omega c^{\dagger}c+Un(n-1) \); \( n=c^{\dagger}c \).
     (c) Fermi-Hubbard Model: \( H=\omega f^{\dagger}f+t(f^{\dagger}+f)+Un(n-1) \); \( n=f^{\dagger}f \); 其中\( Un(n-1) \)一项一般没有贡献.
     Hilbert Space: \[ (a) \vec{\sigma}=(\sigma_{x},\sigma_{y},\sigma_{z}),K=\lbrace \vert \uparrow\rangle,\vert \downarrow\rangle\rbrace;\quad (b) K=\lbrace \vert 0\rangle,\vert 1\rangle,\vert 2\rangle,\cdots\rbrace;\quad (c) K=\lbrace \vert 0\rangle,\vert 1\rangle\rbrace \]
     (a) \( \langle\uparrow\vert H\vert\uparrow\rangle=\langle\uparrow\vert B_{z}\sigma_{z}+B_{x}\sigma_{x}+B_{y}\sigma_{y}\vert \uparrow\rangle=B_{z} \); \( \langle\downarrow\vert H\vert\downarrow\rangle=-B_{z} \); \( \langle\downarrow\vert H\vert\uparrow\rangle=B_{x}+iB_{y} \); \( \langle\uparrow\vert H\vert\downarrow\rangle=B_{x}-iB_{y} \). \[ H= \begin{eqnarray} \left( \begin{array}{cc} B_{z} & B_{x}-iB_{y} \\ B_{x}+iB_{y} & -B_{z} \end{array} \right) \end{eqnarray} \] 本征值\( E=\pm\vert \vec{B}\vert \).
     (b) Bose-Hubbard Model:\( H=T+V \), \(T\)为非对角部分, \( V \)为对角部分. \[ H_{ii}=\langle i\vert V\vert i\rangle= \omega i+ U i(i-1);\quad \langle i\vert T\vert i+1\rangle=\langle i+1\vert T\vert i\rangle=\sqrt{i+1}t,\quad others vanish. \] (c) Fermi-Hubbard: \[ H= \begin{eqnarray} \left( \begin{array}{cc} 0 & t \\ t & \omega \end{array} \right) \end{eqnarray} \] (2) \( N=2 \):
     (a) \( H=\vec{B}\cdot (\vec{\sigma}_{1}+\vec{\sigma}_{2})+J\sigma_{1}\cdot\sigma_{2} \).
     (b) \( H=t(c_{1}^{\dagger}c_{2}+h.c)+\omega (n_{1}+n_{2})+Un_{1}(n_{1}-1)+Un_{2}(n_{2}-1)+Vn_{1}n_{2} \).
     (c) \( H=t(f_{1}^{\dagger}f_{2}+f_{2}^{\dagger}f_{1})+\omega(n_{1}+n_{2})+Vn_{1}n_{2} \)
     Hilbert Space: \[ (a) K=\lbrace \vert \uparrow\uparrow\rangle,\vert \uparrow\downarrow\rangle,\vert \downarrow\uparrow\rangle,\vert \downarrow\downarrow\rangle\rbrace;\quad K=\lbrace\vert n_{1}n_{2}\rangle,0\leq n_{i}\leq N_{c}\rbrace;\quad K=\lbrace\vert n_{1}n_{2}\rangle,n_{i}=\lbrace 0,1\rbrace\rbrace \]
     (a) 对角元：\( \langle\uparrow\uparrow\vert H\vert \uparrow\uparrow\rangle=2B_{z}+J;\langle\uparrow\uparrow\vert H\vert \downarrow\downarrow\rangle=-2B_{z}+J;\langle\uparrow\downarrow\vert H\vert \uparrow\downarrow\rangle=-J;\langle\downarrow\uparrow\vert H\vert \downarrow\uparrow\rangle=-J \).
     非对角元： \( B_{x}(\sigma_{1}^{x}+\sigma_{2}^{x})+B_{y}(\sigma_{1}^{y}+\sigma_{2}^{y})+J(\sigma_{1}^{x}\sigma_{2}^{x}+\sigma_{1}^{y}\sigma_{2}^{y}) \); \( \langle\uparrow\downarrow\vert H\vert \uparrow\uparrow\rangle \); \(\langle\uparrow\uparrow\vert H\vert \downarrow\downarrow\rangle\); \( \langle\uparrow\uparrow\vert H\vert \downarrow\uparrow\rangle \); \( \langle\uparrow\downarrow\vert H\vert \downarrow\uparrow\rangle \).
     (b) Bose-Hubbard: 需要通过索引表将\(i,j\)mapping到\( k \); 对角部分: \( H_{kk}=\omega(i+j)+Ui(i-1)+Uj(j-1)+Vij \).
     非对角部分：\( t\langle nm\vert c_{1}^{\dagger}c_{2}+c_{2}^{\dagger}c_{1}\vert n'm'\rangle;\quad \sqrt{n'}\sqrt{m'+1}\langle nm\vert n'-1,m'+1\rangle=\sqrt{n'}\sqrt{m'+1}\delta_{n,n'-1}\delta_{m,m'+1};\quad \sqrt{n'+1}\sqrt{m'}\langle nm\vert n'+1,m'-1\rangle=\sqrt{n'+1}\sqrt{m'}\delta_{n,n'+1}\delta_{m,m'-1} \).
     (3) \( N=10 \):
     (a) \( H=\vec{B}\cdot (\sum_{i=1}^{N}\vec{\sigma}_{i})+J\sum_{i}\vec{\sigma}_{i}\cdot\vec{\sigma}_{i+1} \)
     (b) \( H=t\sum_{i}(c_{i}^{\dagger}c_{i+1}+h.c)+\omega n_{i}+Un_{i}(n_{i}-1) \).
     (c) \( H=t\sum_{i}(f_{i}^{\dagger}f_{i+1}+h.c)+\omega n_{i}+Vn_{i}n_{i+1} \).
     \[ Hilbert Space: (a) K=\lbrace \vert\downarrow\downarrow,\cdots,\downarrow\rangle,\vert\downarrow\downarrow,\cdots,\downarrow\uparrow\rangle,\cdots\rbrace,2^{10}=1024;\quad (b) K=\lbrace \vert 0,\cdots,0\rangle,\cdots\rbrace,(N_{c}+1)^{d};\quad (c) K=\lbrace \vert 0,\cdots,0\rangle,\vert 0,\cdots,1\rangle,\cdots\rbrace,2^{10}=1024 \] 对于spin model，常考虑总自旋为\(0\)的情况, 对于Fermi-Hubbard model，粒子数守恒，由此可以简化计算. \( K=K_{0}\oplus K_{1}\oplus K_{2}\oplus\cdots \oplus K_{L} \),确定粒子数之后只需要计算一个子空间，填充因子:\( n=N/L \).
     eg: for\( L=5,N=3 \), \( C_{5}^{1}+2C_{5}^{2}+C_{5}^{3}=35 \)
  3. 作业 \( L=5,N=3 \), Bose-Hubbard model本征值. ddl: 1月3日
  4. 课堂笔记25_饶 pdf

• 12.21 周二

  1. 严格对角化
     (1) Bose-Hubbard Model: \( H=-t\sum_{\langle ij\rangle}a_{i}^{\dagger}a_{j}-\mu\sum_{i}n_{i}+Un_{i}(n_{i}-1) \)
     (2) Fermi-Hubbard Model: \( H=-t\sum_{\langle ij\rangle}f_{i}^{\dagger}f_{j}-\mu\sum_{i}n_{i}+\sum_{\langle ij\rangle}Vn_{i}n_{j},\quad n_{i}=f_{i}^{\dagger}f_{i} \)
     (3) Spin Model: \( H=J\sum_{\langle ij\rangle}\vec{S}_{i}\cdot\vec{S}_{j}+h\sum_{i}S_{i}^{z} \), \( \vec{S} \): pauli operator.
     (4) 2 sites: \[ H=-t(a_{1}^{\dagger}a_{2}+h.c)-\mu(n_{1}+n_{2})+Un_{1}(n_{1}-1)+Un_{2}(n_{2}-1) \] \( L=2,N=3 \), \( K=\lbrace \vert 0,3\rangle,\vert 1,2\rangle,\vert 2,1\rangle,\vert 3,0\rangle\rbrace \). \( L=2,N=5 \), \( K=\lbrace \vert 0,5\rangle,\vert 1,4\rangle,\vert 2,3\rangle,\vert 3,2\rangle,\vert 4,1\rangle,\vert 5,0\rangle\rbrace \).
     \( \langle ij\vert H\vert i'j'\rangle \), 对角： \( i=i',j=j' \), 引入\(\delta_{ij}\); 非对角：\( \langle ij\vert a_{1}^{\dagger}a_{2}+h.c\vert i'j'\rangle\Rightarrow \sqrt{j'}\sqrt{i'+1}\delta_{i,i'+1}\delta_{j,j'+1} \)
     \( N=3,L=3 \), \( \lbrace \vert 300\rangle,\vert 030\rangle,\vert 003\rangle,\vert 210\rangle,\vert 201\rangle\cdots\vert 111\rangle\rbrace \)
     非对角：\( H_{ij}=-t\langle i\vert \hat{T}\vert j\rangle=-t\langle i_{1}\cdots i_{L}\vert \hat{T}\vert j_{1}\cdots j_{L}\rangle=-t\sum_{k}\langle i_{1}\cdots i_{L}\vert (a_{1}^{\dagger}a_{2}+h.c)\vert j_{1}\cdots j_{L}\rangle\Rightarrow [ \Pi_{l\neq k,k+1}\delta(i_{l},j_{l}) ]\langle i_{k},i_{k+1}\vert a_{k}^{\dagger}a_{k+1}+h.c\vert j_{k}j_{k+1}\rangle \).
     技巧：(a) 索引表；(b) 分类，比如对角非对角; (c) \( \delta \)函数.
     (5) \( \vert n\rangle \): 对于谐振子\( \vert n\rangle=(1/\sqrt{n!})(a^{\dagger})^{n}\vert 0\rangle \), 第\(n\)能级, Bose-Hubbard model: \(n\)个粒子占据基态.
     (6) Fermion: \( N=1,L=2;K=\lbrace 10\rangle,\vert 01\rangle\rbrace \); \( N=2,L=2,K=\lbrace \vert 1,1\rangle\rbrace \); \( N=3,L=2,K=\lbrace \phi\rbrace \); \( N=1,L=3,K=\lbrace \vert 100\rangle,\vert 010\rangle,\vert 001\rangle\rbrace \); \( N=2,L=3,K=\lbrace \vert 110\rangle,\vert 011\rangle,\vert 101\rangle\rbrace \);\( N=3,L=3,K=\lbrace \vert 111\rangle\rbrace \); \(N=4,L=3,K=\lbrace \phi\rbrace\).
     (7) \( N \)个粒子, \( L \)个格点. \( \vert i_{1},i_{2},\cdots,i_{L}\rangle \), \( \sum_{k}i_{k}=N,i_{k}\in\lbrace 0,1\rbrace \). \[ H_{ii}(\mu,V)=\langle i_{1},\cdots,i_{L}\vert \hat{\mu}+\hat{V}\vert i_{1},\cdots,i_{L}\rangle;\quad -\mu\sum_{i}n_{i}\rightarrow \hat{\mu},Vn_{i}n_{i+1}\rightarrow \hat{V}\] \[ \langle i_{1},i_{2},\cdots,i_{L}\vert \hat{T}\vert j_{1},j_{2},\cdots,j_{L}\rangle \Rightarrow -t\sum_{K}\langle i_{1},i_{2},\cdots,i_{L}\vert f_{k}^{\dagger} f_{k+1}\vert j_{1},j_{2},\cdots,j_{L}\rangle \] \[ f_{k}\vert i_{1},\cdots,i_{L}\rangle=i_{k}\vert i_{1},\cdots,i_{k}-1,\cdots,i_{L}\rangle (-1)^{\sum_{j=1}^{k-1}i_{j}};\quad f_{k+1}^{\dagger}f_{k}\vert i_{1},\cdots,i_{L}\rangle=i_{k}(i_{k+1}-1)(-1)^{i_{k}-1}\vert i_{1},\cdots,i_{k}-1,i_{k+1}+1,\cdots,i_{L}\rangle \] 长程相互作用: eg: \(\langle i_{1},\cdots,i_{L}\vert f_{k+4}^{\dagger}f_{k}\vert j_{1},\cdots,j_{L}\rangle\propto (-1)^{i_{k}-1+i_{k+1}+i_{k+2}+i_{k+3}}\)
     (8) Spin: \[ H=\sum_{\langle ij\rangle}J\vec{S}_{i}\cdot \vec{S}_{j}+h\sum_{i}S_{i}^{z}; \vert \sigma_{1},\cdots,\sigma_{L}\rangle\Rightarrow \vert \sigma\rangle \] \[ H=h\sum_{i}S_{i}^{z}+J\sum_{\langle ij\rangle}S_{i}^{z}S_{j}^{z}+J\sum_{\langle ij\rangle}(S_{i}^{y}S_{j}^{y}+S_{i}^{x}S_{j}^{x}) \] (a) \(\vert \sigma_{1},\cdots,\sigma_{L}\rangle\Rightarrow \vert \sigma\rangle\)索引表.
     (b) 对角：\(H_{\sigma\sigma}=\langle \sigma\vert H\vert \sigma\rangle=h\sum_{i=1}^{L}\sigma_{i}+J\sum_{\langle ij\rangle}(2\sigma_{i}-1)(2\sigma_{j}-1) \)
     (c) 非对角： \[ H_{\sigma s}=\langle\sigma \vert H\vert s\rangle=\sum_{ij}J\langle \sigma_{1}\cdots\sigma_{L}\vert S_{i}^{x}S_{j}^{x}+S_{i}^{y}S_{j}^{y}\vert s_{1}\cdots s_{L}\rangle=\sum_{ij}J\Pi_{l\neq i,j}\delta(\sigma_{l}s_{l})\langle \sigma_{i}\sigma_{j}\vert S_{i}^{x}S_{j}^{x}+S_{i}^{y}S_{j}^{y}\vert s_{i}s_{j}\rangle \]
     (9) 多体局域化： \[ H=\sum_{i} J\vec{S}_{i}\cdot\vec{S}_{i+1}+h_{i}S_{i}^{z} \] 1d Heisenberg model. 一般取\( J=1 \), \( M=0 \)的子空间, \( h_{i}\in U[-W,W] \)
     (a) \( e_{n}=\epsilon_{n+1}-\epsilon_{n} \); (b) \( r=min(e_{n},e_{n+1})/max(e_{n},e_{n+1}) \)
  2. 作业 求\( P(e_{n}),P(r) \)的分布,如果可以的话取\( L=12\sim 14 \).
     ddl:1月4日
  3. 课堂笔记26_饶 pdf

• 12.23 周四

  1. review
     (1) 多体：平均场( Spin, Bose-Hubbard, Fermi-Hubbard ), 多体\( \Rightarrow \)单体\( + \)自洽.
     (2) 严格对角化. (a) 数Hilbert空间; (b) 求矩阵元，分类处理. 技巧：索引表.
     (3) MC, 高维积分.
     (4) 技巧, 去掉比较极端的不太可能出现的情况，缩减子空间大小. 例如\( L=7,N=5,Bose \)的情况, 有3个以上粒子在同一个格点上的情况是可能出现的，但是可能性较小，可以不考虑.
  2. Anderson Localization
     (1) 单体问题， 无序在各种材料中存在. P. W. Anderson于1957年最早提出.
     (2) Phase transition: \( W>W_{c} \), localized, insulator; \( W (3) 比较容易计算; (4) 应用广泛.
     (5) 参考文献： (a) Hoffmann计算物理，讲转移矩阵一章. (b) Thouless: J. Phys. C. 1970. (c) Abrahams. Anderson. PRL. 1979. (d) 李正中， 《固体理论》第\( 12 \)章.
     (6) 方法：转移矩阵.
     \( H=-t\sum_{i}(C_{i}^{\dagger}C_{i+1}+h.c)+V_{i}C_{i}^{\dagger}C_{i} \), 其中\( -t \)为非对角元, \( V_{i} \)为对角元.
     一些工作讨论\( V_{i} \)下的物理: (a) white noise; (b) \( V_{i}=V\cos(\alpha i+\phi) \), 其中\( \alpha \)为无理数. (c) correlated disorder. 下面讨论white noise.
     1d white noise: \( \langle V_{i}\rangle =0 \), \( \langle V_{i}V_{j}\rangle=\sigma^{2}\delta_{ij} \).
     one particle, \( K=\lbrace \vert i\rangle \rbrace=\lbrace C_{i}^{\dagger}\vert 0\rangle , i\in Z \), \( \psi=\sum_{i}\alpha_{i}\vert i\rangle=\sum_{i}\alpha_{i}C_{i}^{\dagger}\vert 0\rangle \). \[ E\psi=H\psi\Rightarrow E\alpha_{i}=-t(\alpha_{i+1}+\alpha_{i-1})+V_{i}\alpha_{i} \] 当\( V_{i}=V=const \), \( \alpha_{j+1}=\alpha_{j-1}+\dfrac{V-E}{t}\alpha_{j} \), \( \Rightarrow x^{2}=1+\dfrac{V-E}{t}x \)，数列求通项的方式来做.
     当\( V_{i} \)为随机数, 且令\( t=1 \): \[ X_{i+1}= \begin{eqnarray} \left( \begin{array}{cc} \alpha_{i+1} \\ \alpha_{i} \end{array} \right)=\left( \begin{array}{cc} V_{i}-E & -1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{cc} \alpha_{i} \\ \alpha_{i-1} \end{array} \right)=T_{i}X_{i} \end{eqnarray} \] \( X_{N+1}=T_{N}T_{N-1}\cdots T_{1}X_{1}\sim e^{-N/\xi} \). QR分解: \( T=QR \), \( R \)为上三角矩阵. \[ T_{1}X_{1}\rightarrow T_{2}Q_{1}R_{1}X_{1}\rightarrow T_{3}Q_{2}R_{2}R_{1}X_{1}\rightarrow\cdots\rightarrow Q_{N}R_{N}R_{N-1}\cdots R_{1}X_{1}=X_{N+1} \] \[ X_{N+1}^{T}X_{N+1}=(X_{1})^{\dagger}R_{1}^{\dagger}R_{2}^{\dagger}\cdots R_{N}^{\dagger}R_{N}\cdots R_{1}X_{1} \] 对于上三角阵：\( (R_{3}R_{2}R_{1})_{kk}=R_{3}^{kk}R_{2}^{kk}R_{1}^{kk} \). \[ X_{N+1}^{\dagger}X_{N+1}=f(N)e^{-2(N+1)/\xi}\Rightarrow \dfrac{\ln f(N)}{N+1}=\dfrac{2}{\xi}+\dfrac{1}{N+1}(X_{1}^{\dagger}R^{\dagger}RX_{1})\Rightarrow \dfrac{2}{\xi}=-\lim_{N\rightarrow \infty}\dfrac{1}{N+1}\ln(X_{1}^{\dagger}R^{\dagger}RX_{1})\] 定理： Oseledec遍历定理：\( \Gamma=\lim_{N\rightarrow\infty}(T_{N}T_{N}^{\dagger})^{1/2N} \)存在，唯一.
     另有：上三角阵的本征值是对角元. \( R=R_{N}\cdots R_{1}\rightarrow \lambda_{k}=(R_{N})_{kk}(R_{N-1})_{kk}\cdots (R_{1})_{kk} \) \[ TT^{\dagger}=P^{\dagger}\lambda^{\dagger}\lambda P;\quad (\lambda^{\dagger}\lambda)_{kk}=\Pi_{j}\vert (R_{j})_{kk}\vert^{2}\Rightarrow f(N)e^{-\dfrac{2N}{\xi}}=\Pi_{j=1}^{N}\vert (R_{j})_{kk}\vert^{2}\Rightarrow \dfrac{1}{\xi}=-\lim_{N\rightarrow\infty}\dfrac{1}{2N}\sum_{j}\ln \vert (R_{j})_{kk}\vert^{2} \]
  3. 课堂笔记27_饶 pdf

• 12.28 周二

  1. review
     (1) 单粒子问题:\( K=\lbrace C_{i}^{\dagger}\vert 0\rangle\rbrace \)
     (2) \( E\alpha_{i}=V_{i}\alpha_{i}-t(\alpha_{i+1}+\alpha_{i-1}) \); \( V_{i} \)为常数, 对应迭代方程; \( V_{i} \)为随机数.
     (3) 转移矩阵, 技巧：QR分解.
  2. Anderson Localization
     (1) 推广到高维:\( H=-t_{i}^{jj'}C_{ij}^{\dagger}C_{i+1,j'}+V_{i}^{jj'}C_{ij}^{\dagger}C_{ij'} \) \[ \begin{eqnarray} \left( \begin{array}{cc} \alpha_{i+1} \\ \alpha_{i} \end{array} \right)=\left( \begin{array}{cc} E-[V_{i}] & -[t_{i}] \\ 1 & 0 \end{array} \right) \left( \begin{array}{cc} \alpha_{i} \\ \alpha_{i-1} \end{array} \right) \end{eqnarray} \] 对于高维情况\( \alpha_{i} \)不再是一个数，而是一个\( d-1 \)维面上所有的点.
     (2) 考虑\( V_{i}=V \), \( H=-t(C_{i}^{\dagger}C_{i+1}+h.c)+VC_{i}^{\dagger}C_{i} \) \[ Fourier transformation: C_{n}=\dfrac{1}{\sqrt{N}}\sum_{k}e^{ikn}C_{k}, \quad C_{n}^{\dagger}=\dfrac{1}{\sqrt{N}}\sum_{k}e^{-ikn}C_{k}^{\dagger};\quad \sum_{n=1}^{N}C_{n}^{\dagger}C_{n}=\sum_{k}C_{k}^{\dagger}C_{k};\quad \sum_{n=1}^{N}C_{n}^{\dagger}C_{n+1}=\sum_{k}(\cos{k}+i\sin{k})C_{k}^{\dagger}C_{k};\quad \sum_{n=1}^{N}C_{n}C_{n+1}^{\dagger}=\sum_{k}(\cos{k}-i\sin{k})C_{k}C_{k}^{\dagger} \] \[ \Rightarrow H=\sum_{k}(-2t\cos{k}+V)C_{k}^{\dagger}C_{k};\quad E(k)=-2t\cos{k}+V \] \[ T_{1}=\cdots=T_{N}=T=\begin{eqnarray} \left( \begin{array}{cc} E-V & -1 \\ 1 & 0 \end{array} \right) \end{eqnarray} \] \[ T=P\lambda P^{-1};\quad T^{N}=P\lambda^{N}P^{-1};\quad \lambda_{11},\lambda_{22}=\dfrac{E}{2}\pm \sqrt{\dfrac{E^{2}}{4}-1},(V=0) \] \(\psi\sim e^{-\dfrac{N}{\lambda}}\); \( \lambda \)复数, extended; \( \lambda \)实数, localized.
     (3) 局域化长度与态密度的关系.
     考虑如下模型： \[ \epsilon_{i}\psi_{i}^{\alpha}-t_{i,i+1}\psi_{i+1}^{\alpha}-t_{i-1,i}\psi_{i-1}^{\alpha}=E_{\alpha}\psi_{i}^{\alpha} \] \(\alpha\)表示不同的本征态，\(t\)表示最近邻格点间的hopping，\(\epsilon\)是对应格点的能量，\(E_{\alpha}\)为对应的本征值. 引入格林函数\(G(E)\), \(G(E)=(EI-H)^{-1}\).
     矩阵\(A\)的逆满足：\(A^{-1}=\dfrac{A^{*}}{det(A)}\), 其中\(A^{*}\)是\(A\)的伴随矩阵. 伴随矩阵的矩阵元是对应的代数余子式, \( det(EI-H)=\prod_{\alpha=1}^{N}(E-E_{\alpha}) \).
     \(G_{1N}\)对应的代数余子式： \[ (-1)^{1+N}\prod_{i=1}^{N-1}-t_{i,i+1}=\prod_{i=1}^{N-1}t_{i,i+1};\quad G_{1N}(E)=\dfrac{\prod_{i=1}^{N-1}t_{i,i+1}}{\prod_{\alpha=1}^{N}(E-E_{\alpha})};\quad G(E)=\dfrac{1}{E-H} \] \(\vert\psi^{\alpha}\rangle\)为\(H\)的本征函数: \[ G(E)=\sum_{\alpha}\dfrac{\vert \psi^{\alpha} \rangle\langle \psi^{\alpha}\vert}{E-E_{\alpha}};\quad G_{1N}(E)=\sum_{\alpha}\dfrac{\langle 1\vert \psi^{\alpha}\rangle\langle\psi^{\alpha}\vert N\rangle}{E-E_{\alpha}} \] 在\(E=E_{\beta}\)附近作围道积分，对应的留数： \[ \langle 1\vert \psi^{\beta}\rangle\langle\psi^{\beta}\vert N\rangle\equiv a_{1}^{\beta}(a_{N}^{\beta})^{*} \] 对\( G_{1N}(E)=\dfrac{\prod_{i=1}^{N-1}t_{i,i+1}}{\prod_{\alpha=1}^{N}(E-E_{\alpha})} \)在\(E_{\beta}\)附近作围道积分，对应的留数为： \[ G_{1N}(E)=\dfrac{\prod_{i=1}^{N-1}t_{i,i+1}}{\prod_{\alpha\neq \beta}(E_{\beta}-E_{\alpha})} \] \[\Rightarrow \ln\vert a_{1}^{\beta}a_{N}^{\beta}\vert=\sum_{i=1}^{N-1}\ln\vert t_{i,i+1}\vert-\sum_{\alpha\neq\beta}\ln\vert E_{\beta}-E_{\alpha}\vert\] 对于一个局域的本征态，假设本征函数取之最大处在\(i\)点： \[ a_{1}^{\beta}\propto e^{-\lambda_{\beta}(i-1)};\quad a_{N}^{\beta}\propto e^{-\lambda_{\beta}(N-i)} \] \[ \lambda_{\beta}=(N-1)^{-1}(-\sum_{i=1}^{N-1}\ln\vert t_{i,i+1}\vert+\sum_{\alpha\neq\beta}\ln\vert E_{\beta}-E_{\alpha}\vert) \] 取\(N\rightarrow\infty\)，写成连续的形式： \[ \lambda=\int\rho(x)\ln\vert E_{\beta}-x\vert dx-\ln\vert t\vert \] 这里\( \lambda^{-1}=\xi \)为局域化长度. (4) 标度率
     (1) \( R\propto L^{2-d} \). For metal, \( g\sim\sigma L^{d-2}\sim (L/\xi)^{d-2} \); for insulator, \( g\propto e^{-L/\xi} \).
     \( \beta \)function: \( g=g_{0}L^{\beta} \), for metal \( \beta=\dfrac{d\ln g}{d\ln L}=d-2 \), for insulator, \( \ln g=\ln \sigma-\dfrac{L}{\xi} \), \( \beta=\dfrac{d\ln g}{d\ln L}=\dfrac{d\ln g}{dL}/\dfrac{d\ln L}{dL}=-\dfrac{L}{\xi} \). 无关联无序的Anderson Localization, 3d有transition， 2d,1d一般认为没有transition, 对任意大小的无序都是局域的.
  3. 作业
     (1) on-site energy\( V_{i}=0 \), \( E(k)=-2t\cos{k} \), 求出态密度\( \rho(E) \), 并用mathematica积分证明\( \xi^{-1}=0 \), \(\xi\)为局域化长度.
     (2) \( \rho(E) \)为lorentz分布\( \dfrac{\gamma/\pi}{x^{2}+\gamma^{2}} \), 求\( \dfrac{1}{\xi(E)} \),\(\xi\)为局域化长度.
     ddl: 1月12日.
  4. 课堂笔记28_饶 pdf
  5. Thouless 1972 pdf

• 1.4 周二

  1. review
     (1) Anderson localization: 转移矩阵. \( C_{i} \): 1d, 1 site; 3d, 一个面. \( -tC_{i}^{\dagger}C_{i+1}\Rightarrow -C_{i}^{\dagger}[t]C_{i+1} \), \( -V_{i}C_{i}^{\dagger}C_{i}\Rightarrow C_{i}^{\dagger}[t]C_{i}\); Thouless.
     (2) 标度率： \( \beta=\dfrac{d\ln g}{d\ln L} \Rightarrow \ln g=\beta \ln L+\ln g_{0},\quad g=g_{0}L^{\beta}\propto\dfrac{1}{R} \).
  2. 超导
     (1) 二次型Hamiltionian, \( y=x^{T}Ax \). For boson: Bogoliubov 变换. For fermion: BCS.
     (2) 表象：\( \rho=0 \), 本质: (a) Meissner effect; (b) cooper pair.
     (3) \( H=-t (C_{1}^{\dagger}C_{2}+h.c)+\epsilon_{1}C_{1}^{\dagger}C_{1}+\epsilon_{2}C_{2}^{\dagger}C_{2} \) \[ H=\begin{eqnarray} \left( \begin{array}{cc} C_{1}^{\dagger} & C_{2}^{\dagger} \end{array} \right) \left( \begin{array}{cc} \epsilon_{1} & -t \\ -t & \epsilon_{2} \end{array} \right) \left( \begin{array}{cc} C_{1} \\ C_{2} \end{array} \right) \end{eqnarray} \] 超导： \[ H=-t(C_{1}^{\dagger}C_{2}+C_{2}^{\dagger}C_{1})+\epsilon_{1}C_{1}^{\dagger}C_{1}+\epsilon_{2}C_{2}^{\dagger}C_{2}+\Delta C_{1}^{\dagger}C_{2}^{\dagger}+\Delta^{*}C_{2}C_{1} \] \( C_{1}\rightarrow x_{1}+ix_{2},\quad C_{1}^{\dagger}\rightarrow x_{1}-ix_{2}, \quad C_{2}\rightarrow x_{3}+ix_{4},\quad C_{2}^{\dagger}\rightarrow x_{3}-ix_{4} \), \( H(x_{1},x_{2},x_{3},x_{4}) \) 二次型. \( kC_{1}^{\dagger}C_{1}(C_{2}+C_{2}^{\dagger}) \), 多体相互作用. \[ H=\begin{eqnarray} \left( \begin{array}{cccc} C_{1}^{\dagger} & C_{2}^{\dagger} & C_{1} & C_{2} \end{array} \right) \left( \begin{array}{cccc} & & & \\ & & & \\ & & & \\ & & & \\ \end{array} \right) \left( \begin{array}{cc} C_{1} \\ C_{2} \\ C_{1}^{\dagger} \\ C_{2}^{\dagger} \end{array} \right) \end{eqnarray} \] 基本要求: (a) Hermite; (b) 等式成立; (c) particle-hole duality.
     下面考虑\( \Delta=0 \)的情况. \[ H=\begin{eqnarray} \left( \begin{array}{cccc} C_{1}^{\dagger} & C_{2}^{\dagger} & C_{1} & C_{2} \end{array} \right) \left( \begin{array}{cccc} x_{1} & x_{2} & 0 & 0 \\ x_{2} & x_{3} & 0 & 0 \\ 0 & 0 & y_{1} & y_{2} \\ 0 & 0 & y_{2} & y_{3} \end{array} \right) \left( \begin{array}{cc} C_{1} \\ C_{2} \\ C_{1}^{\dagger} \\ C_{2}^{\dagger} \end{array} \right) =x_{1}C_{1}^{\dagger}C_{1}+x_{3}C_{2}^{\dagger}C_{2}+y_{1}C_{1}C_{1}^{\dagger}+y_{3}C_{2}C_{2}^{\dagger}+x_{2}C_{1}^{\dagger}C_{2}+x_{2}C_{2}^{\dagger}C_{1}+y_{2}C_{1}C_{2}^{\dagger}+y_{2}C_{2}C_{1}^{\dagger}+const \end{eqnarray} \] \[ H=(x_{1}+y_{1})C_{1}^{\dagger}C_{1}+(x_{3}+y_{3})C_{2}^{\dagger}C_{2}+(x_{2}-y_{2})C_{1}^{\dagger}C_{2}+(x_{2}-y_{2})C_{2}^{\dagger}C_{1}+(y_{1}+y_{3})+const=-t(C_{1}^{\dagger}C_{2}+h.c)+\epsilon_{1}C_{1}^{\dagger}C_{1}+\epsilon_{2}C_{2}^{\dagger}C_{2} \] 考虑\( t=0 \), \( x_{2}=y_{2}=0 \). 可以得到: \( (x_{1},x_{3},y_{1},y_{3})=(\dfrac{1}{2}\epsilon_{1},\dfrac{1}{2}\epsilon_{2},-\dfrac{1}{2}\epsilon_{1},-\dfrac{1}{2}\epsilon_{2}) \).
     另一种情况: \( \epsilon_{1}=\epsilon_{2}=\Delta=0 \); \( x_{1}=x_{3}=y_{1}=y_{3}=0 \), 可以得到\( x_{2}=-\dfrac{t}{2},y_{2}=\dfrac{t}{2} \). \[ H=\dfrac{1}{2}\begin{eqnarray} \left( \begin{array}{cccc} C_{1}^{\dagger} & C_{2}^{\dagger} & C_{1} & C_{2} \end{array} \right) \left( \begin{array}{cccc} \epsilon_{1} & -t & 0 & \Delta \\ -t & \epsilon_{2} & -\Delta & 0 \\ 0 & -\Delta^{*} & -\epsilon_{1} & t \\ \Delta^{*} & 0 & t & -\epsilon_{2} \end{array} \right) \left( \begin{array}{cc} C_{1} \\ C_{2} \\ C_{1}^{\dagger} \\ C_{2}^{\dagger} \end{array} \right)=\dfrac{1}{2}\left( \begin{array}{cc} H_{0} & \Delta \\ \Delta^{\dagger} & -H_{0}^{*} \end{array} \right) \end{eqnarray} \] 对于\( 2L\times 2L \)的情况, \( (H_{0})_{ii}=\epsilon_{i}\), \((H_{0})_{i,i+1}=(H_{0})_{i+1,i}=-t \). \( \Delta_{ij}C_{i}^{\dagger}C_{j}^{\dagger},i < j \), \( H_{i,j+L}=\Delta_{ij} \), \( H_{j,i+L}=-\Delta_{ij} \), \( H_{i,i+L}=0 \).
  3. 课堂笔记29_饶 pdf

• 1.6 周四

  1. Bogoliubov 变换 (1) 1947, on the theory of superfluidity.
     (2) \( H=\sum t_{ij}C_{i}^{\dagger}C_{j}+h.c+\Delta_{ij}C_{i}^{\dagger}C_{j}^{\dagger}+\Delta_{ij}^{*}C_{j}C_{i}=\sum_{i}\epsilon_{i}\gamma_{i}^{\dagger}\gamma_{i} \). \[ C_{i}=u_{ij}\gamma_{j}+v_{ij}\gamma_{j}^{\dagger};\quad C_{i}^{\dagger}=u_{ij}^{\dagger}\gamma_{j}^{\dagger}+v_{ij}^{*}\gamma_{j} \] \[ \psi=\begin{eqnarray} \left( \begin{array}{cc} C \\ C^{\dagger} \end{array} \right)= \left( \begin{array}{cc} U & V \\ V^{*} & U^{*} \end{array} \right) \left( \begin{array}{cc} \gamma \\ \gamma^{\dagger} \end{array} \right) \end{eqnarray} \] 有反对易关系: \[ \lbrace C_{i},C_{j}\rbrace=0;\quad \lbrace C_{i},C_{j}^{\dagger}\rbrace \] 将变换代入: \[ \lbrace C_{i},C_{j}\rbrace=\lbrace u_{ik}\gamma_{k}+v_{ik}\gamma_{k}^{\dagger},u_{jl}\gamma_{l}+v_{jl}\gamma_{l}^{\dagger}\rbrace=u_{ik}v_{jk}+v_{ik}u_{jk}=0\Rightarrow (UV^{T}+VU^{T})_{ij}=0 \] \[ \lbrace C_{i},C_{j}^{\dagger}\rbrace =\lbrace u_{ik}\gamma_{k}+v_{ik}v_{k}^{\dagger},u_{jl}^{*}\gamma_{l}^{\dagger}+v_{jl}^{*}\gamma_{l}\rbrace_{\pm}=u_{ik}u_{jk}^{*}\pm v_{ik}v_{jk}^{*}=(uu^{\dagger}\pm vv^{\dagger})_{ij}=\delta_{ij};\quad uu^{\dagger}\pm vv^{\dagger}=1 \] \[ t_{ij}C_{i}^{\dagger}C_{j}=t_{ij}(u_{ik}^{*}\gamma_{k}^{\dagger}+v_{ik}^{*}\gamma_{k})(u_{jl}\gamma_{l}+v_{jl}\gamma_{l}^{\dagger});\quad \Delta_{ij}C_{i}^{\dagger}C_{j}=\Delta_{ij}(u_{ik}^{*}\gamma_{k}^{\dagger}+v_{ik}^{*}\gamma_{k})(u_{jl}^{*}\gamma_{l}^{\dagger}+v_{jl}^{*}\gamma_{l}) \] 由于\( H=\sum_{j}\epsilon_{j}\gamma_{j}^{\dagger}\gamma_{j} \), 所以\( \gamma_{k}^{\dagger}\gamma_{l}^{\dagger} \)之类的项为\(0\).
     eg: \[ H=\epsilon (a^{\dagger}a+b^{\dagger}b)+\Delta a^{\dagger}b^{\dagger}+\Delta^{*}ba \] \[ H=\begin{eqnarray} \left( \begin{array}{cccc} a^{\dagger} & b^{\dagger} & a & b \end{array} \right)= \left( \begin{array}{cccc} \epsilon & 0 & 0 & \Delta \\ 0 & \epsilon & \Delta & 0 \\ 0 & \Delta & \epsilon & 0 \\ \Delta & 0 & 0 & \epsilon \end{array} \right) \left( \begin{array}{cccc} a \\ b \\ a^{\dagger} \\ b^{\dagger} \end{array} \right) \end{eqnarray} \] 得到的本征值\( \dfrac{1}{2}(\epsilon\pm \Delta) \), 是不对的， 采用另一种方法. \[ i\dot{a}=[a,H]=\dfrac{\partial H}{\partial a^{\dagger}}=\epsilon a+\Delta b^{\dagger};\quad i\dot{b}=\epsilon b+\Delta a^{\dagger} \] \[ i\dot{a}^{\dagger}=-\epsilon a^{\dagger}-\Delta b;\quad i\dot{b}^{\dagger}=-\epsilon b^{\dagger}-\Delta a \] \[ i\dfrac{d}{dt}\begin{eqnarray} \left( \begin{array}{cccc} a \\ b \\ a^{\dagger} \\ b^{\dagger} \end{array} \right)= \left( \begin{array}{cccc} \epsilon & 0 & 0 & \Delta \\ 0 & \epsilon & \Delta & 0 \\ 0 & -\Delta & \epsilon & 0 \\ -\Delta & 0 & 0 & -\epsilon \end{array} \right) \left( \begin{array}{cccc} a \\ b \\ a^{\dagger} \\ b^{\dagger} \end{array} \right) \end{eqnarray} \] 本征值\( \pm\sqrt{\epsilon^{2}-\Delta^{2}} \).
     对于哈密顿量: \[ H=\epsilon (a^{\dagger}a+b^{\dagger}b)+\Delta a^{\dagger}b^{\dagger}+\Delta ba \] 令: \[ a=u\alpha+v\beta^{\dagger};\quad b=x\beta^{\dagger}+y\alpha;\quad a^{\dagger}=u\alpha^{\dagger}+v\beta;\quad b^{\dagger}=x\beta+y\alpha^{\dagger} \] 这里考虑\( u,v,x,y \in R\). \[ [a,a^{\dagger}]=1=[u\alpha+v\beta^{\dagger},u\alpha^{\dagger}+v\beta]=u^{2}-v^{2}=1 \] \[ [a,b^{\dagger}]=0\Rightarrow [u\alpha+v\beta^{\dagger},x\beta+y\alpha^{\dagger}]=0;\quad u y-v x=0 \] 可以得到\(x=u,y=v\). \[ H\Rightarrow \epsilon(u\alpha^{\dagger}+v\beta)(u\alpha+v\beta^{\dagger})+\epsilon(u\beta+v\alpha^{\dagger})(u\beta^{\dagger}+v\alpha)+\Delta (u\alpha^{\dagger}+v\beta)(u\beta^{\dagger}+v\alpha)+\Delta (u\beta+v\alpha^{\dagger})(u\alpha+v\beta^{\dagger})\] 可以得到\( \alpha^{\dagger}\alpha,\beta^{\dagger}\beta \), 对应\( \pm\sqrt{\epsilon^{2}-\Delta^{2}} \), \( \alpha^{\dagger}\beta^{\dagger} \), \( \alpha\beta \)系数为\( 0 \).
     对于\( H=t_{ij}C_{i}^{\dagger}C_{j}+\Delta_{ij}C_{i}^{\dagger}C_{j}^{\dagger}+h.c. =\dfrac{1}{2}\psi^{\dagger}M\psi \). \[ \psi=\begin{eqnarray} \left( \begin{array}{cc} C \\ C^{\dagger} \\ \end{array} \right);\quad M= \left( \begin{array}{cc} t & \Delta \\ \Delta^{\dagger} & t \end{array} \right) \end{eqnarray} \] \( M\rightarrow (M\Lambda) \). For fermion: \( \Lambda=1 \); for boson: \( \Lambda=\sigma_{z} \).
     For fermion: \[ \psi=\begin{eqnarray} \left( \begin{array}{cc} u & v \\ v^{*} & u^{*} \end{array} \right);\quad \psi=T\phi;\quad T\Lambda T^{\dagger}=\Lambda \end{eqnarray} \] \[ \dfrac{1}{2}\psi^{\dagger}M\psi=\dfrac{1}{2}\phi^{\dagger}T^{\dagger}MT\phi=\dfrac{1}{2}\phi^{\dagger}T^{\dagger}MT\Lambda\Lambda\phi=\dfrac{1}{2}\phi^{\dagger}T^{\dagger}(M\Lambda)(T^{\dagger})^{-1}\Lambda \phi=\dfrac{1}{2}\phi^{\dagger}P(M\Lambda)P^{-1}\Lambda \phi \]
  2. 作业 \( 50\times 50 \)格子, \( t=1 \), Fermion, Boson, \( \Delta_{ij} \)随机给出. 求本征值, 波函数. 注意当\( \Delta \)足够大的时候, 系统不稳定.
  3. 课堂笔记30_饶 pdf