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参考书
考核
- 考试：40%，开卷，带电脑，现场编程（8选4）
- 课题报告：20%
- 平时作业：40%
有大四的本科生在7月1号前交project
其他学生在7月30号前交project
- 参考论文(共52篇，选择一篇作为project就行，不用登记)
- 课题报告要求(参看以往拓扑课格式，本课程计算为主)
- 电子版发邮箱或者纸质版交到量子信息实验室办公室517
2.25 周二
- 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 \( \int_a^b |R(x)| dx \)
(1) 过\( (x_0, f(x_0),\ (x_1, f(x_1)),\ \cdots (x_n,f(x_n))\) N+1 个点的插值函数： \[ \tilde{f} (x) = b_0 + b_1(x-x_0) + b_2 (x-x_0)(x-x_1) + \cdots + b_{n}(x-x_0) \cdots (x-x_{n-1}) \]
(2) \(b_0 = f[x_0]=f(x_0)\), \(b_1 =f[x_0,x_1]= \dfrac{f(x_i)-f(x_{x_i+1})}{x_i-x_{i+1}} \), \( b_2 = f[x_0,x_1,x_2] =\dfrac{f[x_i,x_{i+1}]-f[x_{i+1},x_{i+2}]}{x_i-x_{i+2}}\)
(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：
- \(l_i(x_j) = \delta_{ij} \)
- \( \sum_i l_i(x) = 1 \)
- \(l_i(x) \) 是N阶多项式
2.28 周五
- 积分问题
- Newton-Cotes 积分
- Euler Method
- 预估 \( \tilde{y}_{n+1} = y_n + h f(x_n.y_n) \Rightarrow y_{n+1} = y_n + \dfrac{h}{2} [f(x_n,y_n) + f(x_{n+1},\tilde{y}_{n+1})] \)
- 迭代 \(y^{k+1}_{n+1} = y_n +\dfrac{h}{2} [f(x_n,y_n) + f(x_{n+1},y_{n+1}^{k})] \)
- Runge-Kutta Method
- 二阶：\(y_{n+1} = y_n + \dfrac{h}{2} [k_1+k_2],\ k_1 = f(x_n,y_n), k_2=f(x_n+h,y_n+h) \)
- 三阶：\(y_{n+1} =y_n + \dfrac{h}{6} (k_1 + 4 k_2 + k_3),\ k_1=f(x_n,y_n),k_2=f(x_n+\dfrac{h}{2},y_n+\dfrac{h}{2}k_1), k_3 = f(x_n+h,y_n-hk_1+2hk_2)\)
- 四阶：\( y_{n+1}= y_n +\dfrac{h}{6}(k_1+2k_2+2k_3+h_4)\), Ref 张韵华P177-178
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. \(\dot{y} = f(x,y)\)
(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，\(I =\dfrac{h}{6} [f(a)+4f(\dfrac{a+b}{2})+f(b) ] \)
(1) \( \dot{y} = f(x_n,y_n) \Rightarrow y_{n+1} = y_n +h f(x_n,y_n) \)
(2) \( \dot{y} = f(x_n,y_n) \Rightarrow y_{n+1} = y_n + \dfrac{h}{2} [f(x_n,y_n)+f(x_{n+1},y_{n+1})] \)
(1) \[y(x_{n+1}) = y(x_{n}+h) = y(x_n)+ y^\prime(x_n)h + \dfrac{h^2}{2!} y^{\prime \prime}(x_n) + \cdots = y(x_n) + hf(x_n,y_n) + \dfrac{h^2}{2} (f_x + f_y y_n^\prime )\] 比较 \[y_{n+1} = y_n + h[c_1 f(x_n,y_n) + c_2 f(x_n+\alpha h,y_n+bhf(x_n,y_n))] \]待定\(c_1,c_2,\alpha,b\)
(2)
3.6 周五
- 插值问题
- 求根问题
- 作业1 : 求\(AX^2+BX+C=0\)的根，其中\(A,B,C\)都是可逆矩阵，自己给定，\( X \) 为\(N\times N\)的矩阵，求根\(X\)，收敛性如何.
a. Runge现象: 拟合的点数N不是越大越好
b. 多项式插值不能用于外推
(1) a. 非线性方程\(f(x)=0\); b. \(x\rightarrow \vec{x}, f\rightarrow \vec{f}\); c. \(f \rightarrow \) matrix
(2)迭代法: \(f(x)=0 \Rightarrow x_{n+1}=f(x_n)+x_n\)
(3)稳定性: \[x_{n+1}=f(x_n)+x_n \Rightarrow \dot{x}_n=f(x_n),\ \ x=x^{\ast}+z \Rightarrow \dot{z}=f(x^{\ast}+z) = f(x^\ast) + f^\prime(x^\ast)z= Az\] 矩阵A的本征值\(\lambda_k\)， \(\dot{z}_k=\lambda_k z_k,\ z_k(t)=e^{\lambda_k t}z_k(0)\)，\(\lambda<0\)，则收敛. \(\lambda_k\)与lyapunov exponent有关
3.10 周二
- 求根问题
- 矩阵与线性变换 \(\begin{eqnarray} \left \lbrace \begin{array}{cc} \text{矩阵变换} \left \lbrace \begin{array}{ccc} {\rm Schmidt} \text{分解}, \text{正交化处理}\\ {\rm QR} 分解 , A=QR,\ Q^\dagger Q =1, R\text{上三角矩阵}\\ {\rm LU} 分解 ,A=LU,L\text{下三角矩阵},U\text{上三角矩阵} \\ \end{array}\right. \\ \text矩阵对角化 \left \lbrace \begin{array}{cc} 方阵 \left \lbrace \begin{array}{cc} A = A ^\dagger, A=U^\dagger \lambda U,\ U^\dagger U=1, \lambda_i \in \mathbb{R}. \text{若} A,U\in \mathbb{R}, \text{则}U^T U=1, \text{正交矩阵}\\ A \neq A^\dagger, \text{相似变换}A = P^{-1} \lambda P \end{array} \right. \\ 非方阵 A_{n\times m} =\rm SVD(奇异值分解) \end{array} \right. \end{array} \right. \end{eqnarray} \)
-
作业2: Lorenz Attractor
\[\begin{eqnarray}
\left \lbrace
\begin{array}{ccc}
\dot{x}=\sigma (y-x) \\
\dot{y}=\rho x -y -xz \\
\dot{z}= xy-\beta z
\end{array}
\right.
\end{eqnarray}
\]
- 取适合的\(\sigma,\rho,\beta\)，画出运动轨迹
- \(dx/dt=0,\ dy/dt=0,\ dz/dt=0\)，求\(x^\ast,y^\ast,z^\ast\)
- Euler算法迭代求运动轨迹
- \(x=x^\ast+\tilde{x},\ y=y^\ast+\tilde{y},\ z=z^\ast+\tilde{z} \Rightarrow \dot{\vec{\gamma}} = A \vec{\gamma}\)， \(\vec{\gamma}=(x,y,z)\)，求\(A\)的本征值及其稳定性.
- \(x^\ast,\ y^\ast, \ z^\ast\)若取非正中心，稳定性如何？
作业3 : 找一本经典力学和一本线性代数的书，总结主要结果.
(1) Newton Method: \(x_{n+1}=x_n - f(x_n)/f^\prime(x_n) \)
推广(单变量到多变量) \[f \rightarrow \vec{f}=(f_1,\cdots,f_N),\ \vec{x}_{n+1} = \vec{x}_n - J_n^{-1} \vec{f}(x_n),\ J_{ij}= \partial f_i/ \partial x_j\]
(2)Conjugate gradient method \[\vec{x}_{n+1} = \vec{x}_n + \lambda_n \vec{x}_n,\ \lambda_{n+1}= \lambda_n + \beta_n \vec{x}_n\]
3.13 周五
- 经典力学中的计算问题
- 三类问题
- 作业4 : 利用Mathematics的NDSolve函数求解\( \ddot{\theta} = - \sin{\theta} \) (取不同的初始条件)
(1) Refs： Landau 力学；力学与理论力学，科大出版社；经典力学，沈惠川
(2) \(L=L(q,\dot{q}),\ \delta S =0,\ S =\int L dt \Rightarrow \dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{q}} = \dfrac{\partial L}{\partial q}\)
(3) 几类特殊的拉氏量：a. 平动转动，\(L=T-V\)；b. EM Field，\(L=\dfrac{1}{2} mv^2 - e\phi + e \vec{A} \cdot \vec{v} \)； c. 相对论，\(L=-m_0c^2\sqrt{1-v^2/c^2}-V(x)\)
(1) 单摆 \(L=\dfrac{1}{2}ml^2{\dot{\theta}}^2 + mgl \cos{\theta}\)，\(ml^2\ddot{\theta} = -mgl \sin{\theta}\). \(\ \theta \rightarrow 0,\ \ddot{\theta} = -g/l \theta \)
(2)振子 \(T=\dfrac{1}{2}m(\dot{x}_1^2+\dot{x}_2^2+\dot{x}_3^2),\ V = \dfrac{1}{2} k(x_1^2+(x_1-x_2)^2 +x_3^2)\)
(3) RCL Circuit \(V = L di/dt + Ri +\frac{1}{C} \int_0^t idt^\prime =L \ddot{Q}+R\dot{Q}+Q/C \)，与\(m\ddot{x}-\mu m g\dot{x}+kx=0\)一一对应 \[Q\leftrightarrow x, \ i \leftrightarrow \dot{x},\ L \leftrightarrow m ,\ R\leftrightarrow \mu , 1/C \leftrightarrow k, w= 1/\sqrt{k/m} \leftrightarrow 1/\sqrt{LC} \]
3.17 周二
- 经典力学
- RCL电路
- 热力学第一定律\(\leftrightarrow\) 哈密顿力学，\( \mathrm{d} U=-PdV+TdS \leftrightarrow d H=\dot{q}dp-\dot{p}dq\)
- Rigid Body Dynamics
- Parametric resonance \[ \ddot{\theta} + w_0^2(t) \theta = 0, \ w_0(t) = w_0 (1+ \alpha \cos{wt} ),\ |\alpha | \ll 1 \] \[w=2w_0 + \varepsilon \Rightarrow \ddot{\theta} + w_0^2[1+2\alpha \cos{(2w_0+\varepsilon) t}]\theta = 0\] \[\theta= a(t) \cos{ (w_0 +\dfrac{\varepsilon}{2})t } + b(t) \sin{ (w_0+\dfrac{\varepsilon}{2}t) } \Rightarrow \ddot{\theta} = \cdots \] \(a(t),b(t)\)慢变化项，忽略二阶导数\(\ddot{a},\ddot{b}\)，得\(a\sim e^{\mu t},\ b\sim e^{\mu t}, \ a\sim \mu a ,\ b\sim \mu b \)
- 周期问题
- Mathieu eq(1868)， \(y^{\prime\prime} + (a- 2q\cos{2x}) y =0 \)
- Floquet(1883)，\(\dot{x} = A(t) x\)，\(A(t)= A(t+T)\)
- Hill eq (1886)， \(\ddot{y} + f(t) y= 0\)，\(f(t) = f(t+\tau)\)
- Bloch 定理
- \(\dot{x} = A(t) x \Rightarrow \dot{\phi} = A \phi , \ \phi=(x_1,x_2,\cdots)\)
- 解满足 \(\phi(t+T) = \phi(t) \phi^{-1}(0) \phi(T)\)
- \( \phi(t) = P(t) e^{tB} , \ P(t) = P(t+T) \)
- \(\phi(t+T) = \phi(t) e^{TB}\)
- \( {\rm Tr}[TB] = \int_0^T {\rm Tr}[A(s)]ds \ {\rm or } \ {\rm Tr}[B] = T^{-1} \int_0^T {\rm Tr}[A(s)] ds \)
- \({\rm Tr}[\ln{A}] = \ln{[{\rm Det} A]}, \ e^{B} =e^{A_1} e^{A_2} \Rightarrow {\rm Tr}B= {\rm Tr}A_1 + {\rm Tr} A_2\)
-
作业5 :
- 利用Mathematics推导Rigid Body Dynamics的各个方程
- 用NDSolve求解\(\theta(t),\psi(t),\phi(t)\)
- 用NDSolve 研究参数共振问题
- 如何取\(w\)值，使其共振
(1) 牛顿力学 vs Lagrange 力学
(2) 类比思想
\[L= \dfrac{1}{2} I_{\psi} (\dot{\psi} + \dot{\phi} \cos{\theta})^2 + \dfrac{1}{2} I_0 (\dot{\theta}^2 + \dot{\phi}^2 \sin{\theta}^2) - mgl \cos{\theta}\] \[\begin{eqnarray} \dfrac{\partial L}{\partial \dot{\psi}} &=& p_{\psi} = I_{\psi} (\dot{\psi} + \dot{\phi}\cos{\theta}) ={\rm const}, \ \dfrac{\partial L}{\partial \dot{\phi}} &=& p_{\phi} = I_{\psi} (\dot{\psi} + \dot{\phi}\cos{\theta}) \cos{\theta} + I_0 \dot{\phi} \sin{\theta}^2 = {\rm const} \end{eqnarray} \] 用\(p_{\phi},p_{\psi},\theta\)代替\(\dot{\psi},\dot{\phi}\)，得到\(L=\dfrac{1}{2} I_0 \dot{\theta}^2 - V_{\rm eff} (\theta)\)，求运动方程 \(\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{\theta}} = \dfrac{\partial L}{\partial \theta} \Rightarrow I_0 \ddot{\theta} = f(\theta) \)
共振判断 ：a. unstale，\(|\varepsilon | < |\alpha w_0|\)； b. stale，\(|\varepsilon | > |\alpha w_0|\).
(1) 历史发展
3.20 周五
- Bloch方程
- 马丢方程
-
作业7 :
a. 从无穷矩阵看出，\(|k| \leq 1/2\)；
b. 计算Bloch 能带；
c. 扫描a和q，画出稳定区域
作业8 :参考 Compartmental models in epidemiology (wiki) 计算里面所有模型，即以下模型- The SIR model without vital dynamics
- The SIR model with vital dynamics and constant population
- The SIS model
- The MSIR model
- The SEIR model
- The SEIS model
- The MSEIR model
\(-\dfrac{d^2}{dx^2} \phi - 2q\cos{x} \phi = \varepsilon \phi, \ \phi_k(x) = e^{ikx} \sum_n C_n e^{i n x } \Rightarrow (n+k)^2 C_n - q (C_{n+1} + C_{n-1}) =\varepsilon C_ n \) \[\begin{eqnarray} \left( \begin{array}{cccc} \ddots & & & \\ -q & (1+k)^2 & -q & \\ & -q & (0+k)^2 & -q \\ & & -q & (-1+k)^2 & -q \\ & & & & \ddots \end{array} \right) \left( \begin{array}{ccccc} \vdots \\ C_1\\ C_0\\ C_{-1}\\ \vdots \end{array} \right) = \varepsilon \left( \begin{array}{ccccc} \vdots \\ C_1\\ C_0\\ C_{-1}\\ \vdots \end{array} \right) \end{eqnarray} \]
(1)\( -\dfrac{d^2}{dt^2} \phi + (a- 2q \cos t) \phi = 0\)
(2) \( t \Rightarrow x \)，变成Bloch 问题
(3) 不同点：a. \(t \Rightarrow \)演化 \(\Rightarrow\) 给定\(a，q\Rightarrow \phi(t)\) ；b. \(x \Rightarrow \)本征值问题 \(\Rightarrow \varepsilon， \phi(x)\)
3.24 周二
- 量子化
(1) 一次量子化，坐标量子化，\([x,p]=i\hbar\)
(2) 二次量子化( Dirac 1927，refbook)，波函数量子化 \[\begin{eqnarray} \text{哈密顿正则方程} \left\lbrace \begin{array}{cc} \dfrac{d p}{dt} = -\dfrac{\partial H}{\partial q} \\ \dfrac{d q}{d t} = \dfrac{ \partial H}{\partial p} \end{array} \right. \Rightarrow \text{泊松括号形式} \left\lbrace \begin{array}{cc} \dfrac{d p}{dt} =\lbrace q, H\rbrace \\ \dfrac{d q}{d t} =\lbrace p,H\rbrace \end{array} \right. \Rightarrow \left\lbrace \begin{array}{cc} \lbrace q_i , q_j \rbrace=0,\ \lbrace p_i , p_j \rbrace=0 \\ \lbrace q_i , p_j \rbrace= \delta_{i,j} \end{array} \right. \end{eqnarray} \] \[\begin{eqnarray} \left\lbrace \begin{array}{cc} i \dfrac{\partial \psi }{ \partial t} = H \psi \\ \psi = \sum_n c_n(t) |\phi_n \rangle \end{array} \right. \Rightarrow \left\lbrace \begin{array}{cc} E = \langle \psi |H| \psi \rangle = \sum_{nm} t_{nm} c_n^\ast c_m \\ i \dot{c}_n = \sum_m c_m \langle \phi_n |H| \phi_m \rangle =\sum_m t_{nm} c_m \end{array} \right. \Rightarrow \left\lbrace \begin{array}{cc} i \dot{c}_n = \dfrac{\partial E}{\partial c_n^\ast} \\ i\dot{c}_n^\ast = -\dfrac{\partial E}{\partial c_n} \end{array} \right. \end{eqnarray}\] 类比，\(c_n \sim q,\ c_n^\ast \sim p \)，将\(c_n,\ c_n^\ast\)提升到算符\(c_n,\ c_n^\dagger\)
\(\{c_n,c_m\} = \{c^\dagger_n,c_m^\dagger\}=0, \{c_n,c_m^\dagger\}=\delta_{nm} \) 等价描述 \(\{\psi,\psi\} = \{\psi^\dagger,\psi^\dagger\} = 0, \{\psi(x),\psi^\dagger(y)\} = \delta(x-y) \)
波函数量子化和 \([x,p]=i\hbar\)一回事? 可以二维晶格，每一个格子谐振子振动，\( x_n \propto a_n + a_n^\dagger \)来理解
\(H = \sum_n t[c_n^\dagger c_{n+1} + {\rm h.c.}] + \mu c_n^\dagger c_n \)
平移对称性\(\Rightarrow c_n = N^{-1/2}\sum_k c_k e^{i k R_n} \Rightarrow H = \sum_n (2 t \cos{k} + \mu) c_k^\dagger c_k\)
推导技巧：\(\sum_n e^{i k R_n} = N \delta_{k,0}\)
作业10 : \(H = \sum_i t(C_i^\dagger C_{i+1} + {\rm h.c.} ) + \mu C_i^\dagger C_i,\ \mu_i = \mu + \varepsilon_i \)
- 若\(\varepsilon_i\)为高斯分布，\(\langle \varepsilon_i \rangle = 0, \ \langle \varepsilon_i \varepsilon_j \rangle = \sigma^2 \delta_{ij}\)，数值对角化给出局域的波函数.
- \(\varepsilon_i = V\cos(2\pi q i)\)， \(q\)为无理数，验证存在扩展态到局域态相变过程.
3.31 周二
- 量子化
- Fermi-Hubbard model
- Hillbert-Space 的构建：二进制表(\(1,2,3\cdots 2^L\))，判断1的个数是否为N，是则为基矢\( |\psi_i\rangle\)
- \(|\psi_i \rangle= x_1 + x_2 \cdot 2 + \cdots + x_{L} \cdot 2^{L-1}, \ |\psi_j \rangle = y_1 + y_1 \cdot 2 + \cdots + y_{L} \cdot 2^{L-1}\)，满足\( \left| |\psi_i \rangle - |\psi_j \rangle \right|= 2^{\alpha -1}\)的\(\langle \psi_i |C^\dagger_\alpha C_{\alpha+1}|\psi_j\rangle\)不为0
- \(\langle \psi_i |n_\alpha n_{\alpha+1} | \psi_j \rangle= x_\alpha x_{\alpha +1} \delta_{ij} \)
- Bose-Hubbard model
- Spin model
-
作业11 :
计算\(H\)本征值：\(H=H_0+V\)，\(H_0=\hbar w(a^\dagger a + \frac{1}{2}) \)，\(V=\alpha(\frac{\hbar}{2mw})^2 (a+a^\dagger)^4\)，\(H |\phi\rangle = E |\phi\rangle , \ H_0 | n\rangle = \varepsilon_n | n \rangle , \ |\phi \rangle = \sum_n c_n | n \rangle \)
作业12 : 推导Fermi-Hubbard model(spinless)的\(U n_{i}n_{i+1}\)项和Fermi-Hubbard model(spinful)的 \(Un_{i\uparrow}n_{i\downarrow}\) 项
作业13 : 计算Fermi-Hubbard model (spinless)的能谱：a. \(L=10, \ N=5\)；b. \(L=14,\ N=5\).
作业14 : 计算Bose-Hubbard model的能谱：\(L=10, \ N=5, \ Q =3\).
(1) \(H = \dfrac{p^2}{2m}+ \dfrac{1}{2}m w x^2 + \alpha x^4 \rightarrow H =\hbar w (a^\dagger a + \dfrac{1}{2}) + \alpha(\dfrac{\hbar}{2mw})^2 (a+a^\dagger) ^4\)
(2) 位移\(x \rightarrow x\sim a+ a^\dagger,\ H(x) \rightarrow H(a,a^\dagger)\)； 场\(\phi \rightarrow \phi \sim (a, a^\dagger ) ,\ H(\phi) \rightarrow H(a,a^\dagger)\)
(1) \( H = \sum_i ( -t C_i^\dagger C_{i+1} + {\rm h.c.} )+ \mu C_i^\dagger C_i + U n_{i\uparrow}n_{i\downarrow}\)
(2) Spinless F-H model ( \(L\)个格子，\(N\) 个粒子)处理:
(1) \( H = \sum_i ( -t C_i^\dagger C_{i+1} + {\rm h.c.} )+ \mu C_i^\dagger C_i + U n_{i}(n_{i}-1)\)；
(2) 引入 \(Q\) 进制，\(0 \leq Q \leq N\) ，\(Q\)人为截断，越大计算越准确
(1) \(H = \sum_i \sigma_i^x \sigma_{i+1}^x + \sigma_i^y \sigma_{i+1}^y + \sigma_i^z \sigma_{i+1}^z + h \sigma_i^z\)
(2) 方法 一：\(\sigma_i^x \sigma_{i+1}^x \rightarrow I \otimes \cdots \otimes I \otimes\sigma_i^x \otimes \sigma_{i+1}^x \otimes I \otimes \cdots \otimes I \)，\( \sigma_i^z \rightarrow I \otimes \cdots I \otimes\sigma_i^z \otimes I \otimes \cdots \otimes I \)
(3) 方法二：构建基矢\( |\phi_i \rangle= |s_1,s_2\cdots s_L\rangle\)，计算\(\langle \phi_i |H| \phi_j \rangle \)
4.3 周五
- Lanczos method
- \( b \rightarrow b_1 = b/||b||\)
- \(A b_1 = \alpha_1 b_1 + \beta_1 b_2\)，要求 \(b_2 \bot b_1, \ b_1^T b_2 = 0\) \(\Rightarrow \alpha_1 = b_1^T A b_1,\ \beta_1=b_2^T A b_1\)
- \(3\sim N\) 步： \(A b_n = \beta_{n-1}^\ast b_{n-1} + \alpha_n b_n + \beta_n b_{n+1} \) \(\Rightarrow\) \(\beta_{n-1}^\ast = b_{n-1}^T A b_n,\ \alpha_n=b_n^T A b_n ,\ \beta_n = b_{n+1}^T A b_n\)
- 三对角阵
(1) 算法过程:
(2) 缺点：\(b_1\) 的选择； 什么时候cutoff；什么时候收敛.
(3) 改进：krylov subspace iteration method
(1) 三对角阵行列式与连分式 \[ \begin{eqnarray} D_1= {\rm det }(T-\lambda) = \left( \begin{array}{cccc} \alpha_1 -\lambda & \beta_1 & & \\ \beta_1^\ast & \alpha_2 - \lambda & \beta_2 & \\ & \beta_2^\ast & \alpha_3 -\lambda & \\ & & & \ddots \end{array} \right),\ D_2 = \left( \begin{array}{ccc} \alpha_2 - \lambda & \beta_2 & \\ \beta_2^\ast & \alpha_3 -\lambda & \\ & & \ddots \end{array} \right), \ D_3 = \left( \begin{array}{ccc} \alpha_3 - \lambda & \beta_3 & \\ \beta_3^\ast & \alpha_4 -\lambda & \\ & & \ddots \end{array} \right) \end{eqnarray} \] \[D_1 = (\alpha_1 - \lambda) D_2 - |\beta_1|^2 D_3,\ D_2 = (\alpha_2 - \lambda) D_3 - |\beta_2|^2 D_4 , D_4 = D_2/D_3\] \[\dfrac{D_1}{D_2} = \alpha_1 - \lambda - |\beta_1|^2 \dfrac{D_2}{D_3} = \alpha_1 - \lambda - \dfrac{|\beta_1|^2}{ \alpha_2 - \lambda- \dfrac{|\beta_2|^2}{ \alpha_3- \lambda - \cdots}} \]
(2) 特点：\(O(N^2)\)，少内存.
(3) 应用：TB model；差分法；Lanczos \(\rightarrow\) Krylov 空间
4.7 周二
- Cayley 定理
- 平面波法
- 差分法
- \(h \rightarrow 0\)，\(T\) 矩阵元趋于无穷大，为什么结果反而收敛
类比于 TB model \(H = - \frac{\hbar^2} {2m h^2} \sum_i(C_i^\dagger C_{i+1} + {\rm h.c.}) - 2C^\dagger_i C_i\)，
FT 变化，\( \varepsilon_k = (2\cos{kh} - 2) \times (-\frac{\hbar^2} {2m h^2}) \xrightarrow{k\sim 0 展开} \frac{\hbar^2}{2m} (\frac{\pi}{L})^2\)，可以看到\(h\)没有影响结果，是收敛的. - 为什么 \(k\sim 0 \)
\(L\) 很大，\(n\) 很小，考虑低能级. - 为什么可以FT，有周期性?
可以无限个无限深势阱首位连接在一起，构成周期性. -
作业15 :
\(a^2+b^2+c^2+d +Aabcd =B\)，\(A,B,a,b,c,d\) 为正数，
- 数值求解 \(a+b+c+d\)的最大值和最小值
- 从a问结果找出\(a,b,c,d\)对应满足的条件
- 根据得到的条件作解析计算.
作业17 : 差分法求解出下列模型的波函数和能级- \(H = - \dfrac{1}{2} \dfrac{d^2 }{dx^2 } + \dfrac{1}{2} w^2 x^2\)，\(m=1,\hbar =1\)
- 三角形势
\({\rm det} (A -\lambda) = 0 \Rightarrow \lambda^n + b_1 \lambda^{n-1} + b_2 \lambda^{n-2} + \cdots + b_{n-1} \lambda + b_n =0 \)
\(\lambda \rightarrow A\) 矩阵，\(\Rightarrow A^n + b_1 A^{n-1} + \cdots =0 \Rightarrow A^{-1}= -b_n^{-1} ( A^{n-1} + b_1 A^{n-2} + \cdots + b_{n-2} A + b_{n-1})\)
\(A\)的逆矩阵可以用\(A\)的多项式展开
Krylov 空间: \( {\rm Km}(A,b) = \lbrace b,Ab,A^2b \cdots A^{m-1} b \rbrace \)
对于无限深势阱，\(\psi = \sum_n c_n \cos{k_n x}\)，\(k_n = (2n+1) \pi /L\)，与布洛赫函数对比.
(1) FD：Finite-difference (有限差分)，标量场；FDTD：FD time-domain (时域有限差分)，EM 问题
(2) FD： \[\dfrac{df(x)}{dx} = \dfrac{f(x+h)-f(x)}{h} = \dfrac{f(x) - f(x-h)}{h} = \dfrac{f(x+h) -f(x-h)}{2h} \Rightarrow \dfrac{d^2 f}{dx^2} = \dfrac{f(x+h) + f(x-h)－ 2 f(x)}{h^2} ＝f^{\prime \prime } + O(h^2)\]
(3) 薛定谔方程 \[-\dfrac{\hbar^2}{2m} \dfrac{d^2}{dx^2} \psi(x) + V \psi(x) = E \psi(x) \xrightarrow{离散化\ x=na} -\dfrac{\hbar^2}{2m} \dfrac{d^2}{dx^2}(\psi_{n+1} + \psi_{n-1} - 2\psi_{n} )+ V_n \psi_n = E \psi_n \] \[ \begin{eqnarray} V = \left( \begin{array}{ccc} V_1 & & & \\ & \ddots & \\ & & V_n \end{array} \right),\ T = - \dfrac{\hbar^2}{2m h^2} \left( \begin{array}{cccc} -2 & 1 & & \\ 1 & -2 & 1& \\ & 1 & -2 & \\ & & & \ddots \end{array} \right) \end{eqnarray} \]
(4) 含时薛定谔方程 \[ i \dfrac{\psi_{k+1}(x_n) - \psi_k (x_n)}{dt} = H(x_n) \psi_k(x_n) \Rightarrow \psi_{k+1} (x_n) = \psi_k(x_n) - i dt H(x_n) \psi_k(x_n) \] \(k\) (时间离散化)；\(H(x_n)_{N\times N}\)，\(N=L/dx\) (空间离散化)，可以推广到3d
4.10 周五
- FDTD
- \(\Delta t, \ \Delta z\) 取值：\(\Delta t c \sim \Delta z\)，\(\Delta z \sim \frac{1}{5 \sim 10} \lambda\)，\(\lambda\) 波长
- 优点：很好边界性；计算快
- 缺点：计算时间长的话误差大，精度\(O(\Delta t^2),\ O(\Delta z^2)\)
- \(1/2\) 物理上体现了相互作用影响，电磁互感
(1) Ref：Kane S. Yee 1966 原文 (dual lattice method) \[ \begin{eqnarray} \left\lbrace \begin{array}{cc} \partial E_x / \partial t = - \dfrac{1}{\varepsilon_0 } \partial H_y /\partial z \\ \partial H_y /\partial t = \dfrac{1}{\mu_0} \partial E_x /\partial z \end{array} \right. \xrightarrow{f(n\Delta t, k\Delta z)} \left\lbrace \begin{array}{cc} \dfrac{E_x^{n+\frac{1}{2}}(k)-E_x^{n-\frac{1}{2}}(k)}{\Delta t} = - \dfrac{1}{\varepsilon_0 }\dfrac{H_y^{n}(k+\frac{1}{2})-H_y^{n}(k-\frac{1}{2})}{\Delta z} \\ \dfrac{H_y^{n+1}(k+\frac{1}{2})-H_y^n(k+\frac{1}{2})}{\Delta t} = \dfrac{1}{\mu_0} \dfrac{E_x^{n+\frac{1}{2}}(k+1)-E_x^{n+\frac{1}{2}}(k)}{\Delta z} \end{array} \right. \end{eqnarray} \] \[ \begin{eqnarray} \left\lbrace \begin{array}{cc} E_{x}^{n+\frac{1}{2}}(k) = E_x^{n-\frac{1}{2}} (k) + \dfrac{\Delta t}{\varepsilon_0 \Delta z} [H_y^n(k-\frac{1}{2}) - H_y^n(k+\frac{1}{2})]\\ H_y^{n+1}(k+\frac{1}{2}) = H_y^n (k+\frac{1}{2}) + \dfrac{\Delta t}{\mu_0 \Delta z} [E_x^{n+\frac{1}{2}}(k) - E_x^{n+\frac{1}{2}}(k+1)] \end{array} \right. \end{eqnarray} \]
作替换\(E\rightarrow \sqrt{\varepsilon_0/\mu_0} \tilde{E},\ H\rightarrow \sqrt{\mu_0/\varepsilon_0} \tilde{H}\)，化简
4.14 周二
- DFT
- FEM
- 构造\(\int (u^{\prime \prime }v + f v) dx =0\)，\(\forall v(x)\) 都成立
- \(u =\sum_k c_k \psi_k, \ v=\psi_j \)，不要求正交
- \(\sum_k c_k \int \psi_k^{\prime\prime} v dx+ \int f v dx=0 \rightarrow c_k \int \psi_k^{\prime\prime} \psi_j dx + \int f \psi_j dx=0\) ，类似\(\sum_m c_m \langle \psi_n |H| \psi_m \rangle =E c_n\)
- 相似：函数构造，矩阵元\(\psi_k^{\prime \prime} \psi_j \sim \langle \psi_k | \partial^2_x | \psi_j \rangle\)
- 不同：基矢局域，非光滑，不要求正交
- 随机数
- 程序
-
作业18 : 测试加分，减法，乘法，除法，\(\cos\)函数，sqrt 函数，指数函数等时间
作业19 : 用投针法求\(\pi\)，a. 用rand( )发生器；b. \(\pi\) 的精度和\(N\)的关系，精确到\(10^{-3},\ 10^{-4},\ 10^{-5}\)
作业20 : Box-Muller 算法，产生\(U(0,1)\)均匀分布的随机数\(\xi_1,\ \xi_2\)，\(z_1= \sqrt{-2 \ln{\xi_1}} \cos{2\pi \xi_2},\ z_2 = \sqrt{-2 \ln{\xi_1}} \sin{2\pi \xi_2}\)
验证\(z_1,\ z_2\)是独立的Gaussian 分布
\[x_k=\sum_n x_n e^{ikn} = \sum_{n = eve.} x_n e^{ikn} + e^{ik} \sum_{n=odd} x_n e^{ik(n-1)} = \sum_q x_{2q} e^{2ikq} + w_k \sum_q x_{2q+1} e^{i2qk}\]
时间复杂度 \(T(N)= N{\rm ln} N\)，减小重复性计算
求解一定边界条件下\(u(x)\)，\(u^{\prime \prime} + f = 0\)
由\(U(0,1)\)产生其他分布，\(P(x) \xrightarrow{h: \ x \rightarrow y} g(y)\) \[P(x)dx=g(y)dy, \ y=h(x) \Rightarrow P(x) dx = g(y) h^{\prime} (x) dx \Rightarrow P(x) = g(y) h^{\prime}\]
随机数应用：布朗运动，量子耗散，蒙卡，安德森局域，统计物理
ComputationTimeUsed-Tested-By-GM Mathematica-Simplify-NDSolve Test-of-time.f90 Matrix Operators For Numerical Physics
4.17 周五
- 关联的随机数
- 随机微分方程
- 作业21 : 关联的随机数 \(\langle \xi_i \xi_j \rangle =e^{-\alpha |i-j|},\ \langle \xi_i \rangle = 0 \), a. 求出 \(w\) 矩阵；b. 用无关联的随机数 \(\eta\) 来产生有关联的随机数，检验关联函数
关联的随机数 \(\langle \xi_i \xi_j \rangle = f(i-j),\ \langle \xi_i \rangle = 0 \)，用无关联的随机数 \(\eta\) 来产生 \[ \xi_i = \sum_{j} w_{ij} \eta_j ,\ \overline{\xi_i} = 0 \Rightarrow \overline{\eta_j} = 0,\ \langle \xi_i \xi_j \rangle = \sum_{i^\prime j^\prime} w_{i i^\prime } w_{j j^\prime} \langle \eta_{i^\prime} \eta_{j^\prime} \rangle = \sum_{k} w_{ik }w_{jk} \xrightarrow{w=w^T} (w^2)_{ij}\] \[\Rightarrow (w^2)_{ij} = f(i-j),\ w^2 = U^\dagger \lambda U \xrightarrow{\lambda >0 } w= U^\dagger \sqrt{\lambda } U\] Toeplitz matrix \(w^2\)的本征值大多数为正数，极少数为负数，非常小，所以把 \(\lambda\) 中的负数替换为0
随机微分方程的数值解(硕士论文)
布朗运动 \(m\dot{u} = -\eta u + \xi \)，白噪声 \(\langle \xi(t) \xi(t^\prime) \rangle = D k_{\rm B} T \delta(t-t^\prime),\ \langle \xi(t) \rangle = 0\) \[\eta = 0,m=1 \Rightarrow \dot{u} = \xi ,\ u(t)=u(0) +\int_0^t \xi(t^\prime) dt^\prime\] \[\overline{u} = \overline{u(0)} + \int_0^t \overline{(\xi(t^\prime)} dt^\prime = u(0), \ \overline{u^2} = u^2(0) + \int_0^t \langle \xi(t_1) \xi(t_2) \rangle dt_1 dt_2 = u^2(0) + Dk_{\rm B}T\int_0^t\delta(t_1-t_2) dt_1dt_2 = u^2(0)+Dk_{\rm B}T t\] \(u(t)\) 无确定的值，有意义的是平均值，方差
4.21 周二
- 随机微分方程
-
作业22 : 数值求\(\dot{v} =\xi\)
作业23 : 数值求\(\dot{v} =-\eta v + \xi\)
(1) 随机微分方程 = 概率论 + 微积分，\(\int_a^b \xi dt\) 看做一个事件，分布才有意义
定义 \(Y_N=X_1 + X_2 + \cdots + X_N\)，\(\sigma_N^2 = \overline{Y_N^2} - \overline{Y_N}^2 = \sigma^2 N\)，\(\sigma_t^2 = \overline{v_t^2} - \overline{v_t}^2 = At\)，\((N \leftrightarrow t)\)
(2) 求解 \(\dot{v} = \xi \)
| \(\dot{v} = \xi \leftrightarrow dv =\xi dt\) | \(\dot{v} = \sqrt{A} \xi \leftrightarrow dv = \sqrt{A} \xi dt\) |
|---|---|
| \(\langle \xi \rangle = 0, \ \langle \xi(t) \xi(t^\prime)\rangle = A \delta (t-t^\prime)\) | \(\langle \xi \rangle = 0, \ \langle \xi(t) \xi(t^\prime) \rangle = \delta(t-t^\prime)\) |
| Wiener process \(dv = d w_t,\ \langle d w_t \rangle =0, \ \langle dw_t dw_{t^\prime} \rangle = A dt \delta_{tt^\prime}\) | Wiener process \(dv = \sqrt{A}d w_t,\ \langle d w_t \rangle =0, \ \langle dw_t dw_{t^\prime} \rangle = dt \delta_{tt^\prime}\) |
| \(v(t) = \int_0^t dw_t = \sum_i \delta w_t\) | \(v(t) =\sqrt{A} \int_0^t dw_t = \sqrt{A}\sum_i \delta w_t\) |
(3) 求解 \(m\dot{v}= -\eta v + \xi\) \[ v(t) = e^{-\eta/m t} v_0 + \int_0^t e^{-\eta/m (t-t^\prime)} \xi(t^\prime) /mdt^\prime\] Wiener process \[\overline{v}= e^{-\eta/m t} v_0 +\sqrt{A}/m\int_0^t e^{-\eta/m (t-t^\prime)} \overline{dw_{t^\prime}} =e^{-\eta/m t} v_0,\ \overline{v^2} = \overline{v}^2 + A/m^2 \int_0^t e^{-\eta/m (t-t^\prime)} e^{-\eta/m (t-t^{\prime \prime})}\overline{dw_{t^\prime} dw_{t^{\prime\prime}}}= \overline{v}^2 + A/m^2\int_0^t e^{-2\eta/m (t-t^\prime)} dt^\prime \] 一般形式 \(\dot{v} = f(t) \xi(t) \Rightarrow v= v_0 + \int_0^t f(t^\prime) dw_{t^\prime}\)
(4) 数值处理 \(dv=adt +bdw_t, \ a(v,t), \ b=b(v,t)\)，Ref：Numerical Solution Of SDE(第一章) \[方法一\ \ v_{n+1} = v_n + a(v_n,t_n) \Delta_n + b(v_n,t_n) \Delta w_n; \ \ 方法二\ \ v_{n+1} =v_n +a \Delta_n + b \Delta w_n + \frac{1}{2} bb^\prime (\Delta w_n^2 - \Delta_n )\] (5) \(df= f(dw_t) -f(0) = f^\prime + f^{\prime \prime} dw_{t}^2\)，不能忽略二阶导数，break differentiation rules \(\Rightarrow\) New chain rule (Ito's lemma)
4.24 周五
- Ito's lemma
- \(dY(w_t) = Y^\prime(w_t) d w_t + \dfrac{1}{2} Y^{\prime \prime} (w_t) dt\)
- \(dX = a dt+b dw_t\)，\(dY(t,X) = \dfrac{\partial Y}{\partial t} dt+ \dfrac{\partial Y}{\partial X} dX+\dfrac{1}{2} \dfrac{\partial^2 Y}{\partial X^2} dX^2 = (\dfrac{\partial Y}{\partial t}+\dfrac{1}{2}b^2 \dfrac{\partial^2 Y}{\partial X^2}+a \dfrac{\partial Y}{\partial X} ) dt+ b \dfrac{\partial Y}{\partial X} dw_t\)
- Black-Scholes Eq
- 求解 \(dX =a dt + b dw\)
- Ito's lemma：\(dY = (\dfrac{\partial Y}{\partial t} + a\dfrac{\partial Y}{\partial X} + \dfrac{b^2}{2} \dfrac{\partial^2 Y}{\partial X^2} )dt + b \dfrac{\partial Y}{\partial X} dw\)
- 找一个合适的 \(Y\)，使得 \(\dfrac{\partial Y}{\partial X} b = c\)，\(c\) 为常数，得到 \( dY = f(Y) dt + c dw\)
- 类比布朗运动 \(dv = -\eta v dt +\sigma dw\) 求法，\(Y =Y_0 G\)，\(dY_0 =f(Y_0) dt\)
- portfolio
- 作业24 : 数值求Black-Scholes Eq，\(dS = \mu S dt + \sigma S dw\)
参考wiki
注意：\(dS = \mu S dt + \sigma S dw,\ d\ln{S} = \mu dt + \sigma dw,\ \ln{S} = \ln{S_0} + \mu dt +\sigma dw\) 是错的，不符合实际情况，下面利用 Ito's lemma
\[dS = \mu S dt + \sigma S dw, \ dY =\dfrac{\partial Y}{\partial S }dS + \dfrac{1}{2} \dfrac{\partial^2 Y }{\partial S^2} dS^2 = \dfrac{\partial Y}{\partial S } \mu S dt + \dfrac{\partial Y}{\partial S }\sigma S dw + \dfrac{1}{2} \dfrac{\partial^2 Y}{\partial S^2 }\sigma^2 S^2 dt \] 令 \(Y = \ln{S}, \Rightarrow dY = d \ln{S} = (\mu - \dfrac{1}{2} \sigma^2) dt + \sigma dw\)
wiki Black–Scholes–Merton model
4.28 周二
- kubo 文章
- 转移矩阵
- 作业25 : 利用转移矩阵数值求\(H = -\sum_n t C_n^\dagger C_{n+1} + {\rm h.c.} + V_n C_n^\dagger C_n\), \(V_n = V \cos{2\pi q n}\), \(q\)为无理数
kubo_read_p1to10pages
参考Hoffmann Transfer-Matrix Methods and Finite-Size Scaling for Disordered Systems
\[H = -\sum_n t C_n^\dagger C_{n+1} + {\rm h.c.} + V_n C_n^\dagger C_n \Rightarrow E C_n = -t(C_{n+1}+C_{n-1} )+ V_n C_n \Rightarrow C_{n+1} = -C_{n-1} + (V_n-E ) C_n ,\ (t=1)\] \[ \begin{eqnarray} \left( \begin{array}{cc} C_{n+1}\\ C_n \end{array} \right) = \left( \begin{array}{cc} V_n -E& -1 \\ 1 & 0 \end{array} \right) \left( \begin{array}{c} C_{n} \\ C_{n-1} \end{array} \right),\ \left( \begin{array}{cc} C_{n+1}\\ C_n \end{array} \right) = T_N T_{N-1} \cdots T_1 \left( \begin{array}{c} C_{1} \\ C_{0} \end{array} \right) \end{eqnarray}\] \[ T_N T_{N-1} \cdots T_1 \xrightarrow{QR 分解\ T_1 =Q_1 R_1} T_N \cdots T_2 Q_1 R_1 \xrightarrow{\tilde{T_2} = T_2 Q_1} T_N \cdots (\tilde{T_2}) R_1 \xrightarrow{QR 分解 \tilde{T_2} = Q_2 R_2} T_N \cdots Q_2 R_2 R_1\Rightarrow Q_N R_N R_{N-1} \cdots R_2 R_1,\ \ (Q^\dagger Q =1) \] \[\langle \psi_N | \psi_N \rangle = \langle \psi_0 |R_1^\dagger \cdots R_N^\dagger R_N \cdots R_1|\psi_0\rangle = \langle \psi_0 | P^\dagger R_1^\dagger \cdots R_N^\dagger R_N \cdots R_1 P|\psi_0\rangle = \langle \tilde{\psi}_0 | R^\dagger R|\tilde{\psi}_0\rangle, \ (R^\dagger R)_{ii} = \Pi_{k=1}^N \lambda_i^k \Pi_{k=1}^N (\lambda_i^k)^\ast >0, \ (R_k)_{ii} = \lambda_i^k, \ P|\psi_0\rangle=|\tilde{\psi}_0\rangle\] \[\Gamma = \lim\limits_{N \rightarrow \infty} (R^\dagger R)^{1/2N}, \ \Gamma = e^{1/\xi} \Rightarrow \dfrac{1}{\xi_i} = \lim\limits_{N \rightarrow \infty } \dfrac{1}{N} \sum_{k=1}^N \ln|{\lambda_i^k}| ,\ 关联长度 \xi = {\rm max} \lbrace \xi_i \rbrace\]
5.8 周五
- 转移矩阵
- 随机积分
- 作业26 : 自己 找一个函数\(f(x_1,x_2\cdots x_n)\)，\(n=1,\ 2, \ 3,\ 4\)，数值随机积分并与其他方式验证比较结果
(1) 2 clains： 如图， \( E C_{i2} = V_{i2} C_{i2} - 其他五个点\)
randomintegral
RandomCode_GMF90s
(1) 核心：态密度，\(g(\varepsilon) = \dfrac{1}{N} \sum_{i} \delta(\varepsilon-\varepsilon_{i})\) \[\int_a^b f(x)dx = \dfrac{b-a}{N} \sum_i f(x_i) = (b-a) \sum_i P_i f(x_i) ,\ P_i = 1/N ,\ \sum_i P_i =1\] \(P_i f(x_i)\)看做均匀概率乘以函数，推广到任意概率分布函数\(P(x)\) \[\int_a^b f(x) dx = \int_a^b \dfrac{f(x)}{P(x)} P(x) dx = \dfrac{1}{N} \sum_i \int_a^b \dfrac{f(x)}{P(x) }\delta(x-x_i) dx = \dfrac{1}{N} \sum_i \dfrac{f(x_i)}{P(x_i)}\] 推广到2d，3d，任意维度，求和精度与步长无关，与N有关
5.9 周六
- 热统回顾
- 变分MC
- 抽样MC
- 细致平衡
- Metropolis 解：\(W_{i \rightarrow j } = {\rm min} \lbrace 1, \ e^{\beta (E_i -E_j)} \rbrace \)
- Heat-Bath algorithm：费米分布 \(W_{i \rightarrow j } = \dfrac{1}{1+ e^{-\beta(E_i-E_j)}} \)
- Ising(MC求解) 考虑经典二维 Ising 模型 \(H = -J\sum_{\langle i,j \rangle} \sigma_i \sigma_j\)
- 每一个格点随机产生 spin，作为初始构型
- flip spin
- 计算 \(E_{\rm old} ,\ E_{\rm new}\)
- if \( {\rm rand} < e^{-\beta ( E_{\rm new} - E_{\rm old }) } \)，接受新的构型
- 回到步骤二，直到得到稳定的构型
- 作业27：用蒙卡方法计算 Ising model (d=1) ： \(H = -J\sum_{i } \sigma_i \sigma_{i+1}\)，并验证无相变
- 作业28：用蒙卡方法计算 Ising model (d=2) ：\(H = -J\sum_{\langle i,j \rangle} \sigma_i \sigma_j - h \sum_i \sigma_i\)，并验证存在相变
配分函数\( Z= { \rm Tr}{e^{-\beta H}} = \sum_i e^{-\beta \varepsilon_i } = e^{-\beta F}, \ \overline{A} = \sum_i A_i e^{-\beta \varepsilon_i } /Z = \sum_i A_i P_i , \ P_i = e^{-\beta \varepsilon_i }/Z\) \[U = \sum_i E_i e^{-\beta E_i}/Z = -\partial \ln{Z} / \partial \beta, \ C =\partial U/ \partial T = k \beta^2 \partial^2 \ln{Z} / \partial \beta^2, \ S = -k \beta \partial \ln{Z} / \partial \beta + k \ln{Z}, \ F =U -TS\] 统计方法与随机积分对比，\(\int A(\varepsilon) e^{-\beta \varepsilon(x) } /Z dx = \frac{1}{N} \sum_i A(x_i),\ \int f(x) P(x)dx = \frac{1}{N} \sum_i f(x_i)\)
\[E = \dfrac{\langle \phi |H | \phi \rangle }{\langle \phi | \phi \rangle} = \dfrac{\int dx \phi^{\ast} H \phi} {\int dx \phi^\ast \phi} = \dfrac{\int dx \phi^{\ast} {\color{red} \phi} \dfrac{H \phi}{\color{red} \phi }} {\int dx \phi^\ast \phi} = \int dx \dfrac{H \phi}{\phi} \dfrac{\phi^\ast \phi}{\int \phi^\ast \phi dx}\int dx \dfrac{H \phi}{\phi} P(x) = \dfrac{1}{N} \sum_i (\dfrac{H \phi}{\phi})_i\]
\[\overline{A} = \sum_i A_i e^{-\beta \varepsilon_i }/Z = \sum_i \dfrac{A_i e^{-\beta \varepsilon_i }}{P_i} P_i/Z = \sum_i A_i e^{-\beta \varepsilon_i }/(ZP_i) \] if \(e^{-\beta \varepsilon_i }/(ZP_i) =1\)，\(\overline{A} = \dfrac{1}{N} \sum_i A_i\)，这里\(A_i\)服从分布\(P_i\)
(1) 平衡条件 \(N_i W_{i\rightarrow j } = N_j W_{j \rightarrow i }\)，\(N_i = e^{-\beta \varepsilon_i }/Z \)
(2) 找 \(W_{i \rightarrow j }\) 的解
5.12 周二
- 朗之万方程量子化
- quantum damping \[\begin{gather} H = H_0 +H _1 ,\ H_0 = v a^\dagger a + \sum_k v_k b_k^\dagger b_k,\ H_1 = \sum_k g_k (a^\dagger b_k + b_k^\dagger a) \\ i \dot{a} = [a,\ H] = \dfrac{\partial H }{\partial a^\dagger} = v a + \sum_k g_k b_k \Rightarrow \dot{a} = A a + \xi , \ A =-i v , \ \xi = \sum_k g_k b_k \\ \dot{a} = -iv a -i \sum_k g_k b_k , \ \dot{b}_k = -iv_k b_k - i g_k a\\ \dot{a} = -iva -\sum_k g_k^2 \int_0^t dt^\prime a(t^\prime) e^{-i v_k(t-t^\prime) }+ f_a(t), \ f_a(t)= - i \sum_k g_k b_k(0) e^{-iv_k t} \end{gather}\] 如果\(v_k\)变化快，\(a(t^\prime)\)变化慢，\(\dot{a} = - iv a - \Gamma a + f_a(t)\)
- 多体局域化(MBL)
- 作业29：推导Scully 量子光学 chap 9，p272-275
利用 \( m \dot{x} = p \varpropto i (a-a^\dagger) \) \[m \ddot{x} = -\eta \dot{x } +\xi \Rightarrow \dot{a} = -\eta a +\xi, \ \dot{a}^\dagger = -\eta a^\dagger + \xi^\dagger,\ \langle \xi \rangle =0,\ \langle \xi^\dagger (t) \xi(t^\prime) \rangle = D \delta (t-t^\prime),\ \langle \xi^\dagger (t) \xi(t^\prime) \rangle \neq \langle \xi(t^\prime) \ \xi^\dagger (t) \rangle \] \(\xi,\ \xi^\dagger\) 可定义为 \(\xi = \sum_k g_k b_k e^{i kt},\ \xi^\dagger = \sum_k g_k^\ast b_k^\dagger e^{-ikt}\)，无穷个振子表示随机力算符
Ref： Scully 量子光学 chap 9
5.15 周五
- 系统与环境 \[ | \psi(t=0) \rangle = | \psi_s \rangle \otimes | \psi_E\rangle =| \psi_s \rangle \otimes | 0\rangle , \ | \psi(t) \rangle = U |\psi(0)\rangle, \ \rho(t) = U \rho(0) U^\dagger\] 对环境求迹， \[\rho_s (t) = Tr_E U \rho(0) U^\dagger = \sum_k E_k \rho_s(0) E_k^\dagger= \sum_k \rho_k,\ E_k= \langle k |U|0\rangle, \sum_k E_k E_k^\dagger = 1 \Rightarrow Tr(\rho_s(t)) =1\]
- Lindblad eq
- Spin-1/2
- Boson
-
作业30：Spin-1/2，\(\dot{\rho} = -i [w\sigma_z , \rho] + \gamma_+ (2 \sigma_+ \rho \sigma_- - \rho \sigma_- \sigma_+ - \sigma_- \sigma_+ \rho) + \gamma_- (2\sigma_- \rho \sigma_+ - \rho \sigma_+ \sigma_- - \sigma_+ \sigma_- \rho)\)，数值计算\(\rho_{\uparrow\uparrow},\ \rho_{\uparrow \downarrow}, \ \rho_{\downarrow\uparrow},\ \rho_{\downarrow \downarrow}\)
作业31：推导以上boson模型密度矩阵元\(\rho_{nm}\)的运动方程
Ref： The quantum-state diffusion model applied to open systems
\[\begin{gather} | d \psi \rangle = - i H | \psi \rangle dt + \sum_j L_j | \psi \rangle d \xi_j,\ \overline{d\xi}=0,\ \overline{d\xi^\ast d\xi} = 2 dt, \ H^\dagger \neq H\\ \rho(t+dt ) = (| \psi \rangle + | d\psi \rangle )(\langle \psi | + \langle d\psi |) \xrightarrow{作平均} \rho(t) + i [\rho H^\dagger - H \rho] dt + \sum_j L_j \rho L_j^\dagger 2dt\\ H = H_0 - i \Gamma,\ Tr(\rho) =1 \Rightarrow {\rm Lindblad\ eq}\ \dot{\rho} =- i[H,\rho] + \sum_j \gamma_j(2 L_j \rho L_j^\dagger - \rho L_j^\dagger L_j - L_j^\dagger L_j \rho) ,\ L_j \rightarrow \gamma_j L_j \end{gather}\]
\(H _0 = w \sigma_z , \ L_j = \sigma_- ,\ \sigma_+\)，\(\dot{\rho}_{ij} = \sum_{ij} | i \rangle \langle i | \rho |j \rangle \langle j |\) \[\dot{\rho} = -i [w\sigma_z , \rho] + \gamma_+ (2 \sigma_+ \rho \sigma_- - \rho \sigma_- \sigma_+ - \sigma_- \sigma_+ \rho) + \gamma_- (2\sigma_- \rho \sigma_+ - \rho \sigma_+ \sigma_- - \sigma_+ \sigma_- \rho)\]
\(H = w b^\dagger b,\ | n \rangle = | 0 \rangle , | 1 \rangle \cdots ,\ \dot{\rho}_{nm} = \langle n |\dot{\rho}| m \rangle\)，稳态 \(\dot{\rho}_{nm}=0\) \[ \dot{\rho} = -i [H ,\rho] + \gamma_+ [2b^\dagger \rho b - \rho b b^\dagger - b b^\dagger \rho] + \gamma_- [2 b \rho b^\dagger - \rho b^\dagger b - b^\dagger b \rho]\]
5.19 周二
- Bloch 定理
- 作业32：\(H = - \dfrac{\hbar^2}{2m} \dfrac{d^2}{dx^2} +V(x) \)，若势场为Kronig-Penney 势，利用布洛赫定理作数值计算，得出能带.
\(\psi_k = e^{ikx} u_k (x),\ u_k = u_k(x+R) = e^{i kx} \sum_n C_n e^{i G_n x}\) \[H = - \dfrac{\hbar^2}{2m} \dfrac{d^2}{dx^2} + \sum_m V_M e^{i G_M x} , H \psi = E \psi \Rightarrow H_{mn} = \dfrac{\hbar^2}{2m}(k+G_n) \delta_{nm} + V_{m-n}\] 平移性，\(k \leq BZ\)；BZ散射(作微扰处理)产生gap. ( Ref 黄昆固体物理学，Kittel 固体物理导论)
非晶：准晶/非公度，almost periodic systems
多个原子多个轨道交叠理解能带的形成
5.26 周二
- gap 理解
- 在能带中心\(k \sim 0 \Rightarrow\)有效质量\(m^\ast\)
- 布里渊器边界作二阶微扰\(\dfrac{1}{(2\pi/a+k+\delta) - k^2}\)，\(\delta>0, \ 1/\delta\)；\(\delta<0,\ -1/\delta\)，产生gap
- 自由电子+固体物理\(\Rightarrow\) 有效质量带入自由空间运动方程
- 半经典处理
- 布洛赫振荡
- 对称操作
- 对称操作\(T\)不可能导致不同能量态的混合
- \(T\)只能导致\(H\)中简并空间态的混合
- \(T\)维度等于简并个数(排除偶然简并)
- \(T\)的表示，不可约表示
- \(H\) 简并等价\(T\)高维表示，\(H\) 无简并等价\(T\)1d表示
- notation 三种符号
二阶微扰 \[\varepsilon_n = \varepsilon_k + \sum_{k^\prime} \dfrac{\langle k |V| k^\prime \rangle \langle k^\prime| V|k\rangle}{\varepsilon_{k^\prime} - \varepsilon_k } = \varepsilon_k - \sum_n |V_n|^2 /(\varepsilon_{G_n+k} - \varepsilon_k) \xrightarrow{k\sim 0 展开} \dfrac{k^2}{2m} + A k^2 + const = \dfrac{k^2}{2m^\ast}, \ V = \sum_n V_n e^{i G_n x} \]
\[ p = \hbar k,\ \dot{p} = q (v \times B + E) \Rightarrow v = \hbar k /m^\ast = \dfrac{1}{\hbar} \dfrac{\partial}{\partial k } \varepsilon_k, \ \hbar \dot{k} = q (E + \dot{r} \times B)\]
\(H = H_0 (x ) + F x, \ F = -E, \ \varepsilon_k = -2J \cos{k} \)
半经典处理 \(\dot{x} = \dfrac{1}{\hbar} \dfrac{\partial}{\partial k} \varepsilon_k, \ \dot{k} = -e E\)
振荡：\(x(t) = \dfrac{2J}{eE} [\cos{(k_0 -eEt)} - \cos{k_0}],\ k(t) = (k_0 -e Et) mod(2\pi)\)
\([T,H] =0 , \ TH\psi_{l \alpha} = \varepsilon_l T \psi_{l\alpha} = HT \psi_{l\alpha} \Rightarrow T \psi_{l \alpha} = \sum_{\beta} W_{l \alpha}^\beta \psi_{l\beta}\)
5.29 周五
- 能带形成
- floquet 含时演化，波函数稳定与不稳定
- 近自由电子
- Kronig-Penney 势
- \(sp^3\)杂化
- 高对称点简并度与轨道态
- 在\(\Gamma\)点，一重简并，s态；三重简并，p态
- 沿\(\Delta = (1,0,0)\)方向，假设操作P不改变kx空间，那么一重简并，s态，px态；二重简并，py态，pz态
- 沿\(\Sigma = (1,1,1)\)，s态，p态，一重简并
- 对称性减少会导致能级劈裂
6.2 周二
6.5 周五
- 习题课
- 受限系统
6.9 周二
- Tunneling
Tunneling
6.12 周五
- Bogoliubov变换
- 超导理论
-
作业34：将H写成BdG形式，\(H = - \sum_{ij} t_{ij} C^\dagger_i C_j + \Delta_{ij} C^\dagger_{i}C^\dagger_j + {\rm h.c.}\)
作业35：讨论无序\(h_i\)对SC(本征值)的影响，- p-wave \(H = -\sum_i C^\dagger_i C_{i+1} + (\mu+h_i) C_i^\dagger C_i + \Delta C^\dagger_{i} C^\dagger_{i+1} + h.c.\)
- s-wave \( H = -\sum_{i\sigma} C^\dagger_{i\sigma} C_{i+1,\sigma} + \Delta C^\dagger_{i\uparrow}C^\dagger_{i\downarrow}+h.c.+ \sum_{i\sigma} (\mu+h_i) C^\dagger_{i\sigma} C_{i\sigma}\)
6.16 周二
6.19 周五
- Andreev reflection (current)
BTK model