Notes on World of Dreams

A brief summary of Chapter 7: MEAN FIELD THEORY OF PHASE TRANSITIONS

Fri, 23 Apr 2021 14:08:06 +0000

Chapter 7

7.4 Mean Field Theory

the Ising model Hamiltonian,

$$\hat{H}=-J \sum_{\langle i j\rangle} \sigma_{i} \sigma_{j}-H \sum_{i} \sigma_{i}$$

$\text { We will write }\left\langle\sigma_{i}\right\rangle \equiv m$, then we have

$$\begin{aligned} \sigma_{i} \sigma_{j} &=\left(m+\delta \sigma_{i}\right)\left(m+\delta \sigma_{j}\right) \ &=m^{2}+m\left(\delta \sigma_{i}+\delta \sigma_{j}\right)+\delta \sigma_{i} \delta \sigma_{j} \ &=-m^{2}+m\left(\sigma_{i}+\sigma_{j}\right)+\delta \sigma_{i} \delta \sigma_{j} . \end{aligned}$$

neglect the last term on RHS, substitute it in Hamiltonian,

$$\hat{H}{\mathrm{MF}}=\frac{1}{2} N z J m^{2}-(H+z J m) \sum{i} \sigma_{i}$$

define

$$H_{\mathrm{eff}}=H+z J m$$

which indicates the effective “mean field”.

We find the partition function of the canonical ensemble and the free energy that

$$\begin{aligned} Z &=\sum_{state} e^{-\beta \hat{H}} \ &=e^{-\frac{\beta}{2} N z J m^{2}} \cdot(2 \cosh \beta(H+2 J m))^{N} \end{aligned}\ \begin{aligned} F &=-K_{B} T \ln Z \ &=\frac{1}{2} N z J m^{2}-N K_{B} T \ln 2 \cosh \beta(H+z J m) \end{aligned}$$

adimensionalize the free energy

$$f(m, h, \theta)=\frac{1}{2} m^{2}-\theta \ln \left(e^{(m+h) / \theta}+e^{-(m+h) / \theta}\right) .$$

We know that we have to minimize the free energy at equilibrium under constant volume and constant temperature, so we can get the mean field equation by differentiate $f$

$$m=\tanh \left(\frac{m+h}{\theta}\right)$$

discussion:

(i) $h=0$

$$m=\tanh \left(\frac{m}{\theta}\right)$$

if $\theta\ge 1$, one solution $m=0$. If $\theta <1$, three solutions and they are symmetric. It means spontaneous magnetization occurs at the point.

$$\theta_C=1$$

is the mean field transition temperature, and its a second-order transition.

Also we find for $|\theta-1| \ll 1$,

$$m(\theta, h=0)=\pm \sqrt{3}(1-\theta)_{+}^{1 / 2}$$

the exponent $\beta$ equals 1/2.

now we compute the heat capacity. $\text { When } \theta \geqslant 1, m=0, \quad f=-\theta \ln 2 \quad c_{v}=0 \text { . }$

when $\theta<1$,

$$f=\frac{m^{2}}{2}-\theta\ln\left(2 \cosh \frac{m}{\theta}\right)$$ $$c_{V}=-\theta \frac{\partial^{2} f}{\partial \theta^{2}}=\frac{1}{\theta} \cdot \frac{m^{2}(\theta)-m^{4}(\theta)}{\theta-1+m^{2}(\theta)}$$

the calculation is not complex.

$\text { Let } \theta \rightarrow 1 \text { and use } m^{2} \simeq 3 \theta^{2}\left(1-\theta^{2}\right)$, we can obtain $c_V\rightarrow \frac32$

(ii)sets $\frac{\partial f}{\partial m}=0$ and $\frac{\partial^{2} f}{\partial m^{2}}=0$ simultaneously, resulting in

$$h^{*}(\theta)=\sqrt{1-\theta}-\frac{\theta}{2} \ln \left(\frac{1+\sqrt{1-\theta}}{1-\sqrt{1-\theta}}\right) .$$

$\text { The solutions lie at } h=\pm h^{*}(\theta) . \text { For } \theta<\theta_{\mathrm{c}}=1 \text { and } h \in\left[-h^{*}(\theta),+h^{*}(\theta)\right]$, there are three solutions to the mean field equation

Setting $\frac{\partial f}{\partial m}=0$, we obtain

$$\frac{1}{3} m^{3}+(\theta-1) \cdot m-h=0 \text { . }$$

still assumed $h \ll|\theta-1| \ll 1$,

if $\theta >1$ we have

$$m=h /(\theta-1)$$

and we can obtain the Curie-Weiss law, which shows the critical exponent $\gamma=1$

$$\chi(\theta)=\frac{\partial m}{\partial h}=\frac{1}{\theta-1} \propto|\theta-1|^{-\gamma}$$

if $\theta <1$, the solution above becomes a maxima, and the minima is

$$m(\theta, h)=\sqrt{3}(1-\theta)+\frac{h}{2(1-\theta)}$$

the critical exponent $\gamma$ is also 1.

if $\theta=1$,

$$m\left(\theta=\theta_{\mathrm{c}}, h\right)=(3 h)^{1 / 3} \propto h^{1 / \delta}$$

In fact, it turns out that the mean field exponents are exact provided d > du, where du is the upper critical dimension of the theory. For the Ising model, du = 4, and above four dimensions (which is of course unphysical) the mean field exponents are in fact exact.

dynamics

7.5 Variational Density Matrix Method

Suppose we are given a Hamiltonian $\hat{H}$. From this we construct the free energy, $F$ :

$$\begin{aligned} F &=E-T S \ &=\operatorname{Tr}(\varrho \hat{H})+k_{\mathrm{B}} T \operatorname{Tr}(\varrho \ln \varrho) \end{aligned}$$

Let us assume that ̺ is diagonal in the basis of eigenstates of ˆH , i.e.

$$\varrho=\sum_{\gamma} P_{\gamma}|\gamma\rangle\langle\gamma|$$

use it to calculate Ising model

$$\hat{H}=-\sum_{i<j} J_{i j} \sigma_{i} \sigma_{j}-H \sum_{i} \sigma_{i}$$

We now write a trial density matrix which is a product over contributions from independent single sites:

$$\varrho_{N}\left(\sigma_{1}, \sigma_{2}, \ldots\right)=\prod_{i} \varrho\left(\sigma_{i}\right)$$

where

$$\varrho(\sigma)=\left(\frac{1+m}{2}\right) \delta_{\sigma, 1}+\left(\frac{1-m}{2}\right) \delta_{\sigma,-1} .$$

adopt another notation

$$\varrho=\left(\begin{array}{cc} \frac{1+m}{2} & 0 \ 0 & \frac{1-m}{2} \end{array}\right)\ \varrho_{N}=\varrho \otimes \varrho \otimes \cdots \otimes \varrho$$

so that we can use $Tr(A\otimes B)=Tr(A)Tr(B)$ to evaluate F and S:

$$\begin{aligned} E &=\operatorname{Tr}\left(\varrho_{N} \hat{H}\right)=-\sum_{i<j} J_{i j} m^{2}-H \sum_{i} m \ &=-\frac{1}{2} N \hat{J}(0) m^{2}-N H m \end{aligned} \ \begin{aligned} S &=-k_{\mathrm{B}} \operatorname{Tr}\left(\varrho_{N} \ln \varrho_{N}\right)=-N k_{\mathrm{B}} \operatorname{Tr}(\varrho \ln \varrho) \ &=-N k_{\mathrm{B}}\left\\left(\frac{1+m}{2}\right) \ln \left(\frac{1+m}{2}\right)+\left(\frac{1-m}{2}\right) \ln \left(\frac{1-m}{2}\right)\right\ . \end{aligned}$$

and we can obtain the dimensionless free energy per site

$$f(m, h, \theta)=-\frac{1}{2} m^{2}-h m+\theta\left\\left(\frac{1+m}{2}\right) \ln \left(\frac{1+m}{2}\right)+\left(\frac{1-m}{2}\right) \ln \left(\frac{1-m}{2}\right)\right\$$

We extremize $f(m)$ by setting

$$\begin{array}{c} \frac{\partial f}{\partial m}=0=-m-h+\frac{\theta}{2} \ln \left(\frac{1+m}{1-m}\right) . \ m=\tanh \left(\frac{m+h}{\theta}\right) \end{array}$$

which is the same as the result of mean field theory. The discussion next is quite the same. In 7.10 we will see the equivalence of the two methods.

7.6 Landau Theory of Phase Transitions

based on expansion of the free energy of a thermodynamic system in terms of an order parameter

(1) quartic free energy

$$f(m, h, \theta)=f_{0}+\frac{1}{2} a m^{2}+\frac{1}{4} b m^{4}-h m$$

(i) h=0

symmetric

$$\frac{\partial f}{\partial m}=0=a m+b m^{3}$$

unique temperature θc where a(θc) = 0

$$\begin{array}{l} \theta<\theta_{\mathrm{c}} \quad: \quad f(\theta)=f_{0}-\frac{a^{2}}{4 b} \ \theta>\theta_{\mathrm{c}} \quad: \quad f(\theta)=f_{0} \end{array}$$ $$c\left(\theta_{\mathrm{c}}^{+}\right)-c\left(\theta_{\mathrm{c}}^{-}\right)=-\left.\theta_{\mathrm{c}} \frac{\partial^{2}}{\partial \theta^{2}}\right|{\theta=\theta{\mathrm{c}}}\left(\frac{a^{2}}{4 b}\right)=-\frac{\theta_{\mathrm{c}}\left[a^{\prime}\left(\theta_{\mathrm{c}}\right)\right]^{2}}{2 b\left(\theta_{\mathrm{c}}\right)}$$

$\theta_C$二阶相变点，比热突变

(ii)h≠0

asymmetric

$$b m^{3}+a m-h=0 .$$

$f^{\prime \prime}(m)=0$ as well as $f^{\prime}(m)=0$ we can obtain

$$a^{*}(h)=-\frac{3}{2^{2 / 3}} b^{1 / 3}|h|^{2 / 3}$$

$a<a^{*}(h)$, then there will be three real solutions to the mean field equation $f^{\prime}(m)=0$, one of which is a global minimum (the one for which $m \cdot h>0$ ).

For $a>a^{*}(h)$ one minima disappear and there is only a single global minimum

一阶相变

(2) cubic terms

$$f=f_{0}+\frac{1}{2} a m^{2}-\frac{1}{3} y m^{3}+\frac{1}{4} b m^{4}\ \frac{\partial f}{\partial m}=0=a m-y m^{2}+b m^{3}$$

we can obtain

$$\begin{aligned} a>\frac{y^{2}}{4 b} &: 1 \text { real root } m=0 \ \frac{y^{2}}{4 b}>a>\frac{2 y^{2}}{9 b} &: 3 \text { real roots; minimum at } m=0 \ \frac{2 y^{2}}{9 b}>a &: 3 \text { real roots; minimum at } m=\frac{y}{2 b}+\sqrt{\left(\frac{y}{2 b}\right)^{2}-\frac{a}{b}} \end{aligned}$$

thus $y^{2}=\frac{9}{2} a b$ denotes a line of first order transitions

if we impose some dynamics on the system, adimensionalize the free energy

$$m \equiv \frac{y}{b} \cdot u \quad, \quad a \equiv \frac{y^{2}}{b} \cdot r \quad, \quad t \equiv \frac{b}{\Gamma y^{2}} \cdot s \ \varphi(u)=\frac{1}{2} r u^{2}-\frac{1}{3} u^{3}+\frac{1}{4} u^{4} .$$

(3)Sixth order Landau theory : tricritical point

$$f=f_{0}+\frac{1}{2} a m^{2}+\frac{1}{4} b m^{4}+\frac{1}{6} c m^{6}\ \frac{\partial f}{\partial m}=0=m\left(a+b m^{2}+c m^{4}\right)$$

(i)$\text { If } a>0 \text { and } b>0$, $\text { a unique minimum at } m=0 \text { . }$

(ii)$\text { For } a<0$, One of the solutions is $m=0$. The other two are

$$m=\pm \sqrt{-\frac{b}{2 c}+\sqrt{\left(\frac{b}{2 c}\right)^{2}-\frac{a}{c}}}$$

(iii)$a>0 \text { and } b<0$

$$\begin{aligned} &\begin{aligned} b>-2 \sqrt{a c} &: & 1 \text { real root } m=0 \ -2 \sqrt{a c}>b>-\frac{4}{\sqrt{3}} \sqrt{a c} &: & 5 \text { real roots; minimum at } m=0 \end{aligned}\ &-\frac{4}{\sqrt{3}} \sqrt{a c}>b \quad: \quad 5 \text { real roots; minima at } m=\pm \sqrt{-\frac{b}{2 c}+\sqrt{\left(\frac{b}{2 c}\right)^{2}-\frac{a}{c}}} \end{aligned}$$

The point (a, b) = (0, 0), which lies at the confluence of a first order line and a second order line, is known as a tricritical point.

dynamics

$$m \equiv \sqrt{\frac{|b|}{c}} \cdot u \quad, \quad a \equiv \frac{b^{2}}{c} \cdot r \quad, \quad t \equiv \frac{c}{\Gamma b^{2}} \cdot s .\ \varphi(u)=\frac{1}{2} r u^{2} \pm \frac{1}{4} u^{4}+\frac{1}{6} u^{6} .$$

7.11.3 Canted quantum antiferromagnet

Consider the following model for quantum $S=\frac{1}{2}$ spins:

$$\hat{H}=\sum_{\langle i j\rangle}\left[-J\left(\sigma_{i}^{x} \sigma_{j}^{x}+\sigma_{i}^{y} \sigma_{j}^{y}\right)+\Delta \sigma_{i}^{z} \sigma_{j}^{z}\right]+\frac{1}{4} K \sum_{\langle i j k l\rangle} \sigma_{i}^{z} \sigma_{j}^{z} \sigma_{k}^{z} \sigma_{l}^{z}$$

we include a parameter α which describes the canting angle that the spins on these sublattices make with respect to the ˆx-axis.

so the variational density matrix of each site is

$$\begin{array}{l} \varrho_{\mathrm{A}}=\frac{1}{2}+\frac{1}{2} m\left(\sin \alpha \sigma^{x}+\cos \alpha \sigma^{z}\right) \ \varrho_{\mathrm{B}}=\frac{1}{2}+\frac{1}{2} m\left(\sin \alpha \sigma^{x}-\cos \alpha \sigma^{z}\right) \end{array}$$

Finally, the eigenvalues of $\varrho_{\mathrm{A}, \mathrm{B}}$ are still $\lambda_{\pm}=\frac{1}{2}(1 \pm m)$, hence

$$\begin{aligned} s(m) & \equiv-\operatorname{Tr}\left(\varrho_{\mathrm{A}} \ln \varrho_{\mathrm{A}}\right)=-\operatorname{Tr}\left(\varrho_{\mathrm{B}} \ln \varrho_{\mathrm{B}}\right) \ &=-\left[\frac{1+m}{2} \ln \left(\frac{1+m}{2}\right)+\frac{1-m}{2} \ln \left(\frac{1-m}{2}\right)\right] . \end{aligned}$$

the calculation is similar to the 7.5, we can obtain the free energy and adimensionalize it

$$\begin{aligned} F &=\operatorname{Tr}(\varrho \hat{H})+k_{\mathrm{B}} T \operatorname{Tr}(\varrho \ln \varrho) \ &=-2 N\left(J \sin ^{2} \alpha+\Delta \cos ^{2} \alpha\right) m^{2}+\frac{1}{4} N K m^{4} \cos ^{4} \alpha-N k_{\mathrm{B}} T s(m) \end{aligned}\ f(m, \alpha)=-\frac{1}{2} m^{2}+\frac{1}{2}(1-\delta) m^{2} \cos ^{2} \alpha+\frac{1}{4} \kappa m^{4} \cos ^{4} \alpha-\theta s(m)$$

in it, $f \equiv F / 4 N J,\delta \equiv \Delta / J, \kappa \equiv K / 4 J, \text { and } \theta \equiv k_{\mathrm{B}} T / 4 J$

similarly to find its minima, we have

$$\begin{array}{l} \frac{\partial f}{\partial m}=0=-m+(1-\delta) m \cos ^{2} \alpha+\kappa m^{3} \cos ^{4} \alpha+\frac{1}{2} \theta \ln \left(\frac{1+m}{1-m}\right) \ \frac{\partial f}{\partial \alpha}=0=\left(1-\delta+\kappa m^{2} \cos ^{2} \alpha\right) m^{2} \sin \alpha \cos \alpha \end{array}$$

and we can obtain

$$\cos ^{2} \alpha=\left\\begin{array}{ll} 0 & \text { if } \delta<1 \ (\delta-1) / \kappa m^{2} & \text { if } 1 \leq \delta \leq 1+\kappa m^{2} \ 1 & \text { if } \delta \geq 1+\kappa m^{2} \end{array}\right.$$

discussion:

(i)$\delta<1$, $\cos \alpha=0$, $m=\tanh (m / \theta)$, so A and B have no difference in angles. The solution is the same as discussed above.

(ii)$1<\delta<1+\kappa m^{2}$, we have canting with an angle

$$\alpha=\alpha^{*}(m)=\cos ^{-1} \sqrt{\frac{\delta-1}{\kappa m^{2}}}$$

$\text { we can also obtain the relation } m=\tanh (m / \theta)$, and the boundary is

$$\theta_{0}=\frac{m_{0}}{\tanh ^{-1}\left(m_{0}\right)}<1 \quad ; \quad m_{0}=\sqrt{\frac{\delta-1}{\kappa}}$$

(iii)$\theta>\theta_{0} \text { , we have } \delta>1+\kappa m^{2} \text { , and we must take } \cos ^{2} \alpha=1 \text { . }$

$$\delta m-\kappa m^{3}=\frac{\theta}{2} \ln \left(\frac{1+m}{1-m}\right)$$

through this equation we can obtain m.

Windows常用快捷键总结

Mon, 01 Mar 2021 14:32:22 +0000

总结一下好用的Windows快捷键（实际上是我常用的）

Ctrl+C 复制

Ctrl+X 剪切

Ctrl+V 粘贴

Ctrl+Z 撤销

Ctrl+A 全选

Ctrl+S 保存

Ctrl+F 在当前网页或文件夹中查找（相当好用）

Ctrl+Shift+Esc 任务管理器

Alt+Tab 切换任务

Win+D 回到桌面（同触摸屏三指向下）

Win+E 打开文件管理器

Win+L 锁定计算机

Win+R 运行

Win+Ctrl+D 创建新的虚拟桌面

Win+Ctrl+左/右 切换虚拟桌面

PicGo图床+GitHub链接配置方法

Tue, 23 Feb 2021 16:03:12 +0000

最近已经相当习惯于用typora写东西，清晰简洁，也可以直接复制到wordpress发布，各大主流平台也基本都支持markdown的格式。但typora唯一的不方便之处就是图片的链接就十分麻烦，首先是电脑上截屏的图片虽说可以直接粘到typora里，但是截屏的图片是保存在剪贴板里的，链接一段时间后就会失效。其次是直接把md复制到博客中的话图片还要另外上传到服务器中，再重新设置链接，就让人十分烦恼。

今天发现了typora还自带配置图床功能（越发感觉typora的优秀了），于是打算配置一个图床。图床就是将图片上传到服务器上，这样博客和本地都可以通过这个链接访问到图片，不需要麻烦地粘链接啦（当然这需要有网）

下载PicGo

PicGo是typora支持的一个图床软件，在typora上可以直接通过PicGo上传图片并自动生成链接。

去GitHub下载PicGo

https://github.com/Molunerfinn/PicGo/releases

windows下载exe文件 macOS下载dmg文件

安装

创建GitHub仓库

在用户下拉面板 – Your repositories – New
获取token

在 Settings – Developer settings – Personal access tokens

输入Note，勾选repo，再点击下方的 generate token

复制token，保存下来（它只会出现一次）

配置PicGo

点击图床设置-Github图床

仓库名是username/你刚刚设置的名字

分支刚刚生成的默认分支是main，也可以自己新建分支

Token就是刚刚生成的token

存储路径img/就好

确定！ 完成！

在typora中使用PicGo

偏好设置中

将上传服务设为PicGo(app)，路径写刚刚安装的exe的路径就好啦😜

以后打算就用 wordpress+typora+PicGo的写作模式啦

连接信息：要执行请求的操作，WordPress需要访问您网页服务器的权限问题

Tue, 23 Feb 2021 15:19:23 +0000

安装好wordpress后，在进行更新和安装插件时，遇到了 “要执行请求的操作，WordPress需要访问您网页服务器的权限。请输入您的FTP登录凭据以继续。如果您忘记了您的登录凭据（如用户名、密码），请联系您的网站托管商。” 这个问题，

即使输入密码也会显示错误，百度发现这是由于服务器端的权限设置问题，由于我是用的独立服务器，因此需要的是修改网站所在目录属性

`1`	`chmod -R 777 /data/www/wordpress`

其中 /data/www/wordpress 需要改为你自己的网站路径

再执行

`1`	`chown -R www /data/www/wordpress`

www为nginx/apache的user name, 可进入conf文件获知

/data/www/wordpress 同样的为网站路径

其实出现这个的问题就是Apache/Nginx的执行身份非文件属主身份。

Nginx: Job for nginx.service failed because the control process exited 的解决办法

Tue, 23 Feb 2021 13:26:17 +0000

最近在centOS 8.2手动部署nginx的时候发现了如下问题

启动Nginx后

`1`	`service nginx start`

出现了报错

`1`	`Nginx: Job for nginx.service failed because the control process exited`

首先检查一下nginx的配置文件语法是否正确，命令行输入

`1`	`nginx -t`

若有语法错误会显示错误的位置，若语法正确仍然报错

可以尝试下面的方法解决

执行下列命令

1
2

sudo fuser -k 80/tcp
sudo fuser -k 443/tcp

再执行

`1`	`sudo service nginx restart`

解决！

发现这是因为此前服务器上预安装了Apache, 导致端口冲突，因此杀掉这两个进程就好了

参考网址：https://stackoverflow.com/questions/35868976/nginx-job-for-nginx-service-failed-because-the-control-process-exited/51684856#51684856

关于wordpress上传时报错”图像后期处理失败，可能是服务器忙或没有足够的资源。”的解决办法

Thu, 04 Feb 2021 01:32:19 +0000

今天试图上传一个1920×1080的图像时发现上传不上去，并产生了如下报错

我的第一反应是不是租的这个云服务器过于垃圾，于是登入后台查看发现还有很大空间，不至于连一个图片都传不上去。然后百度了一下发现是wordpress新版本的bug(⊙o⊙)？并找到了如下解决办法

经过尝试发现前四种都没有什么卵用╮(╯▽╰)╭ 我又不想压缩我宝贵的图片，于是又找到了下面的办法

将图片通过FTP工具上传到图片默认的WP文件夹中，使用wp插件Add From Server将图片导入媒体库。

成功！

1.首先用FTP客户端连接到主机，把图片传过去，wp默认的文件路径(如我的是) www.lightforever.top/ wp-content/ uploads/ , (不过好像不放在这也行，但是最好还是按照wp默认的目录结构来放叭)

2.打开wp后台，安装插件Add From Server – 从媒体库中打开插件 – 找到自己刚刚上传的图片 – import – 完成\￣︶￣*\)