平均場論(Mean field theory)

一般來說，對於\(H_0[\phi]\)，它的配分函數是 \[\begin{aligned} Z= \int e^{- \beta \{ H_0[\phi] - \int h(x)\phi(x)\, d^dx \}}\, \mathscr{D}[\phi] \end{aligned}\] Basic說或許我們可以將鞍點中的積分函數展開到二階來解，其中鞍點 (saddle point)是長這樣: \[\begin{aligned} \frac{\delta}{\delta \phi(x)} \left \{ H_0[\phi] - \int h(x)\phi(x)\, d^dx \right \} = 0 \end{aligned}\]

\(\phi^4\) theory

先將\(H_0[\phi] - \int h(x)\phi(x)\, d^dx\)進行展開 \[\begin{aligned} H_0[\phi] - \int h(x)\phi(x)\, d^dx = \int \left ( \frac{c}{2}\left \lvert \overset{\rightharpoonup} \nabla \phi \right \rvert ^2 + \frac{at}{2!}\phi^2+\frac{\phi^4}{4!} -h\phi \right)\, d^dx \equiv H \end{aligned}\] 然後讓 \[\begin{aligned} \frac{\delta H}{\delta \phi} \biggr |_{\phi_{saddle}}=0 \end{aligned}\] 就可以得到 \[\begin{aligned} H[\phi]=H[\phi_{saddle}]+\frac{1}{2}\int \int \underbrace {\delta \phi(x) \left[ \frac{\delta^2H}{\delta \phi(x) \delta \phi(x^\prime)} \right]_{saddle} \delta \phi(x^\prime)}_{\delta H}+\cdot \cdot \cdot \, d^dx \, d^dx^\prime \end{aligned}\] 其中\(\delta \phi \equiv \phi - \phi_{saddle}\)，將\(H\)代回\(Z\)就可以得到\(Z\)的形式為 \[\begin{aligned} Z=e^{-\beta F_{MF}} \int e^{- \beta \delta H[\delta \phi]}\, \mathscr{D}[\delta \phi] \end{aligned}\] 其中\(F_{MF}\)指的是平均場的free energy，並且\(F_{MF}=H[\phi_{saddle}]\)，接著進行一些計算 \[\begin{aligned} \frac{\delta^2H}{\delta \phi(x) \delta \phi(x^\prime)} &=\frac{\delta }{\delta \phi(x^\prime)}\left[ -c\nabla^2\phi(x)+at\phi(x)+\frac{a_4}{3!}\phi^3(x) \right]\\ &=\frac{\delta }{\delta \phi(x^\prime)}\int \delta^D(y-x)\left[ -c\nabla_y^2\phi(y)+at\phi(y)+\frac{a_4}{3!}\phi^3(y) \right] \, d^dy\\ &= \int\ \delta^D(y-x)\left[ -c\nabla_y^2+at+\frac{a_4}{2!}\phi^2(y) \right]\delta^D(y-x^\prime) \, d^dy\\ &=\left[ -c\nabla^2+at+\frac{a_4}{2!}\phi^2(x) \right]\delta^D(x-x^\prime) \end{aligned}\] \[\begin{aligned} \delta H [\delta \phi]&=\frac{1}{2}\int \int \delta\phi(x) \left[ -c\nabla^2+at+\frac{a_4}{2!}\phi^2(x) \right]\delta^D(x-x^\prime)\phi(x^\prime)\, d^dx\, d^dx^\prime\\ &=\frac{1}{2} \int \left [ c\left \lvert \overset{\rightharpoonup} \nabla \delta \phi \right \rvert ^2 + \left ( at+\frac{\phi^4}{2}\phi^2_{saddle}\right)\delta \phi^2 \right]\, d^dx\\ &=\frac{1}{2V}\sum_{\overset{\rightharpoonup} k}\underbrace{\left ( ck^2+at+\frac{a_4}{2V}\sum_{k^\prime}\left \lvert \phi_{saddle}(k^\prime) \right \rvert^2\right)}_{k_BTG^{-1}_0(k)}\delta\phi(k)\delta\phi(-k) \end{aligned}\] 最後就可以得到 \[\begin{aligned} \int e^{- \beta \delta H[\delta \phi]}\, \mathscr{D}[\delta \phi]=\prod_{\overset{\rightharpoonup}k}\sqrt{\frac{2\pi V}{G^{-1}_0(k)}} \end{aligned}\] \[\begin{aligned} F=F_{MF}-\underbrace{\frac{k_BT}{2}\sum_{\overset{\rightharpoonup} k}ln\frac{2\pi V}{G^{-1}_0(k)}}_{\mbox{修正為Landau理論}}+\cdot\cdot\cdot \end{aligned}\] 應變函數(response function)\(F_{MF}\) \[\begin{aligned} F_{MF}=\int \left ( \frac{c}{2}\left \lvert \overset{\rightharpoonup} \nabla \phi \right \rvert ^2 + at\phi^2+\frac{a_4}{4!}\phi^4 -h\phi \right)_{\phi=\phi_{MF}}\, d^dx \end{aligned}\] 其中\(\phi_{MF}\)是 \[\begin{aligned} -c\nabla^2\phi+at\phi+\frac{a_4}{3!}\phi^3 -h=0 \end{aligned}\] 接下來看在\(h\to 0\)時\(\phi_{MF}\)的解是甚麼 \[\phi_{MF}\lvert_{h\to 0}= \begin{cases} 0 &,t>0\\ \pm\sqrt{\frac{6a\lvert t\lvert}{a_4}}\equiv\phi_{\pm}&,t<0 \end{cases}\] 因此就可以知道 \[\begin{aligned} \phi_{MF}=\phi_{MF}\lvert_{h\to 0}+\int \chi_T(x-x^\prime)h(x^\prime) \, d^dx^\prime \end{aligned}\] 其中的磁化係數經過計算後可以知道為 \[\begin{aligned} \chi_T(k)=\frac{1}{ck^2+at+\frac{a_4}{2}\phi_{MF}\lvert_{h\to 0}}=\beta G_2(k)\lvert_{MF} \end{aligned}\]

Anomalous Dimensions

我們現在將證明當\(t\to 0\)時，對於\(d>4\)微擾不會導致新的影響。而對於\(d<4\)這個微擾理論是發散的，並且微擾的參數隨著\(T\to T_c\)而變得無限大。我們從配分函數開始: \[\begin{aligned} Z= \int e^{- \beta H}\, \mathscr{D}\phi \end{aligned}\] 其中\(H\)是 \[\begin{aligned} H = \int \left ( \frac{c}{2}\left \lvert \overset{\rightharpoonup} \nabla \phi \right \rvert ^2 + at\phi^2+\frac{a_4}{4!}\phi^4 -h\phi \right)\, d^dx \end{aligned}\] 重新調整階數參數\(\phi\)是一種常規且方便的方法，以便\(\left \lvert \overset{\rightharpoonup} \nabla \phi \right \rvert ^2\)項的係數為\(1/2\)。 這是透過定義來完成的 \[\begin{aligned} \psi&=(\beta c)^{\frac{1}{2}}\phi \\ \frac{r_0}{2}&=\frac{at}{\beta c}\\ \frac{u_0}{4}&=\frac{a_4}{4!\beta c^2} \end{aligned}\] 下面先考慮\(h=0\)的情況，就可以得到 \[\begin{aligned} \beta H = \int \left ( \frac{1}{2}\left \lvert \overset{\rightharpoonup} \nabla \psi \right \rvert ^2 + \frac{r_0}{2}\psi^2+\frac{u_0}{4}\phi^4 \right)\, d^dx \end{aligned}\] 接下來要對\(\beta H\)做些處理，讓它裡面的變數改寫為無因次的變數，共分為兩步驟進行。

第一步

確定方程式中各變數的因次，先確認\([\psi]\)的部分: \[\begin{aligned} \left [\int \left \lvert \overset{\rightharpoonup} \nabla \psi \right \rvert ^2 \, d^dx \right] = 1 \Rightarrow L^{d-2}[\psi]^2=1 \end{aligned}\] 所以 \[\begin{aligned} =L^{1-\frac{d}{2}} \end{aligned}\] \([r_0]\)的部分: \[\begin{aligned} \left [\int r_0 \psi^2 \, d^dx \right] = 1 \Rightarrow L^dL^{2-d}[r_0]=1 \end{aligned}\] 所以 \[\begin{aligned} =L^{-2} \end{aligned}\] \([u_0]\)的部分: \[\begin{aligned} \left [\int u_0 \psi^4 \, d^dx \right] = 1 \Rightarrow L^dL^{4-2d}[u_0]=1 \end{aligned}\] 所以 \[\begin{aligned} =L^{d-4} \end{aligned}\]

第二步

將式子改寫為無因次變數的形式，將算式中有因次的\(\psi\)和\(x\)和\(u_0\)除以它的因次，並取代為: \[\begin{aligned} \varphi&=\frac{\psi}{L^{1-d/2}} \\ r&=\frac{x}{L}\\ \bar{u_0}&=\frac{u_0}{L^{d-4}} \end{aligned}\] 其中\(L=_0^{-1/2}\propto \left \lvert t \right \rvert^{-1/2}\)，最後就可以得到無因次變數的\(\beta H\) \[\begin{aligned} \\ \beta H = \int \left [ \left(\frac{1}{2}\left \lvert \overset{\rightharpoonup} \nabla \varphi \right \rvert ^2 + \frac{1}{2}\varphi^2\right)+\frac{\bar{u_0}}{4}\varphi^4 \right]\, d^dx \\ \end{aligned}\] 可以發現\(\varphi^4\)是裡面最重要的項，而這個項取決於\(\bar{u_0}\)，\(\bar{u_0}\)代表了它的微擾係數，具體的形式為: \[\begin{aligned} \bar{u_0}=u_0r_0^{(d-4)/2} \end{aligned}\] 代入剛剛\(r_0\)正比於\(t\)的關係後，在\(t\to 0\)時可以發現 \[\bar{u_0}= \begin{cases} \infty &,d<4\\ 0 &,d>4 \end{cases}\] 平均場論(Mean field theory)隨著\(T\to T_c^+\)而越來越準確，因此因次分析已經使我們能夠確定正臨界時的大小，和 微擾對它的重要性。