有限温度量子场论-第二章 虚时路径积分（2.1 核心公式 2.2简谐振子模型）

第二章正式进入场论的计算。
平衡态巨正则系综可以表示成一个路径泛函。可以由场论方法非常自然地给出统计关系。

Chapter 2 Imaginary-Time Path Integral

2.1 The core formula in quantum statistic physics

The grand partition function is:

$$\mathcal{Z}=\sum _R{e}^{-(E_R-\sum _{i=1}^M{\mu }_i{N}_i)/T}$$

And for a quantum system with Halmitonian , since the number operators commute with , the (written in energy reprezentation)is:

$$\mathcal{Z}=\sum _R⟨R\left|e^{-(\hat{H}-\sum _{i=1}^M{\mu }_i{\hat{N}}_i)/T}\right|R⟩$$

For interacting system, it is very hard to calculate all eigen states and eigen energy. Generally, we can calculate the trace under any representation:

$$\begin{array}{}\text{(3)}& \mathcal{Z}=\mathrm{Tr}\left[e^{-\beta (\hat{H}-\sum _{i=1}^M{\mu }_i{\hat{N}}_i)}\right]\end{array}$$

Then we can define the density matrix:

$$\rho =[e^{-\beta (\hat{H}-\sum _{i=1}^M{\mu }_i{\hat{N}}_i)}]$$

And for any operator , the thermal average value is:

$$\begin{array}{}\text{(5)}& ⟨\hat{O}⟩=\frac{1}{\mathcal{Z}}\mathrm{Tr}(\hat{O}\rho )\end{array}$$

The equation (3) and (5) are the core formula in quantum statistic physics.

2.2 Thermodynamics of a harmonic oscillator

Harmonic oscillator is a simple but important model in quantum mechanics. It has a simple Halmitonian and energy levels:

$$\hat{H}=\frac{{\hat{P}}^2}{2m}+\frac{1}{2}m{\omega }^2{\hat{X}}^2,E_n=\hslash \omega (n+\frac{1}{2})$$

QM = (0+1)-dimensional QFT: .

The partition function can be easily calculated. First we use the imaginary-time path integral to get the partition func. Then we can compare it to the result from the direct calculation of (3).

In energy representation, can be easily calculated:

$$\begin{array}{rl}\mathcal{Z}& =\mathrm{Tr}\left(e^{-\beta \hat{H}}\right)=\sum _{n=0}^{\mathrm{\infty }}⟨n\left|e^{-\beta \hat{H}}\right|n⟩\\ & =\sum _{n=0}^{\mathrm{\infty }}{e}^{-\beta \hslash \omega (n+\frac{1}{2})}=\frac{e^{-\beta \hslash \omega /2}}{1-e^{-\beta \hslash \omega }}\\ & =\frac{1}{2\sinh \left(\frac{1}{2}\beta \hslash \omega \right)}\end{array}$$

The thermal avg of energy:

$$\begin{array}{rl}E& =\frac{1}{\mathcal{Z}}\mathrm{Tr}\left(\hat{H}{e}^{-\beta \hat{H}}\right)=-\frac{\partial \ln \mathcal{Z}}{\partial \beta }\\ & =\hslash \omega (\frac{1}{2}+\frac{1}{e^{\beta \hslash \omega }-1})=\frac{1}{2}\hslash \omega +\frac{\hslash \omega }{e^{\beta \hslash \omega }-1}\end{array}$$

At low , , .

At high , , .

Path integal: using x-basis.

$$\mathcal{Z}=\int dx⟨x\left|e^{-\beta \hat{H}}\right|x⟩$$

Compare to the propergation amplitude in path integal:

$$\mathcal{A}(t,x;t^{\prime },x^{\prime })=⟨x^{\prime }\left|e^{-i\hat{H}(t^{\prime }-t)/\hslash }\right|x⟩$$

With imaginary time interval we can write the partition func as the form of path integal (divide the time interval to parts: ):

$$\begin{array}{rl}\mathcal{Z}& =\int dx⟨x\left|\prod _{i=1}^N{e}^{-\frac{ϵ\hat{H}}{\hslash }}\right|x⟩\\ & =\int dx⟨x|e^{-\frac{ϵ\hat{H}}{\hslash }}\cdots e^{-\frac{ϵ\hat{H}}{\hslash }}\cdots e^{-\frac{ϵ\hat{H}}{\hslash }}|x⟩.\end{array}$$

Then we do the trick of inserting orthogonal complete basis:

$$\int dx|x⟩⟨x|=\hat{1},⟨x\mid x^{\prime }⟩=\delta (x-x^{\prime })$$$$\int dp|p⟩⟨p|=\hat{1},⟨p\mid p^{\prime }⟩=\delta (p-p^{\prime })$$$$⟨x\mid p⟩=\frac{1}{\sqrt{2\pi \hslash }}{e}^{\frac{ipx}{\hslash }}$$

For each time interval , we insert a basis and a basis:

$$\begin{array}{rl}& ⟨x|e^{-\frac{\mathrm{\Delta }\tau \hat{H}}{\hslash }}\cdots e^{-\frac{\mathrm{\Delta }\tau \hat{H}}{\hslash }}\cdots e^{-\frac{\mathrm{\Delta }\tau \hat{H}}{\hslash }}|x⟩\\ =& \int \prod _{i=1}^{N}d{x}_{i}d{p}_i⟨x_{N+1}\mid p_N⟩⟨p_N\left|e^{-\frac{\mathrm{\Delta }\tau \hat{H}}{\hslash }}\right|x_N⟩\\ & ⟨x_N\mid p_{N-1}⟩⟨p_{N-1}\left|e^{-\frac{\mathrm{\Delta }\tau \hat{H}}{\hslash }}\right|x_{N-1}⟩\cdots \\ & ⟨x_2\mid p_1⟩⟨p_1\left|e^{-\frac{\mathrm{\Delta }\tau \hat{H}}{\hslash }}\right|x_1⟩⟨x_1\mid x_0⟩\end{array}$$

Note here we have the periodic boundary condition(PBC): $x{N+1}=x{0}=x$. This comes from the Boson statistics. And for Fermions we will later see a anti-PBC.

Now we have to consider elements of the type :

$$\begin{array}{rl}& ⟨x_{k+1}\mid p_k⟩⟨p_k\left|e^{-\frac{\mathrm{\Delta }\tau }{\hslash }\hat{H}(\hat{x},\hat{p})}\right|x_k⟩\\ =& \frac{1}{\sqrt{2\pi \hslash }}{e}^{\frac{i{p}_k{x}_{k+1}}{\hslash }}⟨p_k\left|e^{-\frac{\mathrm{\Delta }\tau }{\hslash }H(x_k,p_k)+O\left(\mathrm{\Delta }{\tau }^2\right)}\right|x_k⟩\\ =& \frac{1}{2\pi \hslash }\exp \{-\frac{\mathrm{\Delta }\tau }{\hslash }[\frac{p_k^2}{2m}-i{p}_k\frac{x_{k+1}-x_k}{\mathrm{\Delta }\tau }+V\left(x_k\right)+O(\mathrm{\Delta }\tau )]\}\end{array}$$

Here is defined previously.

So the partition function is:

$$\begin{array}{rl}\mathcal{Z}=& \underset{N\to \mathrm{\infty }}{lim}\int \left[\prod _{i=1}^N\frac{d{x}_{i}d{p}_i}{2\pi \hslash }\right]\\ & {\exp \{-\frac{1}{\hslash }\sum _{k=1}^N\mathrm{\Delta }\tau [\frac{p_k^2}{2m}-i{p}_k\frac{x_{k+1}-x_k}{\mathrm{\Delta }\tau }+V\left(x_k\right)]\}|}_{x_{N+1}=x_1}\end{array}$$

The integral over the momenta is Gaussian, and can be calculated directly:

$$\begin{array}{rl}& {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{d{p}_k}{2\pi \hslash }\exp \{-\frac{\mathrm{\Delta }\tau }{\hslash }[\frac{p_k^2}{2m}-i{p}_k\frac{x_{k+1}-x_k}{\mathrm{\Delta }\tau }]\}\\ =& \sqrt{\frac{m}{2\pi \hslash \mathrm{\Delta }\tau }}\exp [-\frac{m{(x_{k+1}-x_k)}^2}{2\hslash \mathrm{\Delta }\tau }]\end{array}$$

Then the partition func becomes:

$$\begin{array}{rl}\mathcal{Z}=& \underset{N\to \mathrm{\infty }}{lim}\int \prod _{i=1}^N\int \frac{d{x}_i}{\sqrt{2\pi \hslash \mathrm{\Delta }\tau /m}}]\\ & {\exp \{-\frac{1}{\hslash }\sum _{j=1}^N\mathrm{\Delta }\tau [\frac{m}{2}{\left(\frac{x_{k+1}-x_k}{\mathrm{\Delta }\tau }\right)}^2+V\left(x_k\right)]\}|}_{x_{N+1}=x_1}\end{array}$$

Note that here we have a factor which is divergent and independent of :

$$C\equiv {\left(\frac{m}{2\pi \hslash \mathrm{\Delta }\tau }\right)}^{N/2}=\exp [\frac{N}{2}\ln \left(\frac{mN}{2\pi {\hslash }^2\beta }\right)]$$

Whene is big, this factor becomes exponetially big. However we do not need to warry because this could be cancelled out. Finally the is written as a functional integal of :

$$\begin{array}{}\text{(30)}& \mathcal{Z}=C{\int }_{x(\beta \hslash )=x(0)}\mathcal{D}x\exp \{-\frac{1}{\hslash }{\int }_0^{\beta \hslash }d\tau [\frac{m}{2}{\left(\frac{dx}{d\tau }\right)}^2+V(x)]\}\end{array}$$

From quantum mechanical path integral , to finite temperature integral:

• (i) Wick rotation: .
• (ii) $\mathcal{L}{E}=-\mathcal{L}{M}(\tau \equiv i t)=\frac{m}{2}\left(\frac{d x}{d \tau}\right)^{2}+V(x)$.
• (iii) .
• (iv) Require periodicity: .

With these steps the int becomes (30), which is called imaginary-time formalism:

$$\exp (\frac{i}{\hslash }\int dt{\mathcal{L}}_M)\to \exp (-\frac{S_E}{\hslash })\equiv \exp (-\frac{1}{\hslash }{\int }_0^{\beta \hslash }d\tau {\mathcal{L}}_E)$$

E: Euclidean.
M: Minkowski.

For fields with non-zero spin: few modification, still works.

---

To compute (30), write into Fourier sum:

$$\begin{array}{}\text{(33)}& x(\tau )=\frac{1}{\beta }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{x}_n{e}^{i{\omega }_n\tau }\end{array}$$

with periodic conditions, $\omegan = 2\pi n /\beta \hbarx(\tau)x_n^* = x{-n}xxn = a_n +ib_n, a_n = a{-n}, bn = -b{-n}$,

$$\begin{array}{}\text{(34)}& x(\tau )=\frac{1}{\beta }\{a_0+\sum _{n=1}^{\mathrm{\infty }}[(a_n+i{b}_n)e^{i{\omega }_n\tau }+(a_n-i{b}_n)e^{-i{\omega }_n\tau }]\}\end{array}$$

Here, is called the amplitude of Matsubara zero mode.

Generally with the Fourier sum , the quadratic structure int can be calculated:

$$\begin{array}{rl}\frac{1}{\hslash }{\int }_0^{\beta \hslash }d\tau x(\tau )y(\tau )& =\frac{1}{{\beta }^2}\sum _{m,n=-\mathrm{\infty }}^{\mathrm{\infty }}{x}_n{y}_m\frac{1}{\hslash }{\int }_0^{\beta \hslash }d\tau e^{i({\omega }_n+{\omega }_m)\tau }\\ & =\frac{1}{{\beta }^2}\sum _{m,n=-\mathrm{\infty }}^{\mathrm{\infty }}{x}_n{y}_m\beta {\delta }_{n,-m}\\ & =\frac{1}{\beta }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}{x}_n{y}_{-n}\end{array}$$

Use this the int in is:

$$\begin{array}{rl}& -\frac{1}{\hslash }{\int }_0^{\beta \hslash }d\tau \frac{m}{2}[\frac{dx(\tau )}{d\tau }\frac{dx(\tau )}{d\tau }+{\omega }^{2}x(\tau )x(\tau )]\\ =& -\frac{m}{2\beta }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}[i{\omega }_{n}i{\omega }_{-n}+{\omega }^2]x_n{x}_{-n}\\ =& -\frac{m}{2\beta }\sum _{n=-\mathrm{\infty }}^{\mathrm{\infty }}({\omega }_n^2+{\omega }^2)(a_n^2+b_n^2)\\ =& -\frac{m}{2\beta }{\omega }^2{a}_0^2-\frac{m}{\beta }\sum _{n=1}^{\mathrm{\infty }}({\omega }_n^2+{\omega }^2)(a_n^2+b_n^2)\end{array}$$

Next we need to consider the part. With , change the variables to and ,

$$\mathcal{D}x(\tau )=\left|\det \left[\frac{\delta x(\tau )}{\delta x_n}\right]\right|d{a}_0\left[\prod _{n=1}^{\mathrm{\infty }}d{a}_{n}d{b}_n\right]$$

The determinant is hard to compute and could be written into the previous hard-to-compute :

$$C^{\prime }\equiv C\left|\det \left[\frac{\delta x(\tau )}{\delta x_n}\right]\right|$$

Now we have the partition function:

$$\mathcal{Z}=C^{\prime }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}d{a}_0{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\left[\prod _{n=1}^{\mathrm{\infty }}d{a}_{n}d{b}_n\right]\exp [-\frac{m}{2\beta }{\omega }^2{a}_0^2-\frac{m}{\beta }\sum _{n=1}^{\mathrm{\infty }}({\omega }_n^2+{\omega }^2)(a_n^2+b_n^2)]$$

This is a Gaussian integral, and finally:

$$\begin{array}{}\text{(42)}& \mathcal{Z}=C^{\prime }\sqrt{\frac{2\pi \beta }{m{\omega }^2}}\prod _{n=1}^{\mathrm{\infty }}\frac{\pi \beta }{m({\omega }_n^2+{\omega }^2)},{\omega }_n=\frac{2\pi n}{\beta \hslash }\end{array}$$

---

Calculate :