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:

$$\mathcal{Z}=\sum_{R} e^{-\left(E_{R}-\sum_{i=1}^{M} \mu_{i} N_{i}\right) / T}$$

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

$$\mathcal{Z}=\sum_{R}\left\langle R\left|e^{-\left(\hat{H}-\sum_{i=1}^{M} \mu_{i} \hat{N}_{i}\right) / T}\right| R\right\rangle$$

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

$$\mathcal{Z}=\operatorname{Tr}\left[e^{-\beta\left(\hat{H}-\sum_{i=1}^{M} \mu_{i} \hat{N}_{i}\right)}\right]\tag{3}$$

Then we can define the density matrix:

$$\rho = [e^{-\beta\left(\hat{H}-\sum_{i=1}^{M} \mu_{i} \hat{N}_{i}\right)}]$$

And for any operator , the thermal average value is:

$$\langle \hat{O} \rangle = \frac{1}{\mathcal{Z} }\operatorname{Tr}(\hat{O}\rho)\tag{5}$$

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:

$$\hat{H}=\frac{\hat{P}^{2} }{2 m}+\frac{1}{2} m \omega^{2} \hat{X}^{2}, E_{n}=\hbar \omega\left(n+\frac{1}{2}\right)$$

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:

$$\begin{aligned} \mathcal{Z} &=\operatorname{Tr}\left(e^{-\beta \hat{H} }\right)=\sum_{n=0}^{\infty}\left\langle n\left|e^{-\beta \hat{H} }\right| n\right\rangle \\ &=\sum_{n=0}^{\infty} e^{-\beta \hbar \omega\left(n+\frac{1}{2}\right)}=\frac{e^{-\beta \hbar \omega / 2} }{1-e^{-\beta \hbar \omega} } \\ &=\frac{1}{2 \sinh \left(\frac{1}{2} \beta \hbar \omega\right)} \end{aligned}$$

The thermal avg of energy:

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

At low , , .

At high , , .

Path integal: using x-basis.

$$\mathcal{Z}=\int d x\left\langle x\left|e^{-\beta \hat{H} }\right| x\right\rangle$$

Compare to the propergation amplitude in path integal:

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

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

$$\begin{aligned} \mathcal{Z} &=\int d x\left\langle x\left|\prod_{i=1}^{N} e^{-\frac{\epsilon \hat{H} }{\hbar} }\right| x\right\rangle \\ &=\int d x\left\langle x\left|e^{-\frac{\epsilon \hat{H} }{\hbar} } \cdots e^{-\frac{\epsilon \hat{H} }{\hbar} } \cdots e^{-\frac{\epsilon \hat{H} }{\hbar} }\right| x\right\rangle . \end{aligned}$$

Then we do the trick of inserting orthogonal complete basis:

$$\int d x|x\rangle\langle x|=\hat{1}, \quad\left\langle x \mid x^{\prime}\right\rangle=\delta\left(x-x^{\prime}\right)$$ $$\int d p|p\rangle\langle p|=\hat{1}, \quad\left\langle p \mid p^{\prime}\right\rangle=\delta\left(p-p^{\prime}\right)$$ $$\langle x \mid p\rangle=\frac{1}{\sqrt{2 \pi \hbar} } e^{\frac{i p x}{\hbar} }$$

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

$$\begin{aligned} &\left\langle x\left|e^{-\frac{\Delta \tau \hat{H} }{\hbar} } \cdots e^{-\frac{\Delta \tau \hat{H} }{\hbar} } \cdots e^{-\frac{\Delta \tau \hat{H} }{\hbar} }\right| x\right\rangle \\ =& \int \prod_{i=1}^{N} d x_{i} d p_{i}\left\langle x_{N+1} \mid p_{N}\right\rangle\left\langle p_{N}\left|e^{-\frac{\Delta \tau \hat{H} }{\hbar} }\right| x_{N}\right\rangle \\ &\left\langle x_{N} \mid p_{N-1}\right\rangle\left\langle p_{N-1}\left|e^{-\frac{\Delta \tau \hat{H} }{\hbar} }\right| x_{N-1}\right\rangle \cdots \\ &\left\langle x_{2} \mid p_{1}\right\rangle\left\langle p_{1}\left|e^{-\frac{\Delta \tau \hat{H} }{\hbar} }\right| x_{1}\right\rangle\left\langle x_{1} \mid x_{0}\right\rangle \end{aligned}$$

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 :

$$\begin{aligned} &\left\langle x_{k+1} \mid p_{k}\right\rangle\left\langle p_{k}\left|e^{-\frac{\Delta \tau}{\hbar} \hat{H}(\hat{x}, \hat{p})}\right| x_{k}\right\rangle \\ =& \frac{1}{\sqrt{2 \pi \hbar} } e^{\frac{i p_{k} x_{k+1} }{\hbar} }\left\langle p_{k}\left|e^{-\frac{\Delta \tau}{\hbar} H\left(x_{k}, p_{k}\right)+O\left(\Delta \tau^{2}\right)}\right| x_{k}\right\rangle \\ =& \frac{1}{2 \pi \hbar} \exp \left\{-\frac{\Delta \tau}{\hbar}\left[\frac{p_{k}^{2} }{2 m}-i p_{k} \frac{x_{k+1}-x_{k} }{\Delta \tau}+V\left(x_{k}\right)+O(\Delta \tau)\right]\right\} \end{aligned}$$

Here is defined previously.

So the partition function is:

$$\begin{aligned} \mathcal{Z}=& \lim _{N \rightarrow \infty} \int\left[\prod_{i=1}^{N} \frac{d x_{i} d p_{i} }{2 \pi \hbar}\right] \\ &\left.\exp \left\{-\frac{1}{\hbar} \sum_{k=1}^{N} \Delta \tau\left[\frac{p_{k}^{2} }{2 m}-i p_{k} \frac{x_{k+1}-x_{k} }{\Delta \tau}+V\left(x_{k}\right)\right]\right\}\right|_{x_{N+1}=x_{1} } \end{aligned}$$

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

$$\begin{aligned} & \int_{-\infty}^{\infty} \frac{d p_{k} }{2 \pi \hbar} \exp \left\{-\frac{\Delta \tau}{\hbar}\left[\frac{p_{k}^{2} }{2 m}-i p_{k} \frac{x_{k+1}-x_{k} }{\Delta \tau}\right]\right\} \\ =& \sqrt{\frac{m}{2 \pi \hbar \Delta \tau} } \exp \left[-\frac{m\left(x_{k+1}-x_{k}\right)^{2} }{2 \hbar \Delta \tau}\right] \end{aligned}$$

Then the partition func becomes:

$$\begin{aligned} \mathcal{Z}=&\left.\lim _{N \rightarrow \infty} \int \prod_{i=1}^{N} \int \frac{d x_{i} }{\sqrt{2 \pi \hbar \Delta \tau / m} }\right] \\ &\left.\exp \left\{-\frac{1}{\hbar} \sum_{j=1}^{N} \Delta \tau\left[\frac{m}{2}\left(\frac{x_{k+1}-x_{k} }{\Delta \tau}\right)^{2}+V\left(x_{k}\right)\right]\right\}\right|_{x_{N+1}=x_{1} } \end{aligned}$$

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

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

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 :

$$\mathcal{Z}=C \int_{x(\beta \hbar)=x(0)} \mathcal{D} x \exp \left\{-\frac{1}{\hbar} \int_{0}^{\beta \hbar} d \tau\left[\frac{m}{2}\left(\frac{d x}{d \tau}\right)^{2}+V(x)\right]\right\} \tag{30}$$

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:

$$\exp \left(\frac{i}{\hbar} \int d t \mathcal{L}_{M}\right) \rightarrow \exp \left(-\frac{S_{E} }{\hbar}\right) \equiv \exp \left(-\frac{1}{\hbar} \int_{0}^{\beta \hbar} d \tau \mathcal{L}_{E}\right)$$

E: Euclidean.
M: Minkowski.

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

---

To compute (30), write into Fourier sum:

$$x(\tau)=\frac{1}{\beta} \sum_{n=-\infty}^{\infty} x_{n} e^{i \omega_{n} \tau} \tag{33}$$

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

$$x(\tau)=\frac{1}{\beta}\left\{a_{0}+\sum_{n=1}^{\infty}\left[\left(a_{n}+i b_{n}\right) e^{i \omega_{n} \tau}+\left(a_{n}-i b_{n}\right) e^{-i \omega_{n} \tau}\right]\right\} \tag{34}$$

Here, is called the amplitude of Matsubara zero mode.

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

$$\begin{aligned} \frac{1}{\hbar} \int_{0}^{\beta \hbar} d \tau x(\tau) y(\tau) &=\frac{1}{\beta^{2} } \sum_{m, n=-\infty}^{\infty} x_{n} y_{m} \frac{1}{\hbar} \int_{0}^{\beta \hbar} d \tau e^{i\left(\omega_{n}+\omega_{m}\right) \tau} \\ &=\frac{1}{\beta^{2} } \sum_{m, n=-\infty}^{\infty} x_{n} y_{m} \beta \delta_{n,-m} \\ &=\frac{1}{\beta} \sum_{n=-\infty}^{\infty} x_{n} y_{-n} \end{aligned}$$

Use this the int in is:

$$\begin{aligned} &-\frac{1}{\hbar} \int_{0}^{\beta \hbar} d \tau \frac{m}{2}\left[\frac{d x(\tau)}{d \tau} \frac{d x(\tau)}{d \tau}+\omega^{2} x(\tau) x(\tau)\right] \\ =&-\frac{m}{2 \beta} \sum_{n=-\infty}^{\infty}\left[i \omega_{n} i \omega_{-n}+\omega^{2}\right] x_{n} x_{-n} \\ =&-\frac{m}{2 \beta} \sum_{n=-\infty}^{\infty}\left(\omega_{n}^{2}+\omega^{2}\right)\left(a_{n}^{2}+b_{n}^{2}\right) \\ =&-\frac{m}{2 \beta} \omega^{2} a_{0}^{2}-\frac{m}{\beta} \sum_{n=1}^{\infty}\left(\omega_{n}^{2}+\omega^{2}\right)\left(a_{n}^{2}+b_{n}^{2}\right) \end{aligned}$$

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

$$\mathcal{D} x(\tau)=\left|\operatorname{det}\left[\frac{\delta x(\tau)}{\delta x_{n} }\right]\right| d a_{0}\left[\prod_{n=1}^{\infty} d a_{n} d b_{n}\right]$$

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

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

Now we have the partition function:

$$\mathcal{Z}= C^{\prime} \int_{-\infty}^{\infty} d a_{0} \int_{-\infty}^{\infty}\left[\prod_{n=1}^{\infty} d a_{n} d b_{n}\right] \exp \left[-\frac{m}{2 \beta} \omega^{2} a_{0}^{2}\right. \left.-\frac{m}{\beta} \sum_{n=1}^{\infty}\left(\omega_{n}^{2}+\omega^{2}\right)\left(a_{n}^{2}+b_{n}^{2}\right)\right]$$

This is a Gaussian integral, and finally:

$$\mathcal{Z}=C^{\prime} \sqrt{\frac{2 \pi \beta}{m \omega^{2} }} \prod_{n=1}^{\infty} \frac{\pi \beta}{m\left(\omega_{n}^{2}+\omega^{2}\right)}, \quad \omega_{n}=\frac{2 \pi n}{\beta \hbar}\tag{42}$$

---

Calculate :