有限温度量子场论-第一章 复习热力学与统计物理

本学期（2022年春季学期）的“有限温度量子场论”课笔记。授课老师是何联毅。希望能通过坚持记笔记来坚持认真听课~

“有限温QFT”=QFT+统计物理。因为是有限温度，所以系统不局限于基态，因此需要引入配分函数来描述。处理很多物理系统需要用到有限温度下的QFT，例如强关联多体系统。

Chapter 1. Review of Thermodynamics and Statistic Mechanics

1.1 Fundamental Thermodynamics Relations

Closed system with volume in thermodynamic equilibrium (TE), the folowing quantities are conserved:
(i) energy E.
(ii) particle number N.

However in relativistic QFT, N conserv must be modified because we can construct particle pairs from vacuum.
Now we consider some conserved charges, for example, the electrical charge.

In a closed system with volume and conserved charges in TE, the first law would be:

$$dE = TdS-pdV+\sum_{i=1}^M \mu_i dN_i$$

Fundamental for thermodynamics!

Then the energy can be written as a func of variables:

$$E=E(S,V,\{N_i\})$$

This is called as thermodynamically complete equations of states (EoS). This means all other quantities could be calculated as partial derivatives.

Extensive quantities(广延量): .
Intensive quantities(强度量): .

1.2 Microcanonical Ensemble

In statisticle mechanics, a closed system with volume and conserved charges is described by the microcanonical ensemble.
The partition function is related to all the quantities of the system by Boltzmann’s formula:

$$S(E,V,\{N_i\}) = \ln W(E,V,\{N_i\})$$

The partition function can be calculated using the microcanonical density of states .

$$\Delta(E,V,\{N_i\}) = \frac{dW}{dE} = \sum_R^\prime \delta(E-E_R)$$

Here the sum is over all possible micro states with conserved.

With we can get and then get all the quantities. However is very difficult to calculate. So we need the Canonical ensemble.

1.3 Canonical Ensemble

• Microcanonical ensemble: fixed .
• Canonical ensemble: fixed .

Constant : The system is in contact with a heat bath with temperature and is in TE.

• Construct the EoS: From to . F is the free energy. $$F \equiv E-\left.\frac{\partial E}{\partial S}\right|_{V,\left\{N_{i}\right\}} S \equiv E-T S$$

$$d F=d E-T d S-S d T=-S d T-p d V+\sum_{i=1}^{M} \mu_{i} d N_{i}$$

• Partition function: connected to the function of microcanonical ensemble. $$\begin{aligned} Z\left(T, V,\left\{N_{i}\right\}\right) & \equiv \int_{0}^{\infty} d E e^{-E / T} \Delta\left(E, V,\left\{N_{i}\right\}\right) \\ &=\int_{0}^{\infty} d E e^{-E / T} \sum_{R}^{\prime} \delta\left(E-E_{R}\right) \\ &=\sum_{R}^{\prime} e^{-E_{R} / T} \end{aligned}$$

$$F\left(T, V,\left\{N_{i}\right\}\right)=-T \ln Z\left(T, V,\left\{N_{i}\right\}\right)$$

No constraints for energy, but constraints for particle number . Still hard if there are many conserved charges.
No conserved charge -> equal to microcanonical ensemble. (black-body radiation)
(详细推导要看书，讲义没有。)
(以前推过全都忘了呜呜呜，好像还是在David Lee的课上听的。)

1.4 Grand Canonical Ensemble

Mostly used in this course (And research). The easist to calculate!

• Microcanonical ensemble: fixed .
• Canonical ensemble: fixed .
• Grand conanicial ensemble: fixed .

The bath provides not only constant , but also constant . The system exchange both heat and particles.

• Construct the EoS: Grand potential .

$$\Omega \equiv F-\left.\sum_{j=1}^{M} \frac{\partial F}{\partial N_{j}}\right|_{T, V,\left\{N_{i}\right\}_{i \neq j}} N_{j}=F-\sum_{j=1}^{M} \mu_{j} N_{j}$$ $$d\Omega =-S d T-p d V-\sum_{j=1}^{M} N_{j} d \mu_{j}$$

• Grand potential is an extansive quantity: depends linearly on .

$$\left.\frac{\partial \Omega}{\partial V}\right|_{T,\left\{\mu_{i}\right\}}=-p=\frac{\Omega}{V}$$

• Partition function: related to canonical partition function .

$$\begin{aligned} & \mathcal{Z}\left(T, V,\left\{\mu_{i}\right\}\right) \\ \equiv & \sum_{N_{1}} e^{\mu_{1} N_{1} / T} \ldots \sum_{N_{M}} e^{\mu_{M} N_{M} / T} Z\left(T, V, N_{i}\right) \\ =& \sum_{N_{1}} e^{\mu_{1} N_{1} / T} \ldots \sum_{N_{M}} e^{\mu_{M} N_{M} / T} \sum_{R} e^{-E_{R} / T} \\ =& \sum_{N_{1}, N_{2}, \ldots N_{M}} \sum_{R} e^{-\left(E_{R}-\sum_{i=1}^{M} \mu_{i} N_{i}\right) / T} \end{aligned}$$

1.5 Ideal Bose and Fermi Gas

Simple application: ideal (no interaction) Bose and Fermi gases, calculate grand canonical partition function.

Eigen states with energy . No degenerate. The energy for micro state :

$$E_R = \sum_{r=0}^\infty n_r\epsilon_r$$

Conserved charge:

$$N_{R}=\sum_{r=0}^{\infty} q n_{r}$$

Writting down the grand partition function:

$$\begin{aligned} \mathcal{Z}_{0}^{\mathrm{BE} / \mathrm{FD}} &=\sum_{R} e^{-\left(E_{R}-\mu N_{R}\right) / T} \\ &=\sum_{n_{0}=0}^{\infty / 1} \sum_{n_{1}=0}^{\infty / 1} \cdots \exp \left[-\sum_{r=0}^{\infty} \frac{\left(\varepsilon_{r}-\mu\right) n_{r}}{T}\right] \\ &=\sum_{n_{0}=0}^{\infty / 1} e^{-\left(\varepsilon_{0}-\mu\right) n_{0} / T} \sum_{n_{1}=0}^{\infty / 1} e^{-\left(\varepsilon_{1}-\mu\right) n_{1} / T} \cdots \\ &=\prod_{r=0}^{\infty} \sum_{n_{r}=0}^{\infty / 1}\left[e^{-\left(\varepsilon_{r}-\mu\right) / T}\right]^{n_{r}} \end{aligned}$$

• For relativistic ideal bosons, , so . The sum convergent.

  $$\mathcal{Z}_{0}^{\mathrm{BE} }=\prod_{r=0}^{\infty}\left[1-e^{-\left(\varepsilon_{r}-\mu\right) / T}\right]^{-1}$$

• For relativistic ideal fermions, the sum only has two terms (to be occupied or not).

  $$\mathcal{Z}_{0}^{\mathrm{FD} }=\prod_{r=0}^{\infty}\left[1+e^{-\left(\varepsilon_{r}-\mu\right) / T}\right]$$

• The average occupation number of state :

$$\begin{aligned} \left\langle n_{r}^{\mathrm{BE} / \mathrm{FD}}\right\rangle=&\frac{1} { {\mathcal Z}_{0}^{\mathrm{BE} / \mathrm{FD} } } \sum_{R} n_{r} e^{-\sum_{s=0}^{\infty} n_{s}\left(\varepsilon_{s}-\mu\right) / T}\\ =&-\frac{\partial}{\partial x_{r} } \ln \mathcal{Z}_{0}^{\mathrm{BE} / \mathrm{FD} }|_{x_{s, s \neq r} } =\frac{1}{e^{\left(\varepsilon_{r}-\mu\right) / T \mp 1}} \end{aligned}$$

The famous Maxwell-Boltzmann (MB) is the limit (high ) of this result:

$$\ln \mathcal{Z}_{0}^{\mathrm{MB}}=\sum_{r=0}^{\infty} e^{-\left(\varepsilon_{r}-\mu\right) / T}, \quad\left\langle n_{r}^{\mathrm{MB}}\right\rangle=e^{-\left(\varepsilon_{r}-\mu\right) / T}$$

Another approach: 其实就是直接算。其实和在”An introduction to thermal physics”教材里的理想气体那一章的计算一模一样，看了就想起来了对应章节，就不写了。

这里计算一个单粒子问题就能得到气体性质是因为计算的是理想气体，也就是没有粒子间相互作用。如果有相互作用则要复杂许多。