固体比热容

Dulong–Petit

理想固体模型，每个振动自由度能量${\epsilon }^v={\displaystyle \frac{p^2}{2m}+\frac{m{\omega }^2{q}^2}{2}}$，具有两个正平方项，

由能量均分定理，每个正平方项平均能量为 $k_{B}T/2$，$\overline{{\epsilon }^v}=k_{B}T$，

$N$原子固体具有$3N-6\approx 3N$个振动自由度，因此比热容为

$$C=3N{k}_B$$

Einstein

振动同频假设，一维振子能量本征值${\displaystyle E_n=(n+\frac{1}{2})\hslash \omega }$，

配分函数${\displaystyle Z_{1D}=\sum _n{e}^{-\beta E_n}=1/2\sinh \left(\beta \hslash \omega /2\right)}$，$Z_{3D}=(Z_{1D}{)}^3$

$⟨E⟩/N={\displaystyle -\frac{3}{Z_{1D}}\frac{\partial Z_{1D}}{\partial \beta }=3(n_B(\beta \hslash \omega )+\frac{1}{2})\hslash \omega }$，

玻色占有数因子$n_B(x)=1/(e^x-1)$，因此比热容为

$$C=\frac{\partial ⟨E⟩}{\partial T}=3N{k}_B(\beta \hslash \omega {)}^2\frac{e^{\beta \hslash \omega }}{(e^{\beta \hslash \omega }-1{)}^2}$$

爱因斯坦温度$\hslash \omega =k_B{T}_{\text{Einstein}}$

Debye

周期性边界条件${\displaystyle \mathit{k}=\frac{2\pi (n_1,n_2,n_3)}{L}}$，连续化${\displaystyle \sum _{\mathit{k}}\to \frac{L^3}{(2\pi {)}^3}\int \text{d}\mathit{k}}$，
各向同性假设${\displaystyle k=\frac{\omega }{v}}$，状态数密度${\displaystyle {\int }_0^{{\omega }_d}g(\omega )\mathrm{d}\omega =3N}$

$$\begin{gathered} ⟨E⟩=\frac{L^3}{(2\pi {)}^3}\int 3(n_B(\beta \hslash \omega (\mathit{k}))+\frac{1}{2})\hslash \omega (\mathit{k})\text{d}\mathit{k} \\ ={\displaystyle {\int }_0^{{\omega }_d}(n_B(\beta \hslash \omega )+\frac{1}{2})\hslash \omega g(\omega )\mathrm{d}\omega } \end{gathered}$$

德拜频率${\omega }_d^3=6{\pi }^{2}n{v}^3$，状态数密度$g(\omega )={\displaystyle N\frac{9{\omega }^2}{{\omega }_d^3}}$

$$\begin{gathered} ⟨E⟩=\frac{9N\hslash }{{\omega }_d^3}{\int }_0^{{\omega }_d}\frac{{\omega }^3\mathrm{d}\omega }{e^{\beta \hslash \omega }-1}+\mathrm{与}T\mathrm{无}\mathrm{关}\mathrm{常}\mathrm{量} \\ =9N\frac{(k_{B}T{)}^4}{(\hslash {\omega }_d{)}^3}{\int }_0^{x_d}\frac{x^3\mathrm{d}x}{e^x-1}+\mathrm{与}T\mathrm{无}\mathrm{关}\mathrm{常}\mathrm{量} \end{gathered}$$

低温$x_d={\displaystyle \frac{\hslash {\omega }_d}{k_{B}T}\to \mathrm{\infty }}$，${\displaystyle {\int }_0^{x_d}\frac{x^3\mathrm{d}x}{e^x-1}\to 3!\zeta (4)=\frac{{\pi }^4}{15}}$，
德拜温度$\hslash {\omega }_d=k_B{T}_{\text{Debye}}$

$$C=3N{k}_B{\left(\frac{T}{T_{\text{Debye}}}\right)}^3\frac{4{\pi }^4}{5}\propto T^3$$

高温$x_d={\displaystyle \frac{\hslash {\omega }_d}{k_{B}T}\to 0}$，${\displaystyle {\int }_0^{{\omega }_d}\frac{9N\hslash {\omega }^3\mathrm{d}\omega }{{\omega }_d^3(e^{\beta \hslash \omega }-1)}\to =3N{k}_{B}T}$，
比热容回归Dulong–Petit定律

Drude Theory

假设

$1.$ 电子具有特征散射时间 $\tau$，其在时间$\text{d}t$内被散射的概率为${\displaystyle \frac{\text{d}t}{\tau }}$

$2.$ 电子被散射后动量变为零

$3.$ 在两次散射间隔之内电子仅受外部电磁场影响

由上述假设有

$$⟨\mathit{p}(t+\text{d}t)⟩=(1-\text{d}t/\tau )(\mathit{p}(t)+\mathit{F}\text{d}t)$$

$$\frac{\text{d}\mathit{p}}{\text{d}t}=\mathit{F}-\frac{\mathit{p}}{\tau }\mathit{F}=-e(\mathit{E}+\mathit{v}\times \mathit{B})$$

在无外场条件下电子动量期望随时间指数衰减

电磁场

稳态下$\dot{\mathit{p}}=0$，

若只存在电场，$\mathit{p}=-e\tau \mathit{E}$

设电子数密度为$n$，质量为$m$，可求出电导率 $\sigma ={\displaystyle \frac{n{e}^2\tau }{m}}$

若存在电磁场，假设磁场方向沿$\hat{z}$

$$\mathit{E}=\frac{m}{n{e}^2\tau }\mathit{j}+\frac{1}{ne}\mathit{j}\times \mathit{B}=\mathit{\rho }\mathit{j}$$

$$\mathit{\rho }=\left(\begin{array}{ccc}{\displaystyle \frac{m}{n{e}^2\tau }}& {\displaystyle \frac{B}{ne}}& 0\\ -{\displaystyle \frac{B}{ne}}& {\displaystyle \frac{m}{n{e}^2\tau }}& 0\\ 0& 0& {\displaystyle \frac{m}{n{e}^2\tau }}\end{array}\right)$$

$$\mathit{\sigma }={\mathit{\rho }}^{-1}=\left(\begin{array}{ccc}{\displaystyle \frac{n{e}^2\tau }{{\displaystyle m(1+\frac{B^2{e}^2{\tau }^2}{m^2})}}}& -\frac{{\displaystyle n{e}^2\tau }}{{\displaystyle m(\frac{m}{Be\tau }+\frac{Be\tau }{m})}}& 0\\ {\displaystyle \frac{n{e}^2\tau }{m{\displaystyle (\frac{m}{Be\tau }+\frac{Be\tau }{m})}}}& {\displaystyle \frac{n{e}^2\tau }{{\displaystyle m(1+\frac{B^2{e}^2{\tau }^2}{m^2})}}}& 0\\ 0& 0& {\displaystyle \frac{n{e}^2\tau }{m}}\end{array}\right)$$

霍尔系数

$${\displaystyle R_H=\frac{{\rho }_{yx}}{B}=-\frac{1}{ne}}$$

热传导

$${\mathit{j}}_T=-\kappa \mathrm{\nabla }T$$

由

热导率公式，散射长度$\lambda =⟨v⟩\tau$，式中$⟨v{⟩}^2$也可用$⟨v^2⟩$替代

$$\kappa ={\displaystyle \frac{1}{3}n{c}_v⟨v⟩\lambda ={\displaystyle \frac{4}{\pi }\frac{n\tau k_B^{2}T}{m}}}$$

Wiedemann–Franz

$$L=\frac{\kappa }{T\sigma }=\frac{4}{\pi }{\left(\frac{k_B}{e}\right)}^2\approx 0.94\times {10}^{-8}{\text{V}}^2/{\text{K}}^2$$

Sommerfeld Theory

Fermi–Dirac Statistics

占有数因子

$$n_F={\displaystyle \frac{1}{e^{\beta (E-\mu )}+1}}$$

定义费米能 $E_F$ 为 $T=0$ 时化学势，此时占有数因子为阶跃函数，电子均匀分布在自由电子费米球$|k|<k_F$内

$$E_F=\frac{{\hslash }^2{k}_F^2}{2m}{T}_F=\frac{E_F}{k_B}$$

连续化${\displaystyle \sum _{\mathit{k}\sigma }=\frac{2V}{(2\pi {)}^3}\int \text{d}\mathit{k}}$，电子总数

$$N=\frac{2V}{(2\pi {)}^3}{\int }_0^{k_F}\text{d}\mathit{k}=\frac{V{k}_F^3}{3{\pi }^2}$$

$$k_F=(3{\pi }^{2}n{)}^{1/3}{E}_F=\frac{{\hslash }^2(3{\pi }^{2}n{)}^{2/3}}{2m}$$

电子比热容

能量

$$\begin{array}{rl}E& =\frac{2V}{(2\pi {)}^3}{\int }_0^{\mathrm{\infty }}ϵ(\mathit{k})n_F(\mathit{k})\text{d}\mathit{k}\\ \\ & =\frac{2V}{(2\pi {)}^3}{\int }_0^{\mathrm{\infty }}\frac{{\hslash }^2{k}^2}{2m}\cdot \frac{1}{\exp \left({\displaystyle \frac{\beta {\hslash }^2{k}^2}{2m}-\beta \mu }\right)+1}\text{d}\mathit{k}\\ \\ & ={\left(\frac{2m}{{\hslash }^2}\right)}^{3/2}\frac{V}{2{\pi }^2}{\int }_0^{\mathrm{\infty }}\frac{{ϵ}^{3/2}\mathrm{d}ϵ}{e^{\beta (ϵ-\mu )}+1}\\ \\ & \stackrel{T\to 0}{\approx }\frac{3N{k}_{B}T}{2}{\int }_0^{\mathrm{\infty }}\frac{y^{3/2}\mathrm{d}y}{e^{T_F/T(y-1)}+1}\end{array}$$

比热容

$$C=\frac{{\pi }^{2}N{k}_B}{2}\frac{T}{T_F}$$

低温下晶体总比热容

$$C=\alpha T^3+\beta T$$