低温热容

固体在低温下的热容。也许能得到

$$C_V=\alpha T+\beta T^3+higher$$

电子热容

色散关系为

$$\epsilon(\vec k)=\frac{\hbar^2}{2m}\vec k^2,\quad k_1,k_2,k_3=n\frac{2\pi}a$$

所以态密度为

$$\mathcal D(\epsilon)=\frac{\mathrm d}{\mathrm d\epsilon}\int_{\epsilon(\vec k)<\epsilon}\frac V{(2\pi)^3}\mathrm d^3\vec k=\frac V{(2\pi)^3}\frac{\mathrm d}{\mathrm d\epsilon}\left[\frac43\pi\sqrt{\frac{2m\epsilon}{\hbar^2}}^3\right]=V\frac{2\pi(2m)^{3/2}}{(2\pi\hbar)^3}\epsilon^{1/2}\equiv\alpha V\epsilon^{1/2}$$

因此粒子数和能量为

$$N=\int_0^\infty\frac{\mathcal D(\epsilon)}{e^{(\epsilon-\epsilon_F)/k_BT}+1}\mathrm d\epsilon,\quad U=\int_0^\infty\frac{\epsilon \mathcal D(\epsilon)}{e^{(\epsilon-\epsilon_F)/k_BT}+1}\mathrm d\epsilon$$

当$T=0$时计算得

$$N=\frac32\alpha V\epsilon_F^{3/2},\quad U=\frac52\alpha V\epsilon_F^{5/2}$$

假设费米能不变，计算随时间的导数

$$\begin{aligned} \frac{\partial U}{\partial T}=&\int_0^\infty\frac{\epsilon\mathcal D(\epsilon)}{e^{(\epsilon-\epsilon_F)/k_BT}+e^{-(\epsilon-\epsilon_F)/k_BT}+2}\frac{\epsilon-\epsilon_F}{k_BT^2}\mathrm d\epsilon\\ =&\int_{-\epsilon_F/k_BT}^\infty\frac{\alpha V(k_BTu+\epsilon_F)^{3/2}}{e^u+e^{-u}+2}\frac{u}Tk_BT\mathrm du \end{aligned}$$

低温下积分下限可以改写为$-\infty$，而且从分母看只有接近零的部分有贡献

$$\frac{\partial U}{\partial T}\approx\alpha V\epsilon_F^{3/2}k_B\int_{-\infty}^{+\infty}\frac{u+\frac{k_BT}{\epsilon_F}u^2}{e^u+e^{-u}+2}+\mathcal O\left[\left(\frac{k_BT}{\epsilon_F}\right)^2\right]\mathrm du=\alpha V\epsilon_F^{3/2}k_B\frac{k_BT}{\epsilon_F}\frac{\pi^2}3$$

将其写为与体积无关的形式

$$C_V=\frac{2\pi^2}{9}N\frac{k_B^2T}{\epsilon_F}=\frac{2\pi^2m}9N\frac{k_B^2T}{\hbar^2k_F^2}$$

晶格热容

弹性波色散关系为

$$\epsilon(\vec k)=c|\vec k|,\quad k_1,k_2,k_3=n\frac{2\pi}a$$

态密度为

$$\mathcal D(\epsilon)=\frac{\mathrm d}{\mathrm d\epsilon}\int_{\epsilon(\vec k)<\epsilon}\frac V{(2\pi)^3}\mathrm d^3\vec k =\frac V{(2\pi)^3}\frac{\mathrm d}{\mathrm d\epsilon}\left[\frac43\left(\frac{\epsilon}{c}\right)^3\right]=\frac{1}{2\pi^3 c^3}V\epsilon^2$$

总共有$3N$种振动模式，所以

$$3N=\int_0^{k_BT_D}\mathcal D(\epsilon)\mathrm d\epsilon=\frac{Vk_B^3T_D^3}{6\pi^3c^3}$$

内能为

$$U=\int_0^{k_BT_D}\mathcal D(\epsilon)\frac\epsilon{e^{\epsilon/k_BT}-1}\mathrm d\epsilon =\frac{V}{2\pi^3c^3}(k_BT)^4\int_0^{T_D/T}\frac{x^3}{e^x-1}\mathrm dx$$

低温极限将积分上限改为$+\infty$

$$U=\frac{V}{2\pi^3c^3}(k_BT)^4\frac{\pi^4}{15}=\frac{3N}{5T_D^3}\pi^4k_BT^4\Rightarrow C_V=\frac{12\pi^4}{5}k_B\left(T/T_D\right)^3$$

高温极限

$$U=\frac{V}{2\pi^3c^3}(k_BT)^4\frac13(T_D/T)^3=\frac{V}{6\pi^3c^3}k_B^4T_D^3T\Rightarrow C_V=\frac{V}{6\pi^3c^3}k_B^4T_D^3=3Nk_B$$