#### 期刊菜单

Mathematical Description of the Landau Diamagnetism in the High Temperature Limit
DOI: 10.12677/MP.2021.113008, PDF, HTML, XML, 下载: 196  浏览: 303  国家自然科学基金支持

Abstract: In this paper, magnetic susceptibility and total number of particles with their approximate results in the high temperature limit are calculated concretely to the Landau diamagnetism via the choice of a vector potential. The results indicate that magnetic susceptibility and total number of particles are both unrelated to the choice of vector potential, and the Landau diamagnetism will occur in the high temperature limit.

1. 引言

Landau抗磁性最初由L.D. Landau，R. Peierls等人予以考虑 [8] [9] [10]。Landau在1930年用量子力学来考虑这一问题时发现，当KBT比能级间隙还要大时，表面电流消失，即抗磁性 [10]。Lionel Friedman讨论了小尺度的Landau抗磁性，Klaus Richter等通过准经典近似方法得到了任意磁场下的Landau抗磁性 [11] [12]。一个有趣的现象是：中子星内壳层中的电子就是处于高度简并的电子态，Landau反磁矩变得更重要，使得年轻的中子星(温度较高)中有磁星存在 [13]。

2. Landau规范下单电子方程

$H=\frac{1}{2m}{\left(p+\frac{q}{c}A\right)}^{2}-{\mu }_{0}\sigma \cdot H$(1)

$\left(\frac{{\stackrel{^}{p}}_{{x}^{\prime }}^{2}}{2m}+\frac{1}{2}m{\omega }_{0}^{2}{{x}^{\prime }}^{2}\right)f\left({x}^{\prime }\right)={\epsilon }^{\prime }f\left({x}^{\prime }\right)$(2)

${f}_{j}\left({x}^{\prime }\right)=\sqrt{\frac{\alpha }{\sqrt{\pi }{2}^{j}\cdot j!}}{\text{e}}^{\frac{-1}{2}{\alpha }^{2}{{x}^{\prime }}^{2}}{H}_{j}\left(\alpha {x}^{\prime }\right)$${{\epsilon }^{\prime }}_{j}=\left(j+\frac{1}{2}\right)\hslash {\omega }_{0}$$j\in ℕ$(3)

3. 高温极限下配分函数与磁化率

$\mathrm{ln}\Xi =\underset{j=0}{\overset{\infty }{\sum }}\underset{{p}_{z}}{\sum }g\mathrm{ln}\left(1+z{\text{e}}^{-\beta {\epsilon }_{j}}\right)\to \underset{j=0}{\overset{\infty }{\sum }}\frac{g{L}_{z}}{h}{\int }_{-\infty }^{+\infty }\text{d}{p}_{z}\mathrm{ln}\left(1+z{\text{e}}^{-\beta {\epsilon }_{j}}\right).$ (4)

$\begin{array}{l}\mathrm{ln}\Xi \stackrel{\left(1\right)}{=}\underset{j=0}{\overset{\infty }{\sum }}\frac{2g{L}_{z}}{h}{\int }_{0}^{\infty }\text{d}{p}_{z}\underset{n=1}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{n-1}}{n}{z}^{n}{\text{e}}^{-n\beta {\epsilon }_{j}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{\left(2\right)}{=}\underset{j=0}{\overset{\infty }{\sum }}\frac{2g{L}_{z}}{h}\underset{n=1}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{n-1}}{n}{z}^{n}{\int }_{0}^{\infty }\text{d}{p}_{z}{\text{e}}^{-n\beta {\epsilon }_{j}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{\left(3\right)}{=}\underset{j=0}{\overset{\infty }{\sum }}\frac{2g{L}_{z}}{h}\underset{n=1}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{n-1}}{n}{z}^{n}\sqrt{\frac{m\pi }{2n\beta }}{\text{e}}^{-n\beta {{\epsilon }^{\prime }}_{j}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{\left(4\right)}{=}\underset{n=1}{\overset{\infty }{\sum }}\frac{g{L}_{z}}{\lambda }\frac{{\left(-1\right)}^{n-1}}{{n}^{\frac{\text{3}}{\text{2}}}}{z}^{n}\frac{{\text{e}}^{-ny}}{1-{\text{e}}^{-2ny}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\stackrel{\left(5\right)}{=}\underset{n=1}{\overset{\infty }{\sum }}\frac{4\pi mV}{{h}^{2}\beta \lambda }\frac{{\left(-1\right)}^{n-1}}{{n}^{\frac{\text{3}}{\text{2}}}}{z}^{n}\frac{y{\text{e}}^{-ny}}{1-{\text{e}}^{-2ny}}.\end{array}$ (5)

(i) 等号 $\stackrel{\left(3\right)}{=}$ 后的无穷级数(绝对)收敛，因为

$\underset{j=0}{\overset{\infty }{\sum }}\underset{n=1}{\overset{\infty }{\sum }}\frac{{|z|}^{n}}{{n}^{\frac{3}{2}}}{\text{e}}^{-n\beta \hslash {\omega }_{0}\left(j+\frac{1}{2}\right)}\le \underset{j=0}{\overset{\infty }{\sum }}\underset{n=1}{\overset{\infty }{\sum }}\frac{1}{{n}^{\frac{3}{2}}}{\text{e}}^{-j\beta \hslash {\omega }_{0}}=\frac{\zeta \left(\frac{3}{2}\right)}{1-{\text{e}}^{-\beta \hslash {\omega }_{0}}}$(6)

(ii) 级数 $\underset{n=1}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{n-1}{z}^{n}}{{n}^{\frac{3}{2}}}$ 绝对收敛，且 $\frac{{\text{e}}^{-ny}}{1-{\text{e}}^{-2ny}}=\frac{1}{{\text{e}}^{ny}-{\text{e}}^{-ny}}$ 关于n单调递减有界，由Abel判别法知等号 $\stackrel{\left(4\right)}{=}$ 后的无穷级数绝对收敛。由此可知(iii)的交换性成立。

(iii) 等号 $\stackrel{\left(2\right)}{=}$ 后的积分改写成 ${\int }_{0}^{\infty }\text{d}{p}_{z}{\text{e}}^{-n\beta {\epsilon }_{j}}=\underset{m=0}{\overset{\infty }{\sum }}{\int }_{m}^{m+1}\text{d}{p}_{z}{\text{e}}^{-n\beta {\epsilon }_{j}}$，此时转化为二重级数求和问题。类似于(i)中讨论，级数求和 $\underset{n=1}{\overset{\infty }{\sum }}\cdots$$\underset{m=0}{\overset{\infty }{\sum }}\cdots$ 可交换次序。显然当 $|z|<1$ 时级数 $\underset{n=1}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{n}}{n}{z}^{n}{\text{e}}^{-n\beta {\epsilon }_{j}}$ 一致收敛(绝对收敛)，上述定积分 ${\int }_{m}^{m+1}\text{d}{p}_{z}\cdots$ 与求和 $\underset{n=1}{\overset{\infty }{\sum }}\cdots$ 又可交换次序。总之，等号 $\stackrel{\left(1\right)}{=}$ 后的求和号 $\underset{n=1}{\overset{\infty }{\sum }}\cdots$ 与积分号 ${\int }_{0}^{\infty }\text{d}{p}_{z}\cdots$ 可交换次序。

$\text{M}={k}_{B}T{\left(\frac{\partial }{\partial H}\frac{\mathrm{ln}\Xi }{V}\right)}_{T,V,z}=\underset{n=1}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{n}{z}^{n}q{\text{e}}^{-ny}\left[\left(ny+1\right){\text{e}}^{-2ny}+ny-1\right]}{hc\beta \lambda {n}^{\frac{3}{2}}{\left(1-{\text{e}}^{-2ny}\right)}^{2}}.$ (7)

${a}_{n}\left(y\right):=\frac{{\text{e}}^{-ny}\left[\left(ny+1\right){\text{e}}^{-2ny}+ny-1\right]}{{\left(\text{1}-{\text{e}}^{-2ny}\right)}^{2}}$，注意到极限 $\underset{y\to 0+}{\mathrm{lim}}{a}_{n}\left(y\right)=0$，因此对于 ${\epsilon }_{0}=\frac{1}{2}>0$，存在正数 ${\delta }_{\text{0}}={\delta }_{\text{0}}\left(\frac{1}{2}\right)>0$，使得 $|{a}_{n}\left(y\right)|\le \frac{1}{2}$$y\in \left(0,{\delta }_{0}\right)$。当 $y\ge {\delta }_{\text{0}}$ 时有不等式

$|\frac{{\text{e}}^{-ny}\left[\left(ny+1\right){\text{e}}^{-2ny}+ny-1\right]}{{\left(\text{1}-{\text{e}}^{-2ny}\right)}^{2}}|\le \frac{ny{\text{e}}^{-ny}\left({\text{e}}^{-2ny}+1\right)}{{\left(\text{1}-{\text{e}}^{-2ny}\right)}^{2}}+\frac{{\text{e}}^{-ny}}{\text{1}-{\text{e}}^{-2ny}}\le \frac{2}{{\left(\text{1}-{\text{e}}^{-2{\delta }_{0}}\right)}^{2}}+\frac{1}{\text{1}-{\text{e}}^{-2{\delta }_{0}}}.$ (8)

$\chi ={\left(\frac{\partial M}{\partial H}\right)}_{T,V,z}=\underset{n=1}{\overset{\infty }{\sum }}\frac{{\left(-1\right)}^{n-1}{z}^{n}{q}^{2}{\text{e}}^{-ny}\left[\left(ny+2\right){\text{e}}^{-4ny}+6ny{\text{e}}^{-2ny}+ny-2\right]}{4\pi \lambda m{c}^{2}\sqrt{n}{\left(\text{1}-{\text{e}}^{-2ny}\right)}^{3}}.$ (9)

$\begin{array}{l}|\frac{{\text{e}}^{-ny}\left[\left(ny+2\right){\text{e}}^{-4ny}+6ny{\text{e}}^{-2ny}+ny-2\right]}{{\left(\text{1}-{\text{e}}^{-2ny}\right)}^{3}}|\\ \le \frac{ny{\text{e}}^{-ny}\left({\text{e}}^{-4ny}+6{\text{e}}^{-2ny}+1\right)+2{\text{e}}^{-ny}\left(1-{\text{e}}^{-4ny}\right)}{{\left(\text{1}-{\text{e}}^{-2ny}\right)}^{3}}\le \frac{10}{{\left(\text{1}-{\text{e}}^{-2{\delta }_{0}}\right)}^{3}},\text{\hspace{0.17em}}y\ge {\delta }_{0}.\end{array}$ (10)

$〈\stackrel{^}{N}〉={\left(z\frac{\partial \mathrm{ln}\Xi }{\partial z}\right)}_{T,V,H}=\underset{n=1}{\overset{\infty }{\sum }}\frac{4\pi mVy}{\lambda {h}^{2}\beta }\frac{{\left(-1\right)}^{n-1}}{\sqrt{n}}{z}^{n}\frac{{\text{e}}^{-ny}}{1-{\text{e}}^{-2ny}}.$ (11)

4. 近似结果与抗磁性

4.1. 第一种近似

$\left(ax+b\right){\text{e}}^{-cx}\le K,K\ge \mathrm{max}\left\{b,\frac{a}{c}\right\},a,b,c>0,x\ge 0，$ (12)

$\begin{array}{c}\underset{n=2}{\overset{\infty }{\sum }}{\chi }_{n}\le \frac{{z}^{2}{q}^{2}}{4\pi \lambda m{c}^{2}}\underset{n=2}{\overset{\infty }{\sum }}\frac{\left(ny+2\right)\left({\text{e}}^{-5ny}+{\text{e}}^{-ny}\right)+6ny{\text{e}}^{-3ny}}{\sqrt{n}{\left(\text{1}-{\text{e}}^{-2ny}\right)}^{3}}\\ \le \frac{{z}^{2}{q}^{2}}{4\pi \lambda m{c}^{2}}\underset{n=2}{\overset{\infty }{\sum }}\frac{6}{\sqrt{n}{\left(\text{1}-{\text{e}}^{-2ny}\right)}^{3}}.\end{array}$ (13)

$\chi =\left(z{{\chi }^{\prime }}_{1}+{z}^{2}{{\chi }^{\prime }}_{2}\right)+\underset{n=3}{\overset{\infty }{\sum }}{\chi }_{n}\le z\left[{{\chi }^{\prime }}_{1}+z\left({{\chi }^{\prime }}_{2}+{M}_{y}\right)\right]<0,$ (14)

4.2. 第二种近似

$f\left(y\right)=\frac{{\text{e}}^{-y}}{1-{\text{e}}^{-2y}}=\frac{1}{2y}\left(1-\frac{{y}^{2}}{6}\right)+O\left({y}^{3}\right),$ (15)

$1-z+\frac{1}{2}{z}^{2}+O\left({z}^{3}\right)=\left[2z-2{z}^{2}+\frac{4}{3}{z}^{3}+O\left({z}^{4}\right)\right]\cdot \left[\frac{1}{2z}+{c}_{0}+{c}_{1}z+O\left({z}^{2}\right)\right]，$ (16)

$\mathrm{ln}\Xi \approx \frac{\pi mV}{3\lambda {h}^{2}\beta }\left[6{\psi }_{\frac{5}{2}}\left(z\right)-{y}^{2}{\psi }_{\frac{1}{2}}\left(z\right)\right]$(17a)

$\text{M}\approx \frac{-qy}{6\lambda hc\beta }{\psi }_{\frac{1}{2}}\left(z\right)$(17b)

$\chi \approx \frac{-{q}^{2}}{24\pi \lambda m{c}^{2}}{\psi }_{\frac{1}{2}}\left(z\right)$(17c)

$〈\stackrel{^}{N}〉\approx \frac{\pi mV}{3\lambda \beta {h}^{2}}\left[6{\psi }_{\frac{3}{2}}\left(z\right)-{y}^{2}{\psi }_{\frac{-1}{2}}\left(z\right)\right]$(17d)

(i) $x\frac{\text{d}}{\text{d}x}{\psi }_{s}\left(x\right)={\psi }_{s-1}\left(x\right)$$|x|<1$ ；(ii) ${\psi }_{s}\left(1\right)=\left(1-\frac{1}{{2}^{s-1}}\right)\zeta \left(s\right)$${\psi }_{s}\left(-1\right)=-\zeta \left(s\right)$$s>1$

${B}_{0}\left(x\right)=1$${B}_{1}\left(x\right)=x-\frac{1}{2}$

${B}_{2}\left(x\right)={x}^{2}-x+\frac{1}{6}$

${B}_{3}\left(x\right)=x\left(x-1\right)\left(x-\frac{1}{2}\right)$

${B}_{4}\left(x\right)={x}^{4}-2{x}^{3}+{x}^{2}-\frac{1}{30}$

${B}_{5}\left(x\right)=x\left(x-1\right)\left(x-\frac{1}{2}\right)\left({x}^{2}-x-\frac{1}{3}\right)$

${B}_{6}\left(x\right)={x}^{6}-3{x}^{5}+\frac{5}{2}{x}^{4}-\frac{1}{2}{x}^{2}+\frac{1}{42}$

${B}_{7}\left(x\right)=\frac{1}{6}x\left(x-1\right)\left(2x-1\right)\left(3{x}^{4}-6{x}^{3}+3x+1\right)$

${B}_{8}\left(x\right)={x}^{8}-4{x}^{7}+\frac{14}{3}{x}^{6}-\frac{7}{3}{x}^{4}+\frac{2}{3}{x}^{2}-\frac{1}{30}$

${B}_{9}\left(x\right)=\frac{1}{10}x\left(x-1\right)\left(2x-1\right)\left(5{x}^{6}-15{x}^{5}+5{x}^{4}+15{x}^{3}-{x}^{2}-9x-3\right)$

${B}_{10}\left(x\right)={x}^{10}-5{x}^{9}+\frac{15}{2}{x}^{8}-7{x}^{6}+5{x}^{4}-\frac{3}{2}{x}^{2}+\frac{5}{66}$

${B}_{0}=-1$${B}_{1}=\frac{1}{6}$${B}_{2}=\frac{1}{30}$${B}_{3}=\frac{1}{42}$${B}_{4}=\frac{1}{30}$${B}_{5}=\frac{5}{66}$

${B}_{6}=\frac{691}{2730}$${B}_{7}=\frac{7}{6}$${B}_{8}=\frac{3617}{510}$${B}_{9}=\frac{43867}{798}$${B}_{10}=\frac{174611}{330}$

${B}_{2l+1}\left(\frac{1}{2}\right)=0$$l\in ℕ$，因互余宗量关系 ${B}_{n}\left(1-x\right)={\left(-1\right)}^{n}{B}_{n}\left(x\right)$

$\mathrm{ln}\Xi =\underset{l=0}{\overset{\infty }{\sum }}\frac{2\pi mV}{\beta \lambda {h}^{2}}\frac{{\left(2y\right)}^{2l}}{\left(2l\right)!}{B}_{2l}\left(\frac{1}{2}\right){\psi }_{\frac{5}{2}-2l}\left(z\right)$(18a)

$\text{M}=\underset{l=1}{\overset{\infty }{\sum }}\frac{q}{hc\beta \lambda }\frac{{\left(2y\right)}^{2l-1}}{\left(2l-1\right)!}{B}_{2l}\left(\frac{1}{2}\right){\psi }_{\frac{5}{2}-2l}\left(z\right)$(18b)

$\chi =\underset{l=0}{\overset{\infty }{\sum }}\frac{{q}^{2}}{2{c}^{2}m\lambda \pi }\frac{{\left(2y\right)}^{2l}}{\left(2l\right)!}{B}_{2l+2}\left(\frac{1}{2}\right){\psi }_{\frac{1}{2}-2l}\left(z\right)$(18c)

$〈\stackrel{^}{N}〉=\underset{l=0}{\overset{\infty }{\sum }}\frac{2\pi mV}{\beta \lambda {h}^{2}}\frac{{\left(2y\right)}^{2l}}{\left(2l\right)!}{B}_{2l}\left(\frac{1}{2}\right){\psi }_{\frac{3}{2}-2l}\left(z\right)$(18d)

5. 总结和讨论

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