量子统计物理¶

2026-09-202026-10-16

1. density matrix¶

对于混合态的作用算符 $\hat{F}$ ，理论上计算期望得到：

\[ \ev{ \hat{F} } = \sum_i \mel{ \Psi ^i( t ) }{ \hat{F} }{ \Psi ^i( t ) }p_i \]

插入投影算符

\[ \ev{ \hat{F}} = \sum_i\sum_n\sum_m \ip{ \Psi ^i( t ) }{ \psi _n }\mel{ \psi _n }{ \hat{F} }{ \psi _m }\ip{ \psi _m }{ \Psi ^i( t ) }p_i \]

定义矩阵元

\[ F_{nm} = \mel{ \psi _n }{ \hat{F} }{ \psi _m } \]

得到期望值

\[ \ev{ \hat{F} } = \sum_{i, n, m} F_{nm}\ip{ \psi _m }{ \Psi ^i( t ) }\ip{ \Psi ^i( t ) }{ \psi _n }p_i \]

我们定义密度矩阵：

\[ \hat{ \rho } = \sum_i \op{ \Psi ^i( t ) } p_i \]

于是混合态期望可表示为：

\[ \ev{ \hat{F} } = \sum_{n, m}F_{nm}\rho _{nm} = \Tr( \hat{ \rho }\hat{F} ) \]

后面的事情在基础量子就学过：

\[ \Tr( \hat{ \rho } ) = 1\qc \hat{ \rho }^{\dagger} = \hat{ \rho }\qc \Tr( \hat{ \rho }^{2} ) \leq 1 \]

只有纯态是等号成立。作为一个例子，对于

\[ \ket{\Psi ^i} = \cos\alpha_i \ket{\psi _x} + \sin\alpha_i \ket{\psi _y} \]

对应密度矩阵

\[ \hat{ \rho } = \sum_i p_i \pmqty{ \cos ^{2}\alpha_i & \cos\alpha_i \sin\alpha_i \\ \cos\alpha_i \sin\alpha_i & \sin ^{2} \alpha_i } \]

含时 Schrodinger 方程给出

\[ i\hbar \pdv{ t } \ket{\Psi ^i( t )} = \hat{ H } \ket{\Psi ^i( t )} \]

对密度算符求导：

\[ \begin{align} i\hbar \pdv{ t } \hat{ \rho } & = \sum_i p_i \qty [ \qty( i\hbar \pdv{ t } \ket{\Psi ^i( t )} \bra{\Psi ^i( t )} + \ket{\Psi ^i( t )}i\hbar \pdv{ t }\bra {\Psi ^i( t )} ) ] \\ & = \sum_i p_i \qty [ \hat{H}\op{ \Psi ^i( t ) } - \op{ \Psi ^i( t ) }\hat{H} ] \\ & = \comm{ \hat{H} }{ \hat{ \rho } } \end{align} \]

这是 Liouville 方程。

2. ensemble¶

2.1 Classical Statistic¶

相空间的哈密顿量：

\[ \dot{q_i} = \pdv{ H( q, p ) }{ p_i } \qc \dot{p_i} = -\pdv{ H( q, p ) }{ q_i } \]

固定能量

\[ \sum_{i = 1}^{3N} \frac{p_i^{2}}{2m}= E \]

于是对应密度函数

\[ \rho ( p, q ) = \begin{cases} const. & E\leq H( p, q )\leq E+\Delta E \\ 0 & else. \end{cases} \]

这意味着密度函数就是这样一个微小相空间的体积倒数。

2.2 Microcanonical ensemble¶

consider in QM, one can define

\[ p_i = \begin{cases} \Gamma ^{-1}( E, N, V ) & E\leq E_i\leq E+\Delta E \\ 0 & else \end{cases} \]

where $\Gamma$ means the phase space volume of a shallow curve between $E$ and $E+\Delta E$.

define a func

\[ \Delta x = \begin{cases} 1 & 0\leq x\leq \Delta E \\ 0 & else \end{cases} \]

do transform

$$ \begin{align} \hat{ \rho } & = \sum_i p_i\op{ \psi _i }{ \psi _i } \ & = \sum_i \Gamma ^{-1}( E, N, V )\Delta( E_i-E )\op{ \psi _i }{ \psi _i } \ & = \Gamma ^{-1}( E, N, V ) \sum_i\Delta( \hat{H}-E )\op{ \psi _i }{ \psi _i } \ & = \Gamma ^{-1}( E, N, V ) \Delta( \hat{H}-E )

\end{align} $$

我们考虑其对数期望

\[ \begin{align} \ev{ \ln \rho } & = \Tr( \rho \ln \rho ) \\ & = \sum_{m, n}\rho _{mn}( \ln \rho_{mn} ) \end{align} \]

选定

\[ \rho _{mn} = \mel{ \psi _m }{ \sum_i p_i }{ \psi _i }\ip{ \psi _i }{ \psi _n } = p_n\delta_{m, n} \]

因此矩阵为对角，同样 $\ln \rho$ 也对角。于是有：

\[ \begin{align} \ev{ \ln \rho } & = \sum_{m, n}p_n( \ln p_n ) \delta_{m, n} \\ & = \sum_{n}p_n( \ln p_n ) = -\ln\Gamma \end{align} \]

于是可以定义熵，它和态的总数正相关

\[ S = -k_B \ev{ \ln \rho } = k_B \ln\Gamma \]

接下来计算哈密顿量

\[ \begin{align} \ev{ \hat{H} }=\Tr( \rho H ) & =\Gamma ^{-1}\Tr( H\Delta( \hat{H}-E )) \\ & = \Gamma ^{-1} \sum_{m, n}H_{m, n}( \Delta( H-E ) )_{n, m} \end{align} \]

and

\[ \begin{align} \ev{ H } & = \Gamma ^{-1} \sum_{m, n}H_{m, n}( \Delta( H-E ) )_{n, m} \\ \\ & = \Gamma ^{-1} \sum_{m, n}\mel{ \psi _n }{ H }{ \psi _m }\mel{ \psi _m }{( \Delta( H-E ) )_{m, n}}{ \psi _n } \\ & = \Gamma ^{-1} \sum_n E_n\Delta( E_n - E ) \\ & = E \sum_np_n = E \end{align} \]

we found the eqn is self-consisted.

as definition, the equilibrium state def as when

\[ S = S_{max} \]

我们假设有两个体系，分别为 A 和 A‘。外部能量分别为 E 和 E‘。有

\[ E+E'= E^{( 0 )} \]

假定这两个体系之间可以互相传递能量。对于相空间

\[ \Gamma^{( 0 )}( E, E' ) = \Gamma( E )\Gamma'( E^{( 0 )}-E ) \]

当熵最大时，一定有：

\[ \pdv{ \Gamma^{( 0 )}( E, E' ) }{ E } = 0 \]

也就是：

\[ \pdv{ \Gamma( E ) }{ E } \cdot \Gamma'( E' ) = \pdv{ \Gamma'( E' ) }{ E' } \cdot \Gamma( E ) \]

we got

\[ \pdv{ E } \ln\Gamma( E ) = \pdv{ E' } \ln\Gamma'( E' ) = \frac{1}{k_{B}T} \]

带入到熵的定义得到

\[ \qty(\pdv{ S }{ E })_{ N, V } = \frac{1}{T} \]

现在，我们假设系统的体积同样也会发生变化。很容易得到相同的等式：

\[ \begin{gathered} \pdv{ E } \ln\Gamma( E ) = \pdv{ E' } \ln\Gamma'( E' ) = \frac{p}{k_{B}T} \implies \qty(\pdv{ S }{ V })_{ E, V } = \frac{p}{T} \\ \cdots \end{gathered} \]

于是得到基本方程：

\[ \dd{S} = \frac{1}{T}\dd{E} +\frac{p}{T}\dd{V} -\frac{\mu}{T}\dd{N} \]

2.3 Canonical Ensemble¶

前面我们已知

\[ p_i = \frac{1}{Z} e^{ -\beta E_i } \]

c 插入到密度矩阵表达式：

\[ \begin{align} \hat{ \rho } & = \sum_i \op{ \psi _i } \\ & = \frac{1}{Z} \sum_i e^{ -\beta E_i }\op{ \psi _i } \\ & = \frac{1}{Z} \sum_i e^{ -\beta \hat{ H } }\op{ \psi _i } \\ & = \frac{1}{Z}e^{ -\beta \hat{ H } } \end{align} \]

这可以导出经典热力学的很多等式：

\[ E = -\pdv{ }{ \beta } \ln Z \qc F = -k_{B}T\ln Z \]

还有一个特征的熵等式：

\[ \begin{align} -k_B \Tr( \hat{ \rho }\ln \hat{ \rho } ) & = -k_B\sum_i p_i\ln p_i \\ & = \cdots \\ & = \frac{E}{T} - \frac{F}{T} \\ & = S \end{align} \]

这于是可以导出热力学第三定律，也就是 0K 下只有基态

\[ p_0 = 1 \]

导致系统变为纯态。此时 $S=0$ .

和经典统计类似，也可以换成按能量的统计。定义态密度

\[ g( E ) = \dv{ \Gamma( E ) }{ E } \]

通过积分计算

\[ Z = \int_{0}^{\infty} e^{ -\beta E }g( E ) \dd{E}\qc \ev{ f } = \frac{1}{Z} \int_{0}^{\infty} f( E )e^{ -\beta E }g( E ) \dd{E} \]

举例：对于有 $N$ 个量子谐振子形成的系统

\[ \hat{H} = \sum_i \frac{\hbar ^{2}}{2m}\dv [2]{ }{ x } + \frac{1}{2}m\omega ^{2}x^{2} \]

由于每个谐振子都是独立的，我们可以把它拆成独立的哈密顿量

\[ h =\frac{\hbar ^{2}}{2m}\dv [2]{ }{ x } + \frac{1}{2}m\omega ^{2}x^{2} \]

于是就可以拆开能量和波函数

\[ E = \sum_{i = 1}^{N} \varepsilon_i \qc \Psi ( x_1,\cdots , x_N ) = \prod_{i = 1}^{N} \psi _i( x_i ) \]

从经典统计里面，我们拿到公式（这里还没证明）：

\[ Z = z^N\qc z =\sum_{i = 1}^{N} e^{ -\beta\varepsilon_i } \]

然后得到。。。

2.4 Grand Canonical¶

现在我们考虑一个能量和粒子都会交换的体系，此时需要新增一个交换约束

\[ N + N' = N^{( 0 )} \]

根据同样的方法可以得到

\[ \pdv{ N } \ln\Gamma( N ) = \pdv{ N' } \ln\Gamma'( N' ) = -\frac{\mu}{k_{B}T} \]

同样假设外部粒子库的态密度是极高的，因此理论有：

\[ \frac{E}{E^{( 0 )}} = 1- \frac{E'}{E^{( 0 )}} \ll 1\qc \frac{N}{N^{( 0 )}} = 1- \frac{N'}{N^{( 0 )}} \ll 1 \]

对 $\ln \Gamma$ 展开：

\[ \begin{align} \ln \Gamma' & = \ln\Gamma'( E^{( 0 )}, N^{( 0 )} ) + \pdv{ \ln\Gamma' }{ E' } \cdot E + \pdv{ \ln\Gamma' }{ N' } \cdot N \\ & = const. + \frac{\mu}{k_{B}T}N - \frac{1}{k_{B}T}E_i \end{align} \]

由此定义

\[ p_i = e^{ -\beta( E_i - \mu N ) } \]

得到巨配分函数：

\[ \Xi = \sum_{N, i} e^{ -\beta( E_i - \mu N ) } \]

用量子力学的语言表示：

\[ p_i = e^{ -\beta( \hat{H} - \mu \hat{N} ) }\qc \Xi = \Tr( e^{ -\beta( \hat{H} - \mu \hat{N})} ) \]

密度矩阵

\[ \hat{ \rho } = \Xi ^{-1}e^{ -\beta( \hat{H} - \mu \hat{N} ) } \]

计算离子数

\[ \begin{align} \Tr( \hat{ \rho }\hat{N} ) & = \sum_{m, n}\rho _{mn}N_{nm} \\ & = \sum_{N, i}\sum_{N', i'}\rho _{Ni, N'i'}N_{Ni, N'i'} \\ & = \sum_{N, i} \rho _{Ni, Ni}N \end{align} \]

又有对于对应的密度矩阵元

\[ \rho _{Ni, Ni} = \Xi^{-1} e^{ -\beta( E_i-\mu N ) } \]

对应的

\[ \ev{ N } = \Xi^{-1}\sum_{N = 0}^{\infty}Ne^{ -\beta( E_i-\mu N ) } \]

也就是

\[ \ev{ N } = -\pdv{ }{ \alpha } \ln\Xi \]

对于能量：

。。。

另一种通过逸度的定义

\[ \Xi = \sum_{N = 0}^{\infty} z^{N}Z( N, V, T )\qc z = e^{ \beta \mu } \]

之后的大部分推理方法与经典类似，不再赘述。有一个不同的是熵的推导

\[ \begin{align} S & = \cdots \end{align} \]

考虑一个由 $N$ 个能级 $\{n_{1},n_{2},\cdots ,n_N\}$ 组成的巨正则系综，其中个能级的粒子数可能出现 $0 \sim \infty$ 个

\[ \begin{align} Z & = \sum_{n_{1} = 0}^{\infty} \sum_{n_{2} = 0}^{\infty} \cdots \sum_{n_N = 0}^{\infty} e^{ -\beta\sum_{l = 1}^{N} \varepsilon_l } \\ & = \sum_{n_{1},n_{2},\cdots ,n_N}\prod_{l = 1}^{N} e^{ -\beta\varepsilon_l } \\ & = \prod_{l = 1}^{N} \sum_{n_l = 0}^{\infty} e^{ -\beta \varepsilon_l } \end{align} \]

3. Quantum Gases¶

ideal quantum gas¶

对 $\infty$ 个能级分布的 $N$ 个粒子数采取粒子数表象

\[ \ket{n_{1},n_{2},\cdots ,n_\infty} = \ket{N,i} \]

这里的 $i$ 指的是一种抽象的 $N$ 粒子排布方式。考虑巨正则配分函数

\[ \begin{align} \Xi & = \Tr( e^{ -\beta( \hat{H}-\mu \hat{N} ) } ) \\ & = \sum_{N = 0}^{\infty} \sum_i \mel{ N,i }{ e^{ -\beta( \hat{H}-\mu \hat{N} ) } }{ N,i } \\ & = \sum_{n_{1} = 0}^{\infty} \sum_{n_{2} = 0}^{\infty} \cdots \sum_{n_\infty = 0}^{\infty} \mel{ N,i }{ e^{ -\beta( \hat{H}-\mu \hat{N} ) } }{ N,i } \end{align} \]

展开算子

\[ \begin{align} \Xi & = \sum_{n_{1},n_{2},\cdots ,n_\infty} \mel{ N,i }{ \exp( \sum_i \varepsilon_in_i - \mu \sum_i n_i) }{ N,i } \\ & = \prod_{i = 1}^{\infty} \sum_{n_i}\exp( -\beta( \varepsilon_i -\mu)n_i ) \end{align} \]

Bosons

\[ \Psi(\cdots ,x_{i}, \cdots ,x_{j},\cdots ) = \Psi(\cdots ,x_{j}, \cdots ,x_{i},\cdots ) \]

Fermions

\[ \Psi(\cdots ,x_{i}, \cdots ,x_{j},\cdots ) = ( - )\Psi(\cdots ,x_{j}, \cdots ,x_{i},\cdots ) \]

因此我们不能直接拆波函数，必须拆成对称性轮换形式。利用置换算子 $P$ 统计所有交换粒子位置之后的波函数城际

Boson

\[ \Psi_n( x_{1},\cdots ,x_{N} ) = \frac{1}{ \sqrt{ N! }} \sum_P P[ \varphi_{n_{1}}( x_{1} )\cdots \varphi_{n_N}( x_N ) ] \]

Fermions

\[ \Psi_n( x_{1},\cdots ,x_{N} ) =\frac{1}{ \sqrt{ N! }} \sum_P ( - )^{A_P} P[ \varphi_{n_{1}}( x_{1} )\cdots \varphi_{n_N}( x_N ) ] \]

对于 Fermion，还可以用Slater行列式来定义，因为行列式交换行会导致符号反转

\[ \Psi_n( x_{1},\cdots ,x_{N} ) = \frac{1}{ \sqrt{ N! }} \mdet{ \varphi_{n_{1}}( x_{1} ) & \varphi_{n_{1}}( x_{2} ) & \cdots & \varphi_{n_1}( x_N ) \\ \vdots & \vdots & \ddots & \vdots \\ \varphi_{n_{N}}( x_{1} ) & \varphi_{n_{N}}( x_{2} ) & \cdots & \varphi_{n_N}( x_N ) } \]

当存在两行相同时，则行列式变为0. 这就对应Pauli不相容原理

\[ n_i \neq n_j \]

之后量子统计方法和经典相同··

晶格中的统计见固体物理

ideal usal bose gas¶

total particle number $$ \begin{align} N & = \sum_i \frac{1}{e^{ \beta( \varepsilon_i-\mu ) }-1} \ & =g \sum_{\va{k}} \frac{1}{e^{ \beta( \varepsilon-\mu ) }-1} \ & = g \frac{V}{( 2\pi )^{3}} \cdot 4\pi \int_{0}^{\infty} \dd{k} k^{2} \frac{1}{e^{ \beta( \varepsilon-\mu ) }-1} \ & = \frac{gV}{4\pi^{2}} \qty( \frac{2m}{\hbar^{2}} )^{3/2 } \int_{0}^{\infty} \dd{\varepsilon} \frac{ \varepsilon^{1/2 } }{ e^{ \beta( \varepsilon-\mu ) }-1 } \end{align} $$ 对应能量 $$ E = \frac{gV}{4\pi^{2}} \qty( \frac{2m}{\hbar^{2}} )^{3/2 } \int_{0}^{\infty} \dd{\varepsilon} \frac{ \varepsilon^{3/2 } }{ e^{ \beta( \varepsilon-\mu ) }-1 } $$ grand potential $$ \begin{align} \Omega & = \beta ^{-1} \sum_{i = 1}^{\infty} \ln( 1-e^{ \beta( \mu-\varepsilon ) } ) \ & = \beta ^{-1} \frac{gV}{4\pi^{2}} \qty( \frac{2m}{\hbar^{2}} )^{3/2 } \int_{0}^{\infty} \dd{ \varepsilon}\varepsilon^{1/2 } \ln( 1-e^{ \beta( \mu-\varepsilon ) } ) \ & = \frac{gV}{4\pi^{2}} \qty( \frac{2m}{\hbar^{2}} )^{3/2 } \frac{3}{2} \int_{0}^{\infty} \dd{\varepsilon} \frac{ \varepsilon^{3/2 } }{ e^{ \beta( \varepsilon-\mu ) }-1 } \end{align} $$

ideal fermion¶

呃就是改一下 $$ \begin{gathered} N = \frac{gV}{4\pi^{2}} \qty( \frac{2m}{\hbar^{2}} )^{3/2 } \int_{0}^{\infty} \dd{\varepsilon} \frac{ \varepsilon^{1/2 } }{ e^{ \beta( \varepsilon-\mu ) }-1 } \ E = \frac{gV}{4\pi^{2}} \qty( \frac{2m}{\hbar^{2}} )^{3/2 } \int_{0}^{\infty} \dd{\varepsilon} \frac{ \varepsilon^{3/2 } }{ e^{ \beta( \varepsilon-\mu ) }-1 } \ \Omega = -\frac{2}{3} E \end{gathered} $$ fermi distribution on T=0 $$ \frac{1}{e^{ \beta( \varepsilon-\mu ) }+1} = H( \mu-\varepsilon ) $$ Summerfeld expansion $$ \begin{align} \int_{0}^{\infty} \dd{\varepsilon} \frac{ \varepsilon^{3/2 } }{ e^{ \beta( \varepsilon-\mu ) }-1 } & = \int_{0}^{\mu} \dd{\varepsilon} \varepsilon^{3/2 } + \frac{\pi^{2}}{6} ( k_{B}T )^{2} \frac{3}{2}\cdot \mu^{1/2} + \cdots \ & = \frac{2}{5}\mu^{5/2} + \frac{\pi^{2}}{4} ( k_{B}T )^{2} \mu^{1/2} + \cdots \end{align} $$ particle num $$ N = \frac{gV}{4\pi^{2}} \qty( \frac{2m}{\hbar^{2}} )^{3/2 } \qty[ \frac{2}{3}\mu^{3/2} + \frac{\pi^{2}}{8} \mu^{-1/2} ] $$ in T=0 we have $$ N = \frac{gV}{4\pi^{2}} \qty( \frac{2m}{\hbar^{2}} )^{3/2 } \frac{2}{3}\varepsilon_{F}^{3/2} $$ solv $$ \mu = \varepsilon_{F}\qty[ 1-\frac{\pi^{2}}{12} \qty( \frac{k_{B}T}{\varepsilon_{F}} )^{2} ] $$

def magnitization $$ M = -\qty(\pdv{ \Omega }{ B }){ T,V,\mu } $$ 后面可以证明 $$ M = \mu ^{*}( N^+ - N^- ) $$ suspectivity $$ \chi( B ) = \frac{1}{V} \pdv{ M }{ B } $$ 假设对自旋1/2体系，定义能量 $$ \varepsilon = \frac{p^{2}}{2m} - \va{\mu}\cdot \va{B} $$ on T=0 $$ N = N+ + N_- $$

\[ \frac{ \hbar^{2}k_{+}^{2} }{ 2m } - \mu ^{*}B = \frac{ \hbar^{2}k_{-}^{2} }{ 2m } + \mu ^{*}B = \mu $$ for example $$ k_+ = \qty[ \frac{2m}{\hbar^{2}}( \mu+\mu ^{*}B ) ]^{1/2} \]

\[ \begin{align} N_+ = \frac{ Vk_{+}^{3} }{ 6\pi^{2} } \end{align} = \frac{1}{6\pi^{2}} \qty[ \frac{2m}{\hbar^{2}}( \mu+\mu ^{*}B ) ]^{1/2} \]

magnetization $$ M = \mu ^{}( N_{+} - N_- ) = \frac{ \mu ^{}V( 2m )^{3/2} }{6\pi^{2}\hbar\qty[ ( \mu+\mu ^{} }B )^{3/2 } - ( \mu+\mu ^{}B )^{3/2 } ] $$ 计算得到 $$ \chi = \lim_{ B \to 0 } \frac{M}{VB} = \frac{ \mu ^{^2}( 2m )^{3/2 } \varepsilon_F^{1/2} }{ 6\pi^{2}\hbar $$ for arbitrary T $$ \begin{align} \Xi & = \Tr\exp[ -\beta( H-\mu N ) ] \ & =\sum_i \ev{ \exp[ -\beta( \hat{H}-\mu \hat{N} ) ] }{ i } \ & = \sum_{n_{1\uparrow},\cdots ,n_{n\downarrow}} \exp( -\beta( \sum_{\sigma}\sum_i \varepsilon_{i\sigma}n_{i\sigma} - \mu \sum_{\sigma}\sum_i n_{i\sigma}) ) \ & = \prod_{i = 1}^{\infty} \qty[ ( 1+e^{ \beta( \mu-\varepsilon_{j\uparrow} ) } )( 1+e^{ \beta( \mu-\varepsilon_{j\downarrow} ) } ) ] \end{align} $$ we have $$ \ln \Xi =\frac{1}{2}\ln \Xi_{0}( \mu+\mu ^{} }B ) + \frac{1}{2}\ln \Xi_{0}( \mu-\mu ^{}B ) $$ obivous $$ \Omega = \frac{1}{2}\Omega_{0}( \mu+\mu ^{}B ) + \frac{1}{2}\Omega_0 ( \mu-\mu ^{}B ) $$ derivative by B we can prove that $$ M = \mu ^{}( N^+ - N^- ) $$ we can also prove $$ N_{\uparrow} - N_{\downarrow} = \mu ^{} B\qty(\pdv{ N }{ \mu }){ T,V } $$ do taylor expansion around $\Omega_{0}( \mu )$ $$ \Omega( \mu ^{} ) = \Omega_{0}( \mu ) + \pdv{ \Omega_{0}( \mu ) }{ \mu } \mu ^{}B + \frac{1}{2}\pdv[2]{ \Omega_{0}( \mu ) }{ \mu } \qty( \mu^{}B )^{2} $$ 得到 $$ \Omega = \Omega_{0}( \mu ) + \frac{1}{2}( \mu ^{{}B )^{2}\pdv[2]{ \Omega_{0}( \mu ) }{ \mu } $$ 得到的磁化率为 $$ M = -\qty(\pdv{ \Omega }{ B })_{ T,V,\mu } = -\mu ^{})}} B \qty(\pdv[ 2 ]{ \Omega_{0}( \mu ) }{ \mu { T,V } = -\mu ^{*)}} B \qty(\pdv{ N }{ \mu $$

...

Landau 抗磁性¶

Bohr - van Leeuwen Theorem on classical physics (doesn't consider spin) $M=0$

proof on B=0 we have $$ H = \sum_{i = 1}^{N} \frac{ p_i^{2} }{ 2m } + V( q_i ) $$ for kinetic Momentum $$ \tilde{p}i = p_i - \frac{ eA( q_i ) }{ c } $$ on classical stat $$ \begin{align} \Xi( V,\mu,T,B ) & = \sum \ & = \ & = \Xi_{0} \end{align} $$ 于是 }^{\infty} \frac{1}{N!h^{3}} \int\dd[3]{\tilde{p}}\dd[3]{q} e^{ -\beta( H-\mu N ) $M=0$ 又有 $\chi= 0$ ，即没有磁性。

in Landau gauge $$ h=\frac{1}{2m}( p-\frac{ eA }{ c } )^{2} \qc A_x = -By\qc A_y=A_z=0 $$ SE $$ \qty[ \frac{1}{2m} ( p_x + \frac{ eBy }{ c } )^{2} + \frac{1}{2m} ( p_y^{2} + p_z^{2} ) ]\psi_i( r ) = \varepsilon_i\psi_i( r ) $$ B_z is not a good quant #

on B = 0 $$ \varepsilon( \va{p} ) = \frac{1}{2m} ( p_x^{2} + p_y^{2} ) + \frac{1}{2m} p_z^{2} $$ on $B\neq 0$ $$ \varepsilon( j,p_z ) = \mu_BB( j+\frac{1}{2} ) + \frac{1}{2m} p_z^{2} $$ Refer to caption

施加磁场会导致电子被固定到固定的能级上。现在统计每两个圆之间的态密度 $$ \begin{align} g & = \frac{ V^{2/3} }{ ( 2\pi )^{2} } \pi ( k_{max}^{2} ) \ & = \frac{ V^{2/3} 2m }{ 4\pi \hbar^{2} } \cdot \frac{ \hbar^{2} k_{max}^{2} }{ 2m } \ & = \frac{ V^{2/3 }m }{ 2\pi \hbar^{2} }\mu_BB \end{align} $$ 计算得到配分函数 $$ \begin{align} \ln \Xi & = \sum_i \ln( 1+e^{ -\beta( \varepsilon_i-\mu ) } ) \ & = g\sum_{j = 0}^{\infty} \sum_{k_z} \ln( 1+e^{ -\beta( \varepsilon_i-\mu ) } ) \ & = g \sum_{j = 0}^{\infty} \frac{ V^{1/3 } }{ 2\pi } \int_{-\infty}^{\infty} \dd{k_z} \ln( 1+\exp( \beta( \mu_BB( j+\frac{1}{2} ) + \frac{\hbar^{2}k_z -\mu) ) ) \end{align} $$ consider when high T limit }}{2m$\beta \to 0$, using $\ln( 1+x ) = x$ $$ \ln \Xi = \frac{ gV^{1/3 } }{ 2\pi }e^{ -\beta \mu } \sum_{j = 0}^{\infty} e^{ \beta \mu_B B( j+1/2 ) } \cdot \int_{-\infty}^{\infty} e^{ \hbar^{2}k_z $$ obtain $$ \ln \Xi = \frac{ zVeB }{ h^{2}c } \qty( \frac{ 2\pi m }{ \beta } )^{1/2} \qty( 2\sinh( \frac{ \beta e\hbar B }{ 2mc } ) )^{-1} $$ on high T limit $$ \frac{1}{V} \ln \Xi = \frac{z}{\lambda^{3}}\qty[ 1-\frac{1}{6} ( \frac{ e\hbar }{ 2mc } )^{2}( \frac{B}{k_{B}T} )^{2} ] $$ then we can easily obtain}/2m } \dd{k_z

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define $$ F( x ) = \ln( 1+\exp( \beta( \mu_BBx + \frac{\hbar^{2}k_z $$ using Eular - Maclaurin Theorem $$ \frac{1}{2}F( a ) + \sum_{i = 1}^{\infty} F( a+j ) = \int_{a}^{\infty} F( x ) \dd{x} - \frac{1}{12} F'( a ) $$ let }}{2m} -\mu) ) )^{-1$a=1/2$ $$ \frac{1}{2}F( \frac{1}{2} ) + \sum_{j = 1}^{\infty} F( \frac{1}{2} + j ) = \int_{0}^{\infty} F( x ) \dd{x} - \int_{0}^{1/2 } F( x ) \dd{x} - \frac{1}{12}F'( \frac{1}{2} ) $$ using $$ F( x ) =F( 0 ) + xF'( 0 ) $$ finally obtain $$ \sum_{j = 1}^{\infty} F( j+\frac{1}{2} ) = \int_{0}^{\infty} F( x ) \dd{x} + \frac{1}{24}F'(0x ) $$