问题
已知 \(H=H_0 +V_t\)，其中 \(V_t = \hat{B} F_t\) ，为一外场 \(F_t\) 对原有系统造成的微扰，\(\hat{B}\) 是系统中与外场 \(F_t\) 相耦合的力学量。求某力学量 \(\hat{A}\) 在 \(t\) 时刻系综平均值 \(\langle A \rangle_t\) 与 \(0\) 时刻的系综平均值 \(\langle A \rangle_0\) 系综平均值 \(\langle A \rangle_t -\langle A \rangle_0 \) 差的近似结果。
现取一特例，假设外场 \(F_t\) 就是磁场 \(\vec{B_t}\) ，那么系统中与外场 \(\vec{B_t}\) 相耦合的力学量就是系统的总磁矩 \(\vec{m}\) ，总磁矩 \(\vec{m}\) 对应的微观量是系统中单个电子的自旋 \(\vec{S_i}\) ，下标 \(i\) 代表第 \(i\) 个电子。其关系为：
\[ \vec{m} = \sum_i \vec{m_i} = \frac{g_J \mu_B}{\hbar} \sum_i \vec{S_i} \]假设测量量为磁化强度 \(\vec{M} = \vec{m}/V\)
所以对应关系为：
\[ \hat{B} \longleftrightarrow \vec{m} \] \[ F_t \longleftrightarrow \vec{B} \] \[ \hat{A} \longleftrightarrow \vec{M} \]初步路线
写出详细表达式进行具体计算：
\[ \langle \vec{M} \rangle_t -\langle\vec{ M} \rangle_0 = \mathrm{Tr} [(\rho_t - \rho_0)\hat{M}] = \frac{1}{V}\cdot \frac{g_J \mu_B}{\hbar}\sum_i \mathrm{Tr} [(\rho_t - \rho_0)\vec{S_i}] \]接下来求 \(\rho_t - \rho_0\) 。
求 \(t\) 时刻密度矩阵的近似值
Dirac表象中密度矩阵的运动方程：
\[ \dot{\rho_t^D} = \frac{1}{i\hbar} [V_t^D, \rho_t^D] \]假设初值条件为：
\[ \lim_{t\rightarrow -\infty}\rho_t^D(t) =\lim_{t\rightarrow -\infty}\rho_t = \rho_0 = \frac{e^{-\beta H_0}}{\mathrm{Tr}(e^{-\beta H_0})} \]Dirac表象选取 \(t_0=0\) 为时间基点，也就是 \(U(t,t_0)=e^{\frac{1}{i\hbar}(t-t_0)H_0}=e^{\frac{1}{i\hbar}H_0 t} = U(t)\)
所以 \(t\) 时刻密度矩阵可迭代求解：
\[ \rho_t^D = \rho_0 +\frac{1}{i\hbar} \int_{-\infty}^t \mathrm{d}t' [V_{t'}^D, \rho_{t'}^D] =\rho_0 +\frac{1}{i\hbar} \int_{-\infty}^t \mathrm{d}t' [V_{t'}^D, \rho_0] +\cdots \]做近似，只取线性项得：
\[ \rho_t^D \approx \rho_0 +\frac{1}{i\hbar} \int_{-\infty}^t \mathrm{d}t' [V_{t'}^D, \rho_0] \]换到薛定谔表象：
\[ \rho_t \approx \rho_0 +\frac{1}{i\hbar} \int_{-\infty}^t \mathrm{d}t' U(t)[V_{t'}^D, \rho_0]U^{\dagger}(t) \]求解结果
将 \(t\) 时刻密度矩阵的近似结果
\[ \rho_t - \rho_0 \approx \frac{1}{i\hbar} \int_{-\infty}^t \mathrm{d}t' U(t)[V_{t'}^D, \rho_0]U^{\dagger}(t) \]代入到详细表达式进行具体计算：
\[ \mathrm{Tr} [(\rho_t - \rho_0)\vec{S_i}] = \frac{1}{i\hbar} \int_{-\infty}^t \mathrm{d}t' \mathrm{Tr} \left( U(t)[V_{t'}^D, \rho_0]U^{\dagger}(t) \vec{S_i}\right) \]而
\[ V_{t'} = -\vec{m}\cdot \vec{B_{t'}} = -\sum_{\alpha = x,y,z} m^{\alpha} B_{t'}^{\alpha} = -\sum_{\alpha = x,y,z} \sum_i \frac{g_J\mu_B}{\hbar} S_i^{\alpha} B_{t'}^{\alpha} \]所以
\[ \mathrm{Tr} \left( U(t)[V_{t'}^D, \rho_0]U^{\dagger}(t) \vec{S_i}\right) = \mathrm{Tr} \left( [V_{t'}^D, \rho_0] \vec{S_i}(t)\right) = -\sum_{\alpha = x,y,z} \sum_i \frac{g_J\mu_B}{\hbar} \mathrm{Tr}\left( [S_i^{\alpha}(t'),\rho_0] \vec{S_i}(t) \right)B_{t'}^{\alpha} \]而
\[ \mathrm{Tr}\left( [S_i^{\alpha}(t'),\rho_0] \vec{S_i}(t) \right) = \mathrm{Tr}\left( \rho_0 [\vec{S_i}(t),S_i^{\alpha}(t')] \right) = \langle [\vec{S_i}(t),S_i^{\alpha}(t')]\rangle_0 \]所以
\[ \mathrm{Tr} [(\rho_t - \rho_0)\vec{S_i}] = -\sum_{\alpha = x,y,z} \sum_i \frac{g_J\mu_B}{\hbar} \frac{1}{i\hbar} \int_{-\infty}^t \mathrm{d}t' B_{t'}^{\alpha} \langle [\vec{S_i}(t),S_i^{\alpha}(t')]\rangle_0 \] \[ \langle M \rangle_t -\langle M \rangle_0 = \mathrm{Tr} [(\rho_t - \rho_0)\hat{M}] = -\frac{1}{V}\cdot \frac{g_J \mu_B}{\hbar}\sum_{\alpha = x,y,z} \sum_i \frac{g_J\mu_B}{\hbar} \sum_i \frac{1}{i\hbar} \int_{-\infty}^t \mathrm{d}t' B_{t'}^{\alpha}\langle [\vec{S_i}(t),S_i^{\alpha}(t')]\rangle_0 \]注
Wolfgang Nolting, Fundamentals of Many-body Physics, 3.1节笔记
易混淆的量
- magnetetic susceptibility 磁化率： \(\chi\)
- magnetic moment 总磁矩：\(\vec{m}\)
- magnetisation 磁化强度：\(\vec{M}\)