磁致自旋

含有磁缺陷的体系哈密顿量可以写为

$$\left\{\begin{aligned} H&=H_0+H_1\\ H_0&=\sum_{\alpha=\uparrow,\downarrow}\int\mathrm d^3\vec x\ \psi^\dagger_\alpha(\vec x)\left(-\frac{\hbar^2\nabla^2}{2m}\right)\psi_\alpha(\vec x)\\ H_1&=J\vec s(\vec x_i)\cdot\vec S(\vec x_i) \end{aligned}\right.$$

其中大写的$\vec S$为一直的磁缺陷，小写的$\vec s$为自旋

$$\vec s(\vec x)=\frac12\sum_{\alpha,\beta=\uparrow,\downarrow}\psi^\dagger_\alpha(\vec x)[\vec \sigma]_{\alpha\beta}\psi_\beta(\vec x)$$

线性响应

$$\delta\langle\vec s(\vec x,t)\rangle=\frac i\hbar\int_{-\infty}^t\mathrm dt'\ \langle\left[H_{1,H}(t'),\vec s_{H}(\vec x,t)\right]\rangle$$

把具体形式代入得

$$\begin{aligned} \langle s_\mu(\vec x,t)\rangle&=J\frac i\hbar\int_{-\infty}^t\mathrm dt'\ \langle\left[\vec s_{H}(\vec x_i,t'),s_{\mu,H}(\vec x,t)\right]\rangle\cdot\vec S_H(\vec x_i,t)\\ &=J\frac 1\hbar\int_{-\infty}^\infty\mathrm dt'\ \sum_{\nu}D^R_{\mu\nu}(\vec x,t;\vec x_i,t') S_{\nu,H}(\vec x_i,t)\\ &=J\frac 1\hbar\int_{-\infty}^\infty\mathrm dt'\int\frac{\mathrm d^3\vec k\mathrm d\omega}{(2\pi)^4}\ e^{i\vec k\cdot(\vec x-\vec x_i)-i\omega(t-t')}\sum_{\nu}D^R_{\mu\nu}(\vec k,\omega) S_{\nu,H}(\vec x_i,t')\\ \end{aligned}$$

静态极限下外场为$S_{\nu}(\vec x_i)$与时间无关，所以Heisenberg绘景下$S_{\nu,H}(\vec x_i,t’)=e^{iHt’/\hbar}S_\nu(\vec x_i)e^{-iHt’/\hbar}=S_\nu(\vec x_i)$

$$\langle s_\mu(\vec x,t)\rangle=\sum_{\nu}J\frac1\hbar\int\frac{\mathrm d^3\vec k\mathrm d\omega}{(2\pi)^3}e^{i\vec k\cdot(\vec x-\vec x')}\delta(\omega)D^R_{\mu\nu}(\vec k,\omega)S_\nu(\vec x_i) =\sum_{\nu}J\frac1\hbar\int\frac{\mathrm d^3\vec k}{(2\pi)^3}e^{i\vec k\cdot(\vec x-\vec x')}D^R_{\mu\nu}(\vec k,0)S_\nu(\vec x_i)$$

关联函数（编时）

$$D^T_{\mu\nu}(\vec k,\omega)/\hbar=\Pi_{\mu\nu}^\star(\vec k,\omega)+\Pi_{\mu\alpha}^\star(\vec k,\omega)V^{\alpha\beta}(\vec k)\Pi_{\beta\nu}^\star(\vec k,\omega)+\cdots=\Pi_{\mu\alpha}^\star(\vec k,\omega)\left[1-V(\vec k)\Pi^\star(\vec k,\omega)\right]_{\alpha\nu}^{-1}$$

在最低阶近似下

$$\Pi_{\mu\nu}^\star(\vec k,\omega)=-\sum_{\alpha\beta\gamma\delta}\frac{i}\hbar\int\frac{\mathrm d^3\vec q\mathrm d\epsilon}{(2\pi)^4}G^0_{\alpha\sigma}(\vec q,\epsilon)G^0_{\rho\beta}(\vec k+\vec q,\omega+\epsilon)\frac12[\sigma_\mu]_{\alpha\beta}\frac12[\sigma_\nu]_{\rho\sigma}$$

利用$G^0_{\alpha\beta}=G^0\delta_{\alpha\beta}$和$\mathrm{Tr}[\sigma_\mu\sigma_\nu]=2\delta_{\mu\nu}$得

$$\Pi_{\mu\nu}^\star(\vec k,\omega)=\frac{\delta_{\mu\nu}}4\Pi_0(\vec k,\omega),\Pi_0(\vec k,\omega)=-\frac{2i}\hbar\int\frac{\mathrm d^3\vec q\mathrm d\epsilon}{(2\pi)^4}G^0(\vec q,\epsilon)G^0(\vec k+\vec q,\omega+\epsilon)$$

具体的计算比较复杂，所幸考试时会给公式

$$\begin{aligned} \Re\left[\Pi_0(k_F\vec q,\frac{\hbar k_F^2}{m}\nu)\right]=\frac{2mk_F}{\hbar^2}\frac1{4\pi^2}\times\left\{-1+\frac1{2q}\left(1-\left(\frac\nu q-\frac q2\right)^2\right)\ln\left|\frac{\nu-\frac{q^2}2+q}{\nu-\frac {q^2}2-q}\right|\right.\\ \left.-\frac1{2q}\left(1-\left(\frac\nu q+\frac q2\right)^2\right)\ln\left|\frac{\nu+\frac{q^2}2+q}{\nu+\frac {q^2}2-q}\right|\right\} \end{aligned}$$ $$\Im\left[\Pi_0(k_F\vec q,\frac{\hbar k_F^2}m\nu)\right]=\left\{\begin{aligned} &-\frac{mk_F}{\hbar^2}\frac{1}{4\pi q}\left[1-\left(\frac\nu q-\frac q2\right)^2\right] &(q>2\&\left|\frac{q^2}2-\nu\right|\leq q)or(q<2\&|q-\nu|\le\frac{q^2}2\\ &-\frac{mk_F}{\hbar^2}\frac\nu{2\pi q} &(q<2\&0\le\nu\le-\frac{q^2}2+q)\\ &0&(otherwise) \end{aligned}\right.$$

因此

$$\Pi_0(k_F\vec q,0)=\frac{2mk_F}{\hbar^2}\frac{1}{4\pi^2}\times\left\{-1+\left(\frac1q-\frac q4\right)\ln\left|\frac{q-2}{q+2}\right|\right\}$$

所以当$V(\vec k)$为小量时

$$D^T_{\mu\nu}(k_F\vec q,0)=\hbar\Pi^\star_{\mu\nu}(k_F\vec q,0)=\frac{\delta_{\mu\nu}}4\frac{2mk_F}{\hbar}\frac1{4\pi^2}\times\left\{-1+\left(\frac1q-\frac q4\right)\ln\left|\frac{q-2}{q+2}\right|\right\}+\mathcal O(V^2)$$

关联函数（Retard）

$$\Re[D^R(\vec k,\omega)]=\Re[D^T(\vec k,\omega)], \Im[D^R(\vec k,\omega)]=\mathrm{sgn}(\omega)\Im[D^T(\vec k,\omega)]$$

所以

$$D^R_{\mu\nu}(k_F\vec q,0)=\frac{\delta_{\mu\nu}}4\frac{2mk_F}{\hbar}\frac1{4\pi^2}\times\left\{-1+\left(\frac1q-\frac q4\right)\ln\left|\frac{q-2}{q+2}\right|\right\}+\mathcal O(V^2)$$

最终

$$\begin{aligned} \langle s_\mu(\vec x,t)\rangle =&\sum_{\nu}J\frac1\hbar\int\frac{\mathrm d^3\vec k}{(2\pi)^3}e^{i\vec k\cdot(\vec x-\vec x')}D^R_{\mu\nu}(\vec k,0)S_\nu(\vec x_i)\\ =&J\frac1\hbar\int\frac{\mathrm d^3\vec k}{(2\pi)^3}e^{i\vec k\cdot(\vec x-\vec x_i)}\frac{mk_F}{8\pi^2\hbar}\left\{-1+\left(\frac{k_F}k-\frac {k}{4k_F}\right)\ln\left|\frac{k-2k_F}{k+2k_F}\right|\right\}S_\mu(\vec x_i) \end{aligned}$$

所以

$$\langle s_\mu(\vec x)\rangle=JV(\vec x-\vec x_i)S_\mu(\vec x_i)$$ $$V(\vec y)=\frac{mk_F}{8\pi^2\hbar^2}\int\frac{\mathrm d^3\vec k}{(2\pi)^3}e^{i\vec k\cdot\vec y}\left\{-1+\frac{1-x^2}{2x}\ln\left|\frac{x-1}{x+1}\right|\right\}_{x=k/2k_F}$$

势函数

球坐标展开得

$$V(\vec y)=\frac{mk_F}{8\pi^2\hbar^2}\int_0^\infty\frac{2\pi k^2\mathrm dk}{(2\pi)^3}\frac{\sin ky}{ky}\mathrm d\theta\ \left\{-1+\frac{1-x^2}{2x}\ln\left|\frac{x-1}{x+1}\right|\right\}_{x=k/2k_F}$$

换元$u=2k_Fy$，则

$$\begin{aligned} V(\vec y)=&-\frac{mk_F}{8\pi^2\hbar^2(2\pi)^3}\int_0^\infty 2\pi(2k_F)^3x^2\mathrm dx\frac{\sin ux}{ux}\left[1+\frac{1-x^2}{2x}\ln\left|\frac{x+1}{x-1}\right|\right]\\ =&-\frac{mk_F^4}{4\pi^4\hbar^2u}\int_0^\infty\mathrm dx\ \sin ux\left[x+\frac{1-x^2}{2}\ln\left|\frac{x+1}{x-1}\right|\right]\\ =&-\frac{mk_F^4}{4\pi^4\hbar^2u}\lim_{\eta\rightarrow0}\int_0^\infty\mathrm dx\ \sin ux\left[x+(1-x^2)\ln\frac{(x+1)^2+\eta^2}{(x-1)^2+\eta^2}\right]\\ \end{aligned}$$

因为括号内为奇函数，所以

$$V(\vec y)=\frac{imk_F^4}{8\pi^4\hbar^2 u}\lim_{\eta\rightarrow 0}\int_{-\infty}^\infty\mathrm dz\ e^{iuz}\left[z+(1-z^2)\ln\frac{(z+1)^2+\eta^2}{(z-1)^2+\eta^2}\right]$$

做割线$\Re[z]=\pm1,|\Im[z]|\ge\eta$，对上半平面围道积分，值为零，所以

$$V(\vec y)=\frac{imk_F^4}{8\pi^4\hbar^2u}\lim_{\eta\rightarrow 0}\left[\int_{1+\epsilon+i\eta}^{1+\epsilon+i\infty}-\int_{1-\epsilon+i\eta}^{1-\epsilon+i\infty}+(1\rightarrow-1)\right]\mathrm dz\ e^{iuz}\left[z+(1-z^2)\ln\frac{(z+1)^2+\eta^2}{(z-1)^2+\eta^2}\right]$$

显式写出的积分为

$$\int_0^\infty i\mathrm dt\ e^{iu(1+it)}(1-(1+it)^2)\ln\frac{(\epsilon+it)^2+\eta^2}{(-\epsilon+it)^2+\eta^2}=e^{iu}\int_0^\infty\mathrm dt\ (t^2-2it)e^{-ut}(-2\pi i)=-4\pi ie^{iu}\left(\frac{1}{u^3}-\frac{i}{u^2}\right)$$

同理可算另外两项积分，最终为

$$V(\vec y)=\frac{mk_F^4}{2\pi^3\hbar^2u}(\frac{\cos u}{u^3}-\frac{\sin u}{u^2})$$