スピントロニクス理論の基礎/8-8 のバックアップ(No.1)

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スピントロニクス理論の基礎

8-8 相互作用の摂動論的扱い

H=H_0+V

に対して、

(8.96), (8.97)

U=(U_0U_0^\dagger) U=U_0(U_0^\dagger U)\equiv U_0 U_V

として、 U U_0 U_V に分けて書く。

(8.98), (8.99)

&math( i\hbar\frac{\PD}{\PD t}U_V&=i\hbar\frac{\PD U_0^\dagger}{\PD t} U+i\hbar U_0^\dagger\frac{\PD U}{\PD t}\\ &=i\hbar\left(\frac{H_0 U_0}{i\hbar}\right)^\dagger U+i\hbar U_0^\dagger \left(\frac{H U}{i\hbar}\right)\\ &=-U_0^\dagger H_0 U + U_0^\dagger H U\\ &=U_0^\dagger (H-H_0) U\\ &=U_0^\dagger V U\\ &=U_0^\dagger V (U_0 U_0^\dagger) U\\ &=V_{H_0}U_V );

したがって (8.7) と同様にして、

(8.100)

U_V(t,t_0)=Te^{-\frac{i}{\hbar}\int_{t_0}^tdt'V_{H_0}(t')}

これと O_\mathrm H=U^\dagger O U=U_V^\dagger U_0^\dagger O U_0 U_V=U_V^\dagger O_{\mathrm H_0}U_V を用いて、

(8.101)

&math( \overline O(t)&= \frac{1}{Z_0}\trace[U(-i\beta/\hbar+t_0,t_0)U_V^\dagger(t,t_0)O_{\mathrm H_0}(t)U_V(t,t_0)]\\ &=\frac{1}{Z_0}\trace[T_C\,e^{-\frac{i}{\hbar}\int_Cd\tau'V_{\mathrm H_0}(\tau')}O_{\mathrm H_0}(\tau)]\\ &=\big\langle T_C\,e^{-\frac{i}{\hbar}\int_Cd\tau'V_{\mathrm H_0}(\tau')}O_{\mathrm H_0}(\tau)\big\rangle );

同様に、

(8.102)

&math( G(\bm r,t,\bm r',t)= &=-i\big\langle T_C\,e^{-\frac{i}{\hbar}\int_Cd\tau'V_{\mathrm H_0}(\tau')} c_{\mathrm H_0}^\dagger(\bm r',\tau')c_{\mathrm H_0}(\bm r,\tau)\big\rangle );

(8.63) の H H_0+V に置き換え、(8.24A), (8.30A) を用いれば、

(8.103)

&math( &i\hbar\frac{\PD}{\PD \tau}G=\hbar\delta(\tau-\tau')\delta^3(\bm r-\bm r')

  1. i\big\langle T_C\,e^{-\frac{i}{\hbar}\int_Cd\tau'H(\tau')} [H_0+V,c(\bm r,\tau)]c^\dagger(\bm r',\tau')\big\rangle\\ &=\hbar\delta(\tau-\tau')\delta^3(\bm r-\bm r')
  • \left(\frac{\hbar^2}{2m}\nabla^2_{\bm r}+\varepsilon_F\right)G
  1. i\big\langle T_C\,e^{-\frac{i}{\hbar}\int_Cd\tauH(\tau)} [V,c(\bm r,\tau)]c^\dagger(\bm r',\tau')\big\rangle\\ );

&math( &\left(i\hbar\frac{\PD}{\PD \tau}+\frac{\hbar^2}{2m}\nabla^2_{\bm r}+\varepsilon_F\right)G =\hbar\delta(\tau-\tau')\delta^3(\bm r-\bm r')

  1. i\big\langle T_C\,e^{-\frac{i}{\hbar}\int_Cd\tauH(\tau)} [V,c(\bm r,\tau)]c^\dagger(\bm r',\tau')\big\rangle\\ );

この \delta 関数を g_0 で書き換えると、

(8.104)

&math( &\left(i\hbar\frac{\PD}{\PD \tau}+\frac{\hbar^2}{2m}\nabla^2_{\bm r}+\varepsilon_F\right)G(\bm r,t,\bm r',t')\\ &= \left(i\hbar\frac{\PD}{\PD \tau}+\frac{\hbar^2}{2m}\nabla^2_{\bm r}+\varepsilon_F\right)g_0(\bm r,t,\bm r',t')\\ &+\int_Cd\tau_1\int d^3r_1\hbar\delta(\tau-\tau_1)\delta^3(\bm r-\bm r_1)\times \frac{i}{\textcolor{red}{\hbar}}\big\langle T_C\,e^{-\frac{i}{\hbar}\int_Cd\tauH(\tau)} [V,c(\bm r_1,\tau_1)]c^\dagger(\bm r',\tau')\big\rangle\\ &= \left(i\hbar\frac{\PD}{\PD \tau}+\frac{\hbar^2}{2m}\nabla^2_{\bm r}+\varepsilon_F\right)g_0(\bm r,t,\bm r',t')\\ &+\left(i\hbar\frac{\PD}{\PD \tau}+\frac{\hbar^2}{2m}\nabla^2_{\bm r}+\varepsilon_F\right) \int_Cd\tau_1\int d^3r_1g_0(\bm r,t,\bm r',t')\times \frac{i}{\textcolor{red}{\hbar}}\big\langle T_C\,e^{-\frac{i}{\hbar}\int_Cd\tauH(\tau)} [V,c(\bm r_1,\tau_1)]c^\dagger(\bm r',\tau')\big\rangle\\ );


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