弱い周期場

原子配列と一致した，摂動と見なせるほど弱い周期的ポテンシャル $U(\mbox{\boldmath$r$})$を導入する．逆格子ベクトル $\mbox{\boldmath$K$}_{hkl}$を用いて， $U(\mbox{\boldmath$r$})$をフーリエ級数に展開する．

$$U(\mbox{\boldmath$r$}) = \sum_{\mbox{\boldmath$K$}}U_{\mbox{\boldmath$K$}}{\rm e}^{{\rm i}\mbox{\boldmath$K$}\cdot\mbox{\boldmath$r$}},$$ (3.5.48)

$$\mbox{\boldmath$K$} = (h\mbox{\boldmath$a$}^{*},\ k\mbox{\boldmath$b$}^{*},\ l\mbox{\boldmath$c$}^{*}),$$ (3.5.49)

$$U_{\mbox{\boldmath$K$}} = U_{-\mbox{\boldmath$K$}} = U_{\mbox{\boldmath$K$}}^{*}$$ (3.5.50)

固有関数 $\psi_{\mbox{\boldmath$k$}}(\mbox{\boldmath$r$})$の $u_{\mbox{\boldmath$k$}}(\mbox{\boldmath$r$})$もフーリエ級数に展開する．

$$\psi_{\mbox{\boldmath$k$}}(\mbox{\boldmath$r$}) = u_{\mbox{\boldm... ...e}^{{\rm i}(\mbox{\boldmath$k$} + \mbox{\boldmath$K$})\cdot\mbox{\boldmath$r$}}$$ (3.5.51)

上式は，結晶中で電子状態を記述する平面波が周期的ポテンシャル場の中でブラッグ反射を受けることを表す．

(5-1)，(5-2)式をシュレーディンガー方程式に代入する．

$$\left[- {\hbar^{2} \over{2m}}\Delta + U(\mbox{\boldmath$r$})\righ... ... \epsilon_{\mbox{\boldmath$k$}}\psi_{\mbox{\boldmath$k$}}(\mbox{\boldmath$r$}),$$ (3.5.52)

$$- {\hbar^{2} \over{2m}}\Delta\psi_{\mbox{\boldmath$k$}}(\mbox{\bo... ...e}^{{\rm i}(\mbox{\boldmath$k$} + \mbox{\boldmath$K$})\cdot\mbox{\boldmath$r$}}$$ (3.5.53)

$$= \sum{\hbar^{2}(\mbox{\boldmath$k$} + \mbox{\boldmath$K$})^{2} \... ...e}^{{\rm i}(\mbox{\boldmath$k$} + \mbox{\boldmath$K$})\cdot\mbox{\boldmath$r$}}$$ (3.5.54)

$$\equiv \sum_{\mbox{\boldmath$K$}}\epsilon_{\mbox{\boldmath$k$} + ... ...}^{{\rm i}(\mbox{\boldmath$k$} + \mbox{\boldmath$K$})\cdot\mbox{\boldmath$r$}},$$ (3.5.55)

$$\sum_{\mbox{\boldmath$K$}}\epsilon_{\mbox{\boldmath$k$} + \mbox{\... ...e}^{{\rm i}(\mbox{\boldmath$k$} + \mbox{\boldmath$K$})\cdot\mbox{\boldmath$r$}}$$ (3.5.56)

$$= \epsilon_{\mbox{\boldmath$k$}}\sum_{\mbox{\boldmath$K$}}\alpha_... ...k$} + \mbox{\boldmath$K$} + \mbox{\boldmath$K$}')\cdot\mbox{\boldmath$r$}} = 0,$$ (3.5.57)

上式に， ${\rm e}^{-{\rm i}(\mbox{\boldmath$k$} + \mbox{\boldmath$K$}'')\cdot\mbox{\boldmath$r$}}$を掛けて積分する．

$$\sum_{\mbox{\boldmath$K$}}(\epsilon_{\mbox{\boldmath$k$} + \mbox{... ... \mbox{\boldmath$K$}'')\cdot\mbox{\boldmath$r$}}{\rm d}\mbox{\boldmath$r$} = 0,$$ (3.5.58)

$$\sum_{\mbox{\boldmath$K$}}(\epsilon_{\mbox{\boldmath$k$} + \mbox{... ...\delta(\mbox{\boldmath$K$} + \mbox{\boldmath$K$}' - \mbox{\boldmath$K$}'') = 0,$$ (3.5.59)

$$(\epsilon_{\mbox{\boldmath$k$} + \mbox{\boldmath$K$}''}^{0} - \ep... ...' - \mbox{\boldmath$K$}}\alpha_{\mbox{\boldmath$k$} + \mbox{\boldmath$K$}} = 0,$$ (3.5.60)

$$\mbox{\boldmath$K$}''\ \longrightarrow\ \mbox{\boldmath$K$},\ \ \ \mbox{\boldmath$K$}\ \longrightarrow\ \mbox{\boldmath$K$}'$$ (3.5.61)

$$(\epsilon_{\mbox{\boldmath$k$} + \mbox{\boldmath$K$}}^{0} - \epsi... ...- \mbox{\boldmath$K$}'}\alpha_{\mbox{\boldmath$k$} + \mbox{\boldmath$K$}'} = 0,$$ (3.5.62)

上式は，1つの $\mbox{\boldmath$k$}$を指定したときに $\mbox{\boldmath$K$}$の数だけの連立方程式を与える．

第2項の $\mbox{\boldmath$K$}'$の和の中で1つの $U_{\mbox{\boldmath$K$}}$だけが重要である場合を考える．

$$(\epsilon_{\mbox{\boldmath$k$} + \mbox{\boldmath$K$}}^{0} - \epsi... ...\mbox{\boldmath$K$}} + U_{\mbox{\boldmath$K$}}\alpha_{\mbox{\boldmath$k$}} = 0,$$ (3.5.63)

$$(\epsilon_{\mbox{\boldmath$k$}}^{0} - \epsilon_{\mbox{\boldmath$k... ...- \mbox{\boldmath$K$}'}\alpha_{\mbox{\boldmath$k$} + \mbox{\boldmath$K$}'} = 0,$$ (3.5.64)

$$\left( \begin{array}{cc} U_{-\mbox{\boldmath$K$}'} & \epsilo... ...h$K$}}\\ \alpha_{\mbox{\boldmath$k$}} \end{array}\right) = 0,$$ (3.5.65)

$$\vert U_{\mbox{\boldmath$K$}}\vert^{2} - (\epsilon_{\mbox{\boldma... ...\boldmath$k$} + \mbox{\boldmath$K$}}^{0} - \epsilon_{\mbox{\boldmath$k$}}) = 0,$$ (3.5.66)

$$\epsilon_{\mbox{\boldmath$k$}} = {1 \over{2}}\left[\epsilon_{\mbo... ...mbox{\boldmath$K$}}^{0})^{2} + 4\vert U_{\mbox{\boldmath$K$}}\vert^{2}}\right],$$ (3.5.67)

$U_{\mbox{\boldmath$K$}}$の影響の大きいのは

$$\epsilon_{\mbox{\boldmath$k$}}^{0} \simeq \epsilon_{\mbox{\boldmath$k$} + \mbox{\boldmath$K$}}^{0},$$ (3.5.68)

$$\mbox{\boldmath$k$} \simeq \pm\frac{\mbox{\boldmath$K$}}{2},$$ (3.5.69)

$\mbox{\boldmath$k$} = \frac{\mbox{\boldmath$K$}}{2}$では，

$$\epsilon_{\mbox{\boldmath$K$}/2} = \epsilon_{\mbox{\boldmath$K$}/2}^{0} \pm U_{\mbox{\boldmath$K$}},$$ (3.5.70)

図 3.7: 古典的磁石と電子スピン磁石

$\mbox{\boldmath$k$} = \pm\frac{\pi}{a}$での様子を調べる．

$$\epsilon_{\mbox{\boldmath$k$}}^{-} = {1 \over{2}}\left[2\epsilon_... ...t U_{{2\pi}/a}\vert^{2}}\right] = \epsilon_{\pi/a}^{0} - \vert U_{2\pi/a}\vert,$$ (3.5.71)

$${\alpha_{\pi/a} \over{\alpha_{-\pi/a}}} = -{U_{2\pi/a} \over{\vert U_{2\pi/a}\vert}} = 1,\ \ \ (U_{\pm2\pi/a} < 0)$$ (3.5.72)

$$\psi_{\pi/a}^{-}(x) = \sum_{\mbox{\boldmath$K$}}\alpha_{\pi/a + \... ...}{\rm e}^{{\rm i}\frac{\pi}{a}x} = 2\vert\alpha_{\pi/a}\vert\cos\frac{\pi}{a}x,$$ (3.5.73)

$$\epsilon_{\mbox{\boldmath$k$}}^{+} = {1 \over{2}}\left[2\epsilon_... ...ert U_{2\pi/a}\vert^{2}}\right] = \epsilon_{\pi/a}^{0} + \vert U_{2\pi/a}\vert,$$ (3.5.74)

$${\alpha_{\pi/a} \over{\alpha_{-\pi/a}}} = {U_{2\pi/a} \over{\vert U_{2\pi/a}\vert}} = -1,$$ (3.5.75)

$$\psi_{\pi/a}^{+}(x) = \sum_{\mbox{\boldmath$K$}}\alpha_{\pi/a + \... ...a}{\rm e}^{{\rm i}\frac{\pi}{a}x} = 2\vert\alpha_{\pi/a}\vert\sin\frac{\pi}{a}x$$ (3.5.76)

図 3.8: 古典的磁石と電子スピン磁石