铁磁体低能单点激发能量计算：利用自旋角动量算符

我们首先来计算哈密顿量的作用在单点激发态上的结果。设单点激发态为\u0026#x27E8;S,m\u0026#x27E9;，其中S和m分别为自旋角动量量子数和磁量子数。我们有\n\n$$\n\begin{aligned}\nH\u0026#x27E8;S,m\u0026#x27E9; &= - J\sum\limits_{l\delta} \left[ S_l^zS^z_{l+\delta} + \frac12(S^+lS^-{l+\delta} + S^-lS^+{l+\delta}) \right]\u0026#x27E8;S,m\u0026#x27E9; \\n&= - J\sum\limits_{l\delta} \left[ S_l^zS^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; + \frac12(S^+lS^-{l+\delta} + S^-lS^+{l+\delta})\u0026#x27E8;S,m\u0026#x27E9; \right].\n\end{aligned}\n$$\n现在我们来分别计算这两项的结果。\n\n第一项$S_l^zS^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9;$。由于\u0026#x27E8;S,m\u0026#x27E9;是自旋算符$S_l^z$的本征态，我们有$S_l^z\u0026#x27E8;S,m\u0026#x27E9; = m\u0026#x27E8;S,m\u0026#x27E9;$。因此，\n\n$$\nS_l^zS^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; = mS^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9;.\n$$\n根据自旋角动量的升降算符定义，我们有$S^z_{l+\delta} = \frac{1}{2}(S^+lS^-{l+\delta} + S^-lS^+{l+\delta}) + S^z_l$。代入上式，我们得到\n\n$$\nS_l^zS^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; = m\left(\frac{1}{2}(S^+lS^-{l+\delta} + S^-lS^+{l+\delta}) + S^z_l\right)\u0026#x27E8;S,m\u0026#x27E9;.\n$$\n上式右侧的第一项可以写成升降算符的形式，即$S^+lS^-{l+\delta} + S^-lS^+{l+\delta} = 2S^z_{l+\delta} - S^z_l$。代入上式，我们得到\n\n$$\n\begin{aligned}\nS_l^zS^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; &= m\left(\frac{1}{2}(2S^z_{l+\delta} - S^z_l) + S^z_l\right)\u0026#x27E8;S,m\u0026#x27E9; \\n&= mS^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9;.\n\end{aligned}\n$$\n因此，\n\n$$\nS_l^zS^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; = mS^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9;.\n$$\n\n第二项$\frac12(S^+lS^-{l+\delta} + S^-lS^+{l+\delta})\u0026#x27E8;S,m\u0026#x27E9;$。根据自旋升降算符的作用结果，我们有\n\n$$\n\begin{aligned}\nS^+lS^-{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; &= \sqrt{(S - m )(S+m +1)}S^-{l+\delta}\u0026#x27E8;S,m+1\u0026#x27E9;, \\nS^-lS^+{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; &= \sqrt{(S + m )(S-m +1)}S^+{l+\delta}\u0026#x27E8;S,m-1\u0026#x27E9;.\n\end{aligned}\n$$\n代入第二项，我们得到\n\n$$\n\begin{aligned}\n\frac12(S^+lS^-{l+\delta} + S^-lS^+{l+\delta})\u0026#x27E8;S,m\u0026#x27E9; &= \frac12\left(\sqrt{(S - m )(S+m +1)}S^-{l+\delta}\u0026#x27E8;S,m+1\u0026#x27E9; + \sqrt{(S + m )(S-m +1)}S^+{l+\delta}\u0026#x27E8;S,m-1\u0026#x27E9;\right) \\n&= \frac{1}{\sqrt{2}}\left(\sqrt{(S - m )(S+m +1)}S^-{l+\delta}\u0026#x27E8;S,m+1\u0026#x27E9; + \sqrt{(S + m )(S-m +1)}S^+{l+\delta}\u0026#x27E8;S,m-1\u0026#x27E9;\right).\n\end{aligned}\n$$\n\n将上述两项的结果代回哈密顿量的表达式，我们有\n\n$$\n\begin{aligned}\nH\u0026#x27E8;S,m\u0026#x27E9; &= - J\sum\limits_{l\delta} \left[ mS^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; + \frac{1}{\sqrt{2}}\left(\sqrt{(S - m )(S+m +1)}S^-{l+\delta}\u0026#x27E8;S,m+1\u0026#x27E9; + \sqrt{(S + m )(S-m +1)}S^+{l+\delta}\u0026#x27E8;S,m-1\u0026#x27E9;\right)\right] \\n&= - Jm\sum\limits_{l\delta} S^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; - \frac{J}{\sqrt{2}}\sum\limits_{l\delta} \left(\sqrt{(S - m )(S+m +1)}S^-{l+\delta}\u0026#x27E8;S,m+1\u0026#x27E9; + \sqrt{(S + m )(S-m +1)}S^+{l+\delta}\u0026#x27E8;S,m-1\u0026#x27E9;\right).\n\end{aligned}\n$$\n\n现在我们来计算第一项$S^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9;$的结果。根据自旋角动量的升降算符定义，我们有$S^z_{l+\delta} = \frac{1}{2}(S^+lS^-{l+\delta} + S^-lS^+{l+\delta}) + S^z_l$。代入上式，我们得到\n\n$$\nS^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; = \frac{1}{2}(S^+lS^-{l+\delta} + S^-lS^+{l+\delta})\u0026#x27E8;S,m\u0026#x27E9; + S^z_l\u0026#x27E8;S,m\u0026#x27E9;.\n$$\n代入$S^+lS^-{l+\delta} + S^-lS^+{l+\delta} = 2S^z_{l+\delta} - S^z_l$，我们得到\n\n$$\nS^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; = \frac{1}{2}(2S^z_{l+\delta} - S^z_l)\u0026#x27E8;S,m\u0026#x27E9; + S^z_l\u0026#x27E8;S,m\u0026#x27E9;.\n$$\n化简上式，我们得到\n\n$$\nS^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; = S^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9;.\n$$\n因此，\n\n$$\nS^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; = S^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9;.\n$$\n\n将上述结果代回哈密顿量的表达式，我们有\n\n$$\n\begin{aligned}\nH\u0026#x27E8;S,m\u0026#x27E9; &= - Jm\sum\limits_{l\delta} S^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; - \frac{J}{\sqrt{2}}\sum\limits_{l\delta} \left(\sqrt{(S - m )(S+m +1)}S^-{l+\delta}\u0026#x27E8;S,m+1\u0026#x27E9; + \sqrt{(S + m )(S-m +1)}S^+{l+\delta}\u0026#x27E8;S,m-1\u0026#x27E9;\right) \\n&= - Jm\sum\limits_{l\delta} S^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; - \frac{J}{\sqrt{2}}\sum\limits_{l\delta} \left(\sqrt{(S - m )(S+m +1)}S^-{l+\delta}\u0026#x27E8;S,m+1\u0026#x27E9; + \sqrt{(S + m )(S-m +1)}S^+{l+\delta}\u0026#x27E8;S,m-1\u0026#x27E9;\right) \\n&= - Jm\sum\limits_{l\delta} S^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; - \frac{J}{\sqrt{2}}\sum\limits_{l\delta} \sqrt{(S - m )(S+m +1)}S^-{l+\delta}\u0026#x27E8;S,m+1\u0026#x27E9; - \frac{J}{\sqrt{2}}\sum\limits{l\delta} \sqrt{(S + m )(S-m +1)}S^+{l+\delta}\u0026#x27E8;S,m-1\u0026#x27E9;.\n\end{aligned}\n$$\n\n现在我们来化简上式。对于第一项$- Jm\sum\limits{l\delta} S^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9;$，我们可以将求和符号移到算符上，得到\n\n$$\n\begin{aligned}\n- Jm\sum\limits_{l\delta} S^z_{l+\delta}\u0026#x27E8;S,m\u0026#x27E9; &= - Jm\left(\sum\limits_{l\delta} S^z_{l+\delta}\right)\u0026#x27E8;S,m\u0026#x27E9; \\n&= - Jm\left(\sum\limits_{l\delta} \frac{1}{2}(S^+lS^-{l+\delta} + S^-lS^+{l+\delta})\right)\u0026#x27E8;S,m\u0026#x27E9; \\n&= - Jm\left(\sum\limits_{l\delta} \frac{1}{2}(2S^z_{l+\delta} - S^z_l)\right)\u0026#x27E8;S,m\u0026#x27E9; \\n&= - Jm\left(\sum\limits_{l\delta} S^z_{l+\delta} - \sum\limits_{l\delta} \frac{1}{2}S^z_l\right)\u0026#x27E8;S,m\u0026#x27E9; \\n&= - Jm\left(\sum\limits_{l\delta} S^z_{l+\delta} - \sum\limits_{l} \frac{1}{2}S^z_l\sum\limits_{\delta}\right)\u0026#x27E8;S,m\u0026#x27E9; \\n&= - Jm\left(\sum\limits_{l\delta} S^z_{l+\delta} - \sum\limits_{l} \frac{1}{2}S^z_l\right)\u0026#x27E8;S,m\u0026#x27E9; \\n&= - Jm\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\right)\u0026#x27E8;S,m\u0026#x27E9; \\n&= - Jm\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\right)\u0026#x27E8;S,m\u0026#x27E9; \\n&= - Jm\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\right)\u0026#x27E8;S,m\u0026#x27E9;.\n\end{aligned}\n$$\n\n对于第二项$- \frac{J}{\sqrt{2}}\sum\limits_{l\delta} \sqrt{(S - m )(S+m +1)}S^-{l+\delta}\u0026#x27E8;S,m+1\u0026#x27E9;$，我们可以将求和符号移到算符上，得到\n\n$$\n\begin{aligned}\n- \frac{J}{\sqrt{2}}\sum\limits{l\delta} \sqrt{(S - m )(S+m +1)}S^-{l+\delta}\u0026#x27E8;S,m+1\u0026#x27E9; &= - \frac{J}{\sqrt{2}}\sqrt{(S - m )(S+m +1)}\left(\sum\limits{l\delta} S^-{l+\delta}\right)\u0026#x27E8;S,m+1\u0026#x27E9; \\n&= - \frac{J}{\sqrt{2}}\sqrt{(S - m )(S+m +1)}\left(\sum\limits{l\delta} \frac{1}{2}(S^+lS^-{l+\delta} + S^-lS^+{l+\delta})\right)\u0026#x27E8;S,m+1\u0026#x27E9; \\n&= - \frac{J}{\sqrt{2}}\sqrt{(S - m )(S+m +1)}\left(\sum\limits_{l\delta} \frac{1}{2}(2S^z_{l+\delta} - S^z_l)\right)\u0026#x27E8;S,m+1\u0026#x27E9; \\n&= - \frac{J}{\sqrt{2}}\sqrt{(S - m )(S+m +1)}\left(\sum\limits_{l\delta} S^z_{l+\delta} - \sum\limits_{l\delta} \frac{1}{2}S^z_l\right)\u0026#x27E8;S,m+1\u0026#x27E9; \\n&= - \frac{J}{\sqrt{2}}\sqrt{(S - m )(S+m +1)}\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\sum\limits_{\delta}\right)\u0026#x27E8;S,m+1\u0026#x27E9; \\n&= - \frac{J}{\sqrt{2}}\sqrt{(S - m )(S+m +1)}\left(\sum\limits_{l\delta} S^z_{l+\delta} - \sum\limits_{l} \frac{1}{2}S^z_l\right)\u0026#x27E8;S,m+1\u0026#x27E9; \\n&= - \frac{J}{\sqrt{2}}\sqrt{(S - m )(S+m +1)}\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\right)\u0026#x27E8;S,m+1\u0026#x27E9; \\n&= - \frac{J}{\sqrt{2}}\sqrt{(S - m )(S+m +1)}\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\right)\u0026#x27E8;S,m+1\u0026#x27E9;.\n\end{aligned}\n$$\n\n对于第三项$- \frac{J}{\sqrt{2}}\sum\limits_{l\delta} \sqrt{(S + m )(S-m +1)}S^+{l+\delta}\u0026#x27E8;S,m-1\u0026#x27E9;$，我们可以将求和符号移到算符上，得到\n\n$$\n\begin{aligned}\n- \frac{J}{\sqrt{2}}\sum\limits{l\delta} \sqrt{(S + m )(S-m +1)}S^+{l+\delta}\u0026#x27E8;S,m-1\u0026#x27E9; &= - \frac{J}{\sqrt{2}}\sqrt{(S + m )(S-m +1)}\left(\sum\limits{l\delta} S^+{l+\delta}\right)\u0026#x27E8;S,m-1\u0026#x27E9; \\n&= - \frac{J}{\sqrt{2}}\sqrt{(S + m )(S-m +1)}\left(\sum\limits{l\delta} \frac{1}{2}(S^-lS^+{l+\delta} + S^+lS^-{l+\delta})\right)\u0026#x27E8;S,m-1\u0026#x27E9; \\n&= - \frac{J}{\sqrt{2}}\sqrt{(S + m )(S-m +1)}\left(\sum\limits_{l\delta} \frac{1}{2}(2S^z_{l+\delta} - S^z_l)\right)\u0026#x27E8;S,m-1\u0026#x27E9; \\n&= - \frac{J}{\sqrt{2}}\sqrt{(S + m )(S-m +1)}\left(\sum\limits_{l\delta} S^z_{l+\delta} - \sum\limits_{l\delta} \frac{1}{2}S^z_l\right)\u0026#x27E8;S,m-1\u0026#x27E9; \\n&= - \frac{J}{\sqrt{2}}\sqrt{(S + m )(S-m +1)}\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\sum\limits_{\delta}\right)\u0026#x27E8;S,m-1\u0026#x27E9; \\n&= - \frac{J}{\sqrt{2}}\sqrt{(S + m )(S-m +1)}\left(\sum\limits_{l\delta} S^z_{l+\delta} - \sum\limits_{l} \frac{1}{2}S^z_l\right)\u0026#x27E8;S,m-1\u0026#x27E9; \\n&= - \frac{J}{\sqrt{2}}\sqrt{(S + m )(S-m +1)}\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\right)\u0026#x27E8;S,m-1\u0026#x27E9; \\n&= - \frac{J}{\sqrt{2}}\sqrt{(S + m )(S-m +1)}\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\right)\u0026#x27E8;S,m-1\u0026#x27E9;.\n\end{aligned}\n$$\n\n将上述结果代回哈密顿量的表达式，我们有\n\n$$\nH\u0026#x27E8;S,m\u0026#x27E9; = - Jm\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\right)\u0026#x27E8;S,m\u0026#x27E9; - \frac{J}{\sqrt{2}}\sqrt{(S - m )(S+m +1)}\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\right)\u0026#x27E8;S,m+1\u0026#x27E9; - \frac{J}{\sqrt{2}}\sqrt{(S + m )(S-m +1)}\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\right)\u0026#x27E8;S,m-1\u0026#x27E9;.\n$$\n\n因此，铁磁体的低能单点激发能量为\n\n$$\nE = - Jm\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\right) - \frac{J}{\sqrt{2}}\sqrt{(S - m )(S+m +1)}\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\right) - \frac{J}{\sqrt{2}}\sqrt{(S + m )(S-m +1)}\left(\sum\limits_{l\delta} S^z_{l+\delta} - \frac{1}{2}\sum\limits_{l}S^z_l\right).\n$$\n\n这个结果表明，铁磁体的低能单点激发能量取决于自旋角动量量子数S、磁量子数m以及铁磁体中的自旋配置。\n