@jiyanjiang
2015-11-24T15:31:30.000000Z
字数 1702
阅读 1383
固体理论
格点
$$ x_l = \frac{1}{\sqrt N}\sum\limits_{q \in BZ}e^{iqla} x_q $$
$$ x_q = \frac{1}{\sqrt N}\sum\limits_l e^{-iqla}x_l $$
现在引入格点上的动量算符
$$ \left[ x_l ,p_{l'} \right] = i \hbar {\delta}_{l,l'} $$
同时波矢空间中的动量算符
$$ \left[ x_q, p_{q'} \right] = i \hbar {\delta}_{q,q'} $$
可以证明
$$ p_l = \frac{1}{\sqrt N}\sum\limits_{q \in BZ}e^{-iqla} p_q $$
代入
$$ x_l p_{l'} = \sum\limits_q \sum\limits_{q'}\frac{1}{N}e^{iqla-iq'l'a}x_q p_{q'} $$
$$ iqla-iq'l'a=iqla - iql'a + iql'a - i q'l'a $$
代入后:
$$\sum\limits_{q,q'} \frac{1}{N} e^{iq(l-l')a}e^{i(q-q')l'a} x_q p_{q'} = \sum\limits_{q'}\delta_{l,l'}e^{i(q-q')l'a}x_qp_{q'} $$
考虑到:
$$ x_qp_{q'}-p_{q'}x_q = i\hbar \delta_{q,q'} $$
继续化简,
$$[x_l,p_{l'}] = \sum\limits_{q'}\delta_{l,l'}e^{i(q-q')l'a}i \hbar \delta_{q,q'} = i \hbar \delta_{l,l'}$$
QED