量子几何张量【某几何的 Bloch Bands · 3】
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上一篇文章讨论了 Berry phase, Berry curvature 与陈数。本文继续讨论量子几何张量。
Berry connection 使得不同 $\mathbf k$ 处的波函数可以比较。即是说，我们可以通过平行移动将属于不同向量空间 $W^{({\mathbf k})}$ 的向量放入同一向量空间进行比较。
考虑在一般的 non-Abelian 情况下，对于在某过程中从 $\mathbf k$ 移动至 $\mathbf k + \delta \mathbf k$ 的波函数，首先将 $\mathbf k + \delta \mathbf k$ 处的波函数 $\Psi^{(\mathbf k + \delta \mathbf k)}$ 通过绝热过程（平行移动）拉回 $W^{(\mathbf k)}$ 空间
$$ \Psi^{(\mathbf k + \delta \mathbf k)} = \exp\left(i\int_{\Gamma^{-1}}{\mathscr A^{(\mathbf k)}}\right) \tilde \Psi^{(\mathbf k + \delta \mathbf k)} \tag{1} $$从而，波函数变化量可表为
$$ \begin{align} \delta \Psi & = \tilde \Psi^{(\mathbf k + \delta \mathbf k)} - \Psi^{(\mathbf k)}\notag \\ & = \frac{\partial}{\partial k^i}\Psi\, \delta k^i + i\,\mathscr A_i\,\Psi\, \delta k^i \tag{2} \\ \end{align} $$定义协变导数为 $E$ 中的平坦联络在子丛 $F$ 的正交补上的诱导联络
$$ \mathcal D := \nabla^{F^{\perp}} = (1 - Prj_F) \circ \nabla \tag{3} $$从而
$$ \begin{align} \delta \Psi & = \mathcal D_{\frac{\partial}{\partial k^i}} \Psi \, \delta k^i \notag\\ & = \Psi^m \;\mathcal D_{\frac{\partial}{\partial k^i}} u_m \, \delta k^i\qquad (\;\because \mathcal D_{\frac{\partial}{\partial k^i}} \Psi^m = 0\;) \tag{4} \end{align} $$其模长为
$$ ||\delta \Psi||^2 = \Psi^m\, \Psi^n \left \langle \mathcal D_{\frac{\partial}{\partial k^i}} u_m,\;\mathcal D_{\frac{\partial}{\partial k^j}} u_n \right \rangle\, \delta k^i\,\delta k^j \tag{5} $$据此，定义量子几何张量
$$ Q_{mn} = \left \langle \mathcal D u_m|\otimes|\mathcal D u_n \right \rangle= Q_{mn,ij} \,d k^i\otimes d k^j \tag{6} $$其具有类似度规的地位。利用
$$ \mathcal D_{\frac{\partial}{\partial k^i}} u_m= \frac{\partial}{\partial k^i} u_m + i\ {\mathscr A_i\,}^k_m\; u_k \tag{7} $$$$ \left \langle \frac{\partial}{\partial k^i} u_m,\;u_l \right \rangle + \left \langle u_m,\;\frac{\partial}{\partial k^i} u_l \right \rangle = 0 \quad\Rightarrow \quad {\mathscr A_i\,}^m_l = {\mathscr A^{\dagger}_i\,}^l_m \tag{8} $$
整理得到
$$ \begin{align} Q_{mn,ij} = & \left \langle \mathcal D_{\frac{\partial}{\partial k^i}} u_m,\;\mathcal D_{\frac{\partial}{\partial k^j}} u_n \right \rangle \notag \\ = & \left \langle \frac{\partial}{\partial k^i} u_m,\;\frac{\partial}{\partial k^j} u_n \right \rangle - {\mathscr A_j\,}^l_n {\mathscr A_i\,}^m_l \notag \\ = & \left \langle \frac{\partial}{\partial k^i} u_m,\;\frac{\partial}{\partial k^j} u_n \right \rangle - \left \langle\frac{\partial}{\partial k^i} u_m,\; u_l \right \rangle \left \langle u_l,\;\frac{\partial}{\partial k^j} u_n \right \rangle \tag{9} \end{align} $$注意到
$$ {Q^\dagger}_{mn,ij} = Q_{nm,ji} \tag{10} $$可见 $Q_{mn,ij}$ 的实部关于 $i,j$ 与 $m,n$ 对称，而虚部关于 $i,j$ 与 $m,n$ 交错。
1 实部：量子度规 #
度量波函数变化量的表达式
$$ ||\delta \Psi||^2 = \Psi^m\, \Psi^n Q_{mn,ij}\, \delta k^i\,\delta k^j \tag{11} $$其关于指标 $i,j$ 对称，因此 $Q_{mn,ij}$ 的交错部分对度量无贡献。令
$$ g_{mn,ij} = \frac12 (Q_{mn,ij} + Q_{nm,ji}) = Re(Q_{mn,ij}) \tag{12} $$称为量子度规（或 Fubini-Study 度规），从而
$$ ||\delta \Psi||^2 = \Psi^m\, \Psi^n Q_{mn,ij}\, \delta k^i\,\delta k^j = \Psi^m\, \Psi^n g_{mn,ij}\, \delta k^i\,d k^j \tag{13} $$2 虚部：Berry curvature #
量子几何张量的虚部即为其交错部分。在同构的意义下，有
$$ i\;Im(Q_{mn}) = \frac12 \left \langle \mathcal D u_m|\wedge |\mathcal D u_n \right \rangle \tag{14} $$其中
$$ \begin{align} \left \langle \mathcal D u_m|\wedge |\mathcal D u_n \right \rangle & = \left \langle d u_m|\wedge |d u_n\right \rangle - {\mathscr A}^m_k \wedge {\mathscr A}^k_n \notag \\ & = - i\, d{\mathscr A}^m_n - {\mathscr A}^m_k \wedge {\mathscr A}^k_n \notag \\ & = - i\, \mathcal F^m_n \tag{15} \end{align} $$可见
$$ Im(Q_{mn}) = - \frac{1}{2} \mathcal F^m_n \tag{16} $$3 离散 $\mathbf k$ 网络上的计算方式 #
在实际计算中，$\mathbf k$ 的取值离散，此时协变导数利用差分近似给出
$$ \begin{align} \delta k^i\; \mathcal D_{\frac{\partial}{\partial k^i}} u_m & = \left( \frac{\partial}{\partial k^i} u_m - \left\langle u_l^{(\mathbf k)}, \frac{\partial u_m}{\partial k^i}\right\rangle u_l^{(\mathbf k)} \right) \delta k^i \notag \\ & = \pm \left(1 - |u_l^{(\mathbf k)}\rangle \langle u_l^{(\mathbf k)}| \right) \frac{u_m^{(\mathbf k \pm \delta \mathbf k)}}{\left\langle u_m^{(\mathbf k)},\ u_m^{(\mathbf k \pm \delta \mathbf k)} \right\rangle} \tag{17} \end{align} $$从而
$$ \mathcal D_{\frac{\partial}{\partial k^i}} u_m = \frac{1}{2\delta k^i} \left(1 - |u_l^{(\mathbf k)}\rangle \langle u_l^{(\mathbf k)}| \right) \left( \frac{u_m^{(\mathbf k + \delta \mathbf k)}}{\left\langle u_m^{(\mathbf k)},\ u_m^{(\mathbf k + \delta \mathbf k)} \right\rangle} - \frac{u_m^{(\mathbf k - \delta \mathbf k)}}{\left\langle u_m^{(\mathbf k)},\ u_m^{(\mathbf k - \delta \mathbf k)} \right\rangle} \right) \tag{18} $$