本篇主要内容参考Topological insulators and topological superconductors / B. Andrei Bernevig ，夹带些许私货。
考虑基态无简并的含参哈密顿量H ( x a ; λ i ) H(x^a;\lambda^i) H ( x a ; λ i ) λ i \lambda^i λ i x a x^a x a λ i \lambda^i λ i ∣ n ( λ i ) ⟩ |n(\lambda^i)\rangle ∣ n ( λ i ) ⟩ λ i \lambda^i λ i ∣ n ( λ i ) ⟩ |n(\lambda^i)\rangle ∣ n ( λ i ) ⟩ E \mathcal{E} E
(Some math)敏锐的读者可能已经发现，“知道基态”这个说法其实并不严格，因为我们对于量子态的相位其实是未知的。这导致基态也可以取∣ n ′ ( λ i ) ⟩ : = e i θ ( λ i ) ∣ n ( λ i ) ⟩ |n'(\lambda^i)\rangle := e^{i\theta(\lambda^i)}|n(\lambda^i)\rangle ∣ n ′ ( λ i ) ⟩ : = e i θ ( λ i ) ∣ n ( λ i ) ⟩ ∣ n ( λ i ) ⟩ ⟨ n ( λ i ) ∣ |n(\lambda^i)\rangle\langle n(\lambda^i)| ∣ n ( λ i ) ⟩ ⟨ n ( λ i ) ∣ M \mathcal{M} M E \mathcal{E} E M \mathcal{M} M ϕ : ∣ n ( λ i ) ⟩ → ∣ n ( λ i ) ⟩ ⟨ n ( λ i ) ∣ \phi: |n(\lambda^i)\rangle \rightarrow|n(\lambda^i)\rangle\langle n(\lambda^i)| ϕ : ∣ n ( λ i ) ⟩ → ∣ n ( λ i ) ⟩ ⟨ n ( λ i ) ∣ ( E , M , U ( 1 ) , ϕ ) (\mathcal{E},\mathcal{M},U(1),\phi) ( E , M , U ( 1 ) , ϕ )
我们考虑λ i \lambda^i λ i λ i ( t ) \lambda^i(t) λ i ( t ) ∣ ψ ( t ) ⟩ |\psi(t)\rangle ∣ ψ ( t ) ⟩ ∣ ψ ( t ) ⟩ = e − i θ ( t ) ∣ n ( λ i ) ⟩ |\psi(t)\rangle =e^{-i \theta(t)} |n(\lambda^i)\rangle ∣ ψ ( t ) ⟩ = e − i θ ( t ) ∣ n ( λ i ) ⟩ ∣ ψ ( t ) ⟩ |\psi(t)\rangle ∣ ψ ( t ) ⟩ ∣ ψ ( λ i ( t ) ) ⟩ |\psi(\lambda^i (t))\rangle ∣ ψ ( λ i ( t ) ) ⟩ ∣ ψ ( t ) ⟩ |\psi(t)\rangle ∣ ψ ( t ) ⟩
E n ( λ i ( t ) ) ∣ n ( λ i ( t ) ) ⟩ = d t θ ( t ) ∣ n ( λ i ( t ) ) ⟩ + i d t ∣ n ( λ i ( t ) ) ⟩ . E_n(\lambda^i(t))|n(\lambda^i(t))\rangle = d_t\theta(t)|n(\lambda^i(t))\rangle + id_t|n(\lambda^i(t))\rangle.
E n ( λ i ( t ) ) ∣ n ( λ i ( t ) ) ⟩ = d t θ ( t ) ∣ n ( λ i ( t ) ) ⟩ + i d t ∣ n ( λ i ( t ) ) ⟩ . 如果我们将上述方程两边与⟨ n ( λ i ( t ) ) ∣ \langle n(\lambda^i(t))| ⟨ n ( λ i ( t ) ) ∣
θ ( t ) = ∫ E n ( λ i ( t ) ) d t − i ∫ ⟨ n ( λ i ( t ) ) ∣ d t ∣ n ( λ i ( t ) ) ⟩ d t . \theta(t) = \int E_n(\lambda^i(t))dt - i \int \langle n(\lambda^i(t))|d_t|n(\lambda^i(t))\rangle dt.
θ ( t ) = ∫ E n ( λ i ( t ) ) d t − i ∫ ⟨ n ( λ i ( t ) ) ∣ d t ∣ n ( λ i ( t ) ) ⟩ d t . 这里的第一项与路径有关，被称为dynamical phase。而对于第二项我们注意到：
∫ d t ⟨ n ( λ i ( t ) ) ∣ d t ∣ n ( λ i ( t ) ) ⟩ = ∫ d t ⟨ n ( λ i ( t ) ) ∣ d λ j d t ∂ ∂ λ j ∣ n ( λ i ( t ) ) ⟩ = ∫ C ⟨ n ∣ ∂ ∂ λ j ∣ n ⟩ d λ j \int dt \langle n(\lambda^i(t))|d_t|n(\lambda^i(t))\rangle =\int dt \langle n(\lambda^i(t))|\frac{d\lambda^j}{dt}\frac{\partial}{\partial \lambda^j}|n(\lambda^i(t))\rangle = \int_C \langle n|\frac{\partial}{\partial \lambda^j}|n\rangle d\lambda^j
∫ d t ⟨ n ( λ i ( t ) ) ∣ d t ∣ n ( λ i ( t ) ) ⟩ = ∫ d t ⟨ n ( λ i ( t ) ) ∣ d t d λ j ∂ λ j ∂ ∣ n ( λ i ( t ) ) ⟩ = ∫ C ⟨ n ∣ ∂ λ j ∂ ∣ n ⟩ d λ j 我们定义贝里联络A i ( λ ) = i ⟨ n ∣ ∂ ∂ λ i ∣ n ⟩ \mathcal{A}_i(\lambda) =i\langle n|\frac{\partial}{\partial \lambda^i}| n\rangle A i ( λ ) = i ⟨ n ∣ ∂ λ i ∂ ∣ n ⟩
θ ( t ) = ∫ E n ( λ i ( t ) ) d t − ∫ c A i ( λ ) ⋅ d λ i \theta(t) = \int E_n(\lambda^i(t))dt -\int_c \mathcal{A}_i(\lambda)\cdot d\lambda^i
θ ( t ) = ∫ E n ( λ i ( t ) ) d t − ∫ c A i ( λ ) ⋅ d λ i 我们发现上式中的第二项单纯依赖于参数空间的路径并不含时，我们将这一项的贡献称为贝里相位(Berry phase)。这一贡献本身来自于希尔伯特空间的几何效应，在闭合路径下其对应着动力学下从流形M \mathcal{M} M E \mathcal{E} E Derivation of the geometrical phase, Bohm, Arno et. al. 。
从贝里联络的形式中我们容易看出，由于基态波函数全局相位的选取并不是唯一的，一个新的全局相位选择会直接改变贝里联络。如果我们对∣ n ( λ i ( t ) ) ⟩ |n(\lambda^i(t))\rangle ∣ n ( λ i ( t ) ) ⟩ ∣ n ′ ( λ i ( t ) ) ⟩ : = e i ω ( λ i ) ∣ n ( λ i ( t ) ) ⟩ |n'(\lambda^i(t))\rangle:=e^{i\omega(\lambda^i)}|n(\lambda^i(t))\rangle ∣ n ′ ( λ i ( t ) ) ⟩ : = e i ω ( λ i ) ∣ n ( λ i ( t ) ) ⟩
A i ′ ( λ ) = i ⟨ n ′ ∣ ∂ ∂ λ i ∣ n ′ ⟩ = i ⟨ n ∣ ∂ ∂ λ i ∣ n ⟩ − ∂ i ω , \mathcal{A}'_i(\lambda) =i\langle n'|\frac{\partial}{\partial \lambda^i}| n'\rangle = i\langle n|\frac{\partial}{\partial \lambda^i}| n\rangle -\partial_i\omega,
A i ′ ( λ ) = i ⟨ n ′ ∣ ∂ λ i ∂ ∣ n ′ ⟩ = i ⟨ n ∣ ∂ λ i ∂ ∣ n ⟩ − ∂ i ω , 因此
θ ′ ( t ) = ∫ E n ( λ i ( t ) ) d t − ∫ c A i ( λ ) d λ i + ∫ c d ω \theta'(t) = \int E_n(\lambda^i(t))dt -\int_c \mathcal{A}_i(\lambda) d\lambda^i+\int_c d\omega
θ ′ ( t ) = ∫ E n ( λ i ( t ) ) d t − ∫ c A i ( λ ) d λ i + ∫ c d ω 如果我们要求参数空间的路径是闭合路径时，由于∫ c d ω = 0 \int_c d\omega = 0 ∫ c d ω = 0 β : = ∫ c A i ( λ ) d λ i = i ∫ C ⟨ n ( λ i ( t ) ) ∣ ∂ ∂ λ j ∣ n ( λ i ( t ) ) ⟩ d λ j \beta := \int_c \mathcal{A}_i(\lambda)d\lambda^i =i\int_C \langle n(\lambda^i(t))|\frac{\partial}{\partial \lambda^j}|n(\lambda^i(t))\rangle d\lambda^j β : = ∫ c A i ( λ ) d λ i = i ∫ C ⟨ n ( λ i ( t ) ) ∣ ∂ λ j ∂ ∣ n ( λ i ( t ) ) ⟩ d λ j
β = i ∫ S d ( ⟨ n ∣ ∂ ∂ λ j ∣ n ⟩ d λ j ) = i ∫ S d λ k ∧ d λ j ∂ ∂ λ k ( ⟨ n ∣ ∂ ∂ λ j ∣ n ⟩ d λ j ) = i ∫ S d λ k ∧ d λ j ( ∂ ∂ λ k ⟨ n ∣ ) ( ∂ ∂ λ j ∣ n ⟩ ) , \beta = i\int_S d\left(\langle n|\frac{\partial}{\partial \lambda^j}|n\rangle d\lambda^j\right) = i\int_S d\lambda^k\wedge d\lambda^j\frac{\partial}{\partial \lambda^k}\left(\langle n|\frac{\partial}{\partial \lambda^j}|n\rangle d\lambda^j\right) = i \int_S d\lambda^k\wedge d\lambda^j\left(\frac{\partial}{\partial \lambda^k}\langle n|\right)\left(\frac{\partial}{\partial \lambda^j}|n\rangle\right),
β = i ∫ S d ( ⟨ n ∣ ∂ λ j ∂ ∣ n ⟩ d λ j ) = i ∫ S d λ k ∧ d λ j ∂ λ k ∂ ( ⟨ n ∣ ∂ λ j ∂ ∣ n ⟩ d λ j ) = i ∫ S d λ k ∧ d λ j ( ∂ λ k ∂ ⟨ n ∣ ) ( ∂ λ j ∂ ∣ n ⟩ ) , 显然我们可以得到一个等价的表示（真的很显然）：
β = 1 / 2 ∫ S d λ k ∧ d λ j Ω k j , \beta =1/2 \int_S d\lambda^k\wedge d\lambda^j \Omega_{kj},
β = 1 / 2 ∫ S d λ k ∧ d λ j Ω k j , 这里Ω k j \Omega_{kj} Ω k j
Ω k j = i ( ∂ ∂ λ k ⟨ n ∣ ) ( ∂ ∂ λ j ∣ n ⟩ ) − ( k ↔ j ) = ∂ ∂ λ k A j ( λ ) − ∂ ∂ λ j A k ( λ ) \Omega_{kj} = i \left(\frac{\partial}{\partial \lambda^k}\langle n|\right)\left(\frac{\partial}{\partial \lambda^j}|n\rangle\right) - (k\leftrightarrow j) = \frac{\partial}{\partial \lambda^k}\mathcal{A}_j(\lambda)-\frac{\partial}{\partial \lambda^j}\mathcal{A}_k(\lambda)
Ω k j = i ( ∂ λ k ∂ ⟨ n ∣ ) ( ∂ λ j ∂ ∣ n ⟩ ) − ( k ↔ j ) = ∂ λ k ∂ A j ( λ ) − ∂ λ j ∂ A k ( λ ) 当参数空间为3维时，我们可以证明
β = i ∫ S d S ⃗ ⋅ ⟨ ∇ n ∣ × ∣ ∇ n ⟩ = i ∫ S d S i ε i j k ⟨ ∇ j n ∣ ∇ k ∣ n ⟩ \beta = i\int_S d\vec S\cdot\langle\nabla n|\times|\nabla n\rangle = i \int_S dS_i \varepsilon_{ijk}\langle\nabla_j n| \nabla_k| n\rangle
β = i ∫ S d S ⋅ ⟨ ∇ n ∣ × ∣ ∇ n ⟩ = i ∫ S d S i ε i j k ⟨ ∇ j n ∣ ∇ k ∣ n ⟩ 在继续讨论之前，我们注意到上述计算中并没有用到我们演化的量子态是基态的性质，换句话说，只要我们考虑的量子态是本征态且能隙不闭合，我们就能利用绝热定理进行同样的计算。同时我们容易证明⟨ n ∣ ∇ k ∣ n ⟩ \langle n| \nabla_k| n\rangle ⟨ n ∣ ∇ k ∣ n ⟩ i i i − I m -{\rm Im} − I m n n n ∣ n ⟩ |n\rangle ∣ n ⟩
⟨ ∇ j n ∣ ∇ k ∣ n ⟩ = ⟨ ∇ j n ∣ ∑ m ∣ m ⟩ ⟨ m ∣ ∇ k ∣ n ⟩ = ⟨ ∇ j n ∣ ∑ m ≠ n ∣ m ⟩ ⟨ m ∣ ∇ k ∣ n ⟩ + ⟨ ∇ j n ∣ n ⟩ ⟨ n ∣ ∇ k ∣ n ⟩ \langle\nabla_j n| \nabla_k| n\rangle =\langle\nabla_j n|\sum_m|m\rangle\langle m| \nabla_k| n\rangle = \langle\nabla_j n|\sum_{m\neq n}|m\rangle\langle m| \nabla_k| n\rangle +\langle\nabla_j n|n\rangle\langle n| \nabla_k| n\rangle
⟨ ∇ j n ∣ ∇ k ∣ n ⟩ = ⟨ ∇ j n ∣ m ∑ ∣ m ⟩ ⟨ m ∣ ∇ k ∣ n ⟩ = ⟨ ∇ j n ∣ m = n ∑ ∣ m ⟩ ⟨ m ∣ ∇ k ∣ n ⟩ + ⟨ ∇ j n ∣ n ⟩ ⟨ n ∣ ∇ k ∣ n ⟩ 由于上式的第二项为纯虚数乘以纯虚数，所以
I m ⟨ ∇ j n ∣ ∇ k ∣ n ⟩ = I m ∑ m ≠ n ⟨ ∇ j n ∣ m ⟩ ⟨ m ∣ ∇ k ∣ n ⟩ {\rm Im}\langle\nabla_j n| \nabla_k| n\rangle ={\rm Im}\sum_{m\neq n}\langle\nabla_j n|m\rangle\langle m| \nabla_k| n\rangle
I m ⟨ ∇ j n ∣ ∇ k ∣ n ⟩ = I m m = n ∑ ⟨ ∇ j n ∣ m ⟩ ⟨ m ∣ ∇ k ∣ n ⟩ 我们注意到
E n ⟨ m ∣ ∇ i ∣ n ⟩ = ⟨ m ∣ ∇ i ( H ∣ n ⟩ ) = ⟨ m ∣ ∇ i H ∣ n ⟩ + E m ⟨ m ∣ ∇ i ∣ n ⟩ E_n\langle m| \nabla_i| n\rangle = \langle m|\nabla_i(H|n\rangle) = \langle m|\nabla_iH|n\rangle+E_m\langle m| \nabla_i| n\rangle
E n ⟨ m ∣ ∇ i ∣ n ⟩ = ⟨ m ∣ ∇ i ( H ∣ n ⟩ ) = ⟨ m ∣ ∇ i H ∣ n ⟩ + E m ⟨ m ∣ ∇ i ∣ n ⟩ 因此
I m ⟨ ∇ j n ∣ ∇ k ∣ n ⟩ = I m ∑ m ≠ n ⟨ n ∣ ∇ j H ∣ m ⟩ ⟨ m ∣ ∇ k H ∣ n ⟩ ( E m − E n ) 2 {\rm Im}\langle\nabla_j n| \nabla_k| n\rangle = {\rm Im}\sum_{m\neq n}\frac{\langle n|\nabla_jH|m\rangle\langle m|\nabla_kH|n\rangle}{(E_m-E_n)^2}
I m ⟨ ∇ j n ∣ ∇ k ∣ n ⟩ = I m m = n ∑ ( E m − E n ) 2 ⟨ n ∣ ∇ j H ∣ m ⟩ ⟨ m ∣ ∇ k H ∣ n ⟩ β n = − I m ∫ S d S i ε i j k ∑ m ≠ n ⟨ n ∣ ∇ j H ∣ m ⟩ ⟨ m ∣ ∇ k H ∣ n ⟩ ( E m − E n ) 2 \beta_n = -{\rm Im} \int_S dS_i \varepsilon_{ijk}\sum_{m\neq n}\frac{\langle n|\nabla_jH|m\rangle\langle m|\nabla_kH|n\rangle}{(E_m-E_n)^2}
β n = − I m ∫ S d S i ε i j k m = n ∑ ( E m − E n ) 2 ⟨ n ∣ ∇ j H ∣ m ⟩ ⟨ m ∣ ∇ k H ∣ n ⟩ 上述公式由于将求偏导数的操作转移到了哈密顿量上，能避免由于数值计算贝里曲率时本征态规范选取不受控制的情况，因而具有较高的稳定性。同时我们容易看出∑ n β n = 0 \sum_n\beta_n = 0 ∑ n β n = 0
针对(17)式，我们考虑二能级系统。不失一般性，我们假设哈密顿量为H ( λ i ) = d ⃗ ( λ i ) ⋅ σ ⃗ H(\lambda^i) = \vec d(\lambda^i)\cdot\vec\sigma H ( λ i ) = d ( λ i ) ⋅ σ ρ ± = 1 ± d ^ ( λ i ) ⋅ σ ⃗ 2 \rho_\pm = \frac{1\pm\hat d(\lambda^i)\cdot\vec\sigma}{2} ρ ± = 2 1 ± d ^ ( λ i ) ⋅ σ d ^ \hat d d ^ d ⃗ \vec d d E + − E − = 2 ∣ d ⃗ ∣ E_+-E_- = 2|\vec d| E + − E − = 2 ∣ d ∣
⟨ + ∣ ∇ j H ∣ − ⟩ ⟨ − ∣ ∇ k H ∣ + ⟩ = T r [ 1 + d ^ ⋅ σ ⃗ 2 ( ∇ j d ⃗ ) ⋅ σ ⃗ 1 − d ^ ⋅ σ ⃗ 2 ( ∇ k d ⃗ ) ⋅ σ ⃗ ] \langle +|\nabla_jH|-\rangle\langle -|\nabla_kH|+\rangle = {\rm Tr}\left[\frac{1+\hat d\cdot\vec\sigma}{2}(\nabla_j \vec d)\cdot\vec \sigma \frac{1-\hat d\cdot\vec\sigma}{2}(\nabla_k \vec d)\cdot\vec \sigma\right ]
⟨ + ∣ ∇ j H ∣ − ⟩ ⟨ − ∣ ∇ k H ∣ + ⟩ = T r [ 2 1 + d ^ ⋅ σ ( ∇ j d ) ⋅ σ 2 1 − d ^ ⋅ σ ( ∇ k d ) ⋅ σ ] 由于(17)式中只存在虚部的贡献，我们可以得到
I m ⟨ + ∣ ∇ j H ∣ − ⟩ ⟨ − ∣ ∇ k H ∣ + ⟩ = I m 1 4 T r [ − ( ∇ j d ⃗ ) ⋅ σ ⃗ d ^ ⋅ σ ⃗ ( ∇ k d ⃗ ) ⋅ σ ⃗ + d ^ ⋅ σ ⃗ ( ∇ j d ⃗ ) ⋅ σ ⃗ ( ∇ k d ⃗ ) ⋅ σ ⃗ ] = I m 1 4 T r [ d ^ ⋅ σ ⃗ [ ( ∇ j d ⃗ ) ⋅ σ ⃗ , ( ∇ k d ⃗ ) ⋅ σ ⃗ ] ] = I m i 2 T r [ d ^ ⋅ σ ⃗ ( ∇ j d ⃗ ) × ( ∇ k d ⃗ ) ⋅ σ ⃗ ] = d ^ ⋅ ( ∇ j d ⃗ ) × ( ∇ k d ⃗ ) = ∣ d ⃗ ∣ 2 d ^ ⋅ ( ∇ j d ^ ) × ( ∇ k d ^ ) \begin{aligned}
{\rm Im}\langle +|\nabla_jH|-\rangle\langle -|\nabla_kH|+\rangle &={\rm Im} \frac{1}{4}{\rm Tr}\left[-(\nabla_j \vec d)\cdot\vec \sigma\ \hat d\cdot\vec\sigma\ (\nabla_k \vec d)\cdot\vec \sigma+\hat d\cdot\vec\sigma\ (\nabla_j \vec d)\cdot\vec \sigma\ (\nabla_k \vec d)\cdot\vec \sigma\right]\\
&={\rm Im}\frac{1}{4}{\rm Tr}\left[\hat d\cdot\vec\sigma\ [(\nabla_j \vec d)\cdot\vec \sigma,(\nabla_k \vec d)\cdot\vec \sigma]\right]\\
&={\rm Im}\frac{i}{2}{\rm Tr}\left[\hat d\cdot\vec\sigma\ (\nabla_j \vec d)\times(\nabla_k \vec d)\cdot\vec \sigma\right]\\
&=\hat d\cdot (\nabla_j \vec d)\times(\nabla_k \vec d)\\
&=|\vec d|^2\hat d\cdot (\nabla_j \hat d)\times(\nabla_k \hat d)
\end{aligned}
I m ⟨ + ∣ ∇ j H ∣ − ⟩ ⟨ − ∣ ∇ k H ∣ + ⟩ = I m 4 1 T r [ − ( ∇ j d ) ⋅ σ d ^ ⋅ σ ( ∇ k d ) ⋅ σ + d ^ ⋅ σ ( ∇ j d ) ⋅ σ ( ∇ k d ) ⋅ σ ] = I m 4 1 T r [ d ^ ⋅ σ [ ( ∇ j d ) ⋅ σ , ( ∇ k d ) ⋅ σ ] ] = I m 2 i T r [ d ^ ⋅ σ ( ∇ j d ) × ( ∇ k d ) ⋅ σ ] = d ^ ⋅ ( ∇ j d ) × ( ∇ k d ) = ∣ d ∣ 2 d ^ ⋅ ( ∇ j d ^ ) × ( ∇ k d ^ ) 带回(17)式中我们得到
β + = − 1 4 ∫ S d S i ε i j k d ^ ⋅ ( ∇ j d ^ ) × ( ∇ k d ^ ) β − = 1 4 ∫ S d S i ε i j k d ^ ⋅ ( ∇ j d ^ ) × ( ∇ k d ^ ) \beta_+ = -\frac{1}{4}\int_SdS_i\varepsilon_{ijk}\hat d\cdot (\nabla_j \hat d)\times(\nabla_k \hat d)\\
\beta_- = \frac{1}{4}\int_SdS_i\varepsilon_{ijk}\hat d\cdot (\nabla_j \hat d)\times(\nabla_k \hat d)
β + = − 4 1 ∫ S d S i ε i j k d ^ ⋅ ( ∇ j d ^ ) × ( ∇ k d ^ ) β − = 4 1 ∫ S d S i ε i j k d ^ ⋅ ( ∇ j d ^ ) × ( ∇ k d ^ ) 很显然，此时的贝里相位表达式和所谓的Skymion picture的形式便吻合上了。
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