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表面格林函数

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本文介绍了半无限体系的表面格林函数，半无限体系哈密顿量呈块三对角形式，推迟格林函数为(ω+iη−H)GR=I(ω+iη−H)G^R=I，因关注表面格林函数G00G_{00}对大矩阵求逆困难，故介绍了两种迭代求解方法。

半无限体系

半无限体系指系统在一个方向上延伸到无穷远，而在另一个方向上有边界（表面）。

对于半无限体系，其哈密顿量通常可以写成块三对角（block tridiagonal）的形式：

H=(H00H0100⋯H01†H00H010⋯0H01†H00H01⋯00H01†H00⋯⋮⋮⋮⋮⋱)H=\begin{pmatrix} H_{00} & H_{01} & 0 & 0 & \cdots \\ H_{01}^{\dagger} & H_{00} & H_{01} & 0 & \cdots \\ 0 & H_{01}^{\dagger} & H_{00} & H_{01} & \cdots \\ 0 & 0 & H_{01}^{\dagger} & H_{00} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}

H00H_{00}表示单一层的哈密顿量（例如一层原子的内部耦合），H01H_{01}表示相邻层之间的跃迁矩阵，这里只考虑最近邻。

推迟格林函数为 (ω+iη−H)GR=I(ω+iη−H)G^R=I, 可写成:

GR=(ω+iη−H00−H0100⋯−H01†ω+iη−H00−H010⋯0−H01†ω+iη−H00−H01⋯00−H01†ω+iη−H00⋯⋮⋮⋮⋮⋱)−1G^R=\begin{pmatrix} ω+iη-H_{00} & -H_{01} & 0 & 0 & \cdots \\ -H_{01}^{\dagger} & ω+iη-H_{00} & -H_{01} & 0 & \cdots \\ 0 & -H_{01}^{\dagger} & ω+iη-H_{00} & -H_{01} & \cdots \\ 0 & 0 & -H_{01}^{\dagger} & ω+iη-H_{00} & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}^{-1}

我们只关注表面格林函数即GR(1,1)G^R(1,1),对于如此大矩阵求逆是很困难的。

快速迭代求解

(ω−H00−H01−H01†ω−H00−H01−H01†ω−H00⋯⋮⋱)(G00G01G02⋯G10G11G12⋯G20G21G22⋯⋮⋮⋮⋱)=I\begin{pmatrix} \omega - H_{00} & -H_{01} & & \\ -H_{01}^{\dagger} & \omega - H_{00} & -H_{01} & \\ & -H_{01}^{\dagger} & \omega - H_{00} & \cdots \\ & & \vdots & \ddots \end{pmatrix} \begin{pmatrix} G_{00} & G_{01} & G_{02} & \cdots \\ G_{10} & G_{11} & G_{12} & \cdots \\ G_{20} & G_{21} & G_{22} & \cdots \\ \vdots & \vdots & \vdots & \ddots \end{pmatrix}=I

因为只关注GR(1,1)G^R(1,1)也就是G00G_{00},

(ω−H00−H01−H01†ω−H00−H01−H01†ω−H00⋯⋮⋱)(G00G10G20⋮)=(I00⋮)\begin{pmatrix} \omega - H_{00} & -H_{01} & & \\ -H_{01}^{\dagger} & \omega - H_{00} & -H_{01} & \\ & -H_{01}^{\dagger} & \omega - H_{00} & \cdots \\ & & \vdots & \ddots \end{pmatrix} \begin{pmatrix} G_{00} \\ G_{10} \\ G_{20} \\ \vdots \end{pmatrix}= \begin{pmatrix} I \\ 0 \\ 0 \\ \vdots \end{pmatrix}

有如下关系:

(ω−H00)G00=I+H01G10(ω−H00)G10=H01†G00+H01G20⋮(ω−H00)Gn0=H01†Gn−1,0+H01Gn+1,0(\omega - H_{00})G_{00} = I + H_{01}G_{10} \\ (\omega - H_{00})G_{10} = H_{01}^{\dagger}G_{00} + H_{01}G_{20} \\ \vdots \\ (\omega - H_{00})G_{n0} = H_{01}^{\dagger}G_{n - 1,0} + H_{01}G_{n + 1,0}

最后可得如下关系:

G2n,0=(ω−H00)−1(H01†G2n−1,0+H01G2n+1,0)G_{2n,0}=(\omega - H_{00})^{-1}(H_{01}^{\dagger}G_{2n - 1,0} + H_{01}G_{2n + 1,0})

当 n>0时：

[ω−H00−H01(ω−H00)−1H01†−H01†(ω−H00)−1H01]G2n,0=H01†(ω−H00)−1H01†G2n−2,0+H01(ω−H00)−1H01G2n+2,0\left[\omega - H_{00}-H_{01}(\omega - H_{00})^{-1}H_{01}^{\dagger}-H_{01}^{\dagger}(\omega - H_{00})^{-1}H_{01}\right]G_{2n,0} =H_{01}^{\dagger}(\omega - H_{00})^{-1}H_{01}^{\dagger}G_{2n - 2,0}+H_{01}(\omega - H_{00})^{-1}H_{01}G_{2n + 2,0}

当 n=0时:

[ω−H00−H01(ω−H00)−1H01†]G00=I+H01(ω−H00)−1H01†G20\left[\omega - H_{00}-H_{01}(\omega - H_{00})^{-1}H_{01}^{\dagger}\right]G_{00} = I + H_{01}(\omega - H_{00})^{-1}H_{01}^{\dagger}G_{20}

将因子重新写为:

α1=H01(ω−H00)−1H01β1=H01†(ω−H00)−1H01†εs=H00+H01(ω−H00)−1H01†ε1=H00+H01(ω−H00)−1H01†+H01†(ω−H00)−1H01\begin{aligned} \alpha_1 &= H_{01}(\omega - H_{00})^{-1}H_{01} \\ \beta_1 &= H_{01}^{\dagger}(\omega - H_{00})^{-1}H_{01}^{\dagger} \\ \varepsilon^s &= H_{00}+H_{01}(\omega - H_{00})^{-1}H_{01}^{\dagger} \\ \varepsilon_1 &= H_{00}+H_{01}(\omega - H_{00})^{-1}H_{01}^{\dagger}+H_{01}^{\dagger}(\omega - H_{00})^{-1}H_{01} \end{aligned}

参数的迭代关系为:

αi=αi−1(ω−εi−1)−1αi−1βi=βi−1(ω−εi−1)−1βi−1εi=εi−1+αi−1(ω−εi−1)−1βi−1+βi−1(ω−εi−1)−1αi−1εis=εi−1s+αi−1(ω−εi−1)−1βi−1\begin{aligned} \alpha_{i}&=\alpha_{i - 1}(\omega - \varepsilon_{i - 1})^{-1}\alpha_{i - 1}\\ \beta_{i}&=\beta_{i - 1}(\omega - \varepsilon_{i - 1})^{-1}\beta_{i - 1}\\ \varepsilon_{i}&=\varepsilon_{i - 1}+\alpha_{i - 1}(\omega - \varepsilon_{i - 1})^{-1}\beta_{i - 1}+\beta_{i - 1}(\omega - \varepsilon_{i - 1})^{-1}\alpha_{i - 1}\\ \varepsilon_{i}^s&=\varepsilon_{i - 1}^s+\alpha_{i - 1}(\omega - \varepsilon_{i - 1})^{-1}\beta_{i - 1} \end{aligned}

α\alpha和 β\beta将逐渐减小, 取截断可以得到:

得到

(ω−εs)G00=I+α1G20(ω−ε1)G2n,0=β1G2n−2,0+α1G2n+2,0\begin{aligned} (\omega - \varepsilon^s)G_{00} &= I + \alpha_1G_{20}\\ (\omega - \varepsilon_1)G_{2n,0} &= \beta_1G_{2n - 2,0} + \alpha_1G_{2n + 2,0} \end{aligned}

迭代关系为

(ω−εis)G00=I+αiG2i,0(ω−εi)G2i,0=βiG2i(n−1),0+αiG2i(n+1),0\begin{aligned} (\omega - \varepsilon_{i}^{s})G_{00} &= I + \alpha_{i}G_{2^{i},0}\\ (\omega - \varepsilon_{i})G_{2^{i},0} &= \beta_{i}G_{2^{i}(n - 1),0} + \alpha_{i}G_{2^{i}(n + 1),0} \end{aligned}

选择合适的迭代次数i

(ω−εis)G00=IG00=(ω−εis)−1\begin{aligned} (\omega - \varepsilon_{i}^{s})G_{00} = I\\ G_{00} = (\omega - \varepsilon_{i}^{s})^{-1} \end{aligned}

递推矩阵

对于格林函数矩阵

(ω−H00−H01−H01†ω−H00−H01−H01†ω−H00⋯⋮⋱)(G00G10G20⋮)=(I00⋮)\begin{pmatrix} \omega - H_{00} & -H_{01} & & \\ -H_{01}^{\dagger} & \omega - H_{00} & -H_{01} & \\ & -H_{01}^{\dagger} & \omega - H_{00} & \cdots \\ & & \vdots & \ddots \end{pmatrix} \begin{pmatrix} G_{00} \\ G_{10} \\ G_{20} \\ \vdots \end{pmatrix}= \begin{pmatrix} I \\ 0 \\ 0 \\ \vdots \end{pmatrix}

将矩阵乘开可以看到矩阵每一行的格式都是类似的

第一行

(ω−H00)G00=I+H01G10(\omega - H_{00})G_{00} = I + H_{01}G_{10}

第n行

(ω−H00)Gn0=H01†Gn−1,0+H01Gn+1,0 (\omega - H_{00})G_{n0} = H_{01}^{\dagger}G_{n - 1,0} + H_{01}G_{n + 1,0}

写成递推矩阵

(Gn+1,0Gn,0)=(H01−1(ωI−H00)−H01−1H10I0)(Gn,0Gn−1,0)\begin{pmatrix} G_{n + 1,0}\\ G_{n ,0} \end{pmatrix}= \begin{pmatrix} H_{01}^{-1}(\omega I - H_{00}) & -H_{01}^{-1}H_{10}\\ I & 0 \end{pmatrix} \begin{pmatrix} G_{n ,0}\\ G_{n-1,0} \end{pmatrix}

实际上就是转移矩阵T不断幂运算，格林函数的物理意义是两个点的关联程度，而当n不断变大的时候它与第一列的关联程度越来越低，最后一定趋于0，那也就是说右边那个矩阵的幂也是趋于0的，等于说G0G_0是由T的特征值小于1的特征向量构成的，

假设T的特征值小于1的特征向量表示为(S1S2)\begin{pmatrix} S_1\\ S_2 \end{pmatrix}，写成 (G1,0G0,0)=(S1S2)A\begin{pmatrix} G_{1,0}\\ G_{0,0} \end{pmatrix}= \begin{pmatrix} S_1\\ S_2 \end{pmatrix}A。

得到

(ωI−H00)G00−H01G10=IG10=S1S2−1G00\begin{aligned} &(\omega I - H_{00})G_{00} - H_{01}G_{10} = I\\ &G_{10} = S_1S_2^{-1}G_{00}\\ \end{aligned}

化简就是

G00=(ωI−H00−H01S1S2−1)−1G_{00}=(\omega I - H_{00} - H_{01}S_1S_2^{-1})^{-1}

我们假设 G10=γ1G00G_{10}=\gamma_1G_{00} ，上面的式子很容易得到 G00G_{00}

G00=(ωI−H00−H01γ1)−1G_{00}=(\omega I - H_{00} - H_{01}\gamma_1)^{-1}

这里的 γ1\gamma_1实际上就是刚才的 S1S2−1S_1S_2^{-1}。

于是我们只要求解γ1\gamma_1就行了，然后继续假设 G20=γ1G00G_{20}=\gamma_1G_{00} ，很快能得到

γ1=g1H10+g1H01γ2\gamma_1 = g_1H_{10}+g_1H_{01}\gamma_2

其中g1=(ωI−H00)−1g_{1}=(\omega I-H_{00})^{-1}

继续假设 Gn0=γnG00G_{n0}=\gamma_nG_{00}最终

γ2=g2H10g1H10+g2H01γ3γ3=g3H10g2H10g1H10+g3H01γ4γ4=g4H01g3H10g2H10g1H10+g4H01γ5⋅⋅⋅\begin{aligned}&\gamma_{2}=g_2H_{10}g_1H_{10}+g_2H_{01}\gamma_3\\&\gamma_{3}=g_3H_{10}g_2H_{10}g_1H_{10}+g_3H_{01}\gamma_4\\&\gamma_{4}=g_{4}H_{01}g_{3}H_{10}g_{2}H_{10}g_{1}H_{10}+g_{4}H_{01}\gamma_{5}\\&\cdotp\cdotp\cdotp\end{aligned}

可以令γn=0\gamma_n=0，然后一层层带入得到 γ1\gamma_1，就可以得到 G00G_{00}。