物理类

Pulay方法

或者名为DIIS(direct inversion of the iterative subspace)。 原始公式为: pn+1=i=1ncipi p_{n+1}=\sum^n_{i=1}{c_ip_i} 实际上一般还需要加上线性混合: pn+1=i=1nci[βpi,in+(1β)pi,out] p_{n+1}=\sum^n_{i=1}{c_i[\beta p_{i,in}+(1-\beta)p_{i,out}]} 系数cic_i由以下线性方程来解: (01...11b11...b1n............1bn1...bnn)(λc1...cn)=(10...0) \left( \begin{array}{cccc} 0&-1&...&-1\\ -1&b_{11}&...&b_{1n}\\ ...&...&...&...\\ -1&b_{n1}&...&b_{nn} \end{array} \right) \left( \begin{array}{c} \lambda\\c_{1}\\...\\c_{n} \end{array} \right)= \left( \begin{array}{c} -1\\0\\...\\0 \end{array} \right) 其中bi,j=Real(<Δpi,Δpj>),Δpi=pi+1pib_{i,j}=Real(<\Delta p_i,\Delta p_j>), \Delta p_i = p_{i+1}-p_{i}

Gap Function的对称性

ss-wave: Δ(kx,ky)=Δ(kx,ky)=Δ(kx,ky),Δ(kx,ky)=Δ(ky,kx) \Delta(-k_x, k_y) = \Delta(k_x, -k_y) = \Delta(k_x, k_y), \Delta(k_x, k_y) = \Delta(k_y, k_x) dx2y2d_{x^2-y^2}-wave: Δ(kx,ky)=Δ(kx,ky)=Δ(kx,ky),Δ(kx,ky)=Δ(ky,kx) \Delta(-k_x, k_y) = \Delta(k_x, -k_y) = \Delta(k_x, k_y), \Delta(k_x, k_y) = -\Delta(k_y, k_x) dxyd_{xy}-wave: Δ(kx,ky)=Δ(kx,ky)=Δ(kx,ky),Δ(kx,ky)=Δ(ky,kx) \Delta(-k_x, k_y) = \Delta(k_x, -k_y) = -\Delta(k_x, k_y), \Delta(k_x, k_y) = \Delta(k_y, k_x) gg-wave: Δ(kx,ky)=Δ(kx,ky)=Δ(kx,ky),Δ(kx,ky)=Δ(ky,kx) \Delta(-k_x, k_y) = \Delta(k_x, -k_y) = -\Delta(k_x, k_y), \Delta(k_x, k_y) = -\Delta(k_y, k_x)

线性Eliashberg方程的求解

λΔlm(k)=TNVll,mmS(kk)Gll(k)Gmm(k)Δlm(k) \lambda\Delta_{lm}(k)=-\frac{T}{N}V_{ll'',m''m}^{S}(k-k')G_{l''l'}(k')G_{m''m'}(-k')\Delta_{l'm'}(k')

该方程本身高度非线性,不要被名字骗了,如果只要求解最大特征值,可以通过幂法求解,类似于求解一个矩阵最大特征值。

伪代码(Fortran风格):

u=v
while not converge do
    v=Au
    lambda1=abs_max(v)
    u=v/lambda1
end

能带展开的方法

通过实空间中Wannier函数的对称性展开,去掉一些重复的量。 Fe11 to Fe21, corresponds 1 to 6; Fe_1-1\ to\ Fe_2-1,\ corresponds\ 1\ to\ 6; Fe11 to Fe22, corresponds 1 to 7; Fe_1-1\ to\ Fe_2-2,\ corresponds\ 1\ to \ 7; ... ...

将实空间中的Wannier函数变换到展开布里渊区中,在每一个kk点上对一个5*5矩阵进行对角化,得到能带能量的本征值。

一个较好的展开结果如下图:

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