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xirainbow木虫 (正式写手)
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【原创】如何得到力常数
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It has always been my dream to get interatomic force constants from first-principle calculation. I try to get the IFC ever since I began to learn MS and PWscf. In MS, although I found how to get dynamic matrix, I failed. In the manual of PWscf, it claims to be able to get IFC in real space, which attracts me a lot. But almost three months past, no one can explain .fc file to me. Fortunatelly, I find some hints today. I do not comprehend it fully, so I first put it here. "The "search" facility of the mailing list is unfortunately rather crappy, but I was sure this had been already answered, first with 0 bits of information (but accurate nonetheless): http://www.democritos.it/piperma ... ptember/004900.html and then with a finite amount of information:http://www.democritos.it/piperma ... ptember/010099.html Note that the grid of real-space lattice vectors is by construction, periodic! it should be refolded so that it is centered around R=0." "the matrix elements of the IFC are written after these matrices (respectively the dielectric tensor and the effective charges). http://www.democritos.it/piperma ... ptember/004900.html" " Well, let us to recall the notation for the force constants matrix, F_{\alpha,\beta}^{i,j}(R), where \alpha, \beta are atomic numbers, i,j are polarization vectors (x,y,z), and R is the distance between two atoms in units of lattice vectors. Below there are examples how these lines should be treated. 4 4 4 - you used 4x4x4 q-points to generate FC 1 3 1 2 - x, z, 1st atom, 2nd atom 4 3 2 0.1234 - R=i*a+j*b+k*c, where i,j,k are integers from 1 to 4 (you used 4x4x4 mesh), a,b,c are lattice vectors. 0.1234 is F_{1,2}^{x,z}(R(4,3,2)). As QE uses atomic units, F should be in units of Ry/au^2, but not in eV/A^2. http://www.democritos.it/piperma ... ptember/010099.html " " 2 3 4 5.6370000 0.0000000 1.0870000 0.0000000 0.0000000 0.0000000 lattice number of types, number of atoms in the cell, type of lattice,lattice parameters (ibrav = 2) (nat = 3) (ntyp = 4) 1 'Al ' -> atoms 24588.6885119930 2 'B ' -> atoms 9851.88001536857 type #, label, atomic mass in a.u. 1 1 0.0000000 0.0000000 0.0000000 -> positions 2 2 0.5000000 0.2886751 0.5435000 3 2 0.0000000 0.5773503 0.5435000 atom #, type #, positions in the cell in a0 units F T if the file contains epsilon and Z*, F otherwise 8 8 8 grid size n1 n2 n3, defining the grid of q-points in the Brillouin Zone and the side of the R-space supercell used to calculate the Fourier transform. The latter is n1*tau1, n2*tau2, n3*tau3, where tau1, tau2, tau3 are the primitive translations of the lattice 1 1 1 1 indices: i,j,na,nb (polarization1 , polarization 2, atom 1 , atom 2) 1 1 1 1.56778730300E-01 force constant, but for which atoms? m1 m2 m3 force_constant(m1,m2,m3,i,j,na,nb).where m1=1,...,n1, m2=1,...,n2, m3=1,...,n3, define a lattice vector R: R = (m1-1)*tau1 + (m2-1)*tau2 + (m3-1)*tau3. (if m1-1 > n1/2 and so on, refold m1-1 => m1-1-n1 and so on). So: force on atom na in direction i, when moving atom nb in direction j,in a cell that is R far away are the force constants divided by mass ? no How to find the distance in real space (or the number of nearest neighbours) up to which the FC are calculated? see above: approx. n1/2, n2/2, n3/2 unit cells in the three directions http://www.democritos.it/piperma ... 5-April/002408.html 以上内容来自我的博客：http://blog.sina.com.cn/nkasir [ Last edited by xirainbow on 2009-7-17 at 11:35 ] |
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