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【转帖】PWSCF计算H2的振动频率问题
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PWSCF计算H2的振动频率问题zz(2009-05-12 22:15:22)标签：phonon dispersion 分类：第一性原理计算 From the word go, I try to use MS to compute phonon dispersion of layered-material structure (graphene and graphane). Unfortunatelly, the low-wavenumber problem alway fructrates my hope. Just the other day, I found the following website. Maybe, they can give me some hope. 以下内容转载自http://valenhou.blog.edu.cn/2005/133236.html &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& http://www.democritos.it/pipermail/pw_forum/2005-June/002615.html Dear PWscf devepolers and users, I am trying to get phonon frequencies of H2 molecule(gas phase) with PWSCF v.2.0.1. The results for H2 from PWSCF with all other parameters remanining same except different energy cutoffs are given below. I had done for dual= 4, 6 and 8, where "ecutrho= dual* ecutwfc". Results- with box-size 20 bohr, relaxed atomic positions: ================================ ATOMIC_POSITIONS (angstrom) H 1.325979436 0.000000000 0.000000000 H 0.574020564 0.000000000 0.000000000 Phonon frequencies: ================== A1. ecutwfc = 25 Ry, ecutrho = 100 Ry : ************************************************************************** omega( 1) = 2.829060 [THz] = 94.367920 [cm-1] omega( 2) = 2.829060 [THz] = 94.367920 [cm-1] omega( 3) = 3.861495 [THz] = 128.806463 [cm-1] omega( 4) = 4.643165 [THz] = 154.880338 [cm-1] omega( 5) = 4.643165 [THz] = 154.880338 [cm-1] omega( 6) = 129.812814 [THz] = 4330.118153 [cm-1] ************************************************************************** A2. ecutwfc = 25 Ry, ecutrho = 150 Ry : ************************************************************************** omega( 1) = 2.153650 [THz] = 71.838519 [cm-1] omega( 2) = 2.153650 [THz] = 71.838519 [cm-1] omega( 3) = 4.145972 [THz] = 138.295665 [cm-1] omega( 4) = 4.145972 [THz] = 138.295665 [cm-1] omega( 5) = 4.817966 [THz] = 160.711103 [cm-1] omega( 6) = 129.800590 [THz] = 4329.710425 [cm-1] ************************************************************************** A3. ecutwfc = 25 Ry, ecutrho = 200 Ry : ************************************************************************** omega( 1) = -1.588702 [THz] = -52.993762 [cm-1] omega( 2) = -1.588702 [THz] = -52.993762 [cm-1] omega( 3) = 3.878442 [THz] = 129.371746 [cm-1] omega( 4) = 4.580893 [THz] = 152.803167 [cm-1] omega( 5) = 4.580893 [THz] = 152.803167 [cm-1] omega( 6) = 129.718516 [THz] = 4326.972705 [cm-1] ************************************************************************** Thus I got negative and positive non-zero frequencies, where I am expecting 4 zero-frequencies, one frequency close to zero(due to bending) and one frequency in the order of 4560cm-1. Changing the atomic positions and with the same three dual values, the frequencies and their sign vary. Results are given below- Results- in box-size = 20 bohr, relaxed atomic positions: ================================ ATOMIC_POSITIONS (angstrom) H 0.000000000 0.000000000 0.000000000 H 0.434143657 0.434143657 0.434143657 Phonon frequencies: ================== B1. ecutwfc = 25 Ry, ecutrho = 100 Ry : ************************************************************************** omega( 1) = -4.811247 [THz] = -160.486990 [cm-1] omega( 2) = -4.811247 [THz] = -160.486990 [cm-1] omega( 3) = -2.203041 [THz] = -73.486037 [cm-1] omega( 4) = 0.500005 [THz] = 16.678469 [cm-1] omega( 5) = 0.500005 [THz] = 16.678469 [cm-1] omega( 6) = 129.696261 [THz] = 4326.230348 [cm-1] ************************************************************************** B2. ecutwfc = 25 Ry, ecutrho = 150 Ry : ************************************************************************** omega( 1) = 2.357793 [THz] = 78.648043 [cm-1] omega( 2) = 2.357793 [THz] = 78.648043 [cm-1] omega( 3) = 2.816217 [THz] = 93.939507 [cm-1] omega( 4) = 2.816217 [THz] = 93.939507 [cm-1] omega( 5) = 4.476603 [THz] = 149.324395 [cm-1] omega( 6) = 129.796002 [THz] = 4329.557366 [cm-1] ************************************************************************** B3. ecutwfc = 25 Ry, ecutrho = 200 Ry : ************************************************************************** omega( 1) = 2.005189 [THz] = 66.886334 [cm-1] omega( 2) = 2.005189 [THz] = 66.886334 [cm-1] omega( 3) = 3.773415 [THz] = 125.868397 [cm-1] omega( 4) = 4.558315 [THz] = 152.050044 [cm-1] omega( 5) = 4.558315 [THz] = 152.050044 [cm-1] omega( 6) = 129.748872 [THz] = 4327.985258 [cm-1] ************************************************************************** H2 molecule system is so simple and still phonon frequencies obtained are so off! Will you please help me to understand what is the origin of these non-zero(both negative and positive<200 cm-1 )frequencies? How should I get rid of them and have correct frequencies? Best regards, mousumi. ——————————————————————————————————————————————— ——————————————————————————————————————————————— The following is answer from expert of PWscf(编者暗). —————————————————————————————————————————————————————————————————————————————————————————————— I think that there is nothing really wrong in the frequencies you get. The isolated molecule should have 1 non zero frequency, corresponding to the bond-stretching mode and 5 zero frequencies (3 due to global translational symmetry and 2 due to rotational symmetry of a diatomic molecule). In the periodic calculation you are performing the two rotational mode never really vanish due to weak interaction with the periodic images of the molecule. The three translational modes would be zero for a perfect integration of the exchange-correlation energy on the real-space grid used to reperesent the density. This is never achieved in practice (except for exceedingly dense real space grids) Thus some violation of the order of a few percent of the stiffer frequency is common expecially when using gradient corrected XC functionals and/or US pseudopotentials. Your calculation is reasonably accurate, I think, and the violation of the zero-frequency result that you correctly expect is maybe smaller than you think. Think in terms of interatomic force constants: omega_bend = sqrt (PHI/reduced_mass) where PHI is the on site force constant of Hydrogen(1) that by translational symmetry should be equal and opposite to the Hydorgen(1)-Hydrogen(2) interatomic force constant. The fact that this sum rule (Acoustic Sum Rule says \sum_I PHI_{I\alpha,J\beta} = 0 for each J,\alpha,\beta) is violated gives a NON-ZERO translational frequency of the order of omega_trasl = sqrt(Delta PHI/ total_mass) Therefore the relative error in the determination of the interatomic force constants in your calculation is of the order of Delta PHI/PHI = (omega_trals/omega_bend)^2 *total_mass/reduced_mas = approx (150/4300)^2 *4 = 0.5 % which is not that bad, although not wonderful. Translational Acoustic Sum Rule can be enforced (with no noticable effect on the real non-zero frequencies) in the auxiliary code dynmat.x that you can find in the pwtools subdirectory of the espresso distribution. A more refined treatment of Translation and Rotational Acoustic Sum Rules can be enforced in the CVS version of the distribution (and in future web distributions) thanks to a contribution by Nicolas Mounet . best regards, Stefano de Gironcoli Mousumi Upadhyay Kahaly wrote: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 以上内容来自我的博客；blog.sina.com.cn/nkasir |
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