Dear Liang,
> From the results, there are two parts in chi_1_1. (real part and imaginary > part) . No adsorption coefficients Spell correctly: "aBsorption coefficient". Multiply the imaginary part of chi_1_1 by the frequency and plot it. > How can we calculate the adsorption coefficients from polarizability? "aBsorption" S(\hbar \omega) = 2m/( 3 \pi e^2 \hbar) \omega sum_j chi_j_j If you want to analyse the X, Y, and Z components, see above (i.e. multiply chi_j_j by frequency). Regards, Iurii -- Dr. Iurii Timrov Postdoctoral Researcher Swiss Federal Institute of Technology Lausanne (EPFL) Laboratory of Theory and Simulation of Materials (THEOS) CH-1015 Lausanne, Switzerland ________________________________ From: pw_forum-boun...@pwscf.org <pw_forum-boun...@pwscf.org> on behalf of LEUNG Clarence <liangxy...@hotmail.com> Sent: Tuesday, August 1, 2017 12:05 PM To: PWSCF Forum Subject: [Pw_forum] 答复: How to get absorption coefficient Dear Lurii, Thank for your helpful instruction. From the results, there are two parts in chi_1_1. (real part and imaginary part) . No adsorption coefficients How can we calculate the adsorption coefficients from polarizability? The result is shown as follow: #Chi is reported as CHI_(i)_(j) \hbar \omega (Ry) Re(chi) (e^2*a_0^2/Ry) Im(chi) (e^2*a_0^2/Ry) # S(E) satisfies the sum rule chi_1_1= 0.000000000000000E+00 0.189914943334197E+04 0.000000000000000E+00 chi_2_1= 0.000000000000000E+00 -.949575309581559E+03 0.000000000000000E+00 chi_3_1= 0.000000000000000E+00 -.843216803071169E-03 0.000000000000000E+00 chi_1_2= 0.000000000000000E+00 -.949574977107089E+03 0.000000000000000E+00 chi_2_2= 0.000000000000000E+00 0.189915026381885E+04 0.000000000000000E+00 chi_3_2= 0.000000000000000E+00 0.110034487599961E-03 0.000000000000000E+00 chi_1_3= 0.000000000000000E+00 -.838178086194996E-03 0.000000000000000E+00 chi_2_3= 0.000000000000000E+00 0.109036445082823E-03 0.000000000000000E+00 chi_3_3= 0.000000000000000E+00 0.153710402502629E+03 0.000000000000000E+00 S(E)= 0.000000000000000E+00 0.000000000000000E+00 chi_1_1= 0.100000000000000E-03 0.189914977011440E+04 0.269793830860200E-01 chi_2_1= 0.100000000000000E-03 -.949575477967847E+03 -.134896973444010E-01 chi_3_1= 0.100000000000000E-03 -.843216968116708E-03 -.132241869803436E-07 chi_1_2= 0.100000000000000E-03 -.949575145493404E+03 -.134897003736112E-01 chi_2_2= 0.100000000000000E-03 0.189915060059238E+04 0.269794725819193E-01 chi_3_2= 0.100000000000000E-03 0.110034514907608E-03 0.218775411907335E-08 chi_1_3= 0.100000000000000E-03 -.838178244921434E-03 -.131322115320554E-07 chi_2_3= 0.100000000000000E-03 0.109036471525103E-03 0.204656097751975E-08 chi_3_3= 0.100000000000000E-03 0.153710405020382E+03 0.201509717932602E-03 S(E)= 0.100000000000000E-03 0.574659324721647E-06 Thanks. Best regards, LIANG Xiongyi ________________________________ 发件人: pw_forum-boun...@pwscf.org <pw_forum-boun...@pwscf.org> 代表 Timrov Iurii <iurii.tim...@epfl.ch> 发送时间: 2017年8月1日 17:32:37 收件人: PWSCF Forum 主题: Re: [Pw_forum] How to get absorption coefficient Dear LIANG Xiongyi, > First column is energy (Ry), and the second is absorption coefficients, right? Yes > And what is the unit of absorption coefficients? Arbitrary units. What is important is the relative intensity of the peaks and their position. > Can we get the absorption coefficients along x axis and y axis, > respectively? Yes, you can get this information from the chi_i_j matrix: chi_1_1 is the diagonal component along X axis, chi_2_2 is the diagonal component along Y axis, etc. Notice that depending on the symmetry of the molecule under study there might be also non-zero off-diagonal elements in the chi_i_j matrix. In order to obtain the absorption coefficient along a specific direction, do not forget to multiply chi_i_j by frequency. Recall how the S(w) is computed: S(w) ~ w * \sum_j chi_j_j(w). HTH Iurii -- Dr. Iurii Timrov Postdoctoral Researcher Swiss Federal Institute of Technology Lausanne (EPFL) Laboratory of Theory and Simulation of Materials (THEOS) CH-1015 Lausanne, Switzerland ________________________________ From: pw_forum-boun...@pwscf.org <pw_forum-boun...@pwscf.org> on behalf of LEUNG Clarence <liangxy...@hotmail.com> Sent: Tuesday, August 1, 2017 8:34 AM To: PWSCF Forum Subject: [Pw_forum] 答复: How to get absorption coefficient Dear Lurii, Thank you very much. I have collect all S(E) from the out file plot_chi.dat. # S(E) satisfies the sum rule S(E)= 0.000000000000000E+00 0.000000000000000E+00 S(E)= 0.100000000000000E-03 0.574659324721647E-06 S(E)= 0.200000000000000E-03 0.229864029025929E-05 S(E)= 0.300000000000000E-03 0.517195187076012E-05 S(E)= 0.400000000000000E-03 0.919460902323344E-05 S(E)= 0.500000000000000E-03 0.143666326876961E-04 S(E)= 0.600000000000000E-03 0.206880497873775E-04 S(E)= 0.700000000000000E-03 0.281588932289815E-04 S(E)= 0.800000000000000E-03 0.367792019030096E-04 S(E)= 0.900000000000000E-03 0.465490206841405E-04 S(E)= 0.100000000000000E-02 0.574684004316713E-04 First column is energy (Ry), and the second is absorption coefficients, right? And what is the unit of absorption coefficients? Can we get the absorption coefficients along x axis and y axis, respectively? Best regards LIANG Xiongyi ________________________________ 发件人: pw_forum-boun...@pwscf.org <pw_forum-boun...@pwscf.org> 代表 Timrov Iurii <iurii.tim...@epfl.ch> 发送时间: 2017年7月31日 23:33:43 收件人: pw_forum@pwscf.org 主题: Re: [Pw_forum] How to get absorption coefficient Dear Clarence, Before continuing using turboTDDFT, I strongly recommend to read (at least) these two publications: 1. turboTDDFT – A code for the simulation of molecular spectra using the Liouville–Lanczos approach to time-dependent density-functional perturbation theory Original Research Article Authors: Osman Baris Malcioglu, Ralph Gebauer, Dario Rocca, Stefano Baroni Source: Computer Physics Communications Volume: 182 Article Number: 1744 Published: APR 2011 2. turboTDDFT 2.0 - Hybrid functionals and new algorithms within time-dependent density-functional perturbation theory Authors: X. Ge, S. J. Binnie, D. Rocca, R. Gebauer, and S. Baroni Source: Computer Physics Communications Volume: 185 Article Number: 2080 Published: MAR 2014 The full list of publications about the TDDFPT module of Quantum ESPRESSO can be found in qe/TDDFPT/README. The 3x3 matrix chi_i_j is the polarizability (i and j run over the Cartesian components x, y, z, which in the plot_chi.dat file correspond to 1, 2, 3, respectively) - see Eq.(5) in the first reference mentioned above. In the file *.plot_chi.dat in the header you can see what is the meaning of each column, i.e.: second column - energy \hbar \omega (Ry) third column - real part of the polarizability Re(chi) fourth column - imaginary part of the polarizability Im(chi) In the same plot_chi.dat file, there is also the information (only if you performed Lanczos calculations along three Cartesian directions, i.e. ipol=4 - see the first reference above) about S(E) (second column) as a function of the energy (first column), which is the oscillator strength (the absorption coefficient). It is defined as (see the output file produced by turbo_spectrum.x, i.e. *.tddfpt_pp-out): S(\hbar \omega) = 2m/( 3 \pi e^2 \hbar) \omega sum_j chi_j_j HTH Regards, Iurii -- Dr. Iurii Timrov Postdoctoral Researcher Swiss Federal Institute of Technology Lausanne (EPFL) Laboratory of Theory and Simulation of Materials (THEOS) CH-1015 Lausanne, Switzerland +41 21 69 34 881 http://people.epfl.ch/265334 ________________________________ From: pw_forum-boun...@pwscf.org <pw_forum-boun...@pwscf.org> on behalf of LEUNG Clarence <liangxy...@hotmail.com> Sent: Monday, July 31, 2017 4:27 PM To: pw_forum@pwscf.org Subject: [Pw_forum] How to get absorption coefficient Dear QE users, Now, I use the turbo_lanczos.x and turbo_spectrum to get the absorption spectrum. I can get a plot_chi.dat file, as follow: #Chi is reported as CHI_(i)_(j) \hbar \omega (Ry) Re(chi) (e^2*a_0^2/Ry) Im(chi) (e^2*a_0^2/Ry) # S(E) satisfies the sum rule chi_1_1= 0.000000000000000E+00 0.189914943334197E+04 0.000000000000000E+00 chi_2_1= 0.000000000000000E+00 -.949575309581559E+03 0.000000000000000E+00 chi_3_1= 0.000000000000000E+00 -.843216803071169E-03 0.000000000000000E+00 chi_1_2= 0.000000000000000E+00 -.949574977107089E+03 0.000000000000000E+00 chi_2_2= 0.000000000000000E+00 0.189915026381885E+04 0.000000000000000E+00 chi_3_2= 0.000000000000000E+00 0.110034487599961E-03 0.000000000000000E+00 chi_1_3= 0.000000000000000E+00 -.838178086194996E-03 0.000000000000000E+00 chi_2_3= 0.000000000000000E+00 0.109036445082823E-03 0.000000000000000E+00 chi_3_3= 0.000000000000000E+00 0.153710402502629E+03 0.000000000000000E+00 What is meaning of each row and how can I get the absorption coefficients? Many thanks. Clarence City University of Hong Kong
_______________________________________________ Pw_forum mailing list Pw_forum@pwscf.org http://pwscf.org/mailman/listinfo/pw_forum