[Pw_forum] 答复: How to get absorption coefficient
LEUNG Clarence
liangxy123 at hotmail.com
Tue Aug 1 12:05:44 CEST 2017
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-bounces at pwscf.org <pw_forum-bounces at pwscf.org> 代表 Timrov Iurii <iurii.timrov at 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-bounces at pwscf.org <pw_forum-bounces at pwscf.org> on behalf of LEUNG Clarence <liangxy123 at 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-bounces at pwscf.org <pw_forum-bounces at pwscf.org> 代表 Timrov Iurii <iurii.timrov at epfl.ch>
发送时间: 2017年7月31日 23:33:43
收件人: pw_forum at 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-bounces at pwscf.org <pw_forum-bounces at pwscf.org> on behalf of LEUNG Clarence <liangxy123 at hotmail.com>
Sent: Monday, July 31, 2017 4:27 PM
To: pw_forum at 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
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