# [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,

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

________________________________

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

________________________________

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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