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<p><span style="font-family: Calibri, Helvetica, sans-serif, serif, EmojiFont; font-size: 16px;">Dear Lurii,</span></p>
<p><span style="font-family: Calibri, Helvetica, sans-serif, serif, EmojiFont; font-size: 16px;"><br>
</span></p>
<p><span style="font-family: Calibri, Helvetica, sans-serif, serif, EmojiFont; font-size: 16px;">Thank for your helpful instruction.</span></p>
<p><span style="font-family: Calibri, Helvetica, sans-serif, serif, EmojiFont; font-size: 16px;"><br>
</span></p>
<p><span style="font-family: Calibri, Helvetica, sans-serif, serif, EmojiFont; font-size: 16px;">From the results, there are two parts in chi_1_1. (real part and imaginary part) . No <span style="font-family: Calibri, Helvetica, sans-serif, serif, EmojiFont; font-size: 16px;">adsorption
coefficients </span></span></p>
<p><span style="font-family: Calibri, Helvetica, sans-serif, serif, EmojiFont; font-size: 16px;"><br>
</span></p>
<p><font face="Calibri, Helvetica, sans-serif, serif, EmojiFont">How can we calculate the adsorption coefficients from <span>polarizability?</span></font></p>
<p><span style="font-family: Calibri, Helvetica, sans-serif, serif, EmojiFont; font-size: 16px;"><br>
</span></p>
<p><font face="Calibri, Helvetica, sans-serif, serif, EmojiFont">The result is shown as follow:</font></p>
<p><span style="font-family: Calibri, Helvetica, sans-serif, serif, EmojiFont; font-size: 16px;"><br>
</span></p>
<p><span style="font-family: Calibri, Helvetica, sans-serif, serif, EmojiFont; font-size: 16px;"></p>
<div>#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) </div>
<div># S(E) satisfies the sum rule </div>
<div> chi_1_1= 0.000000000000000E+00 0.189914943334197E+04 0.000000000000000E+00</div>
<div> chi_2_1= 0.000000000000000E+00 -.949575309581559E+03 0.000000000000000E+00</div>
<div> chi_3_1= 0.000000000000000E+00 -.843216803071169E-03 0.000000000000000E+00</div>
<div> chi_1_2= 0.000000000000000E+00 -.949574977107089E+03 0.000000000000000E+00</div>
<div> chi_2_2= 0.000000000000000E+00 0.189915026381885E+04 0.000000000000000E+00</div>
<div> chi_3_2= 0.000000000000000E+00 0.110034487599961E-03 0.000000000000000E+00</div>
<div> chi_1_3= 0.000000000000000E+00 -.838178086194996E-03 0.000000000000000E+00</div>
<div> chi_2_3= 0.000000000000000E+00 0.109036445082823E-03 0.000000000000000E+00</div>
<div> chi_3_3= 0.000000000000000E+00 0.153710402502629E+03 0.000000000000000E+00</div>
<div>
<div> S(E)= 0.000000000000000E+00 0.000000000000000E+00</div>
<div> chi_1_1= 0.100000000000000E-03 0.189914977011440E+04 0.269793830860200E-01</div>
<div> chi_2_1= 0.100000000000000E-03 -.949575477967847E+03 -.134896973444010E-01</div>
<div> chi_3_1= 0.100000000000000E-03 -.843216968116708E-03 -.132241869803436E-07</div>
<div> chi_1_2= 0.100000000000000E-03 -.949575145493404E+03 -.134897003736112E-01</div>
<div> chi_2_2= 0.100000000000000E-03 0.189915060059238E+04 0.269794725819193E-01</div>
<div> chi_3_2= 0.100000000000000E-03 0.110034514907608E-03 0.218775411907335E-08</div>
<div> chi_1_3= 0.100000000000000E-03 -.838178244921434E-03 -.131322115320554E-07</div>
<div> chi_2_3= 0.100000000000000E-03 0.109036471525103E-03 0.204656097751975E-08</div>
<div> chi_3_3= 0.100000000000000E-03 0.153710405020382E+03 0.201509717932602E-03</div>
<div> S(E)= 0.100000000000000E-03 0.574659324721647E-06</div>
<br>
</div>
<div>Thanks.</div>
<div><br>
</div>
<div>Best regards,</div>
<div>LIANG Xiongyi</div>
<br>
</span>
<p></p>
</div>
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<div id="divRplyFwdMsg" dir="ltr"><font face="Calibri, sans-serif" style="font-size:11pt" color="#000000"><b>发件人:</b> pw_forum-bounces@pwscf.org <pw_forum-bounces@pwscf.org> 代表 Timrov Iurii <iurii.timrov@epfl.ch><br>
<b>发送时间:</b> 2017年8月1日 17:32:37<br>
<b>收件人:</b> PWSCF Forum<br>
<b>主题:</b> Re: [Pw_forum] How to get absorption coefficient</font>
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<div>
<p>Dear <span style="font-family: Calibri, Helvetica, sans-serif; font-size: 16px; background-color: rgb(255, 255, 255);">LIANG Xiongyi</span>,<br>
</p>
<p><br>
</p>
<div style="font-family: Calibri, Helvetica, sans-serif; font-size: 16px;"><span style="font-size: 12pt;">> First </span>column is energy (Ry), and the second i<span style="font-size: 12pt;">s absorption coefficients, right? </span></div>
<div style="font-family: Calibri, Helvetica, sans-serif; font-size: 16px;"><span style="font-size: 12pt;"><br>
</span></div>
<div style="font-family: Calibri, Helvetica, sans-serif; font-size: 16px;">Yes<br>
</div>
<div style="font-family: Calibri, Helvetica, sans-serif; font-size: 16px;"><span style="font-size: 12pt;"><br>
</span></div>
<div style="font-family: Calibri, Helvetica, sans-serif; font-size: 16px;"><span style="font-size: 12pt;">> And what is the unit of absorption coefficients?</span></div>
<div style="font-family: Calibri, Helvetica, sans-serif; font-size: 16px;"><span style="font-size: 12pt;"><br>
</span></div>
<div style="font-family: Calibri, Helvetica, sans-serif; font-size: 16px;"><span style="font-size: 12pt;">Arbitrary units. What is important is the relative intensity of the peaks and their position.</span></div>
<div style="font-family: Calibri, Helvetica, sans-serif; font-size: 16px;"><span style="font-size: 12pt;"><br>
</span></div>
<div style="font-family: Calibri, Helvetica, sans-serif; font-size: 16px;"><span style="font-size: 12pt;">> Can we get the absorption coefficients along x axis and y axis, respectively?</span><span style="font-size: 12pt;"></span><br>
</div>
<p><br>
</p>
<p>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).<br>
</p>
<p><br>
</p>
<p>HTH </p>
<p><br>
</p>
<p>Iurii<br>
</p>
<p><br>
</p>
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<font color="808080" size="3" face="'Times New Roman', Times, serif">--<br>
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<div id="divRplyFwdMsg" dir="ltr"><font face="Calibri, sans-serif" color="#000000" style="font-size:11pt"><b>From:</b> pw_forum-bounces@pwscf.org <pw_forum-bounces@pwscf.org> on behalf of LEUNG Clarence <liangxy123@hotmail.com><br>
<b>Sent:</b> Tuesday, August 1, 2017 8:34 AM<br>
<b>To:</b> PWSCF Forum<br>
<b>Subject:</b> [Pw_forum] 答复: How to get absorption coefficient</font>
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<p>Dear Lurii,</p>
<p><br>
</p>
<p>Thank you very much.</p>
<p><br>
</p>
<p>I have collect all S(E) from the out file plot_chi.dat.</p>
<p><br>
</p>
<p> </p>
<div># S(E) satisfies the sum rule </div>
<div> S(E)= 0.000000000000000E+00 0.000000000000000E+00</div>
<div> S(E)= 0.100000000000000E-03 0.574659324721647E-06</div>
<div> S(E)= 0.200000000000000E-03 0.229864029025929E-05</div>
<div> S(E)= 0.300000000000000E-03 0.517195187076012E-05</div>
<div> S(E)= 0.400000000000000E-03 0.919460902323344E-05</div>
<div> S(E)= 0.500000000000000E-03 0.143666326876961E-04</div>
<div> S(E)= 0.600000000000000E-03 0.206880497873775E-04</div>
<div> S(E)= 0.700000000000000E-03 0.281588932289815E-04</div>
<div> S(E)= 0.800000000000000E-03 0.367792019030096E-04</div>
<div> S(E)= 0.900000000000000E-03 0.465490206841405E-04</div>
<div> S(E)= 0.100000000000000E-02 0.574684004316713E-04</div>
<div><span style="font-size:12pt"><br>
</span></div>
<div><span style="font-size:12pt">First </span>column is energy (Ry), and the second i<span style="font-size:12pt">s absorption coefficients, right? And what is the unit of absorption coefficients?</span></div>
<div><span style="font-size:12pt"><br>
</span></div>
<div><span style="font-size:12pt">Can we get the absorption coefficients along x axis and y axis, respectively?</span></div>
<div><span style="font-size:12pt"><br>
</span></div>
<div><span style="font-size:12pt">Best regards</span></div>
<div><span style="font-size:12pt">LIANG Xiongyi</span></div>
<p></p>
</div>
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<div id="divRplyFwdMsg" dir="ltr"><font face="Calibri, sans-serif" color="#000000" style="font-size:11pt"><b>发件人:</b> pw_forum-bounces@pwscf.org <pw_forum-bounces@pwscf.org> 代表 Timrov Iurii <iurii.timrov@epfl.ch><br>
<b>发送时间:</b> 2017年7月31日 23:33:43<br>
<b>收件人:</b> pw_forum@pwscf.org<br>
<b>主题:</b> Re: [Pw_forum] How to get absorption coefficient</font>
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<div>
<p>Dear Clarence,</p>
<p><br>
</p>
<p>Before continuing using turboTDDFT, I strongly recommend to read (at least) these two publications:</p>
<p><br>
</p>
<p>1. turboTDDFT – A code for the simulation of molecular spectra using the Liouville–Lanczos approach<br>
to time-dependent density-functional perturbation theory Original Research Article<br>
Authors: Osman Baris Malcioglu, Ralph Gebauer, Dario Rocca, Stefano Baroni<br>
Source: Computer Physics Communications Volume: 182 Article Number: 1744 Published: APR 2011<br>
<br>
2. turboTDDFT 2.0 - Hybrid functionals and new algorithms within time-dependent<br>
density-functional perturbation theory<br>
Authors: X. Ge, S. J. Binnie, D. Rocca, R. Gebauer, and S. Baroni<br>
Source: Computer Physics Communications Volume: 185 Article Number: 2080 Published: MAR 2014</p>
<p><br>
</p>
<p>The full list of publications about the TDDFPT module of Quantum ESPRESSO can be found in qe/TDDFPT/README.<br>
</p>
<p><br>
</p>
<p>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.:</p>
<p>second column - energy \hbar \omega (Ry)</p>
<p>third column - real part of the polarizability Re(chi)</p>
<p>fourth column - imaginary part of the polarizability Im(chi) <br>
</p>
<p><br>
</p>
<p>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):</p>
<p><br>
</p>
<p>S(\hbar \omega) = 2m/( 3 \pi e^2 \hbar) \omega sum_j chi_j_j</p>
<p><br>
</p>
<p>HTH</p>
<p><br>
</p>
<p>Regards,</p>
<p>Iurii<br>
</p>
<p><br>
</p>
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<font size="3" face="'Times New Roman', Times, serif" color="808080">--<br>
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<div id="divRplyFwdMsg" dir="ltr"><font face="Calibri, sans-serif" color="#000000" style="font-size:11pt"><b>From:</b> pw_forum-bounces@pwscf.org <pw_forum-bounces@pwscf.org> on behalf of LEUNG Clarence <liangxy123@hotmail.com><br>
<b>Sent:</b> Monday, July 31, 2017 4:27 PM<br>
<b>To:</b> pw_forum@pwscf.org<br>
<b>Subject:</b> [Pw_forum] How to get absorption coefficient</font>
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<p>Dear QE users,</p>
<p><br>
</p>
<p>Now, I use the turbo_lanczos.x and turbo_spectrum to get the <span>absorption spectrum.</span></p>
<p><span><br>
</span></p>
<p>I can get a plot_chi.dat file, as follow:</p>
<p><br>
</p>
<p></p>
<div>#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) </div>
<div># S(E) satisfies the sum rule </div>
<div> chi_1_1= 0.000000000000000E+00 0.189914943334197E+04 0.000000000000000E+00</div>
<div> chi_2_1= 0.000000000000000E+00 -.949575309581559E+03 0.000000000000000E+00</div>
<div> chi_3_1= 0.000000000000000E+00 -.843216803071169E-03 0.000000000000000E+00</div>
<div> chi_1_2= 0.000000000000000E+00 -.949574977107089E+03 0.000000000000000E+00</div>
<div> chi_2_2= 0.000000000000000E+00 0.189915026381885E+04 0.000000000000000E+00</div>
<div> chi_3_2= 0.000000000000000E+00 0.110034487599961E-03 0.000000000000000E+00</div>
<div> chi_1_3= 0.000000000000000E+00 -.838178086194996E-03 0.000000000000000E+00</div>
<div> chi_2_3= 0.000000000000000E+00 0.109036445082823E-03 0.000000000000000E+00</div>
<div> chi_3_3= 0.000000000000000E+00 0.153710402502629E+03 0.000000000000000E+00</div>
<br>
<p></p>
<p>What is meaning of each row and how can I get the absorption coefficients?</p>
<p><br>
</p>
<p>Many thanks.</p>
<p><br>
</p>
<p>Clarence</p>
<p>City University of Hong Kong</p>
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