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SF TDDFT calculation problem

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Author Post
Member
Registered: Jun 2019
Posts: 1
Dear Q-Chem users,
I am a new Q-Chem user. I want to use Q-Chem to do SF-TDDFT calculation. When I am doing test using examples of TMM diradical in Q-Chem User's Manual. I found the oscillator strength of the first excited state is 1.0000 while the oscillator strength of the other excited state is 0.0000. Then I try to reproduce the calculation of water in "How to analyze SF-DFT calculations: annotated Q-Chem output" (http://iopenshell.usc.edu/howto/sf_dft.pdf). The energy and oscillator strength is far away from results in that documents. I am using Q-Chem 4.3 mpi version. Is there something wrong with input or installation?

This is the input:
$molecule
0 3
C
H 1 rCH
H 1 rCH 2 HCH

rCH 1.0775
HCH 133.29
$end


$rem
job_type SP
UNRESTRICTED true
BASIS 6-31G*
EXCHANGE pbe
SPIN_FLIP 1
CIS_N_ROOTS = 3
STS_MOM = true
wang_ziegler_kernel true
MAX_CIS_CYCLES = 100
THRESH 14
VARTHRESH = FALSE
INCDFT = FALSE
XC_GRID 000150000302
$end

This is a part of output:
Excited state 1: excitation energy (eV) = 1.7543
Total energy for state 1: -38.805333163896
<S**2> : 1.9902
Trans. Mom.: -0.0002 X 0.0000 Y -0.0011 Z
Strength : 1.0000
S( 1) --> S( 1) amplitude = 0.6125 alpha
S( 2) --> S( 2) amplitude = 0.7855 alpha

Excited state 2: excitation energy (eV) = 2.4269
Total energy for state 2: -38.780618294486
<S**2> : 0.0160
Trans. Mom.: -0.0002 X 0.9775 Y 0.0000 Z
Strength : 0.0000
S( 1) --> S( 2) amplitude = 0.3566 alpha
S( 2) --> S( 1) amplitude = 0.9329 alpha

Excited state 3: excitation energy (eV) = 2.8526
Total energy for state 3: -38.764971357505
<S**2> : 0.0486
Trans. Mom.: 0.0000 X 0.0000 Y 0.0000 Z
Strength : -0.1252
S( 1) --> S( 1) amplitude = 0.7864 alpha
S( 2) --> S( 2) amplitude = -0.6151 alpha

Thank you,
Mingyu
Administrator
Registered: Oct 2017
Posts: 22
Dear Mingyu,

You seems to be using a very old version of Q-Chem. I have just run the 5.2 version, and got exactly what the pdf states. I would recommend to update your Q-Chem version.

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