On Wed, Sep 28, 2011 at
11:29 PM, Van Dam, Hubertus J HubertusJJ.vanDam-,-
pnnl.gov <owner-chemistry:ccl.net> wrote:
Sent to CCL by: "Van Dam, Hubertus J" [HubertusJJ.vanDam[*]http://pnnl.gov" target="_blank">pnnl.gov]
Hi Veerapandian,
The output lists every excitation separately (excitations of alpha electrons are
labeled "A" and beta electrons "B"). Hence the excitation
97A -> 99A contributes only once and not twice. So adapting your description
you should use the formula C^2*100 which gives you approximately 61% and your
mystery is resolved.
Best wishes,
Huub
-----Original Message-----
> From: owner-chemistry+hubertus.vandam==pnnl.gov|ccl.net [mailto:owner-chemistry+hubertus.vandam==pnnl.gov|ccl.net] On Behalf Of veera pandian ponnuchamy veera.pandi33
~~ gmail.com
Sent: Tuesday, September 27, 2011 10:44 PM
To: Van Dam, Hubertus J
Subject: CCL: Problem in TDDFT Calculation
Sent to CCL by: "veera pandian ponnuchamy" [veera.pandi33()http://gmail.com" target="_blank">gmail.com]
Hi CCLrs,
I have experienced a surprise case, while
calculating the percentage of contribution in TDDFT calculation.
This is the output
Excited State 1: 2.066-A 1.1028 eV 1124.31 nm
f=0.0003 <S**2>=0.817
97B -> 98B 0.99625
This state for optimization and/or second-order correction.
Total Energy, E(TD-HF/TD-KS) = -1221.52573888
Copying the excited state density for this state as the 1-particle RhoCI
density.
Excited State 2: 2.073-A 2.0309 eV
610.49 nm f=0.0003 <S**2>=0.825
96B -> 98B 0.98993
Excited State 3: 3.476-A 2.0710 eV
598.68 nm f=0.0000 <S**2>=2.771
95A ->103A 0.10782
96A ->102A -0.11015
98A ->100A 0.68284
94B ->105B -0.10218
95B ->103B -0.11202
96B ->101B 0.11454
97B ->100B -0.68191
98A <-100A 0.11802
97B <-100B -0.11806
Excited State 4: 2.852-A 2.5895 eV
478.80 nm f=0.0021 <S**2>=1.783
98A -> 99A 0.99481
Excited State 5: 2.171-A 2.6627 eV
465.63 nm f=0.0193 <S**2>=0.928
97A -> 99A 0.77986
97A ->104A 0.12398
90B -> 98B -0.25513
92B -> 98B 0.47858
93B -> 98B -0.13269
95B -> 98B -0.15685
The above state (excited state 5) is the most probable excited state, here when
I calculate the percentage of contribution using C^2*2*100 formula. I get 121%
how it is possible? or am I doing anything wrong? Your help will be
appreciated.
Thanks in advance...
Yours
Veerapandian.Phttp-:-//www.ccl.net/cgi-bin/ccl/send_ccl_messagehttp://www.ccl.net/chemistry/sub_unsub.shtmlhttp://www.ccl.net/spammers.txt
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