CCL: Problem in TDDFT Calculation



Dear Huub,
      Thank you very much for your answer. As you mentioned my mystery is resolved .

Cheers
veera

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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--
Veerapandian.P
School of Chemistry,
Bharathidasan University,
Tiruchirappalli-620024.