Error: INQRLR: TOO MANY PROPERTIES SPECIFIED

[jd140]

Hello,

I have been running a few quadratic response calcualtions with the imput files being something like:

**DALTON INPUT
.RUN WAVE FUNCTIONS
.RUN RESPONSE
.DIRECT
**INTEGRAL
.NOSUP
.DIPLEN
**WAVE FUNCTIONS
.DFT
CAMB3LYP
*ORBITAL INPUT
.NOSUPSYM
**RESPONSE
.MAXRM
3000
*QUADRA
.MAX IT
1000
.DOUBLE RESIDUE
.DIPLEN
.ROOTS
40
**END OF DALTON INPUT

and I end with the following error in the excitation energies and transition moments calculation (MCTDHF) section:

Code: Select all

*** RSPCTL MICROITERATIONS CONVERGED
 NUMBER OF SPECIFIED PROPERTIES EXCEED THE MAXIMUM ALLOWED
 MXLRQR = 1000 NLRLBL =  1001

  --- SEVERE ERROR, PROGRAM WILL BE ABORTED ---
     Date and time (Linux)  : Mon Mar 24 18:39:59 2014 
     Host name              : compute-00-05.private.dns.zone          

@ MPI MASTER, node no.:     0
@ Reason:  INQRLR: TOO MANY PROPERTIES SPECIFIED

I haven't been able to find anything related in the documentation, couuld somebody help me identifying the cause of this error?
Thanks,
Jean-Christophe

[Joanna]

If in DALTON/include/rspprp.h, you increase
PARAMETER ( MAXLBL = 1000 )
(from 1000 to something higher, since you exceeded the maximum number) and recompile, it should solve the problem.
Good luck!

[jd140]

Hi Joanna,
Thanks, I'll try this.
JCD

[olav]

Hi
Just let me just give you a warning. For every pair of excitations (a,b) you will be solving a linear response equation for frequency wa+wb. This means that given 40 roots you have approximately 1600 equations to solve. This is where you exceed the number 1000. Considering that most of the higher excitations you have obtained with DFT have little to do with reality, I suspect that you will be wasting a lot of computer time.

Regards,
Olav

[jd140]

Hi Olav,
Thanks. I would like to compute such a high number of states because I aim at calculating the excited state absortion spectra for a 0-3eV range (from the excited state, or around 3-6eV relatively to the ground state), and I that's typically how many states I need for this. What does make you think that for such high excitations energies, TD-DFT is going to provide a poor result?
Thanks,
JCD

[lyzhao]

see here:
http://daltonprogram.org/forum/viewtopic.php?f=9&t=191

[janus]

Please see Peter Taylor's comment here:

http://daltonforum.org/viewtopic.php?f=9&t=191

[olav]

Thanks I was just looking for Pete's eloquent reply to a similar question.

Another thing, to save you a lot of computer time there is an .ESA option (excited state absorption) which appears to be undocumented.
Replace

.DOUBLE RESIDUE
.ROOTS
40

with

.ESA
1
40

which is more in line with what you are describing, It means that you only get transition matrix elements <1|x|n>, n=2,40
I agree with Peters comment that the number of meaningful excited state solutions would hardly exceed 10 or so.

Olav

[jd140]

Thanks very much for your having pointed me to this past helpful discussion and for having provided the tip about the undocumented ESA option, which seems super-useful indeed !

[wuyulele]

Thanks I was just looking for Pete's eloquent reply to a similar question.

Another thing, to save you a lot of computer time there is an .ESA option (excited state absorption) which appears to be undocumented.
Replace

.DOUBLE RESIDUE
.ROOTS
40

with

.ESA
1
40

which is more in line with what you are describing, It means that you only get transition matrix elements <1|x|n>, n=2,40
I agree with Peters comment that the number of meaningful excited state solutions would hardly exceed 10 or so.

Olav

Hi Olav,

I tried to compute excited state absorption between S2 and Sn states, so I wrote
.ESA
2
20

but this did not work for me, the ouput file did not print transition moments between excited states. It seems like the .ESA option only works for
.ESA
1
n

Do you have any idea about it?

Thanks

[magnus]

It looks like the ESA option will always be from the lowest excited state to other excited states. The value on the first line is not the number of the state but it denotes the symmetry. So in your case it will use the lowest excited state of symmetry 2 and of course if you don't use or have symmetry in your molecule it will fail. The ESA option is not documented which also means that it is most likely not well tested.

[wuyulele]

It looks like the ESA option will always be from the lowest excited state to other excited states. The value on the first line is not the number of the state but it denotes the symmetry. So in your case it will use the lowest excited state of symmetry 2 and of course if you don't use or have symmetry in your molecule it will fail. The ESA option is not documented which also means that it is most likely not well tested.

Hi magnus,

thanks a lot for your reply! It's very helpful!