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Number of electrons in CCSDT module

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4 years 9 months ago - 4 years 9 months ago #52 by hyungjun96
Dear MRCC users,

I have tried to calculate the electronic structure of Zr atom using CCSDT in MRCC program interfaced to Molpro. Stuttgart ECP is employed during this calculation. (Therefore only 12 electrons are left.)

I have symmetrized HF orbitals and have done CCSDT calculation using these orbitals.

While 4s and 4p orbitals are left as core electrons, 2 electrons in 5s, and 2 electrons in 4d orbitals are distributed, respectively.

When MRCC reads CASSCF MO, there are 4 electrons.
Frozen orbitals: 4 ( 1 1 1 0 1 0 0 0)
Active orbitals: 6 ( 3 0 0 1 0 1 1 0)
Active electrons: 4
Spin quantum number: 1.0

However, the output created by MRCC gives there are only 2 electrons (1alpha and 1beta).
Number of alpha electrons: 1
Number of beta electrons: 1

I don't know why the number of electrons is reduced from 4 to 2. Could you give me a guide to include '4d' electrons in MRCC calculation?

Any comments are welcome.

Yours sincerely,
Last edit: 4 years 9 months ago by hyungjun96.

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4 years 9 months ago #53 by kallay
Could you please send us the output file?
Thank you in advance.

Best regards,
Mihaly Kallay

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4 years 9 months ago - 4 years 9 months ago #54 by hyungjun96
Dear Kallay,

These are my input and output files,

** I cannot attach the files for the unclear reason.

I send the files to your e-mail.

Yours sincerely,
Last edit: 4 years 9 months ago by hyungjun96.

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4 years 9 months ago #55 by kallay
Dear Hyungjun,
The occupation vector and the specification of active orbitals passed over to mrcc is not correct. You should contact the molpro developers about this issue. The quick solution is to specify the directory where mrcc is executed, that is, the corresponding line in the molpro input should be extended, e.g., as
{MRCC,method=CCSDT,spatial=1,HF=0,nacto=0,dir='/scr/mrcc';
Then the scratch files required by mrcc will be saved to that directory. To correct the error edit file fort.56, and replace the last two line by the following ones:
2 2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0
Then execute dmrcc in that directory. Note that in this case an MRCCSDT calculation will be run. If you want to perform a simple CCSDT calculation, the last line of fort.56 must be deleted.

Best regards,
Mihaly Kallay
The following user(s) said Thank You: hyungjun96

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4 years 9 months ago #57 by hyungjun96
Dear Kallay,

Your solution works well. Thank you very much.
I guess the last two line represents the occupation number in each orbital.
Could you give me an explanation about the structure of occupation number for MRCCSDT?(The meaning of the last line, 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0)

I will report this error to molpro developer.

I have one simple additional question, Is it possible to increase CC iteration step?
When I employ larger basis sets, aug-cc-pVTZ-PP, there is a convergence problem.
When I looked up the MRCC part in Molpro manual, there is no keyword related to the iteration.

Yours sincerely,

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4 years 9 months ago #58 by kallay
Dear Hyungjun,
I have just realized the that integral list printed by molpro is also wrong, so you will not get correct results even in this way.
For the moment, I would suggest using ROHF or UHF orbitals, that should work.

Concerning fort.56, the numbers in the third line represent the occupation numbers, the numbers in the fourth line stand for active (1) and inactive (0) orbitals.

Best regards,
Mihaly Kallay

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