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Dear All,
My molecule has Cs symmetry and i want to calculate energy profile along N-H stretching coordinate at the MS-CASPT2//SA2-CASSCF(8,8)/aug-cc-pVDZ level, while imposing the Cs symmetry constraint on the wavefunctions. I would appreciate an example of the input file format for this calculation. The molecule has a total of 72 electrons.
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You may want to reconsider the use of symmetry. With symmetry you'll be forced to use numerical gradients, but without symmetry analytical gradients are available:
* This file could have been saved elsewhere
> file geom.xyz
4
N -0.02309126 -0.00943438 0.00000000
H 0.35905589 -0.94190767 0.00000000
H 0.37178537 0.45144204 0.80448617
H 0.37178537 0.45144204 -0.80448617
> eof
> foreach R in (1.00, 1.01, 1.02, 1.03, 1.04, 1.05)
&GATEWAY
Coord = geom.xyz
Group = NoSym
Basis = cc-pVDZ
RICD
Constraints
d = bond N1 H2
Values
d = $R angstrom
End of constraints
> do while
&SEWARD
&RASSCF
NActEl = 8
Charge = 0
RAS2 = 7
StAverage = 2
RlxRoot = 1
&CASPT2
Multistate = all
IPEA = 0.0
&SLAPAF
> end do
* Save the converged geometry in a new file
> copy $Project.Opt.xyz $MOLCAS_OUTPUT/${Project}_$R.xyz
* And as input for the next scan point
> copy $Project.Opt.xyz geom.xyz
> end foreachIf you insist on using symmetry:
* This file could have been saved elsewhere
> file geom.xyz
4
N -0.02309126 -0.00943438 0.00000000
H 0.35905589 -0.94190767 0.00000000
H 0.37178537 0.45144204 0.80448617
H 0.37178537 0.45144204 -0.80448617
> eof
> foreach R in (1.00, 1.01, 1.02, 1.03, 1.04, 1.05)
&GATEWAY
Coord = geom.xyz
Group = z
Basis = cc-pVDZ
RICD
Constraints
d = bond N1 H2
Values
d = $R angstrom
End of constraints
> do while
&SEWARD
&RASSCF
NActEl = 8
Inactive = 1 0
RAS2 = 5 2
StAverage = 2
RlxRoot = 1
&CASPT2
Multistate = all
IPEA = 0.0
&SLAPAF
> end do
* Save the converged geometry in a new file
> copy $Project.Opt.xyz $MOLCAS_OUTPUT/${Project}_$R.xyz
* And as input for the next scan point
> copy $Project.Opt.xyz geom.xyz
> end foreachFor comparison
R NoSym Cs
1.00 -56.383555700778324 -56.383555922730814
1.01 -56.383927277697623 -56.383927243585440
1.02 -56.384128639020759 -56.384128624844294
1.03 -56.384170556045497 -56.384170535410028
1.04 -56.384063259962922 -56.384063238865402
1.05 -56.383816445018773 -56.383816430207141
1.06 -56.383439305849919 -56.383439268545615
time 125 s 372 sThe larger the molecule, the more expensive numerical gradients will be.
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