Manual specification of double hybrids

Questions
#1

Hello,
I am fairly new to using Q-Chem and quantum chemistry in general. Is there a way to manually specify the ωB2PLYP and ωB2GPPLYP functionals (https://pubs.acs.org/doi/pdf/10.1021/acs.jctc.9b00013) in Q-Chem?

Thanks,

Manual specification of DH-functionals
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#2

It is possible to use Q-Chem for TD-DHDF with a bit of post-processing. However, ωB2PLYP and ωB2GPPLYP require ωB88 exchange, which, as of 5.3.2, is not available in Q-Chem.

#3

The ωB88 exchange functional is actually available in Q-Chem under the name muB88 following the original name from Yanai, Tew, and Handy’s CAM-B3LYP paper.

Here is an example of setting up ωB2PLYP for the formaldehyde molecule (structure taken from the Gordon benchmark set used in https://pubs.acs.org/doi/suppl/10.1021/acs.jctc.9b00013/suppl_file/ct9b00013_si_001.pdf).

$rem
exchange = gen
omega = 300
lrc_dft = 1
hfk_sr_coef =  53000000
hfk_lr_coef = 100000000
correlation = rimp2
basis = aug-cc-pvtz
aux_basis = rimp2-aug-cc-pvtz
scf_convergence = 10
cis_singlets = 1
cis_triplets = 0
cis_n_roots = 20
$end

$xc_functional
k 0.53
x mub88 0.47
c lyp 0.73
$end

$molecule
0 1
C 0.00000000 0.00000000 -0.00545460
O 0.00000000 0.00000000 1.18994350
H -0.93949810 0.00000000 -0.59224440
H 0.93949810 0.00000000 -0.59224440
$end

This job will not scale MP2/(D) correlation, which must be done as a post-processing step. Collect ground-state and excited-state energies:

 SCF   energy in the final basis set =     -114.2321414733
 Total  RIMP2   correlation energy =        -0.4793853081 au
        RIMP2         total energy =      -114.7115267814 au

 Excited state   1: excitation energy (eV) =    4.2287
 Total energy for state  1:                  -114.07673804 au
 Excited state   1: excitation energy (eV) =    3.6294
    Total energy for state   1:   -114.578151315737 au
    CIS(D) correction         :     -0.022027970205 au

and rescale RI-MP2/(D) corrections with a factor of 0.27 for ωB2PLYP (0.36 for ωB2GPPLYP) as follows:

E(GS) = -114.2321414733 + 0.27*(-0.4793853081) = -114.361575506487 (au)
E(ES1) = -114.07673804 + 0.27*(-0.022027970205) = -114.212119625142 (au)
E_ex(ES1) = [-114.212119625142 - (-114.361575506487)]*27.211 = 4.067 (eV)

Repeating this for every state yields the following excitation energies. Here I provide a comparison with excitation energies listed for formaldehyde in Table S4 of SI for 10.1021/acs.jctc.9b00013.

State Energy (a.u.) Excitation energy (eV) Reference result (eV)
11A1 (ground) -114.361575506487
11A2 -114.212119712155 4.0668 4.00
11B2 -114.102288040021 7.0555 7.06
21B2 -114.069871485128 7.9376 7.93
21A1 -114.067004795381 8.0156 7.99
21A2 -114.050872939833 8.4545 8.46
#4

Amazing. Thank you immensely!