Dear Developers,
Is there a way to obtain the solid formation of atomic orbitals used in a calculation? For example, following the H2O example in Psi4,
import psi4
h2o = psi4.geometry("""
O
H 1 0.96
H 1 0.96 2 104.5
symmetry c1
units angstrom
""")
psi4.set_options({'reference': 'rhf', 'basis': 'def2-SVP'})
scf_e, scf_wfn = psi4.energy('scf', molecule=h2o, return_wfn=True)
scf_wfn.get_basisset('ORBITAL').print_detail_out()
will return the following information,
-AO BASIS SET INFORMATION:
Name = DEF2-SVP
Blend = DEF2-SVP
Total number of shells = 12
Number of primitives = 22
Number of AO = 25
Number of SO = 24
Maximum AM = 2
Spherical Harmonics = TRUE
-Contraction Scheme:
Atom Type All Primitives // Shells:
------ ------ --------------------------
1 O 7s 4p 1d // 3s 2p 1d
2 H 4s 1p // 2s 1p
3 H 4s 1p // 2s 1p
==> AO Basis Functions <==
[ DEF2-SVP ]
spherical
****
O 1
S 5 1.00
2266.17677850 -0.00534318
340.87010191 -0.03989004
77.36313517 -0.17853912
21.47964494 -0.46427685
6.65894331 -0.44309745
S 1 1.00
0.80975976 1.00000000
S 1 1.00
0.25530772 1.00000000
P 3 1.00
17.72150432 0.04339457
3.86355054 0.23094121
1.04809209 0.51375311
P 1 1.00
0.27641544 1.00000000
D 1 1.00
1.20000000 1.00000000
****
H 2
S 3 1.00
13.01070100 0.01968216
1.96225720 0.13796524
0.44453796 0.47831935
S 1 1.00
0.12194962 1.00000000
P 1 1.00
0.80000000 1.00000000
****
H 3
S 3 1.00
13.01070100 0.01968216
1.96225720 0.13796524
0.44453796 0.47831935
S 1 1.00
0.12194962 1.00000000
P 1 1.00
0.80000000 1.00000000
****
It is obvious that 25 AOs were used, while scf_wfn.basisset().compute_phi(0, 0, 0).shape
will return (24,)
, which seems to be the number of SO. Is there a way to obtain the details of atomic orbitals?