In this example, we will solve for the eigenroots and eigenvalues of the Time-Dependent Density Functional Theory (TDDFT) Matrix which corresponds to single electron excitation states. Most commonly, this method is used to build up the 0K UV-Vis spectrum of a molecular system.
In order to run a LR-TDDFT calculation with ChronusQ, let's start with a trivial system, LiH+, as an example with open shell character.
First, let's look at the input file structure for a LR-TDDFT job.
[Molecule] charge = 1 mult = 2 geom: Li 0.0 0.0 0.0 H 0.0 0.0 1.2 [QM] reference = Real UB3LYP job = resp [BASIS] basis = 6-31g(d) [RESPONSE] type = residue
In this input we see the familiar
[MOLECULE] section where we define the LiH+ molecule, with a bond length of 1.2 Angstroms, a charge of +1 and a multiplicity of 2, to define the doublet spin state. Next, we use the
B3LYP DFT functional, and specify that this is an unrestricted calculation by setting
reference = Real UB3LYP. By default
Real will be included for a
UB3LYP calculation but we made it explicit anyway.
Next, in the
[QM] section we set
job = resp to specify that this is in general going to be a response calculation. The specific method can now be defined in a new section denoted
[RESPONSE]. Here we set the type of response job to a residue calculation, which gives us the eigenstates of the response matrix corresponding to TDDFT or TDHF.
Important: The first input file above corresponds to a full diagonalization of the response matrix, not an iterative solution.
In order to do an iterative calculation we can use the following input file to solve for the first 3 eigenroots.
[Molecule] charge = 1 mult = 2 geom: Li 0.0 0.0 0.0 H 0.0 0.0 1.2 [QM] reference = Real UB3LYP job = resp [RESPONSE] type = residue dofull = false nroots = 3 [BASIS] basis = 6-31g(d)
As you can see, the two additional keywords are in the
[RESPONSE] section. First, we turn off the full diagonalization using
dofull = false, and next, we specify that we want to solve for 3 eigenroots with
nroots = 3. (In this example we were explicit with the number of roots, but for an iterative diagonalization, the default value is in fact 3 roots.)
Now let's take a look at the output. After the SCF completes, we see that the TDDFT part of the calculation begins. Upon completion we get output data in the following form for the first 3 roots.
RESIDUE RESPONSE RESULTS * RESIDUE EIGENMODES Root 1: W(Eh) = 0.39880715 W(eV) = 10.85209926 f = 0.00500041 MO Contributions: 2A -> 3A -0.99899 Root 2: W(Eh) = 0.41674549 W(eV) = 11.34022649 f = 0.10929524 MO Contributions: 2A -> 4A -0.56282 2A -> 5A -0.82338 Root 3: W(Eh) = 0.41674618 W(eV) = 11.34024540 f = 0.10929521 MO Contributions: 2A -> 4A 0.82338 2A -> 5A -0.56282 . . .
Specifically the energies for each state reported in both Hartrees and eV,
W(eV) respectively. Oscillator strengths are reported on the next line below denoted by
f. Finally, for each root the dominant MO contribution coefficients are printed. For example, in Root 3 we see 2 major transitions
MO Contributions: 2A -> 4A 0.82338 2A -> 5A -0.56282
from the second
\alpha orbital (2A) to the fourth and fifth
\alpha orbitals (4A and 5A).