Linear Response TDDFT

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.

Input File

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.)

Output

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(Eh) and 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).