Questions: Time-Dependent DFT for Excited States

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

You run a TDDFT calculation on a donor-acceptor dye molecule using a GGA functional and find the predicted lowest excitation energy is 1.2 eV below the experimental absorption peak. Which explanation is most likely correct?

AThe GGA functional overestimates the exchange-correlation energy, artificially stabilizing the excited state
BThe calculation is using the wrong basis set, and a larger basis set would correct the error
CThe transition is a charge-transfer excitation, and GGA functionals severely underestimate CT excitation energies due to insufficient long-range exchange
DTDDFT is not applicable to dye molecules because they absorb only in the visible range
Question 2 Multiple Choice

What does a TDDFT linear response calculation directly compute, and what does it NOT compute?

AIt computes the absolute total energy of excited states; it does not compute oscillator strengths
BIt computes excitation energies (energy differences from the ground state) and oscillator strengths; it does not compute the absolute energy of excited states
CIt computes excited-state wavefunctions explicitly; it does not use the ground-state density
DIt computes excited-state geometries; it does not predict absorption wavelengths directly
Question 3 True / False

Standard GGA TDDFT functionals are equally reliable for valence excitations (local transitions) and charge-transfer excitations in large molecules.

TTrue
FFalse
Question 4 True / False

TDDFT requires only a completed ground-state Kohn-Sham DFT calculation as input, making it far less computationally expensive than wavefunction-based excited-state methods.

TTrue
FFalse
Question 5 Short Answer

Why do range-separated hybrid functionals perform significantly better than GGA functionals for charge-transfer excitations in TDDFT, and what physical feature of GGA functionals causes the failure?

Think about your answer, then reveal below.