# Why is time-dependent density-functional theory (TD-DFT) used to describe excited states? [closed]

It seems that TD-DFT is a common choice for a computationally efficient method to describe excited states. But if DFT is a single-configurational method, and hence a ground state method, how does the TD-DFT method treat the excited states?

• I do not have the time to give you that answer. But I do have a recommendation: Try to understand CIS (TDA) first. One can view TD-DFT as an extension to CIS if you like. Here is a good publication that (sort of) explains both: DOI: 10.1021/cr0505627. – AMT Dec 15 '16 at 17:32
• This must be the first time I've heard time-dependent DFT described as 'computationally efficient'! – Jon Custer Dec 16 '16 at 16:15
• Less expensive than, say, CASSCF or multi-configurational coupled cluster? – Yoda Dec 17 '16 at 13:43
• Like most of DFT-related, because it works. – TAR86 Feb 2 '18 at 5:52

TD-DFT is used and developed because there is an excited state theorem, the Runge-Gross theorem, that is analogous to the Hohenberg-Kohn theorem for ground states.

(...) the Runge-Gross theorem shows that for a many-body system evolving from a given initial wavefunction, there exists a one-to-one mapping between the potential (or potentials) in which the system evolves and the density (or densities) of the system.

The Runge–Gross theorem provides the formal foundation of time-dependent density functional theory. It shows that the density can be used as the fundamental variable in describing quantum many-body systems in place of the wavefunction, and that all properties of the system are functionals of the density.

This means that any state of the system can be described by an electron density, not just the ground state. If I understand this correctly, the R-G theorem is more general than the H-K theorem, which can be derived from the R-G theorem by removing the explicit time dependence.

Here is a paywalled link to their original paper.

An interesting comment from the Wikipedia proof is

The proof relies heavily on the assumption that the external potential can be expanded in a Taylor series about the initial time. The proof also assumes that the density vanishes at infinity, making it valid only for finite systems.

In the case of applied fields, such as a slowly varying or oscillating electric field, these may be treated with perturbation theory rather than self-consistently; this is the same thing as the Taylor expansion. A consequence is that TD-DFT may not be valid in the presence of strong applied fields or a strong external potential.

• I don't think you were asking for how TD-DFT works from a computational level, you were only looking for the justification, so I have neglected to make the connection with TD-HF (RPA). – pentavalentcarbon Dec 16 '16 at 14:10

This is a complicated question you ask. You will most likely get an answer like the one provided by azag0. I would be very surprised if this helped you in any way.

I want to rephrase your question: "How does one get excitation energies from TD-DFT?"

I'm gonna assume that this is completely equivalent with the question you've asked, maybe with less of an philosophical touch.

If you ask me, the best path to understanding (in principle) what TD-DFT does is to understand CIS (TDA). You can see TD-DFT as an extension to CIS. This is not how you find it in text books, but the reason why you ask this question is somewhat related to the problem that you don't find it like this in text books. The way you asked the question tells me you have access to publications, maybe you are studying at an university, so I highly recommend a publication by Dreuw and Head-Gordon called "Single-Reference ab Initio Methods for the Calculation of Excited States of Large Molecules", Chem. Rev. 2005, 105 (11), 4009–4037. DOI: 10.1021/cr0505627.

If you do not understand this publication, I would come back and ask simpler questions first, questions one might be able to answer in a simple post on an online board.