Questions: Introduction to Density Functional Theory: From Wavefunctions to Electron Density

5 questions to test your understanding

Score: 0 / 5
Question 1 Multiple Choice

A colleague claims that DFT is fundamentally less rigorous than wavefunction methods because it avoids the exact many-body Schrödinger equation. Which response best refutes this claim?

ADFT is equally rigorous because it uses the same Hartree-Fock equations, just reformulated in terms of density
BThe Hohenberg-Kohn theorems are exact — the ground-state energy is provably a unique functional of the electron density, so DFT's theoretical foundation is as rigorous as any wavefunction method
CDFT is less rigorous, but this is acceptable because computational efficiency outweighs theoretical exactness
DDFT avoids the Schrödinger equation through empirical fitting of density functionals to experimental data
Question 2 Multiple Choice

In Kohn-Sham DFT, the real system of interacting electrons is replaced by a fictitious system of non-interacting electrons with the same density. What is the purpose of this substitution?

ATo avoid the Pauli exclusion principle, which makes multi-electron wavefunctions antisymmetric and hard to compute
BTo reduce the system to a single electron, which can be solved exactly with the hydrogen atom solution
CTo allow the kinetic energy and classical Coulomb energy to be computed straightforwardly, leaving only the unknown exchange-correlation energy to be approximated
DTo convert the wavefunction into an orbital-free representation where no basis set is needed
Question 3 True / False

The Hohenberg-Kohn theorems prove that the exact ground-state energy is a unique functional of the electron density; the only source of error in practical DFT calculations is the approximate exchange-correlation functional.

TTrue
FFalse
Question 4 True / False

A more expensive functional (e.g., a hybrid functional) generally gives more accurate results than a cheaper one (e.g., GGA) for any molecular property.

TTrue
FFalse
Question 5 Short Answer

Why does replacing the 3N-variable many-electron wavefunction with the three-variable electron density provide a computational advantage in DFT? What is the key theoretical result that makes this substitution valid?

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