Page 102 - 71 the abc of dft_opt

204 APPENDIX E. DISCUSSION OF QUESTIONS
Both answers have value, but reflect different priorities. The chemist’s answer is likely
to be more accurate than the physicist’s for problems that are similar to the fitted one.
Errors are likely to be both positive or negative. On the other hand, the physicist’s answer
is likely to be more accurate over a broad range of problems, with systematic errors (but
Appendix E larger than the chemist’s on the fitted systems).

6. In what way will T loc change if spin is included, e.g., for two electrons of opposite
S
Discussion of questions spin in a box?
A detailed answer is given in section 9.2. Here we just doubly occupy each level in the
box. Thus...(Rudy, fill in).
Chapter 1 loc
7. If we add an infinitesimal to n(x) at a point, i.e., &δ(x) as & → 0, how does T
S
change?
1. Why is a Kohn-Sham calculation much faster than a traditional wavefunction calcu-
lation? Again, a detailed answer is given in section 2.2. But we can figure it out easily in this
case. Rudy...
Because a Kohn-Sham calculation requires solving only a one-electron problem (self-
consistently) and occupying N levels, rather than solving a coupled differential equation 8. The simplest density functional approximation is a local one. What form might a cor-
for a function of 3N variables. rection to T loc take?
S
2. If you evaluate the kinetic energy of a Kohn-Sham system, is it equal to the physical A detailed answer is given in section 17.2. But we can make intelligent suggestions here.
kinetic energy? The most obvious is that, since using the density at a point works so robustly, surely
including information on the gradient of the density can help accuracy. Rudy...
No. The kinetic energy operator is the same for the two systems, but their wavefunctions
are different. loc
9. In the chapter and exercises, T does very well. But there are cases where it fails
S
3. Repeat above question for %1/r&, which can be measured in scattering experiments. quite badly. Can you find one, and say why?
3
Yes. %1/r& = ! d r n(r)/r, and so depends only on the one-electron density, which is Consider a particle in a large box, at whose center is a tall but finite barrier. Ignoring the
defined to be the same in both systems. exponentially small tunneling contribution, the lowest molecular orbital is
√
4. Why is the density cubed in the local approximation for T S ? (see section ?? for the φ(x) = (φ A (x) + φ B (x))/ 2 (E.1)
answer).
where φ A (x) is the ground-state wavefunction for the electron to be in the left side,
Performing a dimensional analysis, we note that n ∼ 1/L in one dimension, and since and φ B (x) is that for the right. The exact kinetic energy is then (T A + T B ) = T A by
1
2
p
T ∼ 1/L = dxn (x), p must be 3. 2
!
symmetry. On the other hand, our local functional uses n(x) = (n A (x) + n B (x))/2, so
3
5. Suppose you’d been told that T loc is proportional to ! dxn (x), but were not told the that T S loc = (T A loc + T B loc )/8 = T A loc /4, i.e., it is about 4 times too small!
S
constant of proportionality. If you can do any calculation you like, what procedure 10. Does T loc [n ] work for a single electron in an excited state, with density n (r)?
∗
∗
might you use to determine the constant in T loc ? S
S loc loc
For ANY single excited state of the particle in the box, T [n ] = T [n], so it fails very
∗
A physicist might answer that, since T S becomes more and more accurate as the number S S
badly in this form. Need to show explicit failure for particle in box. Rudy...
of electrons in the box grows, take the limit as N → ∞. We saw in the chapter how that
2
yields the π /6 constant. On the other hand, a practical chemist might answer that, if However, Rudy points out that, since our electrons are not interacting, he can find the
∗
all I want to do is solve problems that look like one electron in a box, I should simply fit energy of the first excited state by writing n (x) = n 2 (x) − n 1 (x), where n i (x) is the
2
loc
loc
∗
the constant to the value for that problem. density with i particle in the box. Then T[n ] = T 2 − T 1 = 17.7/L
203