Page 11 - 71 the abc of dft_opt

1.5. PARTICLE IN A BOX 21 22 CHAPTER 1. INTRODUCTION
The time-independent Schr¨odinger equation has solutions: Exercise 2 Three particles in a box
ˆ
Hφ i = ε i φ i , i = 0, 1, ... (1.9) Calculate the exact energy of three identical fermions in a box of length L. Plot the total
density for one, two, and three particles. Compare the exact energy with that from the local
Thus ε 0 denotes the ground state energy and φ 0 the ground state wave function. Because approximation. What happens as the number of electrons grows? Answer: E = 7π /L ,
2
2
ˆ
the operator H is hermitian the eigenstates can be chosen orthonormal: E loc = 63/L , a 9% underestimate.
2
1
∞
∗
dx φ (x)φ j (x) = δ ij (1.10)
i
These results are gotten by the evaluation of a simple integral over the density, whereas
−∞
Thus each eigenfunction is normalized. The electron probability density is just n(x) = the exact numbers require solving for all the eigenvalues of the box, up to N, the number of
2
|φ 0 (x)| . electrons in it. Thus approximate density functionals produce reliable but inexact results at
Exercise 1 Particle in a one-dimensional box a fraction of the usual cost.
Consider the elementary example of a particle in a box, i.e., V (x) = ∞ everywhere, except To see how remarkable this is, consider another paradigm of textbook quantum mechanics,
0 ≤ x ≤ L, where V = 0. This problem is given in just about all elementary textbooks on namely the one-dimensional harmonic oscillator. In the previous example, the particles were
quantum mechanics. Show that the solution consists of trigonometric functions (standing sine waves; in this one, they are Gaussians, i.e., of a completely different shape. Applying
waves), and making them vanish at the boundaries quantizes the energy, i.e., the same approximate functional, we find merely a 20% overestimate for one particle:
4
5 2
5
φ i = 6 sin(k i x), i = 1, 2, ... (1.11) Exercise 3 Kinetic energy of 1-d harmonic oscillator
L What is the ground-state energy of a 1-d harmonic oscillator? What is its kinetic energy?
2
where k i = πi/L, and the energies are & i = k /2. Check the orthonormality condition, Eq. Calculate its density, and from it extract the local density approximation to its kinetic energy.
i
(1.10). Repeat for two same-spin electrons occupying the lowest two levels of the well.
Next we consider the following approximate density functional for the kinetic energy of
non-interacting electrons in one-dimension. Don’t worry if you never heard of a functional Congratulations. You have performed your first elementary density functional calculations,
before. In fact, chapter ?? discusses functionals in some detail. For now, we need only the and are well on the way to becoming an expert!
fact that a functional is a rule which maps a function onto a number. We use square brackets 14
[..] to indicate a functional dependence. The functional we write down will be the local
approximation to the kinetic energy of non-interacting spinless fermions in one dimension. 12
We do not derive it here, but present it for calculational use, and derive it later in the chapter: 10
3
T loc [n] = 1.645 1 ∞ dx n (x). (1.12) 8
S
−∞
A local functional is one which is a simple integral over a function of its argument. Now, 6 10 particles in a box
since for the particle in a box, all its energy is kinetic, we simply estimate the energy using 4
T loc [n]. Since the electron density is the square of the orbital:
S
2
2
2
n 1 (x) = sin (k 1 x), (1.13) 0
L 0 0.2 0.4 0.6 0.8 1
2
2
2
2
we find T loc = 4.11/L , a 17% underestimate of the true value, π /2L = 4.93/L . What Figure 1.4: Density of 10 particles in a box.
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is so great about that? There is not much more work in finding the exact energy (just take
the second derivative of φ 1 , and divide by φ 1 ) as there is in evaluating the integral. In fact, there’s a simple way to see how this magic has been performed. Imagine the box
But watch what happens if there are two particles, fermions of the same spin (the general with a large number of particles. In Fig. 1.4, we have plotted the case for N = 10. As the
2
2
2
case for which T loc was designed). We find now E = & 1 + & 2 = 5π /2L = 24.7/L . If we number of particles grows, the density becomes more and more uniform (independent of the
S
2
evaluate the approximate kinetic energy again, using n(x) = n 1 (x) + 2 L sin (k 2 x), we find shape of the box!). If we assume it actually becomes uniform as N → ∞, with errors of
2
T loc = 21.8/L , an 11% underestimate. order 1/N or smaller, this gives us the constant in the local approximation. To see this, note
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