Page 15 - 71 the abc of dft_opt

2.2. FUNCTIONAL DERIVATIVES 29 30 CHAPTER 2. FUNCTIONALS
to this increase, i.e., we add an infinitesimal change δr(θ) = &δ(θ − θ 0 ). 2 How does A[r] The first integral is in just the right form for identifying the contribution to the functional
change? derivative, but the second is not. But we can perform an integration by parts to find
 
1 1 2π & 2 ' 1 2π δP r d r #
A[r + δr] − A[r] = dθ (r + &δ(θ − θ 0 )) − r 2 = dθ r&δ(θ − θ 0 ) = & r(θ 0 ) (2.7) = p[r](θ) = √ −  √  (2.12)
2
2 0 0 δr(θ) r + r #2 dθ r + r #2
2
The functional derivative, denoted δA/δr(θ) is just the change in A divided by &, or just r(θ) Note that the end-point term of the integration-by-parts vanishes, because r(θ) is periodic.
in this case. Since this is linear in the change in r, the general definition, for any infinitesimal A little calculus and algebra finally yields
change in r, is just 3 #2
2 ##
2π δA
1 r + 2rr − r r )
A[r + δr] − A[r] = dθ δr(θ), (2.8) p[r](θ) = 2 #2 3/2 (2.13)
0 δr(θ) (r + r )
where the functional derivative δA/δr(θ) is that function of θ which makes this formula Exercise 5 Derivative of a semilocal functional !
∞
exact for any small change in r(θ). Just as the usual derivative df/dx of a function f(x) Show that the functional derivative of a semilocal functional B[n] = −∞ dx b(n, dn/dx),
tells you how much f changes when x changes by a small amount, i.e., f(x + dx) − f(x) = assuming n and its derivatives vanish rapidly as |x| → ∞, is
2
(df/dx) dx + O(dx ), so does the functional derivative, a function, tell you how much a ∂b d < ∂b =
functional changes when its arguments changes by a small “amount” (in this case, a small v B [n](x) = − (2.14)
∂n dx ∂n #
function). Thus
#
δA[r] where n (x) = dn/dx. Use your answer to show
v A [r](θ) ≡ = r(θ), (2.9)
δr(θ) δT VW n ## n # 2
S = − + . (2.15)
where the notation v A [r](θ) emphasizes that the functional derivative of A[r] is a θ-dependent δn(x) 4n 8n 2
functional.
The density functionals we use in practice are three-dimensional:
Exercise 4 Derivative of a local functional Exercise 6 Hartree potential
!
Show that, for any local functional A[n] = dx a(n(x)), the functional derivative is δA/δn(x) = The Hartree (or classical electrostatic) energy of a charge distribution interacting with itself
a (n(x)), where a (n) = da/dn.
#
#
via Coulomb’s law is given by
In general, the way to find a functional derivative is to evaluate the expression A[r+δr]−A[r] 1 1 3 1 n(r) n(r )
#
U[n] = d r d r . (2.16)
3 #
to leading order in δr, and the resulting integral must be cast in the form of a function times 2 |r − r |
#
δr. Often it is necessary to do an integration by parts to identify the functional derivative.
Show that its functional derivative, the Hartree potential, is
For the perimeter, such complications arise. We have
n(r )
1 #
1 2π % v H [n](r) = d r . (2.17)
3 #
# 2
2
P[r + δr] = dθ (r + δr) + (r + δr ) (2.10) #
#
0 |r − r |
where r = dr/dθ and all functions are assumed to have argument θ. To first order in δr, we
#
2.3 Euler-Lagrange equations
find
 
2π rδr + r δr The last piece of functional technology we need for now is how to solve a constrained opti-
1 √ # #
2
P[r + δr] = dθ r + r #2  1 + 
2
0 r + r #2 mization problem, i.e., how to maximize (or minimize) one functional, subject to a constraint
 
2π r r
1 # imposed by another.
= P[r] + dθ  √ δr + √ δr #  (2.11)
2
0 r + r #2 r + r #2 For example, we might want to know what is the maximum area we can enclose inside a
2
loop of string of fixed length l. Thus we need to maximize A[r], subject to the constraint
2 For purists, we take a Gaussian of tiny but finite width, so that as ! → 0, the maximum change becomes arbitrarily small. Then we take the
width to zero! that P[r] = l. There is a well-known method for doing such problems, called the method