Page 18 - 71 the abc of dft_opt

3.3. THREE DIMENSIONS 35 36 CHAPTER 3. ONE ELECTRON
ˆ
and minimize the expectation value of H, subject to the constraint that φ be normalized. 3.4 Differential equations
This leads to a set of linear equations
We have just seen how to construct various trial wavefunctions and find their parameters by
||H − ES|| = 0, (3.6) minimizing the energy. In a few lucky cases, we’ve found the exact solution to the given
problem.
where
1 ∞ But in the real world, where analytic solutions are few and far between, we cannot be
ˆ
H ij = dx φ (x)Hφ j (x) (3.7)
∗
i
−∞ relying on lucky guesses. When we’re given a variational problem to solve, we must find the
is the Hamiltonian matrix, and exact answer, and be sure we’ve done so.
1
∞ Happily, we can do so, by applying the techniques of Chapter 2 to the variational problem of
S ij = dx φ (x)φ j (x) (3.8)
∗
i
−∞ Eq. (3.1). We choose the wavefunction real, and want to minimize E[φ], given the constraint
is the overlap matrix. In general, Eq. (3.6) is a generalized eigenvalue equation. Only if the that the wavefunction be normalized. Thus we must minimize:
basis of trial wavefunctions is chosen as orthonormal (Eq. 1.10) will S = 1, and the equation
be an ordinary eigenvalue equation. F[φ] = T[φ] + V [φ] − µ(N[φ] − 1) (3.11)
The standard textbook problem of chemical bonding is the formation of molecular orbitals where N[φ] = dx φ (x), and µ is a Lagrange multiplier. Taking functional derivatives, we
2
!
+
from atomic orbitals in describing the molecular ion H 2 . find:
+ δT ## δV δN
Exercise 12 1-d H 2 = −φ (x), = 2v(x) φ(x), = 2φ(x) (3.12)
δφ(x) δφ(x) δφ(x)
For a potential V (x) = −δ(x) − δ(x + a), use φ mol (x) = c 1 φ E (x) + c 2 φ E (x + a) as a trial
wavefunction, and calculate the bonding and antibonding energies as a function of atomic Thus
δF
separation a. = −φ (x) + 2(v(x) − µ)φ(x) (3.13)
##
δφ(x)
At the minimum, this should vanish, yielding:
3.3 Three dimensions
1
− φ (x) + v(x) φ(x) = µ φ(x) (3.14)
##
There are only slight complications in the real three-dimensional world. We will consider 2
only spherical 3-d potentials, v(r), where r = |r|. Then the spatial parts of the orbitals are and we can identify µ = E 0 by integrating both sides with φ(x).
characterized by 3 quantum numbers:
This should all make a lot of sense. From the Rayleigh-Ritz variational principle, we have
φ nlm (r) = R nl (r)Y lm (θ, ϕ), (3.9) rededuced the Schr¨odinger equation. Note however that all eigenstates satisfy the Schr¨odinger
equation, but only the ground state satisfies the variatonal principle (all others are only local
where the Y lm ’s are spherical harmonics. Inserting (3.9) into the Schr¨odinger Equation gives
minima).
the radial equation
The general rule is that, given a variational principle, trial wavefunctions can be used to
 
 1 d 2 d l(l + 1)  get approximate answers, but the exact solution can often be found by solving the differential
− r + + v(r) R nl (r) = ε nl R nl (r). (3.10)
2r dr dr 2r
 2 2  equation that results from extremizing the functional, given the constraints.
These are written about in almost all textbooks on quantum mechanics, with emphasis on the
Coulomb problem, v(r) = −Z/r. Less frequently treated is the three-dimensional harmonic 3.5 Virial theorem
2
oscillator problem, v(r) = kr /2.
We close this chapter with a statement, without proof, of the virial theorem in one dimension.
Exercise 13 Three dimensions The proof will be given later in the book. The virial theorem says that, for an eigenstate of
Use the trial wavefunctions of an exponential and a Gaussian in r to deduce the ground-state a given potential,
energy and orbital of (a) the hydrogen atom (Z = 1), and (b) the harmonic oscillator with dV
2T = % x &. (3.15)
k = 1. dx