Page 63 - 71 the abc of dft_opt

16.2. HOOKE’S ATOM 125 126 CHAPTER 16. EXCHANGE-CORRELATION HOLE
Exercise 72 Spin-scaling the hole where R = (r 1 + r 2 )/2, u = r 2 − r 1 , Φ(R) is the ground-state orbital of a 3-d harmonic
Deduce the spin-scaling relation for the exchange hole. oscillator of mass 2, and φ(u) satisfies
 2 
Exercise 73 Holes for one-electron systems  2 ku 1 
u
For the H atom (in 3d), plot P(r, r ), n 2 (r, r ) and n X (r, r ) for several values of r as a  −∇ + 4 + u  φ(u) = & u φ(u) (16.31)
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function of r , keeping r parallel to r.
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Exercise 76 Hooke’s atom exactly
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The correlation hole is everything not in the exchange hole. Since the exchange hole Show that the function φ(u) = C(1 + u/2) exp(−u /4) satisfies the Hooke’s atom equation
satisfies the sum-rule, the correlation hole must integrate to zero: (16.31) with k = 1/4, and find the exact ground-state energy. How big was your error in
1 3 your estimated energy above? Make a rigorous statement about the correlation energy.
d u n C (r, u) = 0. (16.27)
This means the correlation hole has both positive and negative parts, and occasionally the Exercise 77 Hooke’s atom: Exchange and correlation holes
sum of exchange and correlation can be positive. It also has a universal cusp at u = 0, due Use the exact wavefunction for Hooke’s atom at k = 1/4 to plot both n X (r, u) and n C (r, u)
to the singularity in the electron-electron repulsion there. for r = 0 and 1, with u parallel to r. The exact density is:
2
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An alternative way of representing the same information that is often useful is the pair- n(r) = √ 2C exp(−r /2)/r ×
correlation function: # 3 2 √ 2 √ $
7r + r + 8r exp(−r /2)/ 2π + 4(1 + r )erf(r/ 2) (16.32)
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g(x, x ) = P(x, x )/(n(x)n(x )) (16.28)
This pair-correlation function contains the same information as the hole, but can make some where the error function is
results easier to state. For example, at small separations, the Coulomb interaction between 2 1 x 2
erf(x) = √ dy exp(−y ) (16.33)
the electons dominates, leading to the electron-electron cusp in the wavefunction: π 0
√
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dg sph.av. (r, u)/du| u=0 = g sph.av. (r, u = 0), (16.29) and C −1 = 2 5π + 8 π
where the superscript denotes a spherical average in u. Similarly, at large separations, g → 1 Exercise 78 Hooke’s atom: Electron-electron cusp
as u → ∞ in extended systems, due to the screening effect, but not so for finite systems. Repeat the above exercise for the pair-correlation function, both exchange and exchange-
Furthermore, the approach to unity differs between metals and insulators. correlation. Where is the electron-electron cusp? What happens as u → ∞?
16.2 Hooke’s atom 16.3 Transferability of holes
A useful and interesting alternative external potential to the Coulomb attraction of the nucleus The pair density looks very different from one system to the next. But let us consider two
for electrons is the harmonic potential. Hooke’s atom consists of two electrons in a harmonic totally different systems: the 1-d H-atom and the (same-spin) 1-d uniform electron gas. For
well of force constant k, with a Coulomb repulsion. It is useful because it is exactly solvable, any one-electron system, the pair density vanishes, so
so that many ideas can be tested, and also (more importantly), many general concepts can
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be illustrated. n X (r, r ) = −n(r ) (N = 1). (16.34)
Exercise 74 Hooke’s atom: Approximate HF For our 1d H atom, this is just n(x + u), where n(x) = exp(−2|x|), and is given by the solid
Repeat the approximate HF calculation of chapter ?? on Hooke’s atom, using appropriate curve in Fig. 16.2, with x = 0. On the other hand, we can deduce the exchange hole for
orbitals. Make a definite statement about the ground-state and HF energies. the uniform gas from the bulk value of the first-order density matrix. From Eq.(16.7), taking
x, x to be large, we see that the density matrix becomes, in the bulk,
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Exercise 75 Hooke’s atom: Separation of variables
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Show that, for two electrons in external potential v ext (r) = kr /2, and with Coulomb inter- γ unif (u) = n sin(k F u)/(k F u) (16.35)
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action, the ground-state wavefunction may be written as
leading to
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Ψ(r 1 , r 2 ) = Φ(R)φ(u), (16.30) n X (u) = −n sin (k F u)/(k F u) 2 (16.36)