1. Atomic Structure

2. Quantization Due to confinement, like acoustic waves on a string
Same for H-atom
Vacuum
-0.85 eV
-1.51 eV
U(r) = -Zq2/4pe0r
-3.4 eV
-13.6 eV

3. Particle in a Box Results agree with analytical results E ~ n2
Finite wall heights, so waves seep out

4. Generalizing to 3D

5. Recap 2D m Functions Angular nodes promote orthogonality along Q l
Q(j) = exp(imj) Sinusoids m
R(r) = Jn(xmnr/a) Bessel
Functions
Angular nodes promote orthogonality along Q l

6. How about 3D? q f R ^ 3 quantum #s n for radial R l for longitude q
m for latitude f R q f ^

7. Hydrogen Atom F = Rnl(r)/r x Ylm(q,j) For given n, l = 0, 1, 2, …, n-1
(called s, p, d, f, …)
m = -l, -(l-1), -(l-2), … ,l-2, l-1, l
(px, py, pz, dxy, dyz, dxz… “Orbitals”)
2l+1 of multiplets

8. Hydrogen Atom (Variable separation in radial coords)
-(ħ22/2m)F = -ħ2/2m[1/r.∂2(rF)/∂r2] +
+1/2mr2[-ħ2{1/sinq.∂/∂q(sinq∂F/∂q)
+ 1/sin2q.∂2F/∂j2}] L2 ˆ
p2/2m  L2/2I

9. Angular Momentum eigenstates
Lz  Lz ˆ
L2  L2 ˆ
z-component
Quantized mħ
m = -l, -(l-1),...
(l-1), l
Angular Mom
Quantized
l(l+1)ħ2
l = 0, 1, 2,.. (n-1)
Also, Lx, Ly, Lz are not simultaneously measurable

10. Angular space (Orbitals)
Check  they satisfy
L2Ylm = l(l+1)ħ2Ylm
Y00(q,j)
Indep. of
angle
(l=0)
L2 = -ħ2[1/sinq.∂/∂q(sinq.∂/dq)+ 1/sin2q.∂2/∂j2]
Lz = -iħ∂/∂j

11. Angular space (Orbitals)
Check  they satisfy
L2Ylm = l(l+1)ħ2Ylm
E.g. L2cosq = 2ħ2cosq
1st order polynomials
L2 = -ħ2[1/sinq.∂/∂q(sinq.∂/dq)+ 1/sin2q.∂2/∂j2]
Lz = -iħ∂/∂j
l = 1 (“p” Orbitals), m=-1, 0, 1
Px = sinqcosj
Y11(q,j) +
Y1,-1(q,j)
Py = sinqsinj
Y11(q,j) -
Y1,-1(q,j)
Pz = cosq
Y10(q,j)

12. d orbitals (l = 2) dx2-y2 = cos2jsin2q Y22(q,j) + Y2,-2(q,j)
dxy = sin2jsin2q
Y22(q,j) -
Y2,-2(q,j)
dxz = cosjsin2q
Y21(q,j) +
Y2,-1(q,j) x y z x y z x y z
dyz = sinjsin2q
Y21(q,j) -
Y2,-1(q,j)
d3z2-r2 = 3cos2q-1
Y20(q,j) x y z x y z
2nd order polynomials

13. What about radial part? i.e, we use F = R(r)Ylm(q,j)/r Then
Centrifugal
Barrier Uc
i.e, we use F = R(r)Ylm(q,j)/r
Then
-ħ2/2m.d2R/dr2 + [U(r)+l(l+1)ħ2/2mr2]R = ER
Ueff(r)
Ueff U Uc r

14. Why is this centrifugal?
L = mvr (constant)
Fc = mv2/r = L2/mr3
Uc = -∫Fdr = L2/2mr2
L2  L2  l(l+1)ħ2
Ueff(r) = –Zq2/4pe0r + l(l+1)ħ2/2mr2

15. What about R/r? i.e, we use F = R(r)Ylm(q,j)/r Then
-ħ2/2m.d2R/dr2 + [U(r)+l(l+1)ħ2/2mr2]R = ER
Looks like 1D.
Then why R/r?
Normalization over r2dr
∫(R/r)2.r2dr = 1
 ∫R2.dr = 1

16. Hydrogen Atom -ħ2/2m.d2R/dr2 + [U(r)+l(l+1)ħ2/2mr2]R = ER
Centrifugal
Barrier Uc
-ħ2/2m.d2R/dr2 + [U(r)+l(l+1)ħ2/2mr2]R = ER
Effective 1-D problem
with centrifugal barrier
(we know how to solve
this numerically with
Finite difference!)
Ueff(r)
Ueff U Uc r

17. Hydrogen Atom [-(ħ2d2/dr2)/2m + Ueff(r)]R = 0
Ueff(r) = –Zq2/4pe0r + l(l+1)ħ2/2mr2
l = 0, 1, 2, ...
Ueff U Uc r

18. Numerical Solution y yn-1 yn yn+1 xn-1 xn xn+1 Hy = y = -t 2t+U1 -t

19. Matlab code t=1; Nx=101;x=linspace(-5,5,Nx);
q=1.609e-19;e0=4.805e-12;hbar=1.053e-34; m=9.1e-31;
l=0;
%U=[-q./(4*pi*e0*x) + l*(l+1)*hbar^2/(2*m*x.^2*q)];% Coulomb + centrifugal
%Write matrices
T=2*t*eye(Nx)-t*diag(ones(1,Nx-1),1)-t*diag(ones(1,Nx-1),-1); %Kinetic Energy
U=diag(U); %Potential Energy
H=T+U;
[V,D]=eig(H);
[D,ind]=sort(real(diag(D)));
V=V(:,ind);
% Plot
for k=1:5
plot(x,V(:,k)+10*D(k),'r','linewidth',3)
hold on
grid on
end
plot(x,U,'k','linewidth',3); % Zoom if needed
axis([ ])

20. Hydrogen Atom What do finite difference 1D radial solutions look like?
Depends on l
F = Rnl(r)/r x Ylm(q,j)
R(r)/r
P orbitals
(l=1)
Particle stays away
from nucleus
(n-2) nodes
2p eV
3p eV
4p eV
D orbitals
(l=2)
n-3 nodes
3d eV
4d eV
S orbitals
(l=0)
Particle falls in
(No centrif.
barrier)
n-1 nodes
1s eV
2s eV
3s eV
Angular orthogonality between s, p, d due to angular nodes in Ylm
Radial orthogonality within s, p, d set due to (n-l-1) radial nodes in R

21. Hydrogen Atom F = Rnl(r)/r x Ylm(q,j) R(r)/r
Rnl/r ~ rl.e-Zr/na0.L (r/a0)
n-1
l-1
n gives ‘size’, l gives ‘angular momentum’, m gives z component

22. Hydrogen Atom Rnl/r ~ rl.e-Zr/na0.L (r/a0) R(r)/r
Size of atom given by peak of R(r) envelope
r* = nla0/Z

24. So how did Bohr fare? 1. Allowed levels don’t radiate
They are eigenstates, so fixed energy
2. mvr = nħ
LzYlm (q,j) = mħYlm(q,j)
3. mv2/r = Zq2/4pe0r2
Ueff(r) = –Zq2/4pe0r + L2/2mr2
∂Ueff/∂r = 0

25. Hydrogen Energy Levels
Numerical results
1s eV
2s, 2p eV
3s, 3p, 3d eV
Bohr’s results
En = (Z2/n2)E0
E0 = -mq4/8ħ2e0 = eV
E1 = -3.4 eV
E2 = eV
(Grid issues – fine, but enough # points)
Accidentally, energy of 2s, 2p same
Similarly for 3s, 3p, 3d
(ie, E independent of l, dep. only on n)
True only for Hydrogen !

26. Radial Solution for Coulomb electron
[-(ħ2d2/dr2)/2m + Ueff(r)]f = Ef
Ueff(r) = –Zq2/4pe0r + l(l+1)ħ2/2mr2
fnl ~ rl+1e-Zr/na0f(r/a0)
Try f(r) = Spcprp
cp/cp-1 = [(2p+2l)/l-1/a0]/p(p+2l+1)
For series to be finite, need
(2p+2l)/l-1/a0 = 0
E = -ħ2/2ml2 = -(ħ2/2ma02) x 1/(2p+2l)2
= -4E0/4(p+l)2 = -E0/n2
= E0/(n-1/2)2
Twice the energy of 3D Hydrogen !!

27. For most atoms Notice energy depends not just on n, but on n+l
So 4s has lower energy than 3d
(“n+l” rule)