1. What does an atom look like?

2. 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

3. What does L look like? L = r x p
Uncertainty prevents complete specification of
Lx, Ly, Lz
[x, px] = iħ and so on…
(easy to prove from definition of p)
This gives us
[Lx, Ly] = iħLz and so on…
Can choose only one (say, Lz) and also it turns out, L2

4. So what’s the best we can do?
So work only with
L2 = -ħ2{1/sinq.∂/∂q(sinq∂/∂q) + 1/sin2q.∂2/∂j2}
and
Lz = -iħ(x∂/∂y – y∂/∂x)
= -iħ∂/∂j
(like linear momentum in azimuthal coords)

5. The components of L and their eigenstates
Ylm(q,j): eigenstates of L2 and Lz with
eigen-indices l, m

6. Eigenstates of Lz Lz = -iħ∂/∂j LzY = -iħ∂/∂jY = CħY
Solution: Y = f(q)eiCj
Boundary condition: Y(j=0) = Y(j=2p)
C = integer m
Ylm = Plm(q)eimj

7. Range of values for m? L2 = Lx2 + Ly2 + Lz2
Max Lz = l, minimum = -l, with ‘graininess’ ħ
ie,
Lz = mħ, m = -l, -(l-1), -(l-2), …, (l-2), (l-1), l
(2l+1) possibilities

8. Range of values for L2? L2 = Lx2 + Ly2 + Lz2
Naively, we expect L2 = ħ2l2
But the commutator (ie, ordering need) for Lx and Ly
in terms of ħLz makes things a little more complex,
so that we get an extra ħl
L2 = l(l+1)ħ2

9. Angular momentum quantization
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)

10. Hydrogen Atom F = Rnl(r)/r x Ylm(q,j)
n: Principal Quantum Number (Size, r)
l: Azimuthal Quantum Number (Angle q)
m: Magnetic Quantum Number (Angles q, j)
s: Spin Quantum Number (up, down)

11. 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

12. 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]

13. Angular space (Orbitals)
Check  they satisfy
L2Ylm = l(l+1)ħ2Ylm
LzYlm = mħYlm
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)

14. 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

15. Like the modes of a drum 1s 2s
2px
3dxy

16. And like particle in a 2D (q,j) box
Note that particle is ‘free’ in angular direction, as U is independent of angle
f11 (“1s”)
f12 (“2px”)
f22 (“3dxy”)
f21 (“2py”) 16

17. What about radial part? F = Rnl(r)/r x Ylm(q,j)
n: Principal Quantum Number (Size, r)
l: Azimuthal Quantum Number (Angle q)
m: Magnetic Quantum Number (Angles q, j)
s: Spin Quantum Number (up, down)

18. Normalizing the wavefunction
F = Rnl(r)/r x Ylm(q,j) dV
∫F2.r2drdW = 1
dW = sinqdqdj
∫Y2dW = 1
∫(R/r)2.r2dr = 1

19. 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)
Looks like 1D.
Then why R/r?
L = mvr (constant)
Fc = mv2/r = L2/mr3
Uc = -∫Fdr = L2/2mr2
L2  L2  l(l+1)ħ2
Normalization over r2dr
∫(R/r)2.r2dr = 1
∫R2.dr = 1
This looks normalized in 1D ! ˆ

20. 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

21. 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

22. 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

23. 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

24. n-l-1 nodes  Aufbau principle
wikipedia

25. The shell structure of the H-atom
Shows angular and radial nodes for various (nlm) combos
Blue: negative, Red: positive

26. 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 !