
Solution of Schroedinger equation

3 spatial quantum numbers

Probability and ‘Orbitals’

Selection rules for transitions

Zeeman Effects, electron spin

4 quantum numbers

Combining angular momenta
First real test of Schroedinger Equation:
Recall Bohr model:
E n
∝
1
/n
2
, via one quantum number
n
(also, quantised angular momentum L via same n)
L
Successes:
=
Hydrogen line emission spectra
also… size & stability of atoms, Moseley’s Law for Z, etc. n
Problems: Full 3D & spherical symmetry?
Magnetic effects in spectra
Z > 1 atoms’ spectra
Comparison with Bohr model: same expression for
E n
via “principal” quantum number
n
Schroedinger Equation for H atom
Express wavefunction as product
(c.f 2D box)
Ψ
(
r
,
θ
,
φ
)
=
( )
Θ
( )
Φ
Leads to 3 separate equations
−
2
2
mr
2
d dr
⎛
⎝
r
2
dr
⎞
⎠
+
⎛
⎝
2
2
mr
2
+
( )
⎞
⎠
1 sin
θ
d d
θ
⎛
⎝ sin
θ
d
Θ
d
θ
⎞
⎠
+
⎛
⎝
( )
−
m l
2 sin
2
θ
⎞
⎠
( )
=
=
0
d
2
Φ
d
φ
2
+
m l
2
Φ =
0
R solutions are
e
α r
( × r polynomial)
type, Θ are functions of cos & sin θ
Φ solutions are
exp(im l
ϕ
)
type, where
m l
is an integer
Solutions specified by three quantum numbers:
n, l, m l
Schroedinger Equation for H atom (cont.)
Angular momentum  is quantised
Solution shows
L
=
(c.f. angular momentum in lecture 11)
( )
Also Φ must be single valued and
Find
L z
=
m l
(
m l
=
Φ
0,
±
1,
±
= Φ
(
φ +
2
π
)
2...
±
l
)
Note that
L
can be zero
(unlike Bohr model)
And
L z
< L
(unless L=0)
∵
Exists an Uncertainty relation between
L x
L y
L z
Bohr – plane, but must have Δ p normal to plane…
Also seen via
But
⎡⎣
L
2
,
L z
⎤⎦
L x
L y
−
L y
L x
=
⎡⎣
L x
,
L y
⎤⎦
=
i
L z
(try it! – commutator)
=
so no uncertainty between L and L z
& can know both
‘Semiclassical’ vector
pictures for l=2
Schroedinger Equation for H atom (cont.)
Solutions specified by three quantum numbers:
n, l, m l
Energy Levels are
E n
and the same as the Bohr model!
with
n = 1, 2, 3, 4, ….
Solutions show
l = 0, 1, 2, 3… n1
and 
m l
 ≤
l
Notation:
s, p, d, f…
for
l = 0, 1, 2, 3, ….
So, possible states of electron:
(
c.f. 3D box, Lect. 9)
E n
= −
me
4
8
ε
2
o h
2
Note: E n
i.e. l, m
does not depend on
l, m l l states are degenerate –
but holds only for H atom
Electron states in hydrogen atom  orbitals
( )
dr
=
ψ
2
4
π
r
2
dr s
states  spherically symmetric
Radial probability distributions
(a = Bohr radius)
Note for
l = n1 (1s, 2p, 3d…)
there is one maximum, at
r = n
2 a
Angular functions
 orbitals
The (m
l
) subshells combine to give
spherically symmetric (
l
) shells  Spherical
‘harmonics’
(complete shells…) e.g. for 2pstates, cos θ & sin θ dependence
Squared (probability) & add  gives 1
Transition of electrons between orbitals
Selection rules
for transitions:
Δ
l
= ± 1
And
Δ m l
= 0 or ± 1
All others are ‘forbidden’
Conservation overall of angular momentum, photon has angular momentum of
l
determines the magnitude of
L
, m
l
determines the direction
A nonzero magnetic field direction defines z
But B field alters the energy...
N.B.
Magnetic dipole moment
(current in loop x area of loop)
e+ e
µ =
ve
2
π
r
π
r
2 with
mvr
=
µ
B
=
e
2
m
(Bohr Magneton)
So moment =
m l
µ
B
Applied B field shifts energy
(dipolefield interaction)
so
m l
states are split apart by µ
B
B
In nonzero magnetic field B each
l
level splits into 2
l
+1
m
l
sublevels of slightly
different energy, depending
on B
Spectra show three lines
instead of one; only three
because of further selection
rule:
Δ
m l
= ± 1, 0
But…. some transitions
showed more than three 
anomalous
Problem: more than 3 lines and smaller separation(s)
Solution:
“electron spin”
See lecture 11 !
Electron has spin angular momentum
s z
= ±
but
2
Spin angular momentum
S (cf L)
Spin magnetic quantum number
m s
(cf m l
)
µ
z
= −
(
2.000232
)
e
2
m s z
Total combined angular momentum
J
J
= +
S
J
=
( )
j
=
l
±
1
2
Combining
L
and
S
leads to correct number and splitting of emission lines
(first seen by Irishman Thomas Preston)