Lecture 31:
The Hydrogen Atom 2:
Dipole Moments
Phy851 Fall 2009
Electric Dipole Approximation
• The interaction between a hydrogen atom
and an electric field is given to leading order
by the Electric Dipole approximation:
`Semi-Classical’ Approx:
r r
VE = − D ⋅ E (rCM )
•Electric field is classical
•COM motion is classical
• The dipole moment of a pure dipole:
– Vector quantity
– Points from - to +.
– Magnitude is charge _ distance
-
r
r − r−
r+ r
-R
+
r
r r
d = q (r+ − r− )
• For Hydrogen atom this gives:
r
r
D = −eR
Dipole Moment Operator
• The electric dipole moment is an operator in
H(R), which means that its value depends on
the state of the relative motion:
r
r
D = −eR
r r
VE = − D ⋅ E (rCM )
r r
VE = e R ⋅ E (rCM )
• Choosing the z-axis along the electric field
direction gives:
VE =| e | Z E (rCM )
• Expanding onto energy eigenstates gives:
∞
∞ n −1 n ' −1
VE = ∑∑∑∑
l
l'
∑ ∑ nlm (V )
n =1 n '=1 l = 0 l '= 0 m = − l m '= − l '
E nlm ; n 'l 'm '
(VE )nlm;n 'l 'm ' = d nlm;n 'l 'm ' E (rCM )
n' l ' m'
Dipole-Moment Matrix Elements
Z nlm;n 'l 'm ' = nlm R cos Θ n' l' m'
• Separate radial and angular Hilbert spaces:
d nlm;n 'l 'm ' = e nl R n' l'
(R)
lm cos Θ l' m'
(Ω)
• SELECTION RULES:
– Arfken, 3rd ed., 12.213
lm cos Θ l' m'
(Ω)
2
2
 (l + 1) 2 − m 2

l
−
m

= δ m,m '
δ l ,l '+1 +
δ l ,l '−1 
2
2
 4(l + 1) − 1

4l − 1


• The important thing to remember is that
d nlm;n 'l 'm ' ∝ δ m ,m 'δ l ,l '±1
• Electric Dipole Forbidden Transitions
4
3
s
p
d
2
Examples:
1
f
Charged particle in a Magnetic Field
• EM fields are described by both a scalar
potential, _ and vector potential, A
• To include such EM fields, we can make the
transformation:
v
v
v v
P → P − qA( R)
• Here q is the charge and A(R) is the vector
potential
€
• The Hamiltonian of an electron then
becomes:
– Units of B are Gauss (G):
r r 2
r
1 r
H=
P − eA( R) + eΦ( R)
2me
[
]
• This is known as the ‘minimal coupling
Hamiltonian’
€
Vector potential of a uniform B-field
r r
r
• For a uniform B-field, B ( r ) = B0 we have:
r r
1r r
A(r ) = − r × B0
2
• Proof:
r r
r r r
B(r ) = ∇ × A(r )
r r
1r r r
B( r ) = − ∇ × r × B0
2
r r r
r r r
r r r
1 r r r
= − r ∇ ⋅ B0 − B0 ∇ ⋅ r − r ⋅ ∇ B0 + B0 ⋅ ∇ r
2
r
r
1
= − 0 − 3B0 − 0 + B0
2
r
= B0
(
[(
[
€
)
)
( ) (
]
)
)]
(
€
r r r e2 r r
 r e r r 2
e r r r
2
€ P + 2 R × B0  = P + 2 P ⋅ R × B0 + R × B0 ⋅ P + 4 R × B0
[
]
2
r
r
e
 2 2 r r 2
2
= P − eL ⋅ B0 + R B0 − R ⋅ B0 

4
(
€
)
(
)
2
€
An electron in a uniform B-field
• Putting this in the Hamiltonian gives:
r
P2
e r r
e2 2 2
H=
−
L ⋅ B0 +
B0 R⊥ + eΦ( R)
2me 2me
8me
• Choosing B along the z-axis gives:
r
P2
eB0
e 2 B02 2
2
H=
−
Lz +
X + Y ) + eΦ( R)
(
2me 2me
8me
€
e
−
Lz B0
2me
“Paramagnetic term”
e 2 B02 2
2
X
+
Y
(
)
8me
“Diamagnetic term”
•Generates linear Zeeman effect
•Generates quadratic Zeeman effect
Paramagnetic Term: Magnetic Dipole
Interaction
• A loop of current, I, and area, a, creates a
magnetic dipole:
µ= Ia
• The orbital motion of a single electron
constitutes a current
– For a circular orbit we have
ev € a = π r 2
I=−
,
2π r
ev r
Ia=−
2
• An electron therefore has a magnetic dipole
moment associated with its orbital motion
ev r
−
=−
€
2
€
e
e r r
e r r
me vr = −
p×r =
r×p
2me €
2me
2me
r
e r
µ=
L
2me
• The paramagnetic term is therefore the
energy of the orbital dipole moment in the
uniform field:
€
r r
VB = − µ ⋅ B0
eB0
VB = −
Lz
2me
€
Dipole Energy scale
eB0
nlm VB nlm = −
hm
2me
• The energy shift between different m states
is very small compared to Hydrogen level
spacing
• Order of magnitude:
VB
B0
eh
−19 −34 + 30 J
− 23 J
~
= 10
= 10
me
T
T
• Strongest man-made B-fields ~40 T
VB ≤ 10−22 J <<
E1 (2.18 ×10−18 J )
Energy scale of
bare levels
Diamagnetic Term
• An electron in a uniform field will naturally
undergo circular motion in the plane
perpendicular to the field
– Cyclotron motion
• Thus the B-field induces a current
• This leads to an induced magnetic moment,
which must be proportional to B0
r
r
µinduced ∝ B0
• The energy of this magnetic moment in the
uniform B field therefore scales as B2
r
r
E = − µinduced ⋅ B0 ∝ B02
VB 2
e 2 B02 2
=
X +Y 2
8me
(
)
• Order of magnitude:
VB 2
B02
e 2 a02
J
J
~
= 10 −38− 20+30 2 = 10 − 28 2
8me
T
T
VB 2 ≤ 10 − 26 J
<<
VB (10 − 22 J ) <<
E1 (10 −18 J )
– The diamagnetic term can be neglected unless
the B-field is very strong
Zeeman Effect
•
The Hamiltonian of a Hydrogen atom in a uniform
B-field is
– Can neglect diamagnetic term
eB
H = H0 −
Lz
2µ
•
H 0 nlm = En nlm
‘Bare’ Hamiltonian
Eigenstates are unchanged
H n, l, m = E n, l, m
•
Energy eigenvalues now depend on m:
En , m
•
h 2 1 eB
=−
−
m
2
2
2 µa0 n
2µ
The additional term is called the Zeeman shift
– We already know that it will be no larger than 10-22
J~10-4eV
– E.g. 100 G field:
• EZeeman~10-25 J
• EZeeman/EI ~ 10-25+18 ~ 10-7
•
To get the correct Zeeman shift, we will also need
to include spin.
– We will do this next semester using perturbation
theory and the Wigner-Ekert Theorem