Atomic Structure and the Fine structure constant α
Niels Bohr
Erwin Schrödinger
Lecture Notes Fundamental Constants 2015; W. Ubachs
Wolfgang Pauli
Paul Dirac
The Old Bohr Model
An electron is held in orbit by the Coulomb force: (equals centripetal force)
FCentripetal = FCoulomb
mv 2
Ze 2
=
rn
4πε 0 rn2
Bohrs postulate:
Quantization of angular momentum
h
L = mvr = n
= n
2π
n 2 h 2ε 0 n 2
rn =
= r1
2
πmZe
Z
Ze 2 r n 2  2
v r =
= 2
4πε 0 m
m
2 2
h 2ε 0
−10
r1 =
m
=
0
.
529
×
10
2
πme
The size of the orbit is quantized, and we know the size of an atom !
Lecture Notes Fundamental Constants 2015; W. Ubachs
The Old Bohr Model: Energy Quantisation
Ze 2
Z2
1 2
En = mv −
= − 2 R∞
2
4πε 0 rn
n
Quantisation of energy
2
 e  me

R∞ = 
 2 2
πε
4
0

The Rydberg constant is the scale unit
of energies in the atom
Z2
Z2
En = − 2 R∞ ⇒ − 2
n
2n
Energies in the atom in atomic units
1 Hartree = 2 Rydberg
2
En = −
2
2
Z
Z
2
2
R
mc
=
−
α
∞
n2
2n 2
dimensionless energy
Lecture Notes Fundamental Constants 2015; W. Ubachs
with
α=
e2
4πε 0 c
The Old Bohr Model; velocity of the electron
Velocity in Bohr orbit
vn =1 = Zαc
α=
e2
4πε 0 c
Lecture Notes Fundamental Constants 2015; W. Ubachs
Limit on the number of elements ?
Classical argument
Schrodinger Equation; Radial part: special case l=0

 2 d 2 dR 
2
(
)
−
+
(
)
+


+
1
r
V
r

 R = ER
2mr 2 dr dr 
2mr 2

Find a solution for 
=0
2 
2  Ze 2
−
R = ER
 R"+ R '  −
2µ 
r  4πε 0 r
Physical intuition; no density for r → ∞
trial:
R(r ) = Ae −r / a
A
R
R' = − e −r / a = −
a
a
A
R
R" = 2 e −r / a = 2
a
a
2  1
2  Ze 2
−
=E
 − −
2m  a 2 ar  4πε 0 r
must hold for all values of r
Lecture Notes Fundamental Constants 2015; W. Ubachs
Prefactor for 1/r:
2
Ze 2
−
=0
ma 4πε 0
4πε 0 
Solution for the
a=
length scale paramater
Ze 2 m
2
1
a = a0
Z
4πε 0 2
with a0 = 2
e me
Bohr radius
Solutions for the energy
2
2
 me
2
2 e

= − Z 
E=−
2
2ma
4
πε
0  2

E = − Z 2 R∞
Ground state in the
Bohr model (n=1)
Quantum mechanics: same result
The effect of the proton-mass in the atom
Velocity vectors:

v1 =
M 
v
m+M
M 

v2 = −
v
m+M
Relative coordinates:
Relative velocity

 dr
v=
dt
  
r = r1 − r2
Centre of Mass
1
1
1
K = m1v12 + m2v22 = µv 2
2
2
2
Position vectors:
Angular momentum
M 
r
m+M

r2 = −
m1v12 m2v22 µv 2
=
=
F=
r1
r2
r
Quantisation of angular momentum:
L = µvr = n
h
= n
2π
Kinetic energy


mr1 + Mr2 = 0

r1 =
Centripetal force
L = m1v1r1 + m2v2 r2 = µvr
With reduced mass
m 
r
m+M
Lecture Notes Fundamental Constants 2015; W. Ubachs
µ=
mM
m+M
Problem is similar, but
m µ
r
relative coordinate
Reduced mass in the old Bohr model  isotope shifts
Results
Quantisation of radius in orbit:
n 2 4πε 0 2 n 2 me
a0
rn =
=
2
Z e µ
Z µ
1. Isotope shift
on an atomic transition
Energy levels in the Bohr model:
Z2  µ 
En = − 2   R∞
n  me 
Rydberg constant:
 µ 
RH =   R∞
 me 
Lecture Notes Fundamental Constants 2015; W. Ubachs
2. Effect of proton/electron mass
ratio on the energy levels
µ red
me
=
mM
M
M /m
µ
/m =
=
=
m+M
m + M 1+ M / m 1+ µ
Conclusion: the atoms are not a good
probe to detect a variation of µ
General conclusions on atoms and atomic structure
En = −
2
2
Z
Z
2
2
mc
R
−
α
=
∞
2n 2
n2
Note units (different units in this equation):
R∞ = −
EI
= 1.0973731568549(83) ×107 m −1
hc
dimensionless energy
Conclusion 1: All atoms have the Rydberg as a scale for energy;
they cannot be used to detect a variation of α
µ red
M /m
µ
=
=
m 1+ M / m 1+ µ
Conclusion 2: the atoms are not a good
probe to detect a variation of µ
Lecture Notes Fundamental Constants 2015; W. Ubachs
Relativistic effects in atoms
Electron spin
No classical analogue for this phenomenon
s=
1

2
Origin of the spin-concept
-Stern-Gerlach experiment;
space quantization
Pauli:
There is an additional “two-valuedness”
in the spectra of atoms, behaving like
an angular momentum
Goudsmit and Uhlenbeck
This may be interpreted/represented
as an angular momentum
Lecture Notes Fundamental Constants 2015; W. Ubachs
-Theory: the periodic system requires
an additional two-valuedness
Electron spin as an angular momentum operator
In analogy with the orbital angular momentum
of the electron


L
µL = −gLµB

1
s= 
2
Spin is an angular momentum, so it
should satisfy
S 2 s, ms =  2 s(s + 1) s, ms
gL = 1
A spin (intrinsic) angular momentum can be

defined:

µS = − g S µB
S

a) in relativistic Dirac theory
S z s , m s = m s s , m s
gS = 2
1
1
s = , ms = ±
2
2
b) in quantum electrodynamics
g S = 2.00232...
Note: the spin of the electron cannot be explained from a classically “spinning” electronic charge
e
Electron radius
2
=
m
c
from EM-energy: e
4πε 0 re
2
Lecture Notes Fundamental Constants 2015; W. Ubachs
Angular momentum L = Iω = 2 m r 2 v = 1 
e
e e
from spin
5
re 2
Spin-orbit interaction
Frame of nucleus:

v
Frame of electron:
-e
+Ze
+Ze
-e

−v
The moving charged nucleus induces
a magnetic field at the location of the
electron, via Biot-Savart’s law
 µ0 Ze(− v )× r
B=
4π
r3
1

 
ε
=
µ
0 0
Use L = mr × v ;
c2


Ze
L
Bint =
Then
4πε 0 me c 2 r 3
Spin of electron is a magnet with dipole

µS = − ge
µB 

The dipole orients in the B-field with energy
2
 


VLS = − µ S ⋅ B =
Ze
4πε 0 me2c 2 r 3
S ⋅L
A fully relativistic derivation
 
(Thomas Precession) yields VLS = ζ (r )S ⋅ L
with
2
ζ (r ) =
Ze
1
8πε 0 me2c 2 r 3 nl
Use:
1
r
3
=
2
a n ( + 1 / 2 )( + 1)
3 3
2
 Zαmc 
 3

 n  n ( + 1 / 2 )( + 1)
3
Lecture Notes Fundamental Constants 2015; W. Ubachs
S
=
Fine structure in spectra due to Spin-orbit interaction
In first order correction to energy
for state lsjm j
 j ( j + 1) − ( + 1) − s(s + 1) 
ESO = α Z hcR 

 2n3( + 1 / 2 )( + 1) 
2 4
ESO = lsjm j VSL lsjm j
 
= lsjm j ζ nl L ⋅ S lsjm j
Evaluate the dot-product
 2
 
2
2
2
J = L+S
Then the full interaction energy is:
= L + S + 2L ⋅ S
S-states
 = 0, j = s
P-states
 = 1, j =  ± 1 / 2
ESO = 0
Then
(
)
 
1
L ⋅ S sjm j = J 2 − L2 − S 2 sjm j
2
1
=  2 { j ( j + 1) − ( + 1) − s(s + 1)} sjm j
2
ESO =
α 2 Z 4 hcR
2n3
Show that the “centre-of-gravity”
does not shift
Lecture Notes Fundamental Constants 2015; W. Ubachs
Kinetic Relativistic effects in atomic hydrogen
Relativistic kinetic energy
rel
Ekin
mc
2
=
2 2
2 4
2
p c + m c − mc =
2
2 2
2
1 + p / m c − mc =


p
p
mc 1 +
−
+ 
2 2
4 4
8m c

 2m c
2
2
4
First relativistic correction term
4
K rel = −
p
8me3c 2
To be used in perturbation analysis:


p=− ∇
i
operator does not
change wave function
Lecture Notes Fundamental Constants 2015; W. Ubachs
K rel = Ψnjm −
−
Z 4α 2
2n3
p4
8me3c 3
Ψnjm =
1
3
− 
 2 + 1 8n 
(hc )R
Relativistic effects in atomic hydrogen: SO + Kinetic
Relativistic energy levels:
Enj = En −
Z 4α 2
2n 3
(hc )R
2
3 
− 
 2 j + 1 4n 
Fine structure splitting ~ Z4α2
Also the outcome
of the Dirac equation
(cα ⋅ p + βmc )ψ = ih ∂∂ψt
2
P.A.M. Dirac
Lecture Notes Fundamental Constants 2015; W. Ubachs
j=1/2
levels
degenerate
Hyperfine structure in atomic hydrogen: 21 cm
Nucleus has a spin as well, and therefore
a magnetic moment


I
;
µI = gI µN

µN =
e
2M p
Interaction with electron spin, that may have density
at the site of the nucleus (Fermi contact term)
(
    1 2
I ⋅S = I ⋅J = F − J2 − I2
2
)
Splitting : F=1 ↔ F=0 1.42 GHz
F=1
F=0
Magnetic dipole transition
Lecture Notes Fundamental Constants 2015; W. Ubachs
or λ = 21 cm
Scaling: g pα 2 / µ
Alkali Doublets
VSL =
2
 
S ⋅L
Ze
4πε 0 2m 2c 2 r 3
Selection rules:
with
(
  1 2
S ⋅ L = J − L2 − S 2
2
)
Na doublet
∆j = 0,±1
∆ = ±1
∆s = 0
np
2P
3/2
2P
1/2
ns
Lecture Notes Fundamental Constants 2015; W. Ubachs
2S
1/2
ESO =
α 2 Z 4 hcR
2n3
The Alkali Doublet Method
Lecture Notes Fundamental Constants 2015; W. Ubachs
The Many Multiplet Method
1.
3.
2.
1. Strong transitions
2. Weak, narrow transitions
3. Hyperfine transitions
Lecture Notes Fundamental Constants 2015; W. Ubachs
The Many Multiplet Method
Z2
En = − 2 R∞
n
2
 e 2  me

R∞ = 
2

 4πε 0  2
Lecture Notes Fundamental Constants 2015; W. Ubachs
Relativistic corrections in the Many Multiplet Method
Relativistic correction to energy level
me 4 Z 2 (Zα )
∆n = −
2 2
n3
2
 2
3 

− 
 2 j + 1 4n 
Further include Many body effects
∆ n ≅ En
(Zα )2 
ν

1

(
)
−
C
Z
j
l
,
,
 j +1/ 2


(note: atomic units different)
2
(
Zα )
∆ n ≅ En
ν ( j + 1 / 2)
with:
En is the Rydberg energy scaling
ν is effective quantum number
In many cases: C (Z , j , l ) ≅ 0.6
These effects separate light atoms (low Z)
from heavy atoms (high Z)
Lecture Notes Fundamental Constants 2015; W. Ubachs
Many Multiplet Method
Dependence of the energy levels on α:
(two values for different times)
Advantages of MM-Method:
in simplified form:
1) Many atoms can be “used” simultaneously
with:
 α

x=
 α lab




2
“q” given in frequency/energy units
Lecture Notes Fundamental Constants 2015; W. Ubachs
2) Transition frequencies can be used
(not just splittings)
3) Combine heavy and light atoms
Results
All allowed E1 transitions
Negative signs for:
d→p and p→s
Lecture Notes Fundamental Constants 2015; W. Ubachs
Quasar Lines
Lecture Notes Fundamental Constants 2015; W. Ubachs


1
T = T0 1 −
3/ 2 
(
)
+
1
z


Lecture Notes Fundamental Constants 2015; W. Ubachs
“Quasar
Absorptie
Quasar
absorption
spectra
Spectra”
Quasar
To Earth
Lyman limit
Lyα
Lyβ
Lyαem
SiII CII
SiII CIV
SiIV
Lyβem
NVem
CIVem
SiIVem
Lecture Notes Fundamental Constants 2015; W. Ubachs
On weak and strong lines
E2
E2 − E1 = hν
Cuν
A
Buν
E1
Einstein coefficients
C=B
A 8πhν 3
=
B
c3
Dipole strength
Lifetime
2
πe 2
B=
µ
2 ij
3ε 0
1
τ=
A
Heisenberg uncertainty
Γ=
1
2πτ
Strong lines  broadened
Weak lines  narrow
Lecture Notes Fundamental Constants 2015; W. Ubachs
Similar calculations for “laboratory lines”
Clock
transitions
Ion traps
Lecture Notes Fundamental Constants 2015; W. Ubachs
Optical lattice clock
“Accidental degeneracies”
Level A: q/(hc)= 6x103 cm-1
Level B: q/(hc)= -24x103 cm-1
Dy
atom
∆q~ 30x 103 cm-1 ~ 9x105 GHz
 α 
 α 
δν = ∆q  = 2∆q 
α 
α 
 α 
= 1.8 × 1015   Hz
α 
2
Look for “rate of change”
τΑ=7.9 µs
τΒ=200 µs
ΓA~ 2x104 Hz ; Line split~ 10-4
Lecture Notes Fundamental Constants 2015; W. Ubachs
 α 
−15
  ~ 10
α 
per year
δν = 1.8Hz
per year
∆ν(A-B) ~ 235 MHz
Precision ~ 10-8
Cingoz et al,
Phys. Rev. Lett. 98, 040801 (2007)
Modern Clock Comparisons
Further parametrization:
f = const ⋅ Ry ⋅ F (α )
Constraints from various experiments
d ln f d ln Ry
d ln α
=
+ A⋅
dt
dt
dt
d ln F
A=
d ln α
Lecture Notes Fundamental Constants 2015; W. Ubachs
Cf: Peik, Nucl. Phys B Supp. 203 (2010) 18
Functional dependence on fundamental constants
Lecture Notes Fundamental Constants 2015; W. Ubachs