Download Complex Atoms I L-S Coupling - Astrophysik Uni

Atomic Spectra in Astrophysics
Potsdam University : Wi 2016-17 : Dr. Lidia Oskinova
[email protected]
01
Ioniztion of hydrogen and Saha equaiton
Ionization stages: Saha law for hydrogen
ne n p
(2πme kT )3/2 − hνkT1c
=
e
nH
h3
Lets define the ionization fraction as
ne
x = n p +nH
Using
n p = ne and n p + nH = n one can re-write the Saha equation
x2
1−x
1
3/2 − 13.6eV
kT
T
e
n p +nH
∝
Ionization fraction is determined by two parameters: temperature and density
Temperature of the Universe T(z)=2.728(1+z)
Density (n p +n H )(z)=1.6(1+z) 3 m -3
T=4000K x=0.6
T=3800K x=0.29
T=3000K x=0.00017
Hydrogen Atom
Direct observation of H electron orbital (Stodolna et al. 2013, Phys. Rev. Lett.
110, 213001)
The Schrödinger Equation for a H-like atom
The Hamiltonian operator of H-like system
Ĥ =
Atomic Units
−~2 2
∇
2µ
Ze2
4πǫ0 r
−
Electron mass, m e =1.66 10 -27 kg
Electron charge, e=1.6 10 -19 C
Bohr radius, a 0 =
4πǫ0 ~2
me2
= 5.29 10 -11 m
Dirack constant, h/2π = 1 a.u.
Ĥ =
1 2
− 2µ ∇
−
Z
r
For a system with energy E and wavefunction ψ:
For H-like atom
h
1 2
− 2µ ∇ −
Z
r
i
Ĥψ = Eψ
− E ψ(~r) = 0
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04
Wavefunctions and separating the variables
m1 m2
m1 +m2
µ=
r = (r, θ, φ)
Coordinates ~
Reduced mass:
For H-like atoms one can solve
Schrödinger eq. analytically by
separating the variables.
Radial solutions Laguerre polinomials
Angular solutions spherical harmonics
ψ(r, θ, φ) = Rnl (r)Ylm (θ, φ)
R(r)
solutions exists only if main quantum number n=1,2,..., ∞
Y(τ,ψ):
orbital quantum number l = 0,1,2,..., n-1
and
magnetic quantum number m l =-l, -l+1,..l-1, l (2l+1 values)
and
spin quantum number m s =+1/2,-1/2
05
Quantum numbers
n determines the energy of the atom
l describes the electron angular momentum:
~[l(l + 1)]1/2
0 1 2 3 4 5 6 7 ...
s p d f g h i k l ...
m l the magnetic quntum number: determines level splitting in the
presence of magnetic field.
m~ is the projection of angular
momentum on the z-axis
~[s(s + 1)]1/2 . √
3/2~
for H-like atoms angular momentum
s spin. The electron angular momentum is
Electron spin is 1/2
s z projection of spin angular momentum. It can have s, -s+1,...,s1, s values. For one electron system only -1/2, +1/2.
06
Energy levels and Rydberg constant
For bound states the solution of Schrödinger equation
En =
Z2
−R n2
The constant R takes different values for different H- like ions
RH =
µ
R
me ∞
The rydberg constant for a system of infinite nuclear mass
R∞ = 109737.31 cm-1
For hydrogen measured experimentaly
RH = 109677.581 cm-1
07
Angular momentum coupling in the Hydrogen atom
Sources of angular momentum:
electron orbital angular momentum l
electron spin angular momentum s
nuclear spin angular momentum i - determined by spin coupling
of the various nucleons: protons and neutrons have spin 1/2.
Only the total angular momentum is conserved. Individual
momenta must be combined. This is done using a coupling
scheme. Different schemes are possible, but ususally the
strongest couplings are considered first.
l + ~s = ~j
Hydrogen the usual coupling scheme: ~
Next, combine electron and nuclear spin: ~j + ~i =
Rules for vector addition:
f~
|a − b| ≤ c ≤ a + b
In quantum mechanics the result is quantized with a unity step.
l1 = 2, l2 = 3, ~L = l~1 + l~2 → L = 1, 2, 3, 4, 5
08
The fine structure of hydrogen
Fine structure in the energy split according to the value of j.
For Hydrogen s=1/2. Lets take l=0,1,2,3.
(2S+1)
Configuration l
s
ns
0 1/2
LJ
j
1/2
H atom
ns1/2
Term
n2 S
Level
n2 S1/2
np
1 1/2
1/2, 3/2
np1/2 , np3/2
n2 Po
n2 Po1/2 , n2 Po3/2
nd
2 1/2
3/2, 5/2
nd3/2 , nd5/2
n2 D
n2 D3/2 , n2 D5/2
nf
3 1/2
5/2, 7/2
nf5/2 , nf7/2
n2 Fo
n2 Fo5/2 , n2 Fo7/2
E.g. n=2 give rise to three fine strucutre levels
Which line is formed by spontaneus transitions from this level?
22 S 1/2 , 22 Po1/2 , 22 Po3/2
09
Selectrion Rules
Selection rules are derived rigorously using quintum mechanics.
Strong transitions are driven by electric dipoles. Weaker
transitions could also be driven by quadrupoles.
For electric dipole transitions in hydrogen
∆n any
∆l = +-1
∆s = 0 (for H is always satisfied as s=1/2)
∆j = 0, +-1
∆m j = 0, +-1
Lets consider Hα: n=2-3, ∆l = +-1
2s-3p, 2p-3s, 2p-3d are
allowed, 2s-3s, 2p-3p, 2s-3d are not allowed.
Considering fine structure, further constrains are put by ∆j = 0, +-1
rule
Complex Atoms
11
Non-relativistic Schrödinger Equation
hP 2
N
−~
i=1
2
−
∇
2me i
Ze2
4πǫ0 ri
+
PN−1 PN
i=1
e2
j=1+1 4πǫ0 |~
ri −~
r j|
i
− E × ψ(~
r1 , .., r~N ) = 0
The first sum is a kinetic energy operator for the motion of each
electron and the Coulomb attraction between this electron and the
nucleus.
The second term: electron-electron repulsion term. Because of
this term the equation cannot be solved analytically even for N=2.
Therefore, to understand such systems it is necessary to introduce
approximations
12
Central Field model
The easiest approximation
reduce the problem to a single
particle situation. I.e. a potential does not depend on the angular
position of electrons around the nucleus. Central field potential the force acting on each electron only depends on its distance
from the nucleus. The Schrödinger eq. can be written for ith
electron
where
h 2
i
−~
( 2me ∇2i + Vi (ri ) φi (ri ) = E i φi (ri )
Vi (ri ) is the angle-independent cental potential of each
electron.
The total energy of the system is then
The solutions,
E=
P
i
Ei
φi (ri ) , are called orbitals. Importnant to
remember that this is an approximation.
13
By analogy with H, each atomic orbital can be written as the
product of a radial and an angular function.
φi (~ri ) = Rni li (ri )Yli mi (θi , φi )
The angular part does is independent of the other electrons.
To specify central potential
Vi (ri ) , one usually considers a
potion of an electron in the central field of all the other electrons.
Vi (ri ) =
Ze2
4πǫ0 ri
+
P
j,i
D
e2
4πǫ0 |~
ri −~
r j|
E
where <> represent average. Hence the radial part,
Rni li (ri )
depends on all other electrons and cannot be obtained
analytically.
Usually it is done iteratively - the self-consistent field (SCF) problem
14
It is standard to use the H-atom orbital labels: n, l, m. The angular
behaviour, given by and m is the same. However, radial part is
not the same as in case of H.
The total wavefunction of the system
ψ(~
r1 , ..., r~N ) = φ1 (~
r1 ), ..., φN (r~N )
Lets consider spin and generalize orbital to spin- orbital
Φi ( j) = φi (~
r j , σ j ) , where Φi ( j) means that spin-orbital i is a
function of the space-spin coordinates of electron j.
The space- spin wavefunction for the total system then
Ψ(1, ..., N) = Φ(1), ..., Φ(N)
Indistinguishable Particles
Consider a system with two identical particles. If the wavefunction
of these particles is
Ψ(1, 2) , and the particles are
indistinguishable, what property their wavefunction must have?
|Ψ|2 . Then, if particles are
indistinguishable |Ψ(1, 2)|2 = |Ψ(2, 1)|2 .
Physically observable is
Symmetric solution
Ψ(1, 2)| = +Ψ(2, 1) , antisymmetric
Ψ(1, 2)| = −Ψ(2, 1) . Pauli Principle:
Wavefunctions are antisymmetric with respect to interchange of identical Fermions
Fermions - any particles with half-integer spin (electrons, protons, neutrons).
A two-electron wavefunction which obeys the Pauli Principle can be written as
Ψ(1, 2) =
√1
2
[Φa (1)Φb (2) − Φa (2)Φb (1)] = −Ψ(2, 1)
1
where √ is to preserve normalization.
2
15
The Pauli exclusion principle
Consider a two- electron wavefunction. If the two spin- orbitals,
Φa = Φb , then Ψ(1, 2) = 0 . This solution is not allowed.
Hence solutions which have the two particle occupying the same
spin-orbital are ecluded. The Pauli exclusion principle:
No two electrons can occupy the same spin-orbital
This exclusion is the key to atomic structure.
It also provides the degeneracy pressure which holds up the
gravitational collaps of white dwarfs and neutron stars.
16
Electron Configurations
For a H-like atom, the energy is determined only by principle quntum number:
E(1s)<E(2s)=E(2p)<E(3s)=E(3s)=E(3p)=E(3d)<E(4s)...
For complex atoms the degeneracy on l is lifted.
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18
Following the Pauli exclusion principle, each orbital labeled nl
actually consists of orbitals with l+1 different m value, each with
two possible values s. Thus the nl orbital can hold a maximum
2(2l+1) electrons.
Write possible combinations of magnetic and spin quantum
numbers for p electron
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Following the Pauli exclusion principle, each orbital labeled nl
actually consists of orbitals with l+1 different m value, each with
two possible values s. Thus the nl orbital can hold a maximum
2(2l+1) electrons.
An atomic configuration is given by distributing electrons among
orbitals. First fill the atomic orbitals in energy order from the
lowest energy orbitals upwards.
Write the ground state configuration for carbon.
Write the ground state configuration for neon.
20
Following the Pauli exclusion principle, each orbital labeled nl actually
consists of orbitals with l+1 different m value, each with two possible
values s. Thus the nl orbital can hold a maximum 2(2l+1) electrons.
An atomic configuration is given by distributing electrons among
orbitals. First fill the atomic orbitals in energy order from the lowest
energy orbitals upwards.
The ground state configuration for carbon 1s 2 2s 2 2p 2
The 2p orbital is not full.
The ground state configuration for neon 1s 2 2s 2 2p 6
The 2p orbital has closed shell configuration. A closed shell or subshell makes no conribution to the total orbital or spin momentum, L or
S.
Atomic ions which have the same number of electrons form
isoelectronic series. E.g. neon-like iron is
21
Following the Pauli exclusion principle, each orbital labeled nl actually
consists of orbitals with l+1 different m value, each with two possible
values s. Thus the nl orbital can hold a maximum 2(2l+1) electrons.
An atomic configuration is given by distributing electrons among
orbitals. First fill the atomic orbitals in energy order from the lowest
energy orbitals upwards.
The ground state configuration for carbon 1s 2 2s 2 2p 2
The 2p orbital is not full.
The ground state configuration for neon 1s 2 2s 2 2p 6
The 2p orbital has closed shell configuration. A closed shell or subshell makes no conribution to the total orbital or spin momentum, L or
S.
Atomic ions which have the same number of electrons form
isoelectronic series. E.g. neon-like iron is Fe XVII (Fe 16+ )
22
The Periodic Table Dmitry
Mendeleyev (1834-1907)
The structure of peridic tableelectron configurations of
elements
Closed shells: n=1 orbitals Kshells
Closed shells: n=2 orbitals Lshells
Closed shells: n=3 orbitals
M-shells etc.
Atoms with the same electron
configuration outside a closed shell
share similar chemical and optical
properties.
Li, Na, K - a single electron outside
a closed shell. Alkali metals.
Excited states - when one of the
outermost electrons jumps to
higher orbital. He: 1s2s, 1s2p,
1s3s.
Angular Momentum in Complex Atoms
Complex atoms have more than one electron, hence several
sources of angular momentum.Ignoring nuclear spin, the total
conserved angular momentum, J is the sum of spin plus orbital
momenta of all electrons.
There are two coupling schemes of summing the individual
electron angular momenta.
L-S or Russell-Saunders coupling
P~
~
The total orbital angular momentum L = i li and the total electron angular
P
~
si
momentum S = i ~
These are then added to give
Pauli Principle
J~ = ~L + S~
closed shells and sub-shells (e.g. 1s 2 or 2p 6 ), have both L=0
and S=0. Hence, it is necessary to consider only electrons in open or partlyfilled shells.
23
2
2
24
Consider OIII with the configuration 1s 2s 2p3d
Which electrons contribute to angular momentum? What are their l
and s?
Find L and S. Find J. Write possible terms.
(2S+1)
LJ
2
2
Consider OIII with the configuration 1s 2s 2p3d
For the 2p electron: l 1 =1and s 1 =1/2.
For the 3d electron: l 2 =2and s 2 =1/2.
~ = l~1 + l~2 → L = 1, 2, 3;
L
S~ = s~1 + s~2 → S = 0, 1;
L S
1 0
(2S+1)
J
1
LJ
1
1
0, 1, 2
J~ = ~L + S~
2
0
2
2
1
1, 2, 3
3
0
3
3
1
2, 3, 4
Level
1 o
P1
3
Po0 ,3 Po1 ,3 Po2
1
3
Do2
Do1 ,3 Do2 ,3 Do3
1
3
F 3o
F 2o ,3 F 3o ,3 F 4o
Twelve levels arise from 1s 2 2s 2 2p3d configuration of OIII.
25
j-j coupling
P
li + ~si → J~ = i ~ji
Alternative scheme: 1) ~ji = ~
Consider again OIII and show that the result is the same.
Why to schemes are needed? We considered non-relativistic Schrödinger
equation. In this case levels with the same L and S have the same energy. In
practice, relativistic effects split this degeneracy (these effects are introduced
as a perturbation of the non-relativistic treatment).
For light ions (lighter than Fe), relativisitc effects are weak. L-S coupling
scheme is the most appropriate.
For heavy ions (heavier than Fe), relativisitc effects are stron. j-j coupling
scheme is the most appropriate.
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Specroscopic Notations
The standard notation is called spectroscopic notation and works
with L-S coupling
A term is a state of a configuration with a specific value of L and S:
(2S+1)
L
(o)
A state with S=0 is a singlet (2S+1=1); a state with S=1/2 is a doublet; a state
with S=1 is a triplet.
o means odd parity. When parity is even, no superscript. A level is a term with a
specific value of J:
(2S+1)
L
(o)
J
Each term can be split on 2J+1 sub-leveles called states which are designated
by the total magnetic quantum number M J = -J, -J+1, ..., J-1, J. These states
are degenerate in the absence of an eternal field.
The splitting of levels into states in a magnetic field Zeeman effect.
Parity of the Wavefunction
28
The parity of the wavefunction is dermined by how the wavefunction behaves
upon inversion. Inversion, for an atom is is replacing vector r with -r. Probability
distribution, i.e., square of the wavefunction must be unchanged, hence
~ 1 , ..., −r
~ N ) . Even parity states are given by + ψ , odd
ψ(~
r1 , ..., r~N ) = ±ψ(−r
arity states are given by − ψ .
In practice, the parity is given by
(−1)l1 +l2 +...+lN .
As closed shells and sub-shells have an even number of electrons, it is
necessary to consider only active electrons.
For our example OIII with the configuration 1s 2 2s 2 2p3d we must consider only
2p and 3d electrons. What is the parity?
The parity is importnant since it leads to a rigorous dipole slection rule, the
Laporte rule: All electric dipole transitions connect states of opposite parity.
Strong transitions can only link configurations with even to those with odd parity
and vice versa