Download Elementary Treatment The ground state of hydrogen atom has been

LECTURE.8
Prof.R.Parthasarathy
Atomic Structure - Elementary Treatment
The ground state of hydrogen atom has been solved exactly in the nonrelativistic treatment. The ground state of helium atom has been handled
in variational scheme. What about heavier atoms? Here we assume that
each of the atomic electrons moves in a spherically symmetric potential V (r)
produced by the nucleus and all the other electrons. This is ’central field
2
approximation’. We have seen in hydrogen-like atom, the potential − Zer
produced infinite number of energy levels. We ’expect’ similar situation for
V (r) as well. However, there is going to be a difference! We have seen in
hydrogen-like atoms, the energy is proportional to n−2 and independent of
`, so that for given n, the states having different `’s are degenerate. For
example, the states 21m and 200 have the same energy. This degeneracy is
removed in a non-Coulomb central field. How? When many electrons are
present, the nucleus is screened by electrons. We want to show that for given
n, the states of lower ` have lower energy. Let us introduce the screening
2
by the screened Coulomb potential V (r) = Zer e−λr , where λ is a positive
2
p2
constant. Then, H = 2m
− Zer e−λr . This is the hamiltonian felt by one of
the electrons. Expanding the exponential and neglecting the constant term
Ze2 λ, we have
1
p2
Ze2 1 2 2
−
− Ze λ r + Ze2 λ3 r2 − · · ·
2m
r
2
6
= H0 + H1 + H2 + · · · ,
H =
(1)
where
1
1
H1 = − Ze2 λ2 r ; H2 = Ze2 λ3 r2 .
2
6
(2)
In (1), H0 is the hamiltonian for hydrogen-like atom whose Schrödinger equa2 2
tion has been solved exactly. The energy eigenvalues are En = − Z2ae0 n12 and
the eigenfunctions are ψn`m (r, θ, φ).
For small λ, the term H1 is relevant and can be considered as a perturbation. The first order correction to the energy is < H1 >n`m . For a given
1
n, this perturbation cannot connect different ` states. So non-degenerate
perturbation theory is a good approximation! We have calculated the expectation value of r in our treatment of hydrogen atom. So,
Ze2 λ2
< r >n`m ,
2
3
`(` + 1) e 2 λ2
a0 n2 −
= −
.
2
2
2n2
< H1 >n`m = −
(3)
From (3), we get
E200 = −|E20 | − 3e2 λ2 a0 ,
E21m = −|E20 | − 2.5e2 λ2 a0 ,
(4)
2 2
where |E20 | is the unperturbed energy in n = 2 state Z8ae0 . Clearly the 200
state has lower energy that 21m state. Thus, the first order correction not
only removes the ` degeneracy but also gives the result that lower angular
momentum states have lower energy.
Identical Particles
We have seen the quantum mechanics of a single particle in a potential. When we have more than one particle, say electron, because of the
absense of determinism in quantum mechanics , it is impossible to label
the electrons. They are indistinguishable, when both the electrons are having same quantum numbers. So, physical observation must be invariant
under their exchange. As an example, consider helium atom. There are
two electrons. Let us neglect their mutual interaction. What is the ground
state? Let us denote the lowest single particle state by |s >. Then the amplitude for finding electron 1 (label it temporarily!) and 2 in this state
is < 1|s > and < 2|s >. Knowing the electron has spin 21 , we have
< 1|s+ >, < 1|s− > ; < 2|s+ >, < 2|s− >. Then the possible two-particle
amplitudes are:
< 1|s+ >< 2|s+ >, < 1|s+ >< 2|s− >, < 1|s− >< 2|s+ >, < 1|s− >< 2|s− >,(5)
or their linear combinations. Experimentally, only one linear combination
appears! i.e.,
< 1|s+ >< 2|s− > − < 1|s− >< 2|s+ >
2
(6)
which changes sign when 1 ↔ 2. This is a result of the exclusion principle
of Pauli- ”the amplitude for a system containing several electrons must be
antisymmetric under the exchange of any two electrons”. Equivalently, no
two electrons are in the same single particle state. For if any two electrons
are in the same state, the totally antisymmetric amplitude vanishes. Given
no two electrons can be in the same single particle state, can we get the
antisymmetry?
Let us proceed further to a general situation. Let
ΦI = < 1|a >< 2|b >: ΦII = < 1|b >< 2|a > .
(7)
Since 1 and 2 are indistigushable, we have two linear combinations;
ΦS = ΦI + ΦII , Bosons,
ΦAS = ΦI − ΦII . F ermions.
(8)
Consider a system of two identical particles 1 and 2 governed by a hamiltonian
H(12). Let there be a stationary state |I >, which can be characterized by
labelling 1 by an index α and 2 by β, where α, β refer to position, spin
projection etc. Then |I >= |1α 2β > will be in the Hilbert space H1,2 =
H1 ⊗H2 . Since 1 and 2 are identical, the state |II >= |2α , 1β > will also be a
stationary state with same energy as |I > since H(12) = H(21). Introducing
P12 the ’permutation operator’,
P12 |I >= |II >, P12 |II >= |I > , [P12 , H] = 0.
(9)
So P12 and H have simulataneous eigenfunctions. Let P12 |ψ >= η|ψ >. Then
P12 P12 |ψ >= η 2 |ψ >= |ψ > and so η = ±1. Then, the case, P12 |ψ >= |ψ >
corresponds to ’symmetric state’ representing Bosons while the case with
P12 |ψ >= −|ψ > corresponds to ’antisymmetric state’ representing Fermions.
For a two electron system, the totally antisymmetric wave function will be
|1α 2β > −|2α 1β > = ψα (1)ψβ (2) − ψα (2)ψβ (1),
ψα (1) ψα (2)
=
ψβ (1) ψβ (2)
(10)
For a three electron system, the totally antisymmetric wave function is
ψα (1) ψα (2) ψα (3)
ψβ (1) ψβ (2) ψβ (3)
ψγ (1) ψγ (2) ψγ (3)
3
(11)
Antisymmetry in (11) under 1 ↔ 2, 2 ↔ 3, 3 ↔ 1 is evident using the
property of determinants!
Symmetric wave functions will represent Bosons, particles of intrinsic spin
0, 1, 2 etc and obey Bose-Einstein statistics. Antisymmetric wavefunctions
represent Fermions, particles of intrinsic spin 12 , 23 , etc and obey Fermi-Dirac
statistics. This connection ”spin-statistics theorem” has been verified. But
the source or proof is not well understood. The above wave function is not
normalized. A normalized antisymmetric wave function is given by Slater
determinant. For the three-fermion system,
ψα (1) ψα (2) ψα (3)
1
ψαβγ (1, 2, 3) = √ ψβ (1) ψβ (2) ψβ (3)
3! ψ (1) ψ (2) ψ (3)
γ
γ
γ
(12)
Is this procedure always correct? No. When the electrons are non-interacting,
it is a correct method. When the interaction among them is weak, the above
state can be used for perturbative calculations.
Ground State Configurations
The quantum numbers are:
n = 1, 2, 3, · · · principal quantum number
` = 0, 1, 2, · · · (n − 1), orbital ang.momentum qtm number
m = −` to + ` in steps of 1, z − component of ang.mom qtm number
1
,
s =
2
1 1
ms =
,− ,
2 2
Hydrogen
Z = 1. Ground state: n = 1, ` = 0, m = 0, ms = ± 12 The single electron
configuration is (1s)1 . (notation: (n`)occupants . The chemical valency is just
the number of electrons in the last shell which here is incomplete and so the
valency is 1.
Helium
4
Z = 2. The two electrons occupy n = 1, ` = 0 shell and have ms = ± 21
so as to respect Pauli principle. The electronic configuration is (1s)2 . With
two electrons in 1s shell, it is completely occupied. The valency is zero and
this element is chemically inert. Helium is an inert gas.
Lithium
Z = 3. First the shell 1s is completely filled by two of the three electrons.
The third electron can go to n = 2, ` = 0 with ms = 12 or − 12 . The electronic
configuration is (1s)2 (2s)1 . The valency is due to the electron in the 2s shell
and so it is 1. This element is chemically active.
Berylliium
Z = 4. The electronic configuration is (1s)2 (2s)2 . It has a closed sub-shell
since the 2p state is empty. The normal valency is 2.
We give the electronic configurations of some light elements.
Element(Z)
Hydrogen(1)
Helium(2)
Lithium(3)
Beryllium(4)
Boron(5)
Carbon(6)
N itrogen(7)
Oxygen(8)
F luorine(9)
N eon(10)
Sodium(11)
M agnesium(12)
Aluminum(13)
Silicon(14)
P hospherous(15)
Sulphur(16)
Chlorine(17)
Argon(18)
ElectronicConf iguration
(1s)1
(1s)2
(1s)2 (2s)1
(1s)2 (2s)2
(1s)2 (2s)2 (2p)1
(1s)2 (2s)2 (2p)2
(1s)2 (2s)2 (2p)3
(1s)2 (2s)2 (2p)4
(1s)2 (2s)2 (2p)5
(1s)2 (2s)2 (2p)6
(1s)2 (2s)2 (2p)6 (3s)1
(1s)2 (2s)2 (2p)6 (3s)2
(1s)2 (2s)2 (2p)6 (3s)2 (3p)1
(1s)2 (2s)2 (2p)6 (3s)1(3p)2
(1s)1 (2s)2 (2p)6 (3s)2 (3p)3
(1s)2 (2s)2 (2p)6 (3s)2 (3p)4
(1s)2 (2s)2 (2p)6 (3s)2 (3p)5
(1s)2 (2s)2 (2p)6 (3s)2 (3p)8 .
5
Upto Z = 18, the filling up of the energy levels with the understanding
that for a given n, lower ` levels are filled first works. The removal of the
` degenercay being explained by the screening of the nuclear charge by the
remaining electrons. This sequence changes now! After Argon, we have
Potassium Z = 19. The nineteenth electron should go to the 3d state. But it
goes to 4s state! After filling the 4s subshell, 3d starts getting electrons. For
example, the electronic configuration of potassium (K,Z=19) is (Ar) + (4s)1 ,
that of Calcium (Ca, Z=20) is (Ar) + (4s)2 , that of Scandium (Sc, Z=21) is
(Ar) + (4s)2 (3d)1 and so on till, for Zinc (Zn, Z=30), it is (Ar) + (4s)2 (3d)10 .
Now that the 3d subshell is completely filled, the next electron in Galium
goes to 4p state, namely, Ga(Z=31) has (Ar) + (4s)2 (3d)10 (4p)1 . Can we
understand this qualitatively? From the first order correction (3), it follows
E320
E400
Z 2 e2
− 5.25λ2 e2 a0 ,
= −
18a0
Z 2 e2
− 12λ2 e2 a0 .
= −
32a0
(13)
If the screening is strong, then it is plausible to understand that the 4s level
has lower energy than 3d level.
Alkali atoms
The alkali atoms are Li, Na, K etc. From the electronic configuration
table, we note that the ground state configuration of an alkali atom consists
of filled shells followed by a single s electron. The filled shell configuration
corresponds to inert gas configuration and so stable. Then, the excited states
of the alkali atom involve only the valence s electron. Thus, the alkali atom
can be treated to a good approximation in terms of a model in which a single elctron moves in a spherically symmetric non-Coulomb potential energy
V (r), non-Coulomb due to screening. In the absense of external fields, the
∇2
+ V (r). To this we add the spin-orbit interaction.
hamiltonian could be − 2m
First of all we consider hydrogen atom. The energy levels of hydrogen
atom obtained using non-relativistic Schrödinger equation depend on the
principal quantum number n and there is degeneracy for `. Classically, a
~
particle of mass m, charge −e moving with velocity ~v in an electric field E
6
~ v
~ = E×~
~ is derivable
experiences a magnetic field B
and if the electric field E
c2
~ = − ~r dV so that
from a potential V (r), E
r dr
~ = − 1 dV ~r × ~v ,
B
c2 r dr
1 dV ~
`,
= − 2
c rm dr
~ and a magnetic
where ~` = ~r × p~. The energy of the interaction between B
1
dV
~ so that Hs.o = −m
~ = 2
m
~ · ~`. For an electron,
dipole m
~ is −m
~ ·B
~ ·B
mc r dr
e
m
~ = − m ~s so that
Hs.o = −
e
m 2 c2 r
dV ~
` · ~s.
dr
A correct calculation (Dirac) leads to
Hs.o = −
dV ~
e
` · ~s.
2m2 c2 r dr
For an electron in hydrogen-like atom, V =
Hs.o =
Ze
r
(14)
and so
Ze2 ~
` · ~s.
2m2 c2 r3
(15)
We know from hydrogen atom wave functions,
< n`m|
1
Z3
|n`m
>
=
.
r3
n3 a30 `(` + 21 )(` + 1)
(16)
From the time-independent perturbation theory, the first order correction to
the energy levels of hydrogen-like atom due to the spin-orbit term is
< n`m|Hs.o |n`m > =
Z 4 e2
< ~` · ~s > .
2m2 c2 a30 `(` + 12 )(` + 1)
(17)
The expression < ~` · ~s > can be easily evaluated in the coupled basis since
~` · ~s = 1 {j 2 − `2 − s2 }.
2
7
Thus the first order correction is
Es.o =
Z 4 e2
1 2
h̄ {j(j + 1) − `(` + 1) − s(s + 1)}.
(18)
1
3 3
2
2
2m c a0 n `(` + 2 )(` + 1) 2
The above result can be expressed also as
Es.o
Z 2 |En0 |α2
=
n(2` + 1)
1
`+1
f or j = ` +
1
2
− 1`
f or j = ` −
1
2
(19)
where |En0 | is the unperturbed energy.
This can be easily generalized to alkali atoms. With this, let us evaluate
the energies for 3p 23 and 3p 12 .
1
,
18
−1
= −E30 + Z 2 α2 |E30 |
,
9
E3p 3 = −E30 + Z 2 α2 |E30 |
2
E3p 1
2
(20)
so that
1
E3p 3 − E3p 1 = Z 2 |E30 |α2 .
2
2
6
(21)
Let us consider sodium atom. Experimentally, there are two dominant
yellow doublet. Their wavelengths are: 5890Å and 5896Å. They correspond
to the atomic transitions 3p 32 → 3s 21 and 3p 12 → 3s 12 . For Coulomb potentials, such transitions cannot occur as the energy depends only on n. For
non-Coulombic potentials, such transitions can occur. As the non-Coulombic
potential is not known (at this level), we cannot calculate the above transition energies. However, we adopt a model: The spectrum of sodium atom is
determined by the valence electron in 3s 21 state. The closed shell configuration upto 10 electrons corresponds to Argon the inert gas. Here the nuclear
charge (+11 e) is completely shielded by the 10 atomic electrons, so that the
valence electron ”sees” the nuclear charge of one unit. In this model, the
difference of the energies of 3p 23 and 3p 12 is given by (16) with |E30 | being
8
simply 13.6
eV . Using the value of the fine structure constant α as
9
Z = 11, this energy difference (16) becomes,
= 1.623 × 10−3 eV.
E3p 3 − E3p 1 = 4E
2
2
1
137
and
(22)
For the transition E3p 3 → E3s 1 , the wavelength of the radiation is λ1 =
2
2
hc
and
for
the
other
transition,
λ2 = E 1 hc
. Hence,
−E 1
E 3 −E 1
3p 2
3s 2
3p 2
4λ = λ1 λ2
3s 2
4E
.
hc
(23)
Using the values of h = 6.626 × 10−34 J.s and c = 3 × 1010 cm/s, we find,
4λ = 4.54Å while the experimental value is 6Å. The simple model is able
to explain the sodium doublet. It is possible to calculate the intensities of
these lines using time-dependent perturbation theory.
Doublet of Sodium Yellow Lines
5890Å 5896Å
Zeeman Effect
We are considering an atom with one valence electron placed in a magnetic
field. The result is that the energy levels are split into several components,
Zeeman splitting. The interaction energy responsible for these splittings
consists of two parts: (1) due to the spin of the electron and (2) due to the
orbital motion.
The electron spin angular momentum is 12 h̄ and has projections ± 12 h̄. So
eh̄
the eletron has a component of magnetic moment ∓ 2mc
in the direction in
which the component of spin angular momentum is ± 12 h̄. As the energy
9
~ (external) is −m
~ the
of a particle of magnetic moment m
~ in a field H
~ · H,
interaction energy of the electron is
e
e ~
~ =
~s · H
H · 2~s.
mc
2mc
(24)
(In 1925 Uhlenbeck and Goudsmit (Nature 117 (1926) 264) discovered electron possessing an intrinsic angular momentum whose component is restricted
to ± 21 h̄. Pauli in 1927, described it by his matrices!)
The effect of external magnetic field on the orbital motion of the electron
is found now. The hamiltonian for a particle of charge −e in the e.m vector
~ (H
~ =∇
~ × A)
~ is
potential A
H =
2
1
e~ 2
1 2
e
~+A
~ · p~) + e A
~ 2 . (25)
(~p + A)
=
p~ +
(~p · A
2
2m
c
2m
2mc
2mc
~ are operators. Then,
In here, p~, A
~
~ · (Aψ),
~
(~p · A)ψ
= −ih̄∇
~ · A)ψ
~ − ih̄A
~ · (∇ψ),
~
= −ih̄(∇
~ · p~)ψ,
= (A
(26)
~ ·A
~ = 0 (Coulomb gauge) is used in the penultimate
where the gauge ∇
e ~
e2 ~ 2
A · p~ + 2mc
step. So the energy due to the magnetic field is mc
2 A . For a
1
~ = (H
~ × ~r). Show this! Then, we see
uniform magnetic field, we know A
2
~ 2 = 1 (H
~ 2 r2 − (H
~ · ~r)2 ). We first neglect A
~ 2 term. Then, the interaction
A
4
can be written as:
e ~
e
~ × ~r) · p~,
A · p~ =
(H
mc
2mc
e
=
ijk Hj rk pi ,
2mc
e
=
Hj jki rk pi ,
2mc
e ~
=
H · (~r × p~),
2mc
e ~ ~ ~
=
H · L, L = ~r × p~.
2mc
10
(27)
Combining with the spin contribution, the total effect of the external
magnetic field on the atomic electron is
e ~
e ~ ~
H0 =
H · 2~s +
H · L,
2mc
2mc
e ~ ~
=
H · (L + 2~s),
2mc
e ~ ~ ~
~ + ~s.
=
H · (J + S), J~ = L
(28)
2mc
If the applied magnetic field is along the z-axis, then the interaction is
e
H(Jz + sz ). In (28), we have combined the orbital and spin angular mo2mc
menta. So, the wave functions should be appropriately constructed. From the
angular momentum wave functions |`m > and spin wave functions | 12 ms >
we have the following:
1
1
C(` j; mms M ) |`m > | ms >,
2
2
ms
1
1
1
11
= C(` j; m m + ) |`m > |
>
2
2
2
22
1
1
1
1 1
+ C(` j; m, − , m − ) |`m > | − > .
2
2
2
2 2
|jM > =
X
(29)
This for j = ` + 21 , becomes
s
`+M +
1
|` + M > =
2
2` + 1
s
`−M +
+
2` + 1
and for j = ` −
1
2
1
2
|`M −
1
1 1
> | , >
2
2 2
|`M +
1
1 1
> | , − >,
2
2 2
(30)
1
2
s
` − M + 12
1
1
11
|` − M > = −
|`, M − > |
>
2
2`
+
1
2
2
2
s
` + M + 21
1
1 1
+
|`, M + > | , − > .
2` + 1
2
2 2
Now it is easy to find
1
eH
1
eh̄H M (2` + 1 + 1)
< ` + , M|
(Jz + sz )|` + , M > =
,
2
2mc
2
2mc
2` + 1
11
(31)
1
eH
1
eh̄H M (2` − 1 + 1)
< ` − , M|
(Jz + sz )|` − , M > =
, so that
2
2mc
2
2mc
2` + 1
1
eH
1
eh̄H M (2` ± 1 + 1)
< ` ± , M|
(Jz + sz )|` ± , M 0 > =
δM M 0 .(32)
2
2mc
2
2mc
2` + 1
The factor
2`±1+1
2`+1
is known as Landé g-factor.
Weak Field Case
When the external magnetic field is weak, the spin-orbit interactrion cannot be neglected. This interaction necessitates the j-j coupling. In this state,
we have evaluated the matrix elements of the magnetic interaction in (32).
To this we add the spin-orbit coupling. Denoting the radial matrix element
as ζn` , we have this as 21 ζn` {j(j + 1) − `(` + 1) − s(s + 1)}.
Strong Field Case
When the external magnetic field is strong, the spin-orbit interaction
is small and in this case, we use (n, `, m` , ms ) functions (without angular
momentum coupling). Then,
eH
eh̄H
1
1
(Lz + 2sz )|n`m` ms > =
(m` + 2ms )
< n`m` ms |
2
2mc
2
2mc
eh̄H
(M + ms ).
=
2mc
(33)
Each level is split symmetrically into 2` + 3, as m` + 2ms takes ` + 1, `, · · · −
(` + 1) which totals to 2` + 3. This case is known as Paschen-Back effect. We
now include the spin-orbit interaction. The states are labelled by n, `, m` , ms .
The diagonal matrix element < n`m` ms |ξ ~` · ~s|n`m` ms > cannot connect x, y
components of angular momenta and so this is ζn` m` ms .
Transition Case
We will use the fundamental system of eigen functions labelled by n`m` ms
following E.U.Condon and G.H.Shortley, The Theory of Atomic Spectra. In
eH
this scheme, the matrix of (`z + 2sz ) is diagonal, where = 2mc
. We shall
consider the matrix for a given n`, say 2p and 2s pertaining to alkali atom
sodium. We shall denote the radial integral of ξ by ζ.
12
1
1
−
2s States: The eigen functions are Y00 χ 21 ; Y00 χ 1 2 . The matrix element of ~`·~s
2
2
can be omitted since it does not shift them with respect to each other. The
shift due to the magnetic interaction (`z + 2sz ) is ±h̄ for m` = 0; ms = ± 12 .
1
−1
2p States: The eigen functions Y11 χ 21 ; Y1−1 χ 1 2 have m = ± 32 . For them the
2
2
additional energies are
1
ζ {j(j + 1) − `(` + 1) − s(s + 1)} + < `z + 2sz >,
2
which for j =
3
2
(34)
and m = ± 32 gives 21 ζ ± 2h̄.
1
1
−1
−1
2
2
2
2
The wave functions Y10 χ 21 , Y1−1 χ 21 , Y11 χ 1 2 , Y10 χ 1 2 combine in pairs according to m` + ms = m = ± 21 . These are:
1
−1
1
−1
2
2
2
2
{Y10 χ 21 , Y11 χ 1 2 ; Y1−1 χ 21 , Y10 χ 1 2 }.
(35)
One such pair can be considered. Let
1
|ψ1 > = Y10 χ 21 ,
2
|ψ2 > =
−1
Y11 χ 1 2 ,
(36)
2
corresponding to m = 12 . Now we use ’degenerate perturbation theory, as
these states are degenerate. We want to find the matrix elements of
H 0 = ξ(r)~` · ~s + (`z + 2sz ).
(37)
First, the easy one. It is seen that
< ψ1 |(`z + 2sz )|ψ1
< ψ1 |(`z + 2sz )|ψ2
< ψ2 |(`z + 2sz )|ψ1
< ψ2 |(`z + 2sz )|ψ2
>
>
>
>
=
=
=
=
h̄,
0,
0,
0.
(38)
Next, we need < ξ(r)~` · ~s > between these states. Writing
~` · ~s = 1 {`+ s− + `− s+ } + `z sz ,
2
13
(39)
and using
`± |`, m` > =
q
(` ∓ m` )(` ± m` + 1) |`, m` ± 1 >,
(40)
we find
< ψ1 |ξ(r)~` · ~s|ψ1 > = 0,
1
< ψ1 |ξ(r)~` · ~s|ψ2 > = √ ζh̄2 ,
2
1
< ψ2 |ξ(r)~` · ~s|ψ1 > = √ ζh̄2 ,
2
1
< ψ2 |ξ(r)~` · ~s|ψ2 > = − ζh̄2 .
2
(41)
Thus in the spirit of degenerate perturbation theory, we end up with
M =
√1 ζh̄2 h̄
2
,
√1 ζh̄2 − 1 ζh̄2
2
2
(42)
whose eigenvalues are
1
1
λ =
{h̄ − ζh̄2 ±
2
2
s
9
(2 h̄2 + ζh̄3 + ζ 2 h̄4 )}.
4
(43)
Here we recall that ζ represents the spin-orbit term contribution and the
magnetic perturbation.
2
Then, ζh̄
→ 0 corresponds to the Strong Field Case. In this case, you
h̄
can see the two eigenvalues are:
1
λ+ = h̄ ; λ− = − ζh̄2 .
2
Similarly,
h̄
ζh̄2
(44)
→ 0 corresponds to the Weak Field Case and then in this case,
1
2
1
λ+ = ζh̄2 + h̄ ; λ− = −ζh̄2 + h̄.
2
3
3
(45)
We recall in the Strong Field Case, we had the energy shift as h̄(m +
ms ) + ζh̄2 m` ms which for m = 12 , m` = m − ms , ms = ± 12 gives: ms = 12 ; h̄
14
and ms = − 12 ; − 12 ζh̄2 agreeing with (44). In the Weak Field Case, we had
+ 21 ζh̄2 (j(j + 1) − `(` + 1) − s(s + 1)) which is
the energy shift as: h̄m 2`±1+1
2`+1
h̄m 2`±1+1
+ 21 ζh̄2 {` or − (` + 1)} for j = ` + 12 or j = ` − 12 . This, for ` = 1
2`+1
agrees with (45).
3
2
2
1
2
p3
2
············
− 21
− 23
2
1
2
p1
2
2
1
2
s1
2
············
− 21
············
− 21
Weak Field Zeeman splitting
15
1+
0+
−1+
2
p 1−
0−
−1−
0+
2
············
s
−
0
Strong Field Paschen-Back splitting
16