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Introduction to Atomic Physics

Presentation · June 2023

DOI: 10.13140/RG.2.2.34491.59687

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Introduction to Atomic Physics

Badis Ydri

Annaba University - Algeria

June 1, 2023
Content
Bohr’s atom and fine structure of the hydrogen atom 3
Angular momentum and spherical harmonics 3
The Stern-Gerlach experiment and discovery of the spin 6
The qubit model of the spin 11
Addition of angular momenta 13
Central problem and Bohr’s atom 15
Bohr’s hydrogen atom as a harmonic oscillator problem 19
Time-independent Perturbation theory 23
The spin-orbit coupling and fine structure of hydrogen 25
The hyperfine structure (21 cm cosmology) 30
Stark effect and Lamb shift 33
The Zeeman effect 36
The interaction between Bohr’s atom and radiation and quantum jumps 39
Time-dependent perturbation theory 39
Fermi’s Golden rule 41
Harmonic perturbations and detailed balance 42
Stimulated emission (LASER) and absorption by electromagnetic radiation 44
Spontaneous emission and blackbody radiation 49
Exact solution for two-level systems (MASER and NMR) 51
Hydrogen atom quantum computation 54
Angular momentum and spherical harmonics
I In quantum mechanics the angular momentum ~ r ×~
L=~ p is given by the operators
L̂1 = x̂2 p̂3 − x̂3 p̂2 , L̂2 = x̂3 p̂1 − x̂1 p̂3 , L̂3 = x̂1 p̂2 − x̂2 p̂1 . (1)
I By using the fundamental commutation relations [x̂i , p̂j ] = i~δij we compute the commutation relations

[L̂1 , L̂2 ] = i~L̂3 , [L̂3 , L̂1 ] = i~L̂2 , [L̂2 , L̂3 ] = i~L̂1 . (2)

I The square of the total angular momentum L̂2 = L̂2 + L̂2 + L̂2 commutes with the components L̂i , viz
1 2 3
2 2 2 2 2
L̂ = L̂1 + L̂2 + L̂3 , [L̂ , L̂i ] = 0. (3)

I Thus, we can diagonalize L̂2 and one of the components of the angular momentum operator. We consider
2
L̂3 |f i = µ|f i , L̂ |f i = λ|f i. (4)

We must always have µ2 ≤ λ.

I Next, we define the raising and lowering operators by

L̂± = L̂1 ± i L̂2 . (5)

Thus, L̂± |f i is an eigenvector of L̂3 with eigenvalue µ ± ~.

I In other words, starting from |f i we obtain via the application of L̂+ the eigenvectors with eigenvalues
µ + n~ where n is a positive integer. We must always have (µ + n~)2 ≤ λ and hence there is a maximum
value of n.
I The corresponding eigenvector is the highest weight state denoted by |fh i = |lli with corresponding
eigenvalue denoted by +~l.
I This highest weight state must also satisfy L̂+ |lli = 0 and L̂2 |lli = ~2 l(l + 1)|lli. Thus, λ = ~2 l(l + 1).
We have then
2 2
L̂3 |lli = ~l|lli , L̂+ |lli = 0 , L̂ |lli = ~ l(l + 1)|lli. (6)
I In the same manner we can show that there exists a lowest weight state |l − li with eigenvalue equals
−~l and which satisfies L̂− |l − li = 0. We have then
2 2
L̂3 |l − li = −~l|l − li , L̂− |l − li = 0 , L̂ |l − li = ~ l(l + 1)|l − li. (7)
Angular momentum and spherical harmonics
I In general, the eigenvalues of L̂3 will be denoted by ~m where m takes the values between −l and +l in N
integer steps. Thus, l = −l + N or equivalently l = N/2. In other words, there are 2l + 1 states in total
and l can be either integer (corresponding to ordinary angular momentum) or half-integer (corresponding
to spin). The eigenvectors will be denoted by |lmi such that
2 2
L̂ |lmi = ~ l(l + 1)|lmi , L3 |lmi = ~m|lmi
1 3
l = 0, , 1, , ... , m = −l, −l + 1, ..., l − 1, l. (8)
2 2
I In spherical coordinates the components of the angular momentum operator and the square of the total
angular momentum are given by

~ ∂ ∂
~ ∂ ∂
~ ∂
L̂1 = − sin φ − cot θ cos φ , L̂2 = cos φ − cot θ sin φ , L̂3 = . (9)
i ∂θ ∂φ i ∂θ ∂φ i ∂φ

1 ∂ ∂ 1 ∂2
 
2 2
L̂ = −~ (sin θ )+ . (10)
sin θ ∂θ ∂θ sin2 θ ∂φ2

I The common eigenfunctions of L̂2 and L̂3 are Y m (θ, φ) = hθ|hφ|lmi. They satisfy
l

1 ∂ ∂ 1 ∂2 ~ ∂
 
2 m 2 m m m
−~ (sin θ )+ Yl = ~ l(l + 1)Yl , Yl = ~mYl . (11)
sin θ ∂θ ∂θ 2
sin θ ∂φ 2 i ∂φ
I The explicit solution can be obtained by separation of variables. We write Y m (θ, φ) = Θm (θ) exp(imφ)
l l
where m is an integer, viz m = 0, ±1, ±2, ....
I We obtain the differential equation (with x = cos θ)

d dΘm m2
 
2 l m
(1 − x ) + [l(l + 1) − ]Θl = 0. (12)
dx dx 1− x2
This is the Legendre equation.
Angular momentum and spherical harmonics
I The solution is given in terms of the associated Legendre polynomial P m (x) by the equation
l
|m|  d |m|
m m m 2 2
Θl (θ) = APl (x) , Pl (x) = (1 − x ) Pl (x). (13)
dx
I Hence, the associated Legendre polynomial P m (x) is given in terms of the Legendre polynomial Pl (x)
l
which in turn is given by the so-called Rodrigues formula
1 d l 2
 
l
Pl (x) = (x − 1) . (14)
l
2 l! dx
From this equation it is clear that l is a positive integer and Pl (x) is a polynomial of degree l in x = cos θ.
Hence, if |m| > l then Plm (x) = 0 and the allowed values of l and m are thus given by
l = 0, 1, 2, ... , m = −l, −l + 1, ..., 0, ...., l − 1, l. (15)
As before we have 2l + 1 states for every value of l. However, l is here always an integer.
I The complete solution is therefore given by
m m imφ
Yl (θ, φ) = APl (cos θ)e . (16)
I We will impose the normalization condition
Z 2π Z π
m 2
sin θdθdφ |Yl (θ, φ)| = 1. (17)
0 0

We find
s
2l + 1 (l − |m|)!
A= . (18)
4π (l + |m|)!

m
 = (−1) , m ≥ 0 ,  = 1 , m ≤ 0. (19)

We can also check the orthonormalization condition

Z 2π Z π
m ∗ s
sin θdθdφ [Yl (θ, φ)] Yt (θ, φ) = δlt δms . (20)
0 0
The Stern-Gerlach experiment and discovery of the spin
I The Stern-Gerlach experiment is one of the greatest experiments in physics, atomic physics and quantum
mechanics.
I This experiment plays also a major role in the foundation and philosophy of quantum mechanics. The
Stern-Gerlach apparatus can be used to prepare the initial state of the system, i.e. it allows us to select
any desired polarization for the initial state.
I More importantly, the Stern-Gerlach experiment provides an almost idealized conceptual model for the
process of quantum measurement as we will discuss below.
I It was designed originally to decide between Larmor’s classical theory and Sommerfeld’s old quantum
theory describing the motion of charged particles in magnetic fields. As it turns out, both theories are in
fact wrong and the correct description is given by quantum mechanics.
I More precisely, this experiment was designed to test the so-called ”space quantization”, i.e. the
quantization of the angular momentum suggested originally by Bohr.
I It led immediately to the discovery of the ”spin” which is a purely quantum property of the electron.
I In the Bohr-Sommerfeld model we imagine the atom as a positively charged dense nucleus with negatively
charged electrons moving in orbits around the center.
I In order for the electrons to not spiral towards the center, as they radiate energy in the electric field of the
nucleus, Bohr restricted their motion to specific orbits, called orbitals, determined by some integer (the
principal quantum number n). This is called orbital quantization and it allowed Bohr to explain why atoms
emit and absorb radiation only under a discrete form, i.e. at a discrete set of electromagnetic frequencies.
In fact, Bohr was also successful in deriving the celebrated Rydberg formula for these frequencies.
I Bohr’s model relied on the fundamental assumption that the angular moment of the electron must be
quantized. We write this in the form
I
pϕ dϕ = nϕ h.

Thus, in the ground state of the atom the electron must have only two values of the angular momentum
along any direction in space corresponding to nϕ = ±1.
I This prediction was called space quantization and Stern devised his experiment to test it directly. In fact,
Stern was very sceptical about this prediction and he went in his scepticism as far as to raise a wager to
quite physics if Bohr’s quantization turns out to be correct. Stern was shortly joined by Gerlach who was
instrumental in the success of the experiment. Yet, it is Stern who won the Nobel prize in physics on this
work (called the molecular ray method) in 1943.
I As we will see this one assumption, which seemed to enrage Stern, turns out to be wrong yet the beams of
atoms prepared in the ground states (such as the silver and the hydrogen atoms used in the experiments)
are observed to suffer a splitting into two streams in a non-uniform magnetic field as predicted by Bohr
(but for the wrong reason).
The Stern-Gerlach experiment and discovery of the spin
I The orbital angular momentum is indeed quantized but not according to the above rule and the
Stern-Gerlach experiment which sought to discover this space quantization ended up discovering the spin
quantum number and vindicating the new theory of quantum mechanics.
I We can also add to the above quantization of the angular momentum, following Sommerfeld, the following
radial quantization
I
pr dr = nr h.

I This will extend Bohr’s original circular orbits to elliptical orbits. These quantization conditions govern the
so-called Bohr’s correspondence principle which states that in the limit of large quantum numbers nϕ and
nr the predictions of the old quantum theory should reduce to those of classical physics.
I In the Stern-Gerlach experiment (1922) a beam of hot silver atoms (characterized by a single unpaired
electron) is sent through a non-uniform magnetic field. After going through the magnets the beam reaches
a detector plate. The atoms, as we will see, are characterized by a dipole magnetic moment and thus as
they move in the non-uniform magnetic field they will experience a torque which acts differently on the two
ends of the dipoles leading to a net force on the atoms. This causes the atoms to deflect differently, in the
non-uniform magnetic field, according to the magnitude of their magnetic moments.
I This experiment was repeated with hydrogen atoms (which are characterized by a single electron) by
Phipps and Taylor in 1927.
I Both silver and hydrogen are neutral atoms found in the ground state l = 0 (as opposed to the assumption
of the Bohr-Sommerfeld old quantum mechanics which predicts l = 1).
I Yet, in both cases the Stern-Gerlach experiment shows that the beam of atoms splits in fact into two
beams, as originally thought by Bohr and Sommerfeld but not for the right reasons, signaling therefore the
existence of another quantum property.
I This quantum property is the so-called spin angular momentum with value s = 1/2 which was first
proposed by two graduate students (Goudsmit and Uhlenbeck) and then confirmed later by Dirac in his
quantum relativistic theory of the electron.
I In the case of the hydrogen atom the extra spin angular momentum is associated with its single electron
whereas in the case of the silver atom the extra spin quantum number is associated with its unpaired
outermost electron.
I Thus, the Stern-Gerlach experiment invalidates both classical mechanics and the old quantum theory. The
Stern-Gerlach experiment is a major vindication of the new quantum theory of Heisenberg, Schrodinger,
Dirac, Pauli and Bohr.
The Stern-Gerlach experiment and discovery of the spin
I In general if a beam of atoms with total angular momentum J~ = ~ L+S~ is prepared in the orbital ground
state l = 0 then allowed to go through the Stern-Gerlach appartus it will be observed to split into 2s + 1
beams corresponding to the eigenvalues ms of the spin angular momentum Sz which are given by
ms = s, s − 1, ..., −s. In other words, if we measure the angle between the spin angular momentum and
the z−axis there can only be 2s + 1 possible values.
I Every atom acts as an electromagnet, i.e. it is characterized by a magnetic dipole moment µ ~ . Thus, it will
experience a torque µ~ ×B ~ in a magnetic field B
~ which makes the dipole moment µ ~ want to align with the
~ giving therefore a minimum energy. The corresponding potential energy is
direction of the magnetic field B
given by

H = −~ ~
µ.B.
I In the simplest case of the hydrogen atom we have a single electron moving around its orbit and thus
creating a tiny loop of electric current I . The magnetic moment of this current is µl = I .A where A is the
area of the loop. If v is the speed of the electron and r is the radius of the orbit then 2πr is the distance
traveled by the electron in a single period, i.e. v /2πr is the inverse period and ev /2πr is precisely the
current, viz I = ev /2πr where e is the charge of the electron. The angular momentum of the electron is
L = mrv . From all these considerations we have
e e
µl = L⇒µ
~l = − ~
L.
2m 2m
We also note that the magnetic moment and the angular momentum are in opposite directions because of
the negative charge of the electron.
I Similarly, the magnetic dipole moment of a spinning charge is given in terms of its spin angular momentum
~ by the equation (with q = −e)
S
gq gq
µ
~s = ~⇒H =−
S ~ B.
S. ~
2m 2m
The factor g is the gyromagnetic ratio which is given by g = 2 for the electron.
I In a non-uniform magnetic field there exists, in addition to the above torque, a force on the magnetic
dipole given by

~ = ∇(~
F ~
~ µ.B).
The Stern-Gerlach experiment and discovery of the spin
I Let us imagine that the beam is directed in the direction of the y -axis and that the non-uniform magnetic
field is given by

~ = −αx iˆ + (B0 + αz)k̂ , ∇

B ~B~ = 0.
I The parameter α provides the non-uniformity. B0 represents a uniform magnetic field giving rise to the
so-called Larmor precession. This effect is given by the following expectation values of the spin angular
momentum components

~ ~ ~
hSx i = sin α cos γB0 t , hSy i = − cos α sin γB0 t , hSz i = cos α.
2 2 2
~ precesses around the z-axis at a constant angle α with a frequency ω = B0 t (Larmor
I Thus, hSi
frequency). The above result is an instance of Ehrenfest’s theorem, viz

d
~ = h~
hSi ~
µ × Bi.
dt
I We return to the non-uniform magnetic field and we compute the corresponding force. We find the result

~ = γα(−Sx iˆ + Sz k̂).
F
I But we know from Larmor precession that Sx oscillates rapidly and averages to zero leaving only the
z-component of the force, viz

~ = γαSz k̂.
F
I For a spin 1/2 particle the beam is either deflected up for the eigenvalue m = +~/2 of Sz or down for the
eigenvalue m = −~/2.
I This splitting into two beams was observed by Gerlach for the first time in February 1922.
I The correct explanation of the Stern-Gerlach experiment is thus not in terms of Bohr-Sommerfeld old
quantum theory which suggested that the ground state of the silver atom is characterized by a quantized
angular momentum equal l = 1.
I This is because we know that this ground state is in fact characterized by l = 0 and thus does not lead to
any splitting (and even if it was characterized by l = 1 it would have given a splitting into three beames
corresponding to l = +1, 0, −1 and not two beams corresponding to l = +1, −1).
The Stern-Gerlach experiment and discovery of the spin
I The correct explanation is given in terms of the spin angular momentum of the outermost unpaired
electron. The total angular momentum J~ = ~ L+S ~ with l = 0 and s = 1/2 leads to j = 1/2 and thus a
splitting into two beams.
I In general the beam is split into 2s + 1 components corresponding to the 2s + 1 values of the force
associated themselves with the 2s + 1 values of the spin, viz
~ = γα~mk̂ , m = s, s − 1, ..., −s.
F
I This is the basic physics behind the Stern-Gerlach experiment.
I Thus, the Stern-Gerlach experiment is an experimental proof for the spin quantum number and not for
space quantization (although we also know today that orbital angular momentum is indeed quantized).
I More importantly, Stern-Gerlach experiment is an experimental proof for the new theory of quantum
mechanics.

Stern-Gerlach experiment. Source: https://physicsworld.com/

Discovery of the spin. Source: https://plato.stanford.edu/

The qubit model of the spin
I In summary, the Stern-Gerlach experiment comes with two main results:
I ) First: The magnetic dipole moment of the atoms is quantized not continuous (discrete set of angles
instead of a continuous distribution).
II ) Second: In the ground state of atoms we have zero orbital angular momentum l = 0 which corresponds to
zero magnetic dipole moment µ = 0 giving rise therefore to no deflection at all. Yet, we observe two peaks.
I Conclusion: There must exist a new physical quantity making an extra contribution to the magnetic dipole
moment (spin). This contribution has no relation with the rotational motion of the electron.
I The spin can be characterized by a single bit or more precisely by a single quantum bit or qubit as follows.
1) The output of the Stern-Georlach apparatus consists of two beams | + Z i (up) and | − Z i (down). We
call this the z−Stern-Gerlach apparatus (since it measures the spin or the qubit in the ẑ-direction as it is
oriented in the ẑ-direction).
2) We place now an x-Stern-Gerlach apparatus oriented in the x̂-direction (and thus measures the spin or
qubit in the x̂-direction) in series with the original ẑ-Stern-Gerlach apparatus.
3) Let us block the beam |Z −i. Thus, all transmitted atoms have magnetic dipole moment which is up. And
classically since the magnetic dipole moment is oriented in the ẑ-direction there will be no deflection in a
magnetic field oriented in the x̂-direction. In other words, one should have one central peak. However,
again we observe two peaks labeled | + X i, | − X i.
4) Let us again block the beam | − X i and place a third z-Stern-Gerlach apparatus in the way of the beam
| + X i. We expect to see one peak corresponding to the fact that the atoms in the beam | + X i retained
their | + Z i orientation. But again we observe a splitting into two beams | + Z i and | − Z i.
5) The conclusion is that the state | + Z i contains equal amounts of | + X i and | − X i and the state
| + X i contains equal amounts of | + Z i and | − Z i.
I In terms of the qubit computational states |0i and |1i we have then the following correspondence :

| + Z i ≡ |0i , | − Z i ≡ 1i.

1 1
| + X i ≡ |+i = √ (|0i + |1i) , | − X i ≡ |−i √ (|0i − |1i).
2 2
I Thus, the z-Stern-Gerlach apparatus measures the spin quantum number or the qubit in the computational
basis {|0i, |1i}. Similarly, the x-Stern-Gerlach apparatus measures the spin quantum number or the qubit
in the computational basis {|+i, |−i}.
I The spin can thus be completely captured by the qubit model.
The qubit model of the spin

Source: Nielsen and Chuang

Addition of angular momenta
~ in addition to the orbital angular momentum ~
I In the presence of a non-zero spin angular momentum S, L,
the total angular momentum J~ becomes given by

J~ = ~ ~
L + S. (21)

I We have two sets of compatible operators: {~ L2 , L3 , S

~2 , S3 } and {~
L2 , S
~2 , J~2 , J3 }.
I The eigenvectors |lsjj3 i of J~2 , J3 , ~
L2 and S
~2 can be given as linear combinations of the eigenvectors
|lmi|sσi of ~ L2 , L3 , S
~2 and S3 with coefficients C lmsσ known as the Clebsch-Gordan coefficients.
jj3
I Explicitly, we have

~2 2
L |lmi|sσi = ~ l(l + 1)|lmi|sσi , L3 |lmi|sσi = ~m|lmi|sσi
~2 |lmi|sσi = ~2 s(s + 1)|lmi|sσi , S3 |lmi|sσi = ~σ|lmi|sσi.
S (22)

~2 2 ~2 |lsjj3 i = ~s(s + 1)|lsjj3 i

L |lsjj3 i = ~ l(l + 1)|lsjj3 i , S
2 2
J~ |lsjj3 i = ~ j(j + 1)|lsjj3 i , J3 |lsjj3 i = ~j3 |lsjj3 i. (23)
I For spin s = 1/2 we have explicitly

lj3 − 1 1 1 1 11 lj3 + 1 1 − 1 1 1 1
|lsjj3 i = Cjj 2 2 2 |lj3 − i| i + Cjj 2 2 2 |lj3 + i| − i
3 2 22 3 2 2 2
1 1
= A|lj3 − i|+i + B|lj3 + i|−i. (24)
2 2
I By using the condition J~2 = ~
L2 + S
~2 + 2L3 S3 + L+ S− + L− S+ we can compute the two constraints
s
1 1 1 1 1
 
2 2 2
J~ |lj3 − i|+i = ~ l(l + 1) + j3 + |lj3 − i|+i + ~ l(l + 1) − j32 + |lj3 + i|−i. (25)
2 4 2 4 2

s
1 1 1 1 1
 
2 2 2
J~ |lj3 + i|−i = ~ l(l + 1) − j3 + |lj3 + i|−i + ~ l(l + 1) − j32 + |lj3 − i|+i. (26)
2 4 2 4 2
Addition of angular momenta

I Using now the condition J~2 |lsjj3 i = ~2 j(j + 1)|lsjj3 i and comparing with the above two constraints leads
to two equations in the unknowns A and B which are equivalent. The first equation reads

s
1 1
 
A l(l + 1) + j3 + +B l(l + 1) − j32 + = j(j + 1)A. (27)
4 4

I We must also have |A|2 + |B|2 = 1. Explicitly, we get the solution

 r r

l+ 1 +j l+ 1 −j3


2 3 2
, j = l + 1/2


 2l+1 2l+1
(A, B) = r r . (28)
l+ 1 −j l+ 1 +j3

3


 2 , − 2 j = l − 1/2
2l+1 2l+1


Central problem and Bohr’s atom
I The three dimensional Schrodinger’s equation in position basis in spherical coordinates for a central
potential V (r ) reads

∂ ~2 ∂ ∂ L̂2
 
2
i~ Ψ = − (r )+ + V (r ) Ψ. (29)
∂t 2mr 2 ∂r ∂r 2mr 2
I We solve this equation by separation of variables. First, we write

iEn t
−
Ψ = Ψ(t, ~
r ) = ψnlm (~
r )e ~ . (30)
I The wave function ψnlm (~
r ) solves the differential equation

~2 ∂ ∂ L̂2
 
2
En ψnlm = − (r )+ + V (r ) ψnlm . (31)
2mr 2 ∂r ∂r 2mr 2
I Second, we separate the variables further as follows

m unl (r ) m
r ) = Rnl (r )Yl (θ, φ) =
ψnlm (~ Yl (θ, φ). (32)
r
I The radial wave function unl (r ) satisfies the differential equation

~2 d 2 unl ~2 l(l + 1)
 
− + V (r ) + unl = En unl . (33)
2m dr 2 2m r2
This is Schrodinger equation in one dimension with an effective potential given by

~2 l(l + 1)
Veff (r ) = V (r ) + . (34)
2m r2
The second term is called the centrifugal term. It tends to push the particle away from the origin. The
normalization condition reads
Z ∞ Z ∞
2 2 2
r dr |Rnl (r )| = dr |unl (r )| = 1. (35)
0 0
Central problem and Bohr’s atom
I In the hydrogen atom the proton is much heavier than the electron, i.e. mp  me .
I In the limit mp /me −→ 0 the proton is identified with the center of mass of the system which is at rest at
the origin while the electron is identified with the reduced mass, i.e. ~
r is the vector position of the electron.
I The Coulomb potential corresponding to the hydrogen atom is given by

e2 1
V (r ) = − . (36)
4π0 r
I We are looking for bound states and thus E < 0. We define κn and ρ by
√
−2mEn
κn = , ρ = κn r . (37)
~
I Next, we remove the asymptotic behaviors at ρ −→ ∞ and ρ −→ 0 which are given by ρl+1 and e −ρ
respectively by considering the following ansatz
l+1 −ρ
unl (r ) = ρ e vnl (ρ). (38)
I The function vnl (ρ) is found to satisfy the differential equation

d 2 vnl (ρ) dvnl me 2

 
ρ + 2(l + 1 − ρ) + ρ0n − 2(l + 1) vnl = 0 , ρ0n = . (39)
dρ2 dρ 2π0 ~2 κn
I We consider now a solution given by the power series

∞
j
X
vnl (ρ) = cj ρ . (40)
j=0

I By inserting this power series into the above differential equation we obtain

∞  
j
X
(j + 1)(j + 2l + 2)cj+1 + (ρ0n − 2(j + l + 1))cj ρ = 0. (41)
j=0
Central problem and Bohr’s atom
I In other words, we must have the recursion relation

2(j + l + 1) − ρ0n
cj+1 = cj . (42)
(j + 1)(j + 2l + 2)
I For large j we have cj+1 ' 2cj /(j + 1).
I By assuming that this result is exact we obtain cj = 2j c0 /j!.
I By substituting in we obtain the function vnl (ρ) = c0 e 2ρ .
I In other words, the radial wave function is given by un (ρ) = c0 ρl+1 e ρ which has the wrong asymptotic
behavior at ρ −→ ∞.
I This means that the series must terminate and as a consequence there must exist a maximum value jmax
of j such that cjmax +1 = 0.
I In summary , we have

jmax
j
X
vnl (ρ) = cj ρ ⇒ cj = 0 , j > jmax ⇒
j=0

jmax  
j
X
(j + 1)(j + 2l + 2)cj+1 + (ρ0n − 2(j + l + 1))cj ρ = 0 ⇒
j=0

2(j + l + 1) − ρ0n
cj+1 = cj , j < jmax
(j + 1)(j + 2l + 2)
and
cjmax +1 = 0 ⇒ 2(jmax + l + 1) − ρ0n = 0. (43)

I In other words, jmax must satisfy 2(jmax + l + 1) − ρ0n = 0. Instead of jmax we work with the integer
n defined by n = jmax + l + 1. Thus, we have (with En ≡ En0 )

me 2 ~2 me 2
2
1 1

0
ρ0n = 2n ⇒ κn = ⇒ En =− . (44)
4π0 ~2 n 2m 4π0 ~2 n2
Central problem and Bohr’s atom
I This is the celebrated Bohr quantization formula of the energy levels of the hydrogen atom given explicitly
by

e2 α2
2
m 1 13.6eV

0 2
En =− =− mc = −
2~2 4π0 n2 2n2 n2
e2 1
α= = . (45)
4π0 ~c 137.036
The constant α is known as the fine structure constant. This celebrated number determines the coupling
constant of electromagnetic interactions.
I The complete spatial wave functions read

ρl+1 −ρ m
ψnlm (r , θ, φ) = e vnl (ρ)Yl (θ, φ). (46)
r
I The ground state corresponds to n = 1 with energy E1 = −13.6 eV . Clearly, in this case jmax = 0 and
as a consequence l = 0, vnl (ρ) = c0 and R10 = c0 κ1 e −κ1 r . The normalization condition fixes c0 . The
constant κ1 is the inverse of the so-called Bohr radius, viz
p
−2mE1 1
κ1 = = . (47)
~ a
I In general the function vnl (ρ) is a polynomial of degree jmax = n − l − 1 in ρ. It is equal, modulo a
2l+1
normalization, to the associated Laguerre polynomial Ln−l−1 (2ρ).
I The associated Laguerre polynomials Lp (x) are defined in terms of the Laguerre polynomials Lq (x) by
q−p
p q
d d
 
p p x −x q
Lq−p (x) = (−1) Lq (x) , Lq (x) = e (e x ). (48)
dx dx
I Clearly, for a fixed value of n the quantum number l can only take the values 0, 1,...,n − 1.
I For every fixed value of l the quantum number m can take the 2l + 1 values −l,−l + 1,....,l − 1,l.
Pn−1
I Thus, for a fixed value of n there is
l=0
(2l + 1) = n2 states. This is the degree of degeneracy of the
energy level En .
Bohr’s hydrogen atom as a harmonic oscillator problem
I Schrodinger equation for the hydrogen atom can be rewritten in the form

2 λ 4
~ +
(4∇ − α )ψ = 0
r
8 4 32π0 E
λ= , α =− . (49)
a e2a
I We introduce the stereographic complex coordinates ξA and ξB by the transformation
2 2 2
x + iy = 2ξA ξ̄B , z = ξA ξ̄A − ξB ξ̄B ⇒ r = x + y + z = ξA ξ̄A + ξB ξ̄B . (50)
I Schrodinger equation becomes (with ∂A = ∂/∂ξA , ∂¯A = ∂/∂ ξ̄A , etc)
 
¯ ¯ 4
4∂A ∂A + 4∂B ∂B + λ − α (ξA ξ̄A + ξB ξ̄B ) ψ = 0. (51)

I This is precisely the Schrodinger equation for a four-dimensional harmonic oscillator. Indeed, we can make
the change of variables
ξA = q1 + iq2 , ξB = q3 + iq4 . (52)
I The above Schrodinger equation takes then the form

~2 1 0
 
2 2 2 2 2 2 2 2 2
− (∂1 + ∂2 + ∂3 + ∂4 ) + mω (q1 + q2 + q3 + q4 ) ψ=E ψ
2m 2
0
2mE 2 mω
λ= , α = . (53)
~2 ~
This is indeed the Schrodinger equation for a four-dimensional harmonic oscillator.
I The fundamental commutation relations are immediately given by (with µ, ν = 1, ..., 4)

[q̂µ , p̂ν ] = i~δµν . (54)

I This four-dimensional harmonic oscillator can also be regarded as a coupled system of two two-dimensional
harmonic oscillators.
Bohr’s hydrogen atom as a harmonic oscillator problem
I Indeed, by separating the variables as ψ = ψA (q1 , q2 )ψB (q3 , q4 ), we obtain the Schrodinger equations
for the two-dimensional harmonic oscillators A and B which are characterized by the complex coordinates
ξA = q1 + iq2 and ξB = q3 + iq4 , viz

~2 1 0
 
2 2 2 2 2
− (∂1 + ∂2 ) + mω (q1 + q2 ) ψA = EA ψ A
2m 2
~2 1 0
 
2 2 2 2 2
− (∂3 + ∂4 ) + mω (q3 + q4 ) ψB = EB ψB
2m 2
0 0 0
E = EA + EB . (55)
I However, these two-dimensional harmonic oscillators A and B are not decoupled. This is seen as follows.
1) The complex coordinates ξA and ξB (rotation group U(2)) determine uniquely the Cartesian coordinates
x, y and z (rotation group O(3)). But, x, y and z determine ξA and ξB only to within an arbitrary value
of the angle σ defined by
σ = argξA ξB . (56)
2) Indeed, we have

√ θ i σ+φ √ θ i σ−φ
ξA = r cos e 2 , ξB = r sin e 2 . (57)
2 2
3) The angle σ is clearly unphysical and hence the wave function ψ must be independent thereof, viz
∂ψ
=0 ⇒ (ξ̄A ∂¯A − ξA ∂A )ψ = −(ξ̄B ∂¯B − ξB ∂B )ψ
∂σ
⇒ (q1 ∂2 − q2 ∂1 )ψ = −(q3 ∂4 − q4 ∂3 )ψ
⇒ LA ψ = −LB ψ. (58)
Thus the two-dimensional harmonic oscillators A and B have equal but opposite angular momenta.
I In fact, equation (58) is the constraint which reduces the number of independent variables form 4
(harmonic oscillator coordinates qµ ) back to 3 (Cartesian coordinates x, y and z).
I Bohr’s hydrogen atom is equivalent to the two equations (55) and (58) which describe a system of two
coupled two-dimensional harmonic oscillators. The coupling is such that their total angular momentum is
zero.
Bohr’s hydrogen atom as a harmonic oscillator problem 0 0 0 0
I The two-dimensional harmonic oscillators A and B have energy levels E = E
A nA lA and EB = EnB lB which
must be quantized as follows
0 0 0 0
EA = En l = ~ω(2nA + |lA | + 1) = ~ω(NA + 1) , EB = En l = ~ω(2nB + |lB | + 1) = ~ω(NB + 1). (59)
AA B B

The quantum numbers nA and nB take the values 0, 1, 2,.... The degrees of degeneracy of the energy
0 0
levels EA and EB are NA + 1 and NB + 1 respectively.
I The quantum numbers lA and lB take the values 0, ±1, ±2,.... They are precisely the angular momentum
quantum numbers associated with the angular momentum operators LA and LB , i.e.

LA = ~la , LB = ~lB . (60)

I The angular momenta L̂A = ij q̂i p̂j = −i~ij q̂i ∂j and L̂B = ab q̂a p̂b = −i~ab q̂a ∂b satisfy also (with
i, j = 1, 2 and a, b = 3, 4)

[L̂A , q̂i ] = −i~ij q̂j , [L̂A , p̂i ] = −i~ij p̂j , [L̂B , q̂a ] = −i~ab q̂b , [L̂B , p̂a ] = −i~ab p̂b . (61)
0
I Equation (58) imposes the constraint lA = −lB = l whereas equation (55) gives the energy levels E as

0 0 0 2~2 ~2 ~2 8
 
2
EA + EB = 2~ω(nA + nB + |l| + 1) = E ⇒ α (nA + nB + |l| + 1) = λ=
m 2m 2m a
2 2 1
⇒ α = , n = nA + nB + |l| + 1
a n
32π0 E 4 1
⇒ − =
e2a a2 n 2
 2 2
m e 1
⇒ E =− . (62)
2~2 4π0 n2
These are precisely Bohr’s energy levels, i.e. E ≡ En . It is not difficult to check that the degeneracy of
each energy level n is indeed n2 .
I For more detail see Cornish 1983 where it is also noted that this construction is equivalent to solving the
hydrogen atom in parabolic coordinates (Condon and Shortley 1953).
Bohr’s hydrogen atom as a harmonic oscillator problem
I The underlying physical reason behind the quantization of the energy levels in Bohr’s atom consists
therefore of the following facts.
1 The radial wave function unl (r ) must vanish at infinity causing its series expansion around the origin to
become cutoff.
2 Thus, the probability of finding the electron at any fixed location in space is finite.
3 And, the total probability of finding the electron somewhere in space is finite equals precisely 1.
4 All this is a reflection of the fact that the electron behaves in fact as a wave according to the wave/particle
duality.
5 Indeed, the Heisenberg uncertainty principle can be written down in a straightforward way by going to the
harmonic oscillator formulation of the problem in which Bohr’s hydrogen atom can be described by a
system of two coupled two-dimensional harmonic oscillators where the coupling is such that their total
angular momentum is zero.
6 Effectively, we formulate the problem in a six-dimensional phase space: q1 , q2 , q3 , q4 and LA , LB . We
have then to quantize this phase space by postulating the extended fundamental commutation relations

[q̂i , p̂j ] = i~δij , [q̂a , p̂b ] = i~δab . (63)

[L̂A , q̂i ] = −i~ij q̂j , [L̂A , p̂i ] = −i~ij p̂j , [L̂B , q̂a ] = −i~ab q̂b , [L̂B , p̂a ] = −i~ab p̂b . (64)

These commutation relations provide the desired Heisenberg uncertainty principle. In particular, the second
set (64) reduces the number of degrees of freedom from 6 to 4.
7 In addition to the above commutation relations (63) and (64) which describe a quantized four-dimensional
phase space, we must also have the constraint

L̂A + L̂B = 0. (65)

In fact, without this constraint the system genuinely describes a four-dimensional harmonic oscillator or
equivalently a noncommutative space R4θ . In other words, this constraint reduces the number of degrees of
freedom from 4 (harmonic oscillator coordinates qµ ) back to 3 (Cartesian coordinates x, y and z).
8 The hydrogen atom should then be viewed as providing or defining a novel quantized phase space, i.e. a
noncommutative space in three dimensions.
Time-independent perturbation theory
I The unperturbed problem is given in terms of a Hamiltonian H 0 by
0 0 0 0 0 0
H |ψn i = En |ψn i , hψn |ψm i = δnm . (66)
I Now, let H be some other Hamiltonian which can be written as
0 1
H = H + λH . (67)

The Hamiltonian λH 1 is called the perturbation where λ is a small control parameter.

I The perturbed problem is defined by

H|ψn i = En |ψn i. (68)

I Non-degenerate perturbation theory: We assume that the eigenvalues En are non-degenerate. The
first-order non-degenerate pertrubation theory is given by

0 1 1 0 1 0
En = En + λEn , En = hψn |H |ψn i. (69)

I Degenerate perturbation theory: Let us suppose now that there are two unperturbed states |ψ 0 i and
a
|ψb0 i which share the same unperturbed energy E 0 , i.e. we have
0 0 0 0 0 0 0 0
H |ψa i = E |ψa i , H |ψb i = E |ψb i , hψa |ψb i = 0. (70)

I In fact, any linear combination of |ψ 0 i and |ψ 0 i is also an eigenstate of H 0 with the energy E 0 :
a b
|ψ i = α|ψa i + β|ψb i. Generally, the perturbation λH 1 will lift this degeneracy.
0 0 0
I The first-order degenerate perturbation theory requires then the diagonalization of the pertrubation matrix
0 1 0
Wij = hψi |H |ψj i. (71)

I Indeed, the two solutions are given by

1
 q 
0 0 1 1
E± =E + λE± , E± = Waa + Wbb ± (Waa − Wbb )2 + 4|Wab |2 . (72)
2
I In the case of an n-fold degenerate energy level E 0 the first-order corrections E 1 will be given by the
eigenvalues of the n × n pertrubation matrix Wij = hψi0 |H 1 |ψj0 i.
Time-independent perturbation theory
I Degenerate perturbation theory in the presence of a symmetry: Let A be a hermitian operator which
commutes with H 0 and H 1 . We assume that |ψa0 i and |ψb0 i are also eigenvectors of A with distinct
eigenvalues, viz
0 0 0 0
A|ψa i = µ|ψa i , A|ψb i = ν|ψb i , µ 6= ν. (73)

I Since [A, H 1 ] = 0 we compute

0 1 0 0 1 0 0 1 0
0 = hψa |[A, H ]|ψb i = hψa |AH |ψb i − hψa |H A|ψb i
= (µ − ν)Wab . (74)
I Since µ 6= ν we conclude that Wab = 0. In other words, we have the two solutions
1 1
E+ = Waa , E− = Wbb . (75)

I This is the result we would obtain using non-degenerate first-order perturbation theory. The plus sign
corresponds to α = 1 and β = 0, i.e. |ψ 0 i = |ψa0 i whereas the minus sign corresponds to α = 0 and
β = 1, i.e. |ψ 0 i = |ψb0 i.
I In summary, if we can find a hermitian operator A which commutes with H we can choose the common
eigenvectors as the unperturbed states and use ordinary first-order perturbation theory.

Source: https://slideplayer.com/slide/15719370/
The spin-orbit coupling and fine structure of hydrogen
The fine structure of Bohr’s atom, which is of order α4 mc 2 , consists of two effects:
1) The relativistic correction given by the perturbation

1 (pi2 )2
λHr =− . (76)
8m3 c 2
This is the relativistic correction to the kinetic energy.
2) This spin-orbit coupling which is due to the interaction between the magnetic field of the proton and the
magnetic dipole moment of the electron (in the electron rest frame).
Alternatively, the spin-orbit coupling can be given by the interaction between the electric field of the proton
and the electric dipole moment of the electron in the laboratory frame.
The spin-orbit correction, in the electron rest frame, is given by the perturbation

1 1
λHso = −( )~ ~int .
µe . B (77)
2
The factor of 1/2 is due to the fact that the electron rest frame is not inertial (Thomas precession).
~
~ e is the magnetic dipole moment of the electron which is proportional to its spin S
In the above equation µ
where the proportionality factor is called the gyromagnetic ratio and is given classically by ge = q/2m.
Explicitly, we have
q
µ
~ e = 2. ~ , q = −e.
S (78)
2m
The extra factor of 2 can only be calculated using the Dirac equation.
We remark that this correction to the gyromagnetic ratio cancels exactly Thomas precession.
In the electron rest frame the proton circles around the electron and as a consequence it creates a
magnetic field B~ given by the Biot-Savart law, viz
µ0 µ0 e e e
Bint = I = = = L. (79)
2r 2r T 20 rc 2 T 4π0 r 3 mc 2

The external magnetic field B~int is in fact in the direction of the orbital angular momentum ~ r ×~
L = m~ v
of the electron (in the proton rest frame). Thus, we must have

e
~int =
B ~
L. (80)
4π0 r 3 mc 2
The spin-orbit coupling and fine structure of hydrogen
I The relativistic perturbation is spherically symmetric and thus it commutes with the square L2 of the
i
angular momentum and with its third component L3 (Li ≡ L̂i ).
I In other words, n, l and m are ”good” quantum numbers and we can use the basis {|ψ 0 i} of the
nlm
unperturbed Hamiltonian H 0 .
I If the spin is included then {|Ψ0 0
nlmsσ i ≡ |ψnlm i|sσi} is the appropriate basis, i.e. n, l, s, m and σ are
the ”good” quantum numbers.
I As we will show, the ”good” quantum numbers in the case of the spin-orbit coupling are in fact given by n,
l, s, j and j3 where j and j3 define the basis of the total angular momentum J~ = ~ L + S.~ Hence, we need to
use the basis {|Ω0nlsjj i} which can be expanded in the basis {|Ψ0nlmsσ i}.
3
I Thus, in order to calculate the relativistic and the spin-orbit corrections we will use first-order
non-degenerate perturbation theory with the unperturbed states |Ψ0nlmsσ i and {|Ω0nlsjj i} respectively, viz
3

1 0 1 0 1 0 1 0
λEr = λhΨnlmsσ |Hr |Ψnlmsσ i , λEso = λhΩnlsjj |Hso |Ωnlsjj i. (81)
3 3

I The relativistic correction can be computed using first-order non-degenerate perturbation theory with the
states |Ψ0nlmsσ i since [Hr1 , ~
L] = 0. This is given explicitly by

1 0 1 0
λEr = λhΨnlmsσ |Hr |Ψnlmsσ i
0 1 0
= λhψnlm |Hr |ψnlm i
1 0 2 2 0 2 0 0 0
= − hψnlm |(p̂i ) |ψnlm i , p̂ |ψnlm i = 2m(En − V )|ψnlm i
8m3 c 2
1 0 0 2 0
= − hψnlm |(En − V ) |ψnlm i
2mc 2
1 0 1 1 1 α mc 1 α2 m 2 c 2 1
 
0 2 2 2 2
= − (En ) + 2~cαEn h i +~ c α h i ,h i = , h i=
2mc 2 r r2 r n2 ~ r2 ~2 n3 (l + 21 )

(En0 )2 4n
 
= − −3 . (82)
2mc 2 l + 12
The spin-orbit coupling and fine structure of hydrogen
I The spin-orbit interaction of the hydrogen atom is described by the perturbation

1 e2 1
λHso = ~~
S.L
8π0 m2 c 2 r 3
e2 1 2 2
J~ − ~ ~2 , J~ = ~ ~


= L −S L + S. (83)
16π0 m2 c 2 r 3
I We remark immediately that this perturbation does not commute with L3 and S3 separately. But it
commutes with J3 = L3 + S3 where J~ is the total angular momentum.
I Thus, λH 1 commutes with ~ L2 , S
~2 , J~2 and J3 and as a consequence the ”good” quantum numbers are l,
so
s, j, j3 and n (and not l, s, m, σ and n). In fact, the unperturbed Hamiltonian H 0 itself commutes with
L2 , S
~ ~2 , J~2 and J3 .
I Thus, the unperturbed eigenvectors will be taken to be the common eigenvectors of H 0 , ~ L2 , S
~2 , J~2 and
J3 . These are given by
0
|Ωnlsjj i = |Rnl i|jj3 i. (84)
3

This should be compared with |Ψ0nlmsσ i = |ψnlm 0

i|sσi = |Rnl i|lmi|sσi which are the common
eigenvectors of H 0 , L̂2 , Lˆ3 , Ŝ 2 , Ŝ3 . The unperturbed energies corresponding to either |Ω0nlsjj i or
3
|Ψ0nlmsσ i are still given by Bohr’ energies.
L2 , S
I The angular momentum states |jj3 i are the eigenvectors of ~ ~2 , J~2 and J3 with eigenvalues ~2 j(j + 1)
and ~j3 of J~2 and J3 respectively where j = l ± 1/2 and m = j, j − 1, ..., −j. They are given explicitly by
X lmsσ
|jj3 i = Cjj |lmi|sσi
3
m,σ

lj3 − 1 s 1 1 1 lj3 + 1 s 1 1 1
= Cjj 2 2 |lj3 − i|s i+ Cjj 2 2 |lj3 + i|s − i. (85)
3 2 2 3 2 2
The coefficients Cjjlmsσ are the Clebsch-Gordon coefficients which satisfy among other things Cjjlmsσ = 0
3 3
unless j3 = m + σ.
The spin-orbit coupling and fine structure of hydrogen
I We have shown that J~2 and J3 commute with the total Hamiltonian. Thus, we can compute the spin-orbit
correction to Bohr’s energy levels using ordinary first-order non-degenerate perturbation theory with the
common eigenvectors |Ω0nlsjj i as follows
3

1 0 1 0 e2 1 ~2 3 1
λEso = hΩnlsjj |λHso |Ωnlsjj i = (j(j + 1) − − l(l + 1))hRnl | |Rnl i
3 3 8π0 m2 c 2 2 4 r3
α~3 3 1 1 α3 m 3 c 3
= (j(j + 1) − − l(l + 1))h i, h i=
4m2 c 4 ~3 n3 l(l + 1)(l + 21 )
r3 r3

(En0 )2 n 3
 
= j(j + 1) − − l(l + 1) . (86)
mc 2 l(l + 1)(l + 12 ) 4
I Explicitly, we compute


(En0 )2 2n 1 j = l + 21


 2mc 2 l+ 1 l+1

1 2
λEso = . (87)
(En0 )2 2n 1 1
− j =l−


 2mc 2 l+ 1 l
 2
2
I In summary, the fine-structure correction of the hydrogen atom is given by

0 2
 (En ) (3 − 4n ) j = l + 1 (En0 )2

4n

1 1 2mc 2 l+1 2 1 1
λEr + λEso = 0 2 ⇒ λEr + λEso = (3 − 1
). (88)
(En ) 4n 1 2mc 2 j +
2 (3 − l ) j =l− 2


 2
2mc

I The energy levels of the hydrogen atom become

α2 n 3
 
0 1 1 0
Enj = En + λEr + λEso = En 1+ ( − ) . (89)
n2 j + 12 4

The degeneracy in l is lifted. But, there is degeneracy in j.

I This fine structure correction is predicted more naturally and very efficiently by the Dirac equation.
The spin-orbit coupling and fine structure of hydrogen

The fine structure of hydrogen. Source: Griffiths

The hyperfine structure (21 cm cosmology)
I The magnetic dipole moment µ ~ by the
~ of a particle of mass m and charge q is given in terms of its spin S
relation
gq
µ
~ = ~
S. (90)
2m
Here, g is the gyromagentic ratio of the particle. For the electron and proton we have g = gp = 5.59 and
g = g2 = 2 respectively.
I According to classical electrodynamics the above magnetic dipole generates a magnetic field given by the
relation
µ0 2µ0 3
~ =


B 3(~
µ.ˆ r −µ
r )ˆ ~ + µ
~ δ (~
r ). (91)
4πr 3 3
δ 3 (~
r ) is Dirac’s delta function. Recall also that the speed of light is given by
1 8
c = √ = 2.998 × 10 m/s. (92)
µ0 0

I The Hamiltonian λHhf = −~ ~p describing the motion of the electron in the magnetic field B
µe . B ~p due to
the proton’s magnetic dipole moment is given by
1 ~p
λHhf = −~
µe .B
µ0 2µ0
 
3
= − 3(~
µp .ˆ
r )(~ r) − µ
µe .ˆ µe −
~ p .~ µ
~ p .~
µe δ (~
r)
4πr 3 3
µ0 gp ge e 2 2µ0 gp ge e 2
 
= ~
3(Sp .ˆ ~ ~ ~
r ) − Sp .Se +
r )(Se .ˆ S ~e δ 3 (~
~p .S r ). (93)
3
4πr 4mp me 3 4mp me
I This is called the hyperfine perturbation.
I From the above expression it is expected that the ”good” quantum numbers in this case are s and σ where
~ is now the total spin, viz S
S ~=S ~p + S~e . Indeed, the first term in H 1 vanishes by rotational invariance
hf
(its expectation value vanishes in spherically symmetric orbitals) whereas the second term, proportional to
the delta function, commutes obviously with S ~2 and S3 .
I We can then use first-order non-degenerate pertrubation theory with the states |ψ 0 i|sσi where |sσi
nlm
~2 and S3 , i.e. S
are the eigenvectors of S ~2 |sσi = ~2 s(s + 1)|sσi and S3 |sσi = ~σ|sσi.
The hyperfine structure (21 cm cosmology)
I The states |sσi are given explicitly by

1
S = 0 , |00i = √ (|+i|−i − |−i|+i) : singlet. (94)
2

1
S = 1 , |11i = |+i|+i , |10i = √ (|+i|−i + |−i|+i) , |1 − 1i = |−i|−i : triplet. (95)
2
I The first-order hyperfine correction to Bohr’s energy levels E 0 = −13.6eV /n2 is then given by
n
1 0 1 0
λEhf = hψnlm |hsσ|λHhf |ψnlm i|sσi
µ0 gp ge e 2 ~p .ˆ
(S r )(S~e .ˆ
r) ~p .S
S ~e  2µ0 gp ge e 2

= 3h i−h i + hS ~e δ 3 (~
~p .S r )i. (96)
4π 4mp me r 3 r 3 3 4mp me
I We consider only spherically symmetric orbitals, i.e. we assume that l = m = 0. In these orbitals the only
non-trivial angular integral is of the form
4π
Z
sin θdθdφ(~
v .ˆ
r )(~
w .ˆ
r) = (~
v .~
w ). (97)
3
I We can immediately show that

~p .ˆ
(S ~e .ˆ
r )(S r) ~p .S
S ~e
3h i−h i = 0. (98)
r3 r3
I The first-order hyperfine correction reduces to

1 2µ0 gp ge e 2 3 2µ0 gp ge e 2 0 2 ~ ~
λEhf = ~ ~
hSp .Se δ (~
r )i = |ψn00 | hS p .Se i. (99)
3 4mp me 3 4mp me
I We consider the most important case which is the ground state itself n = 1 and l = m = 0. And we use
the result (a is Bohr’s radius)

2 1 4π0 ~2 −11
|ψ100 (0)| = , a= = 5.292 × 10 m. (100)
πa3 me e 2
The hyperfine structure (21 cm cosmology)
I The first-order hyperfine correction to the ground state energy E 0 is then given by
1

1 4gp ~2 2gp ~2 3~2

λEhf = ~p .S
hS ~e i = ~2 −
h(S )i
3mp me2 c 2 a4 3mp me2 c 2 a4 2
2gp ~4 3
= (s(s + 1) − )
3mp me2 c 2 a4 2

 g p ~4
− S = 0 : singlet


mp me2 c 2 a4
= . (101)
g p ~4
S = 1 : triplet


 3m m2 c 2 a4

p e

I The transition between the triplet and the singlet states is then given by the following energy, frequency
and wavelength

1 1 4gp ~4 −6
∆Ehf = λ(Ehf |S=1 − Ehf |S=0 ) = = 5.88 × 10 eV
3mp me2 c 2 a4
∆Ehf
ν = = 1421MHz
h
c
λ = = 21cm. (102)
ν
I This radio microwave radiation is important in cosmology especially in the study of the epoch of
reionization (EoR). The so-called 21cm cosmology aims at studying the universe during its first 1 billion
years by observing redshifted 21 cm line of atomic hydrogen. In particular, this will allow us to understand
the formation of the first stars and galaxies and thermal evolution of the universe.
I Indeed, at the time of recombination neutral hydrogen atoms are formed and the CMB radiation was
released. From this epoch (around 380 thousand years after the big bang), through the epoch of
reionization of hydrogen in stars (around 400 million years after the big bang), up to the epoch of full
reionization (around 1 billion years after the big bang) the above 21 cm microwave radiation was the
dominant form of radiation. The 21 cm line is observed as an emission line in the CMB for temperature
T > TCMB and redshift z < 10 and as absorption line for T < TCMB and 30 < z < 150.
Stark effect and Lamb shift

Stark effect of the level n = 2. Source: Sakurai

I Here, we discuss two more corrections to the energy levels of Bohr’s atom: Stark effect and Lamb Shift.
I Stark effect: In Stark effect we subject the hydrogen atom to a uniform electric field in the z-direction.
The corresponding potential energy of the electron is given by
V = −eEext z = −eEext r cos θ. (103)
I The first energy level E1 of the hydrogen atom is non-degenerate (by ignoring spin) so we can apply
non-degenerate perturbation theory to compute its first-order correction. We find that the ground state
energy is not affected by this perturbation.
I By employing the Wigner-Eckart theorem we can compute the second-order correction to the ground state.
We find the first-order energy correction

1 2 9a3
∆E1 = − αEext , α= . (104)
2 2
α is the polarizability of the hydrogen atom.
I The first excited energy level E2 of the hydrogen atom is four-fold degenerate: ψ200 , ψ21−1 , ψ210 and
ψ21+1 so we need to use degenerate perturbation theory. By employing symmetry principles we find that
the only non-zero matrix elements of the pertubation matrix are
hψ200 |V |ψ210 i = hψ210 |V |ψ200 i = 3eaEext . (105)
The first-order energy corrections and the corresponding eigenvectors are given by
1
∆E2± = ±3eaEext , |±i = √ (|ψ200 i ± |ψ210 i). (106)
2
Stark effect and Lamb shift
I In summary, the uniform electric field lifts the degeneracy between the two sub-levels 2s and 2p.
I However, if we include the fine structure correction of the hydrogen atom, then the degeneracy between
the two sub-levels 2s and 2p is partially lifted as 2s1/2 and 2p1/2 become separated from 2p3/2 . The
persistence of the degeneracy between 2s1/2 and 2p1/2 justifies the use of degenerate pertrubation theory.
I As it turns out, the remaining degeneracy between 2s1/2 and 2p1/2 is also lifted by the Lamb sift. If the
electric field perturbation matrix is much larger than the Lamb shift then the use of degenerate
perturbation is still justified. In the other extreme, if the the Lamb shift is much larger than the electric
field perturbation matrix then we must use non-degenerate pertrubation theory (the correction will turn out
to be quadratic as in the case of the ground state energy level).
I The time-dependent Stark effect: If we consider now a time-dependent perturbation given by a
time-dependent electric field E ~ext (t) in the z-direction then transitions between the n = 1 and n = 2
levels of the hydrogen become possible.
I In fact, we can verify that we can only access the state |ψ210 i from the ground state |Ψ100 i by means of
this perturbation and as a consequence the system (hydrogen atom+time-dependent electric field) behaves
effectively as a two-level system.
I The time-dependent perturbations, quantum transitions (quantum jumps) and two-level systems will be
discussed in great detail in the next chapter.
I Lamb shift: The two sub-levels 2s1/2 and 2p1/2 of the hydrogen atom were believed to be degenerate
before the experimental discovery of the Lamb shift by Lamb and Retherford in 1947.
I This effect is due to the quantization of the electromagnetic field and it causes the sub-level 2s1/2 to be
higher in energy than the sub-level 2p1/2 .
I In fact, the discovery of the Lamb shift provided a major stimulus for the development of the theory of
quantum electrodynamics (QED) in particular and for the development of quantum field theory (QFT) in
general.
I In the context of quantum field theory Lamb shift is a radiative correction due to the Feynman loop
diagrams depicted below.
1 The first diagram (vacuum polarization) is responsible the production of electron-positron virtual pairs
around the proton causing a screening of its charge.
2 The second diagram (electron mass renormalization) causes a change in the energy of the electron as it
propagates in the vacuum fluctuations of the quantum electromagnetic field.
3 The third diagram (vertex correction) causes a modification of the electron’s magnetic moment called the
anomalous magnetic moment.
Stark effect and Lamb shift

I The Lamb shift is very difficult to compute and it is given by


5 2 1
α mc 4n3 k(n,


0)

l =0
∆ELamb = 5 2 1 1 . (107)
α mc 4n3 k(n, 0) ± π(j+ 1 )(l+ 1 )
 l 6= 0 , j = l ± 21
2 2

I Thus, the Lamb shift lifts the degeneracy in the quantum number l. The effect is more pronounced for
l = 0 (the numerical factor k(n, 0) changes from 12.7 for n = 1 to 13.2 for n −→ ∞).

Feynman diagrams responsible for the Lamb shift. Source: Griffiths

The Zeeman effect
I The magnetic dipole moments of the electron due to the angular momentum ~ L and the spin S~ are given
~ l = q~
respectively by µ L/2m and µ ~
~ s = 2q S/2m.
~ext is then given by the perturbation
I The Zeeman effect of Bohr’s atom in an external magnetic field B

1 e
λHZ = −(~
µl + µ ~ext =
~ s ).B (~ ~ B
L + 2S). ~ext . (108)
2m
~ext should be compared with the internal magnetic field B
I This external magnetic field B ~int responsible for
the spin-orbit coupling. We will assume that B~ext is in the z-direction.
I The total Hamiltonian including the relativistic and spin-orbit corrections of Bohr’s atom is then given by

0 1 1 1 0 1 e e
H + λHr + λHso + λHZ = H + λHr + ~B
S. ~int + (~ ~ B
L + 2S). ~ext . (109)
m 2m
I Here, we distinguish two limiting regimes. In the limit of weak external magnetic field the common
eigenvectors are |Ω0nlsjj i as in the original Bohr’s atom. While in limit of strong external magnetic field
3
the common eigenvectors are in fact given by |Ψ0nlmsσ i.
I In both these regimes we can use first-order non-degenerate perturbation theory.
I In the intermediate region we need to use first-order degenerate perturbation theory (direct and explicit
diagonalization of the pertrubation).
I Weak external magnetic field: In other words, Bext  Bint and thus the spin-orbit interaction
dominates compared with the Zeeman effect. In other words, the Zeeman effect should be treated as a
perturbation and the ”good” quantum numbers are still given by l, s, j, j3 and n.
This means that the first-order correction is computed using first-order non-degenerate perturbation theory
with the states |Ω0nlsjj i, viz
3

1 0 1 0 e
λEZ = hΩnlsjj |HZ |Ωnlsjj i = ~ext hΩ0 ~ ~ 0
3 3
B nlsjj |(L + 2S)|Ωnlsjj i.
3 3
(110)
2m
I This can be computed directly using the explicit expression of the states |Ω0
nlsjj3 i. This will allow us to
confirm that Zeeman effect in the limit of weak external magnetic field is rotational invariant, i.e. the
Hamiltonian effectively commutes with J~2 and J3 and as a consequence j and j3 are the ”good” quantum
numbers.
I This can also be verified using the Wigner-Eckart theorem (or more precisely the projection theorem) as
follows.
The Zeeman effect
I The total angular momentum J~ is conserved. Thus, we decompose the angular momentum ~ L and the spin
~ into the sum of their parallel and perpendicular components along the fixed direction of J.
S ~ We have then

~
L.J~ J~2 − S
~2 + ~
L2
h~
Li = h~
Lk i + h~
L⊥ i = h ~ ~
Ji + hL⊥ i = h ~ + h~
Ji L⊥ i. (111)
J~2 2J~2

~ J~
S. J~2 + S
~2 − ~
L2
~ = hS
hSi ~ i + hS
~⊥ i = h ~ ~
Ji + hS⊥ i = h ~ + hS
Ji ~⊥ i. (112)
k ~2 ~2
J 2J
I The angular momentum ~ ~ precess about the direction of J.
L and the spin S ~ Thus, the time average of their
perpendicular components ~
L⊥ and S~⊥ vanish. We end up with the result

J~2 − S
~2 + ~
L2 J~2 + S
~2 − ~
L2
h~
Li = h ~ , hSi
Ji ~ =h ~
Ji. (113)
2J~2 2J~2
I Equivalently, we have

J~2 + S
~2 − ~
L2
 
h(~ ~
L + 2S)i = h 1+ ~ = gJ hJi
Ji ~
2J~2

j(j + 1) + 34 − l(l + 1)
gJ = 1+ . (114)
2j(j + 1)
I The factor gJ is called Lande g -factor.
I Thus, despite the fact that the total magnetic moment of the electron µ ~ J is not in the direction of the
total angular momentum J~ its expectation value h~ µJ i is in the direction of the expectation value of the
total angular momentum hJi.~
I Although J~ does not commute with the perpendicular components ~ L⊥ and S ~⊥ it clearly commutes with
the parallel components ~Lk and S ~ . Thus, despite the fact that the perpendicular part of the Hamiltonian
k
does not commute with J it nevertheless vanishes by virtue of the fact that the time average of ~
~ L⊥ and
~
S⊥ vanish. As a consequence, the Hamiltonian effectively reduces to its parallel part which commutes with
the total angular momentum J~ and hence the ”good” quantum numbers are indeed j and j3 and we have
rotational invariance.
The Zeeman effect
I Strong external magnetic field: In other words, Bext  Bint and thus Zeeman effect dominates
compared with the spin-orbit coupling. The ”good” quantum numbers in this case are given by l, m, s, σ
and n but now the degeneracy in m and σ is lifted because of the external magnetic field.
I Indeed, in this case the unperturbed Hamiltonian and its corresponding Bohr’s energy levels are given by

0 1 0 e
Hunper = H + λHZ = H + (~ ~ B
L + 2S). ~ext . (115)
2m

0 13.6eV
Enmσ = − + µB Bext (m + 2σ). (116)
n2
I Clearly, |Ψ0 0
nlmsσ i are eigenvectors of Hunper while |Ωnlsjj3 i are no longer eigenvectors.
I The unit of the quantized magnetic moment is given by Bohr magneton µB which is defined by
µB = e~/2m.
I For a strong external magnetic field the fine structure correction is given as before by the relativistic and
spin-orbit corrections, viz

1 1 e
λHr + λHso = λHr + ~B
S. ~int . (117)
m
I This perturbation commutes with ~ L2 and L3 and thus we can use first-order non-degenerate perturbation
theory with the states |Ψ0nlmsσ i.
I The relativistic correction comes out as before.
I But, the spin-orbit correction can be found to be proportional to mσ since

0 1 0 0 ~~ 0 0 0
h|Ψnlmsσ |Hso |Ψnlmsσ i ∝ hΨnlmsσ |S.L|Ψnlmsσ i = hΨnlmsσ |S3 L3 |Ψnlmsσ i. (118)
I The fine structure correction in the presence of a strong external magnetic field is then given by

(En0 )2 4n 4nmσ
 
1 1
λEr + λEso =− −3− . (119)
2mc 2 l + 12 l(l + 1)(l + 21 )
Time-dependent perturbation theory
I Quantum jumps are transitions between Bohr’s energy levels which correspond to the process of emission
or absorption of radiation by atoms.
I These jumps are only possible in the case of time-dependent potentials (non-stationary problems).
I We will then consider a time-dependent Hamiltonian H of the form

H = H0 + V (t). (120)
I Let En and |ni be the eigenvalues and the eigenstates of the time-independent unperturbed Hamiltonian
H0 , viz

H0 |ni = En |ni. (121)

I We will expand the state vector |ψ(t)i for t > 0 as

iEn t
X − X
|ψ(t)i = cn (t)e ~ |ni , |ψ(t)iI = cn (t)|ni. (122)
n n

|ψ(t)iI is the state vector in the interaction or Dirac picture, namely |ψ(t)iI = exp(iH0 t/~)|ψ(t)i.
I The initial state vector at time t = 0 is assumed to be given by |ii, i.e. we must have
X
|ψ(0)i = cn (0)|ni ≡ |ii ⇒ cn (0) = δni . (123)
n

I The Schrodinger equation for |ψ(t)i leads to a coupled system of first-order differential equations for the
probability amplitudes cn (t) = hn|ψ(t)iI . Explicitly, we have

dcm (t) X iωmn t

i~ = cn (t)e V mn (t). (124)
dt n

Em − En
ωmn = . (125)
~

Vmn (t) = hm|V (t)|ni. (126)

Time-dependent perturbation theory
I Dyson series is a formal perturbative solution of this system given by

(0) (1) (2)

cn (t) = cn (t) + cn (t) + cn (t) + .... (127)
I At 0th, 1st and 2nd orders of perturbation theory we have

(0)
cn (t) = δni . (128)

−i
 Z t
(1) iωni t1
cn (t) = dt1 e Vni (t1 ). (129)
~ t0

2 Z t
−i
 Z t
(2) 1 X iω t iω t
cn (t) = dt1 dt2 e nm 1 e mi 2 Vnm (t1 )Vmi (t2 ). (130)
~ t0 t0 m

I The transition probability from the initial state i to the final state n is then given by

2 (0) (1) (2) 2

Pi−→n (t) = |cn (t)| = |cn (t) + cn (t) + cn (t) + ...| . (131)
I As an example, let us consider the step function perturbation which is of the form
(
0 t <0
V (t) = . (132)
V t ≥0

I For n 6= i we compute the first-order transition probability at time t:

(1) 2 4|Vni |2 2 (En − Ei )t

Pi−→n (t) = |cn | = sin . (133)
(En − Ei )2 2~
I For a fixed value of t we study the probability Pi−→n (t) as a function of ωni , viz Pi−→n (t) ≡ P(ωni ).
We obtain the following resonance behavior:
1) The maximum of this probability is at ωni = 0 where it is proportional to t 2 .
2) There are smaller peaks at ωni = (2n + 1)π/t, n = 1, 2....
3) The probability vanishes at ωni = 2nπ/t, n = 1, 2....
4) Thus, the width of the probability P(ωni ) is 1/t.
Fermi’s golden rule
I In summary, the probability Pi−→n (t) = |cn(1) |2 for very large times is always negligible except for states
|ni with energies around En satisfying the Heisenberg uncertainty relation, viz

|En − Ei |t ∼ 2π~. (134)

I This uncertainty relation means that for t −→ 0 we obtain a broader peak and thus transitions with
non-conservation of energy are probable (virtual processes occur at very short times) whereas for t −→ ∞
we obtain a narrower peak and the transitions with En ' Ei are the most probable.
I The total probability for the transition from the initial state |ii to a group of final states |ni with energies
centered around En is

(1) 2
X
Pi−→[n] (t) = |cn | . (135)
En 'Ei

I We assume that the final states constitute a continuum. We introduce a density of final states ρ(E ). In
other words, ρ(E )dE is the number of states with energy between E and E + dE . We can then replace the
above total probability with the integral

4|Vni |2 (En − Ei )t
Z Z
(1) 2 2
Pi−→[n] (t) = dEn ρ(En )|cn | = dEn ρ(En ) sin . (136)
(En − Ei )2 2~

I For very large times we can use the identity sin2 tx/tx 2 = πδ(x) to obtain

2πt
Z
2
Pi−→[n] (t) = dEn ρ(En )|Vni | δ(En − Ei )
~
2πt
 
= ρ(En )|Vni |2 . (137)
~ En =Ei

|Vni |2 is the average of |Vni |2 (the matrix elements Vni are similar for similar final states |ni).
I The transition rate is the transition probability per unit time. We obtain Fermi’s golden rule which is given
explicitly by
d X 2π
 
(1) 2 2
wi−→[n] = |cn | = ρ(En )|Vni | . (138)
dt E 'E ~ En =Ei
n i
Harmonic perturbations and detailed balance
I We consider now the monochromatic harmonic perturbation consisting of a single frequency ω given by
iωt + −iωt
V (t) = Ve +V e . (139)

Again V and V + depend only implicitly on time.

I This perturbation corresponds typically to a monochromatic, polarized and coherent electromagnetic wave.
I At the initial time t = 0 only the eigenstate |ii of H0 is populated.
I We compute immediately the first-order amplitude

−i
Z t
(1) iω t
cn (t) = dt1 e ni Vni (t)
~ 0
i(ωni −ω)t 
1 1 − e i(ωni +ω)t +1−e

= Vni + Vni . (140)
~ ωni + ω ωni − ω
I This should be contrasted with the result obtained for the step function perturbation given by

1 1 − e iωni t
 
(1)
cn (t) = Vni . (141)
~ ωni
I In other words, the only change is

ωni −→ ωni ± ω. (142)

I Hence, for large times t the probability Pi−→n (t) = |cn(1) |2 is important only in the following two
mutually exclusive cases
ωni + ω = 0 ⇔ En = Ei − ~ω : stimulated emission. (143)

ωni − ω = 0 ⇔ En = Ei + ~ω : absorption. (144)

I Clearly, the energy of the system alone is not conserved but it is the combination system plus external
perturbation V (t) which conserves energy.
I The second case (144) is only possible if En > Ei , i.e. if En is an excited state. It corresponds therefore to
absorption where the system receives an energy ~ω from the perturbation V (t).
I The first case (143) is only possible if En < Ei , i.e. if Ei is an excited state. It corresponds therefore to
stimulated emission where the system gives up an energy ~ω to the perturbation V (t).
Harmonic perturbations and detailed balance
I The emission is termed ”stimulated” because it is caused by the perturbation -as opposed of being
spontaneous-.
I For example when we shine a light on an atom in an excited state Ei it can make a transition to the lower
state En .
I In other words, the single incident photon becomes two outgoing photons with the same frequency. This is
the principle of amplification underlying LASER (light amplification by stimulated emission of radiation).
I In LASER all atoms need to be prepared in the upper or excited state Ei (population inversion).
I This system is then triggered by a single incident photon which generates two photons with the same
frequency and these two photons themselves generate four photons and these four generate eight and so
on. This chain reaction produces therefore, virtually instantly, a very large number of photons with the
same frequency.
I Stimulated emission by electromagnetic interaction was first predicted by Einstein in 1917.
I The Fermi’s Golden rule in this case reads

2π
 
stim−emis 2
wi−→[n] = ρ(En )|Vni | . (145)
~ En =Ei −~ω

2π
 
abso + 2
wi−→[n] = ρ(En )|Vni | . (146)
~ En =Ei +~ω

I From the result |Vni |2 = |V + |2 we get the so-called ”detailed balance” which expresses the symmetry
in
between absorption and stimulated emission. This reads

stim−emis abso
wi−→[n] wn−→[i]
= . (147)
ρ(En ) ρ(Ei )
Stimulated emission (LASER) and absorption by radiation
I The most important case of harmonic perturbations is interaction with electromagnetic radiation.
I We will consider a monochromatic, polarized and coherent electromagnetic plane wave characterized by a
single frequency ω and a single polarization vector ~
 and is propagating in the direction n̂.
I We consider thus the field of a monochromatic, polarized and coherent electromagnetic plane wave given
by the following potentials
ω 2π
~ = 2ˆ
φ=0, A A0 cos(~ x − ωt) , ~
k.~ k = k n̂ , k = = . (148)
c λ
The corresponding electric and magnetic fields are given in terms of the scalar potential φ and the vector
~ by E
potential A ~ = −∇φ ~
~ − ∂ A/∂t and B ~ =∇ ~ × A.~
I We will impose the Coulomb gauge condition

~A
∇ ~ = 0 ⇒ ˆn̂ = 0. (149)
~ and B
I The Hamiltonian of a charge q moving under the influence of the electric and magnetic fields E ~ is
given by
1
H = (~ ~ 2 + qφ , φ = 0
p − q A)
2m
p2
~ q q2 A
~2
= − ~p +
A~ ~ 2  |A|
, |A| ~
2m m 2m
~2
p
= + V (t). (150)
2m
I The electromagnetic harmonic perturbation is given explicitly by

iωt + −iωt qA0 ~

−i k.~
x
V (t) = Ve +V e , V =− ˆ~
pe . (151)
m
I The term e iωt V corresponds to stimulated emission whereas the term e −iωt V + corresponds to
absorption, viz
(
e iωt V stimulated emission
V (t) = . (152)
e −iωt V + absorption
Stimulated emission (LASER) and absorption by radiation
I We treat now ~ x and ~
p as quantum operators satisfying the commutation relations [xi , pj ] = i~δij .
I The first-order amplitude cn(1) (t) and the probability Pi−→n (t) = |cn(1) |2 for stimulated emission and
absorption are given respectively by
−i
Z t
(1) iω t
cn (t) = dt1 e ni Vni (t)
~ 0
i(ωni −ω)t 
1 1 − e i(ωni +ω)t +1−e

= Vni + Vni . (153)
~ ωni + ω ωni − ω

stim−emis (1) 2 4|Vni |2 2 (ωni + ω)t

Pi−→n (t) = |cn | = sin , ω ' −ωni . (154)
~2 (ωni + ω)2 2

+ 2
abso (1) 2 4|Vni | 2 (ωni − ω)t
Pi−→n (t) = |cn | = sin , ω ' ωni . (155)
~2 (ωni − ω)2 2
I We compute (with q = e)

2 2
|V |2 = e A0 |ˆ p e −ik n̂~ x
|ii|2

hn|~ stimulated emission

ni m2 . (156)
e 2 A2
|V + |2 = hn|e ik n̂~
0 |ˆ x~
p |ii|2

 absorption
ni m2

+ 2
This satisfies detailed balance |Vni |2 = |Vin | .

The process of stimulated emission (i) versus absorption (ii). Source: Sakurai
Stimulated emission (LASER) and absorption by radiation
~ x ~ x
p , e i k.~
I We can also use in the above equation the facts [~ ke i k.~
] = −i~~ and ˆ.~
k = 0 to write

2 2
|V |2 = e A0 |ˆ p e −ik n̂~ x
|ii|2

hn|~ stimulated emission

ni m2 . (157)
e 2 A2
|V + |2 = 0 |ˆ p e ik n̂~
hn|~ x
|ii|2

 absorption
ni m2

I We can now make the approximation that the wavelength of the radiation is much larger than the size of
the atoms, viz |k n̂~
x | = 2π|n̂~
x /λ|  1.
I Hence, we can approximate the exponential by 1. This is the electric dipole approximation. The matrix
elements for stimulated emission and absorption become equal in this approximation, viz

2 + 2 e 2 A20 2
|Vni | = |Vni | = |ˆ
hn|~
p |ii| . (158)
m2
I From the identity [x, H0 ] = i~px /m we calculate that hn|~
p |ii = imωni hn|~
x |ii. Hence, we obtain

2 + 2 2 2 ~ 2
|Vni | = |Vni | = A0 ωni |ˆ
hn|P|ii| . (159)

~ is precisely the electric dipole moment defined by

I The vector P

~ = e~
P x. (160)

~ and the magnetic field of the above monochromatic plane wave are given by
I The electric field E
~ = −ˆ
E x − ωt) and B
(2ωA0 ) sin(k n̂~ ~ = −(n̂ × ˆ)(2ωA0 /c) sin(k n̂~x − ωt).
I The energy density (energy per unit volume) in this electromagnetic wave is then given by the formula
u = (0 E 2 + B 2 /µ0 )/2 = 0 E 2 .
I Thus, the average over a full cycle is u = 20 ω 2 A2 . Hence, we get (we also use the fact that ωni = ±ω)
0

2 + 2 u 2
|Vni | = |Vni | = |ˆ ~
hn|P|ii| . (161)
20
Stimulated emission (LASER) and absorption by radiation
I The probability for stimulated emission and absorption takes now the form

t2 u sin (ω0 − ω)t/2 2 2

 
ω ~ 2
Pi−0→n (t) = |ˆ
hn|P|ii| t . (162)
~2 20 (ω0 − ω)t/2

(
ωin stimulated emission
ω0 = . (163)
ωni absorption

I For a non-monochromatic radiation we need to replace the energy density u by ρ(ω)dω which is the
energy density in the frequency range dω centered about ω. Then, we integrate over all frequencies. We
obtain then

t2 1 sin (ω0 − ω)t/2 2

Z ∞  
ω0 ~ 2
Pi−→n (t) = |ˆ
hn|P|ii| ρ(ω) dω
~2 20 0 (ω0 − ω)t/2
Z ω0 t
t 1 2 2 2x sin2 x
= |ˆ ~
hn|P|ii| ρ(ω0 − ) dx
~2 0 −∞ t x2
sin2 x
Z +∞
t 1 2
= |ˆ ~
hn|P|ii| ρ(ω0 ) dx , t −→ ∞
~2 0 −∞ x2
t π 2
= |ˆ ~
hn|P|ii| ρ(ω0 ). (164)
~2 0
I The transition rate is (transition probability per unit time) is then given by

ω d ω 1 π 2
Ri−0→n = Pi−0→n (t) = |ˆ ~
hn|P|ii| ρ(ω0 ). (165)
dt ~2 0
I This is the result for non-monochromatic radiation.
I But, we are interested in non-monochromatic, unpolarized and incoherent radiation.
I Thus, we need to take the average of the above expression over all incident directions n̂ and all polarization
directions ˆ.
Stimulated emission (LASER) and absorption by radiation
I First, note that the vectors n̂ and ˆ are perpendicular.
I We will work in spherical coordinates.
I We choose the z axis along the propagation direction n̂. Thus, ˆ is in the xy plane.
I We choose the y axis such that the vector hn|P|ii~ lies in the zy plane. The x axis is then fixed. The angle
~
between n̂ and hn|P|ii is θ and the angle between the x axis and ˆ is φ. We have then

~
ˆ = cos φiˆ + sin φjˆ , hn|P|ii ~
= |hn|P|ii|(cos ˆ
θ k̂ + sin θ j). (166)

I Obviously, we have

~ 2 2 2 2
|hn|P|ii| = hn|Px |ii + hn|Py |ii + hn|Pz |ii . (167)

I The desired average is given by the integral

1 π 1
Z
ω ~ 2 2 2
Ri−0→n = |hn|P|ii| ρ(ω0 ) sin φ sin θ sin θdθdφ
~2 0 4π
π 2
= ~
|hn|P|ii| ρ(ω0 ). (168)
30 ~2

I We get then Fermi’s Golden rule for stimulated emission and absorption, viz

stim−emis ω ≡ω π 2
wi−→[n] ≡ Ri−0→n in = ~
|hn|P|ii| ρ(ωin ). (169)
30 ~ 2

And

abso ω ≡ω π 2
wi−→[n] ≡ Ri−0→n ni = ~
|hn|P|ii| ρ(ωni ). (170)
30 ~ 2

We have precise detailed balance between absorption and stimulated emission.

Spontaneous emission and black body radiation
I We consider a gas of Bohr’s atoms in thermal equilibrium with radiation at temperature T .
I We assume that the atoms function effectively as two-state systems, i.e. only two states |ψa i (ground
state) and |ψb i (excited state) can be accessed by the atoms.
I Let Na be the number of atoms in the ground state |ψa i and Nb be the number of atoms in the excited
state |ψb i.
I Let ω0 be the transition frequency between the two states, viz ~ω0 = Eb − Ea .
I Let A be the spontaneous emission rate. Thus, ANb is the number of atoms per unit time transitioning
from |ψb i to |ψa i via spontaneous emission.
I The stimulated emission rate and the absorption rate must be proportional to the energy density ρ(ω0 ) of
the electromagnetic radiation.
I Thus, let Bba ρ(ω0 ) be the stimulated emission rate, i.e. Bba ρ(ω0 )Nb is the number of atoms per unit
time transitioning from |ψb i to |ψa i via stimulated emission.
I Similarly, let Bab ρ(ω0 ) be the absorption rate, i.e. Bab ρ(ω0 )Na is the number of atoms per unit time
transitioning from |ψa i to |ψb i via absorption.
I The rate of change of the number of particle Nb is then given by

dNb
= −Nb A − Nb Bba ρ(ω0 ) + Na Bab ρ(ω0 ). (171)
dt
I The condition of thermal equilibrium gives immediately

dNb A
= 0 ⇒ ρ(ω0 ) = N . (172)
dt aB − B
Nb ab ba

I From the condition of thermal equilibrium we also know (from Boltzmann distribution) that

e −Ea /kB T
~ω0
Na kB T
= =e . (173)
Nb e −Eb /kB T
I Thus, we obtain the rate of change

A
ρ(ω0 ) = ~ω0
. (174)
e kB T Bab − Bba
Spontaneous emission and black body radiation
I Again, from the condition of thermal equilibrium, the gas of Bohr’s atom in thermal equilibrium with the
radiation should behave as a blackbody. Namely, we must have

~ ω03
ρ(ω0 ) = . (175)
π2 c 3 ~ω0
e kB T −1
I By comparing the above two equations we get the condition of detailed balance between stimulated
emission and absorption and the spontaneous emission rate in terms of the rates of stimulated emission
and absorption, viz
Bab = Bba . (176)
And
~ω03
A= Bba . (177)
π2 c 3
I But we have already found that
π 2
Bab = Bba = ~
|hb|P|ai| . (178)
30 ~2
I Thus, the spontaneous emission rate is given by

ω03 2
A= ~
|hb|P|ai| . (179)
3~π0 c3

Source: Griffiths
Exact solution for two-level systems (MASER and NMR)
I We assume again a time-dependent Hamiltonian H of the form H = H0 + V (t).
I However, we will assume that the time-independent unperturbed Hamiltonian H0 has only two states |ψ1 i
and |ψ2 i with energies E1 and E2 respectively where E2 > E1 . We have

E2 − E1
H0 |ψ1 i = E1 |ψ1 i , H0 |ψ2 i = E2 |ψ2 i , ω0 = > 0. (180)
~
I We will expand the state vector |ψ(t)i for t > 0 as

iE1 t iE2 t
− −
|ψ(t)i = c1 (t)e ~ |ψ1 i + c2 (t)e ~ |ψ2 i. (181)
I The Schrodinger equation for |ψ(t)i leads to a coupled system of two first-order differential equations
given explicitly by

dc1 (t) X iω1n t −iω0 t

i~ = cn (t)e V 1n = c1 (t)V11 + c2 (t)e V 12 . (182)
dt n

dc2 (t) X iω2n t iω0 t

i~ = cn (t)e V2n = c1 (t)e V 21 + c2 (t)V22 . (183)
dt n

I The system starts initially in the state |1i. In other words, we use the initial conditions

c1 (0) = 1 , c2 (0) = 0. (184)

I We will assume that the time-dependent potential is sinusoidal given by

∗ iωt
V11 = V22 = 0 , V12 = V21 = γe . (185)
I We find immediately the solution

ω−ω0  ω − ω0

i γ −i ω−ω0 t
i t
c1 = e 2 cos ωr t −i sin ωr t , c2 = − e 2 sin ωr t. (186)
2ωr ωr ~
Exact solution for two-level systems (MASER and NMR)
I In the above solution ωr is the so-called Rabi flopping frequency ωr defined by

2 (ω − ω0 )2 γ2
ωr = + . (187)
4 ~2
I So there are three different frequencies Ω in this problem:

ω0 natural

Ω= ω driving . (188)

ω
r Rabi

The two-level system with natural frequency ω0 oscillates really with Rabi frequency ωr in response to the
perturbing potential which is sinusoidal with frequency ω.
I The probability of finding the system at time t in the state |2i is then given by

2 γ 2 /~2 2 γ 2 /~2 1
|c2 | = sin ωr t = (1 − cos 2ωr t). (189)
ωr2 ωr2 2

This is Rabi’s formula.

I The probability of finding the system in the state |2i oscillates in time with a frequency 2ωr .
I The amplitude of oscillation is maximum at ω = ω0 . This is a resonance behavior.
I The resonance condition is therefore defined by
γ
ω = ω0 , ωr = . (190)
~
I The transition probability at resonance becomes

2 1
|c2 | = (1 − cos 2ωr t). (191)
2
I From time t = 0 to t = π~/2γ the probability of finding the system in |2i increases until it reaches 1.
Thus, during this time the system absorbs energy from the perturbation until at time t = π~/2γ only the
state |2i is populated.
I From t = π~/2γ to t = π~/γ the probability |c2 |2 decreases and thus the system gives up energy to the
perturbation until at at time t = π~/γ only the state |1i is populated.
I This emission-absorption cycle continues indefinitely.
Exact solution for two-level systems (MASER and NMR)
I The time-dependent sinusoidal pertrubation V (t) is causing absorption transitions |1i −→ |2i and
emission transitions |2i −→ |1i.
I Away from resonance the maximum of the transition probability |c2 |2 is given by

2 γ2
2 1 γ 2 2 ~2
|c2 | = sin ωr t ≤ |c2 |max = . (192)
ωr2 ~2 (ω−ω0 )2 γ2
+
4 ~2

I This is a resonance curve with a peak at ω = ω0 . The frequencies corresponding to a probability

|c2 |2 = 1/2 are ω = ω0 ± 2γ/~. Therefore, the width at half maxima is 4γ/~. Clearly, the widths
become smaller, i.e. we get narrower resonance peaks for weaker potentials.
I Away from resonance the emission-absorption cycle also takes place but populating/depopulating the state
|2i/|1i occurs only partially since |c2 |2max < 1.
I Among the applications of this two-level system we only mention nuclear magnetic resonance (NMR) and
microwave amplification by stimulated emission of radiation (MASER).

Resonance behavior. Source: Sakurai

Hydrogen atom quantum computation
I The above two-level systems are a very important ingredient for quantum computation as they play the
role of single quantum bits or qubits. The paradigm for a two-level system is a single spin 1/2 particle.
I But, in quantum computation, we also need in a fundamental way two-qubit systems.
I The two-qubit systems used in quantum logic gates can be given by two-particle systems where the
particles are spin 1/2 particles.
I As examples of two-particle systems we consider the hydrogen atom and positronium in their ground states
and nuclear magnetic resonance or NMR which is the most advanced existing technology for quantum
computation today.
I The spin-dependent Hamiltonian of the hydrogen atom in the presence of a uniform magnetic field is given
by
eB (e − ) gp eB (p) −
H = Sz − ~(e ) .S
S z + J0 S ~(p) , me  mp
me 2mp
eB (e − ) (e − ) (p)
= Sz ~
+ J0 S ~
.S . (193)
me
I A much more symmetric problem is the positronium (a bound state of an electron and a positron). The
spin-dependent Hamiltonian of positronium in the presence of a uniform magnetic field is given by
eB (e − ) eB (e + ) (e − ) (e + )
H = Sz − Sz ~
+ J0 S ~
.S . (194)
me me
I Both systems are examples of two-particle systems where the particles are spin 1/2 particles and with
Hamiltonian operators of the generic form

H = −~ ~−µ
µA . B ~ + J0 S
~ X .B ~A .S
~X
q
= ~A .B
−γA S ~ − γX S
~X .B
~ + J0 S
~A .S
~X , γ = g (1 − σ)
2m
S
= −γA B~MA − γX B~MX + J0 S ~X , M = z
~A .S
~
γB
= ~A .S
−hνA MA − hνX MX + J0 S ~X , ν = . (195)
2π
Hydrogen atom quantum computation
I A very important example described by the above Hamiltonian is NMR.
I Indeed, for NMR spectra in which molecules contain two spin 1/2 protons A and X with very different
chemical shifts we have the gyromagnetic ratios γA = gp e(1 − σA )/2mp and γX = gp e(1 − σX )/2mp .
This system is called an AX system.
I The quantity σ is called the shielding constant of the proton. It expresses the fact that the external
~ induces electronic currents which themselves give rise to additional magnetic fields at the
magnetic field B
~loc = B
protons. The local magnetic field at the protons is in fact given by B ~ + δB~ = (1 − σ)B. ~
I The Larmor frequency is then different for protons in different environments since it is given by

e~
hν = gp BµN (1 − σ) , µN = . (196)
2mp

µN is the nuclear magneton. This should be compared with Bohr’s magneton µB which is the
fundamental quantum of magnetic moment given by µB = e~/2me .
I The so-called chemical shift is an empirical quantity δ which measures the difference between the Larmor
frequency of the proton in question and some standard frequency, viz
ν − ν0 σ0 − σ
δ= = . (197)
ν0 1 − σ0
As the shielding σ decreases the chemical sift δ increases, i.e. a large chemical shift corresponds to strong
deshielding.
I The hydrogen atom is formally a special case of the AX system. Explicitly, we have

− e
A=e ⇒ ge = 2 , σ e = 0 , γ e = −
me
gp e
X = p ⇒ gp = 5.586 , σp = 0 , γp =
2mp
γe B
νe = − , hνe = −γe ~B = ge µB B , He − = hνe Me
2π
γp B
νp = , hνp = γp ~B = gp µN B , Hp = −hνp Mp . (198)
2π
I The ensuing discussion should then hold at least formally.
Hydrogen atom quantum computation
I The Hamiltonian for a single spin (proton) is given by H = −γB~M with eigenvalues E = −γB~m in the
eigenstates |mi of Sz = ~M. The resonance condition for a single proton is then immediately given by

∆E = Eβ − Eα = γB~ ≡ hν , α ≡ +1/2 , β ≡ −1/2. (199)

I In other words, resonant absorption occurs when the frequency of the electromagnetic radiation matches
Larmor frequency. Because of the different chemical shifts of the protons A and X the corresponding
resonance frequencies are different, viz

gp BµN (1 − σA )
γA B~ = hνA ≡ hν ⇒ ν = νA =
h
gp BµN (1 − σX )
γX B~ = hνX ≡ hν ⇒ ν = νX = . (200)
h

I For the two-spin system we treat H0 = −γA B~MA − γX B~MX as a non-perturbed Hamiltonian and
V = JS ~A .S
~X as a perturbation. The unperturbed energy levels are immediately given by

EmA mX = −γB~mA − γB~mX = −hνA mA − hνX mX . (201)

I We use non-degenerate perturbation theory. In the basis {|mA i|mX i} we compute the first-order
corrections

2
∆EmA mX = hV i = J0 ~ mA mX = hJmA mB . (202)

I The total energy of the state |mA i|mX i is then given by (we assume that J > 0)

EmA mX + ∆EmA mX = −hνA mA − hνX mX + hJmA mB . (203)

Hydrogen atom quantum computation
I Here, we have two resonances:
1) Resonance A: In which proton A makes a transition while proton X remains unchanged, i.e. ∆mA = −1,
∆mX = 0. There are two absorption transitions αA −→ βA with ∆mA = −1, ∆mX = 0 which are
αA αX −→ βA αX and αA βX −→ βA βX . The energies of the transitions are given by
hJ 1
∆EA = −hνA ∆mA + hJmX ∆mA = −hνA ∆mA ± ∆mA , mX = ±
2 2
hJ
= hνA ∓ , ∆mA = −1 : αA −→ βA . (204)
2
The A resonance consists therefore of a doublet separated by J centered on the chemical shift of A.
2) Resonance X : In which proton X makes a transition while proton A remains unchanged: ∆mA = 0,
∆mX = −1. There are two absorption transitions αX −→ βX with ∆mA = 0, ∆mX = −1 which are
αA αX −→ αA βX and βA αX −→ βA βX . The energies of the transitions are given by
hJ 1
∆EX = −hνX ∆mX + hJmA ∆mX = −hνX ∆mX ± ∆mX , mA = ±
2 2
hJ
= hνX ∓ , ∆mX = −1 : αX −→ βX . (205)
2
The X resonance consists therefore of a doublet separated by J centered on the chemical shift of X .

The A and X resonances. Source: Atkins

Hydrogen atom quantum computation
I The above two-spin system is what is used in NMR quantum computation. The two spins A and X play
the role of two qubits Qubit 1 and Qubit 2.
I In NMR quantum computation the states |+i and |−i correspond to the computational states |0i and
|1i respectively, viz

1
α≡+ , |+i = |0i , lower energy
2
1
β ≡− , |−i = |1i , higher energy. (206)
2

I The frequencies νA = ω1 /2π and νX = ω2 /2π are the resonance frequencies that drive the transitions
between the levels of Qubit 1 and Qubit 2 respectively. We have

ω1
νA = , Qubit 1
2π
ω2
νX = , Qubit 2. (207)
2π

I The states of Qubit 1 are then denoted by |0i1 and |1i1 and the states of Qubit 2 are denoted by |0i2
and |1i2 .
I A radio electromagnetic radiation of frequency ω1 will change the spin of Qubit 1 as |0i1 −→ |1i1
(absorption) or |1i1 −→ |0i1 (stimulated emission). Similarly, for Qubit 2 with a radio electromagnetic
radiation of frequency ω2 . We are assuming that spontaneous emission can be neglected.
I We need now to use one of the qubits to control the other one in order to build quantum logic gates
allowing us to perform arbitrary operations on the two-qubit system.
I As it turn out, it is sufficient to build the controlled-NOT or CNOT gate as every other gate is in fact a
combination of CNOT gates and single qubit rotations. The CNOT gate is said to be universal.
I The CNOT gate changes the logical value of Qubit 2 if and only if the logical value of Qubit 1 is 1.
Hydrogen atom quantum computation

The truth table of the CNOT gate. Source: Foot

I As we have shown the two-qubit system is characterized by four states |00i = | + +i, |01i = | + −i,
|10i = | − +i and | − −i = |11i and two resonances A and X .
1 Resonance A:

x0 hJ
αA αX = |00i → βA αX = |10i , ∆EA = ~ω1 = hνA −
2
x1 hJ
αA βX = |01i → βA βX = |11i , ∆EA = ~ω1 = hνA + . (208)
2

2 Resonance X :

0x hJ
αA αX = |00i → αA βX = |01i , ∆EX = ~ω2 = hνX −
2
1x hJ
βA αX = |10i → βA βX = |11i , ∆EX = ~ω2 = hνX + . (209)
2

I We observe immediately that a radio electromagnetic radiation of frequency ω 1x will switch the state of
2
Qubit 2 if and only if Qubit 1 is in the state |1i1 , i.e. as |10i −→ |11i (absorption), |11i −→ |10i
(stimulated emission). The other two states |00i and |01i remain unchanged. This is precisely the CNOT
gate.
 Hydrogen atom quantum computation

The absorption spectrum. Source: Foot

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