Alexandra Gabriela S ERBAN [604249]

Universitatea din Bucures ,ti
Facultatea de Fizic a
Alexandra Gabriela S ERBAN
Interaction of the Electromagnetic Field with the
Atomic Systems
Bachelor Thesis
Scientic Adviser
Prof. dr. Virgil B ARAN
Bucharest, 2019

Contents
Introduction 1
1 Electromagnetic eld 2
1.1 Maxwell's Equations for the Electromagnetic Field . . . . . . . . . . 2
1.2 The Electromagnetic Potentials . . . . . . . . . . . . . . . . . . . . . 3
1.2.1 Equations for the electromagnetic potentials . . . . . . . . . . 4
1.2.2 Lorenz Gauge . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.2.3 Coulomb Gauge . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.3 Plane Waves Expansion . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4 Quantization in the Plane Waves Basis . . . . . . . . . . . . . . . . . 9
2 Time Dependent Perturbation Theory 21
2.1 Interaction picture . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.2 Transition probability . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
Bibliography 26

Introduction
INTRO
jkjmi=q
2j+1
4R
d~ nDj
m(~ n)ay
(~k)j0i
ay
(~k)j0i=1X
j=1jX
m=jr
2j+ 1
4Dj
m(~ n)jkjmi
jktm
kzi=q
kt
2Rdk
2(1)m
eim
k~kE
~kE
=X
jmr
2j+ 1
4Dj
m(~ n)by
jm(~k)j0i
by
jm(~k) =q
2j+1
4R
d~ nDj
m(~ n)ay
(~k)
jktm
kzi=X
jr
2j+ 1
4(1)m
d(j)
m
()by
jm
(~k)j0i
1

Chapter 1
Electromagnetic eld
1.1 Maxwell's Equations for the Electromagnetic
Field
Consider~E(~ r;t) and~B(~ r;t) the time varying classical electric and magnetic elds
that satisfy Maxwell's equations:
8
>>>>>>>><
>>>>>>>>:~r
~E(~ r;t) =f(~ r;t)
"0~r
~B(~ r;t) = 0
~r~E(~ r;t) =@~B(~ r;t)
@t
~r~B(~ r;t) =0~j(~ r;t) +0"0@~E(~ r;t)
@t(1.1)
Where,
f= density of free charges
~j= density of currents
at a point where the electric and magnetic elds are evaluated. And,
"0= electric permittivity
0= magnetic permeability
are constants that determine the properties of the vacuum.
We dene,
c=1
p"00; c = 299792458 m/s
as the speed of light in vacuum.
2

1.2 The Electromagnetic Potentials
Due to the fact that the divergence of ~Bis zero, the magnetic eld must be the curl
of a vector eld.
~r
~B= 0
=)~B=~r~A (1.2)
Starting from the following equation,
~r~E=@~B
@t
=)~r~E=@
@t(~r~A) =~r
@~A
@t!
=)~r~E+
~r@~A
@t!
= 0
=)~r
~E+@~A
@t!
= 0 (1.3)
By using the following identity,
~r~r= 0
whereis an arbitrary scalar function, it is obtained that the expression in the
brackets from equation 1.3 can be written as the gradient of a scalar potential,
~E+@~A
@t=~r
=)~E=@~A
@t~r (1.4)
Maxwell's equations should be independent of the particular choice of the po-
tentials. The potentials ~A(~ r;t) and(~ r;t) are not uniquely dened. They do not
have a unique physical signicance since innite families of potentials represent the
same eld. Only the elds have a unique meaning.
A simultaneous change of the potentials with an arbitrary scalar function (~ r;t),
continuous and with the derivatives continuous, leaves the elds invariant.
(~ r;t)!(~ r;t) +@
@t (~ r;t)
~A(~ r;t)!~A(~ r;t)~r (~ r;t)
3

1.2.1 Equations for the electromagnetic potentials
The expressions of the electric and magnetic elds as a function of the scalar and
vector potentials are,
8
<
:~B=~r~A
~E=@~A
@t~r(1.5)
From the four coupled Maxwell's equations, the following results are obtained:
~r
~E=f
"0=)~r

@~A
@t~r!
=f
"0
=) @
@t(~r
~A)4=f
"0(1.6)
~r
~B= 0 =)~r
~r~A= 0
~r~E=@~B
@t=)~r
@~A
@t~r!
=@
@t(~r~A)
~r~B=0~j+0"0@~E
@t
=)~r (~r~A) =0~j+0"0@
@t
@~A
@t~r!
=)~r(~r
~A)4~A=0~j1
c2@2~A
@t2~r1
c2@
@t
=)1
c2@2~A
@t24~A=0~j~r
~r
~A+1
c2@
@t
=)~A=0~j~r
~r
~A+1
c2@
@t
(1.7)
Maxwell's equations and all the associated physical quantities, when expressed
in terms of the electromagnetic potentials, must be invariant under gauge transfor-
mations, which are transformations between physically equivalent sets of potentials.
4

1.2.2 Lorenz Gauge
~r
~A+1
c2@
@t= 0 (1.8)
Replacing the condition above back in equation 1.7, it is obtained the D'Alambertian
of the vector potential,
=)~A=0~j (1.9)
~r
~A+1
c2@
@t= 0 =)~r
~A=1
c2@
@t
Replacing the divergence of ~Aback in equation 1.6, it is obtained the D'Alambertian
of the scalar potential,
@
@t
1
c2@
@t
4=f
"0
=)1
c2@2
@t24=f
"0
=)=f
"0(1.10)
1.2.3 Coulomb Gauge
~r
~A= 0 (1.11)
The choice above is known as the transverse gauge.
=)8
>><
>>:~A=0~j~r1
c2@
@t
4=f
"0
1.3 Plane Waves Expansion
In order to obtain the wave equation, the Maxwell's equations for free space are
considered. This implies that there are no free charges and no free currents.
(
f(~ r;t) = 0
~j(~ r;t) = 0
5

In the Coulomb Gauge 1.11, it is considered the Lorenz Gauge 1.8 and the
following result is obtained,
~A= 0 =)1
c2@2~A
@t24~A= 0 (1.12)
which represents the wave equation.
The wave equation has the general solution of the form of an innite set of plane
waves propagating in the ~kdirection.
~A(~ r;t) =1p
VX
~k;~A(~k)ei(!kt~k
~ r)(1.13)
The factorp
Vat the denominator was introduced as a normalization condition
taking into account that the electromagnetic eld is conned to a large volume
V=L3with periodic boundary conditions.
~A(x+L;y;z;t ) =~A(x;y+L;z;t ) =~A(x;y;z +L;t) =~A(x;y;z;t )
eikxL=eikyL=eikzL= 1
kx=2
Lnx; ky=2
Lny; kz=2
Lnz
whereni(i=x;y;z ) is an integer number, positive, negative or zero.
The meaning of the parameters in 1.13 is,
= index of polarization
~k= wave propagating direction~k=!~k
c
~A(~k) = amplitude of the frequency !k
Transversality condition
From the condition of the Coulomb Gauge 1.11 it is obtained that,
~k
~A(~k) = 0
which means that the amplitude vectors of the eld are normal to the propagation
direction.
6

Due to the fact that ~A(~k) and~kare orthogonal, the amplitude can be specied
in terms of components along two mutually perpendicular directions transverse to
~k.
The unit vectors along these directions are denoted by ~ (~k) and they represent
the real linear unit vectors of the eld polarization. These unit vectors satisfy the
following conditions:
~ i(~k)
~ j(~k) =ij
~ (~k)
~k= 0
~ 1(~k)~ 2(~k) =^k
The vector potential has to be real so it will be written as a linear combination
of …?
~A(~ r;t) =1
21p
VX
~k;h
~A(~k)ei(!kt~k
~ r)+~A
(~k)ei(!kt~k
~ r)i
(1.14)
=)~A(~ r;t) =1
21p
VX
~k;h
~ (~k)A(~k)ei(!kt~k
~ r)+~ (~k)A
(~k)ei(!kt~k
~ r)i
The physical meaning of this expansion of the vector potential is that the wave
which propagates in the opposite direction also has to be introduced, so the wave
which has the wavevector ~k. Obviously, ~ 1(~k) and~ 2(~k) are orthogonal on ~k, but
they do not form a …. (triedru drept). In order to obtain a … (triedru drept) the
following conditions should be imposed,
(
~ 1(~k) =~ 2(~k)
~ 2(~k) =~ 1(~k)
Energy of the electromagnetic eld
Considering that the electromagnetic wave is conned in a space of volume V,
the energy of the electromagnetic eld is given by the Hamiltonian:
H=1
2Z
VdV
"0~E2
+1
0~B2
="0
2Z
VdV~E2
+1
0"0~B2
="0
2Z
VdV~E2
+c2~B2
7

which is given by the integral of the energy density over the volume occupied
by the eld.1
2"0~E2
represents the energy density of the electric eld and1
20~B2
represents the energy density of the magnetic eld.
We observe that ( c~B+i~E)(c~Bi~E) =~E2
+c2~B2
.
=)H="0
2Z
VdV(c~B+i~E)(c~Bi~E) (1.15)
8

1.4 Quantization in the Plane Waves Basis
The starting point in the quantization of the electromagnetic eld is the expansion
of the vector potential 1.14.
~A(~ r;t) =1
21p
VX
~k;
~ (~k)A(~k)ei(!kt~k
~ r)+~ (~k)A
(~k)ei(!kt~k
~ r)
=1
21p
VX
~k
~ 1(~k)A1(~k)ei~k
~ rei!kt+~ 1(~k)A
1(~k)ei~k
~ rei!kt+
+~ 2(~k)A2(~k)ei~k
~ rei!kt+~ 2(~k)A
2(~k)ei~k
~ rei!kt
=1
21p
VX
~k
~ 1(~k)A1(~k)ei~k
~ rei!kt+~ 1(~k)A
1(~k)ei~k
~ rei!kt+
+~ 2(~k)A2(~k)ei~k
~ rei!kt+~ 2(~k)A
2(~k)ei~k
~ rei!kt
=1
2X
~kei~k
~ r
p
V
~ 1(~k)A1(~k)ei!kt+~ 2(~k)A
1(~k)ei!kt+~ 2(~k)A2(~k)ei!kt+~ 1(~k)A
2(~k)ei!kt
=1
2X
~kei~k
~ r
p
Vh
~ 1(~k)
A1(~k)ei!kt+A
2(~k)ei!kt
+~ 2(~k)
A
1(~k)ei!kt+A2(~k)ei!kti
=)~A(~ r;t) =1
2X
~kh
~ 1
C1(~k) +C
2(~k)
+~ 2
C2(~k) +C
1(~k)iei~k
~ r
p
V(1.16)
The electric eld ~E(~ r;t) and the magnetic eld ~B(~ r;t) are calculated with the
form of the vector potential ~Aobtained above.
~E(~ r;t) =@
@t~A(~ r;t) =1
2X
~k(i!k)"
~ 1(~k)h
C1(~k)C
2(~k)i
+~ 2(~k)h
C2(~k)C
1(~k)i#
ei~k
~ r
p
V
=)~E(~ r;t) =i
2X
~k!k"
~ 1(~k)h
C1(~k)C
2(~k)i
+~ 2(~k)h
C2(~k)C
1(~k)i#
ei~k
~ r
p
V
~B(~ r;t) =~r~A(~ r;t) =i
2X
~k"
~k~ 1(~k)h
C1(~k) +C
2(~k)i
+
~k~ 2(~k)h
C2(~k) +C
1(~k)i#
ei~k
~ r
p
V
=)c~B(~ r;t) =ci
2X
~kh
~k~ 2(~k)
C1(~k) +C
2(~k)
~k~ 1(~k)
C2(~k) +C
1(~k)iei~k
~ r
p
V
=)c~B(~ r;t) =i
2X
~k!kh
~ 2(~k)
C1(~k) +C
2(~k)
~ 1(~k)
C2(~k) +C
1(~k)iei~k
~ r
p
V
9

Now, the terms which appear in the formula for the energy of the electromagnetic
eld 1.15 have to be calculated.
c~Bi~E=i
2X
~k!kh
~ 2(~k)
C1(~k) +C
2(~k)
~ 1(~k)
C2(~k) +C
1(~k)iei~k
~ r
p
V+
+i
2X
~k!kh
~ 1(~k)
C1(~k)C
2(~k)
+~ 2(~k)
C2(~k)C
1(~k)iei~k
~ r
p
V
=1
2X
~k!kh
i~ 2(~k)
C1(~k) +C
2(~k)
i~ 1(~k)
C2(~k) +C
1(~k)
+
+~ 1(~k)
C1(~k)C
2(~k)
+~ 2(~k)
C2(~k)C
1(~k)iei~k
~ r
p
V
=1
2X
~k!k
i~ 2(~k)C1(~k) +i~ 2(~k)C
2(~k)i~ 1(~k)C2(~k)i~ 1(~k)C
1(~k)+
+~ 1(~k)C1(~k)~ 1(~k)C
2(~k) +~ 2(~k)C2(~k)~ 2(~k)C
1(~k)ei~k
~ r
p
V
=1
2X
~k!kh
~ 1(~k) +i~ 2(~k)
C1(~k)iC2(~k)

~ 1(~k)i~ 2(~k)
iC
1(~k) +C
2(~k)iei~k
~ r
p
V
The following vectors are introduced,
(
~ +(~k) =~ 1(~k) +i~ 2(~k)
~ (~k) =~ 1(~k)i~ 2(~k)(1.17)
=)c~Bi~E=1
2X
~k!kh
~ +(~k)
C1(~k)iC2(~k)
~ (~k)
iC
1(~k) +C
2(~k)iei~k
~ r
p
V
By taking the complex conjugate of the term above, it is obtained that,
=)c~B+i~E=1
2X
~k!kh
~ (~k)
C
1(~k)+iC
2(~k)
~ +(~k)
iC1(~k)+C2(~k)iei~k
~ r
p
V
(1.18)
10

The calculation of the product ( c~Bi~E)(c~B+i~E) is,
(c~Bi~E)(c~B+i~E) =1
4VX
~k!~kh
~ +(~k)
C1(~k)iC2(~k)
~ (~k)
iC
1(~k) +C
2(~k)i
ei~k
~ r


X
~k0!~k0h
~ (~k0)
C
1(~k0) +iC
2(~k0)
~ +(~k0)
iC1(~k0) +C2(~k0)i
ei~k0
~ r
=1
4VX
~kX
~k0!~k!~k0h
~ +(~k)
C1(~k)iC2(~k)
~ (~k)
iC
1(~k) +C
2(~k)i



h
~ (~k0)
C
1(~k0) +iC
2(~k0)
~ +(~k0)
iC1(~k0) +C2(~k0)i
ei(~k~k0)
~ r
=1
4VX
~kX
~k0!~k!~k0h
2
C1(~k)iC2(~k)
C
1(~k0) +iC
2(~k0)
+
+ 2
iC
1(~k) +C
2(~k)
iC1(~k0) +C2(~k0)i
ei(~k~k0)
~ r
=1
2VX
~kX
~k0!~k!~k0h
C1(~k)iC2(~k)
C
1(~k0) +iC
2(~k0)
+
+
iC
1(~k) +C
2(~k)
iC1(~k0) +C2(~k0)i
ei(~k~k0)
~ r
The obtained results above are replaced back in equation 1.15.
H="0
2Z
VdV(c~B+i~E)(c~Bi~E)
="0
21
2Z
VdVX
~kX
~k0!~k!~k0h
C1(~k)iC2(~k)
C
1(~k0) +iC
2(~k0)
+
+
iC
1(~k) +C
2(~k)
iC1(~k0) +C2(~k0)iei(~k~k0)
~ r
V
="0
4X
~k!2
~kh
C1(~k)iC2(~k)
C
1(~k0) +iC
2(~k0)
+
iC
1(~k) +C
2(~k)
iC1(~k0) +C2(~k0)i
="0
4X
~k!2
~k
C1(~k)C
1(~k) +iC1(~k)C
2(~k)iC2(~k)C
1(~k) +C2(~k)C
2(~k)+
+C
1(~k)C1(~k) +iC
1(~k)C2(~k)iC
2(~k)C1(~k) +C
2(~k)C2(~k)
="0
4X
~k!2
~k
2C1(~k)C
1(~k) + 2C2(~k)C
2(~k)
="0
2X
~k!2
~k
C
1(~k)C1(~k) +C
2(~k)C2(~k)
=)H="0
2X
~k;!2
~kC
(~k)C(~k) (1.19)
11

A comparison with the case of the classical harmonic oscillator is appropriate.
The linear combinations are introduced,
8
><
>:C=x+ip
m!
C=xip
m!(1.20)
Thus, the classical Hamiltonian becomes,
Hosc=p2
2m+m!2×2
2
=m!2
2
x2+p2
m2!2
=m!2
2
x+ip
m!
xip
m!
=m!2
2CC(1.21)
The analogy is complete when as "mass" of the electromagnetic osciallation,
the electrical permittivity of the vaccum "0is considered. Therefore, the eld of
the electromagnetic radiation behaves, from a dynamic point of view, like a system
composed of an innity of linear harmonic oscialltors.
Passing now to the quantum theory imposes no diculty as it comes very natural.
For each oscillator, the position and linear momentum operators are replaced with
operators. Then, the coecients C(~k) andC
(~k) also become operators, together
with the vector potential. The mass is also replaced with its equivalent "0, the
electrical permittivity of the vacuum.
8
>><
>>:C(~k) =Q(~k) +iP(~k)
"0!k
Cy
(~k) =Q(~k)iP(~k)
"0!k(1.22)
The fundamental commutation relations are considered,
h
Q(~k);P0(~k0)i
=i~0kk0
h
Q(~k);Q0(~k0)i
= 0
h
P(~k);P0(~k0)i
= 0(1.23)
12

The commutation relation between the coecients C(~k) andCy
(~k) is calculated.

C(~k);Cy
0(~k0)
=C(~k)Cy
0(~k0)Cy
0(~k0)C(~k)
=
Q(~k) +iP(~k)
"0!k
Q0(~k0)iP0(~k0)
"0!k0


Q0(~k0)iP0(~k0)
"0!k0
Q(~k) +iP(~k)
"0!k
=Q(~k)Q0(~k0)i
"0!k0Q(~k)P0(~k0)+
+i
"0!kP(~k)Q0(~k0) +1
"0!k!k0P(~k)P0(~k0)
=h
Q(~k);Q0(~k0)i
i
"0!k0h
Q(~k);P0(~k0)i

i
"0!kh
Q0(~k0);P(~k)i
+1
"0!k!k0h
P(~k);P0(~k0)i
=i
"0!k0i~0kk0i
"0!ki~0kk0
=~
"0!k00kk0+~
"0!k0kk0
If,
~k6=~k0=)!k6=!k0;butkk0= 0
~k=~k0=)!k=!k0andkk0= 1
h
C(~k);Cy
0(~k0)i
=2~
"0!k0kk0 (1.24)
Making the analogy with the classical case, the rising and lowering operators are
dened,
8
>>>><
>>>>:a(~k) =r
"0!k
2~
Q(~k) +iP(~k)
"0!k!
ay
(~k) =r
"0!k
2~
Q(~k)iP(~k)
"0!k! (1.25)
Analogously to the previous proof, the commutation relation between the low-
ering and rising operator si obtained very easy.
h
a(~k);ay
0(~k0)i
=0kk0 (1.26)
13

The number operator is dened as the product between the rising and the low-
ering operator.
N(~k) =ay
(~k)a(~k)
=r"0!k
2~
Q(~k)iP(~k)
"0!kr"0!k
2~
Q(~k) +iP(~k)
"0!k
="0!k
2~
Q2
(~k) +i
"0!kQ(~k)P(~k)i
"0!kP(~k)Q(~k) +1
"2
0!2
kP2
(~k)
="0!k
2~
Q2
(~k) +1
"2
0!2
kP2
(~k) +i
"0!k
Q(~k);P(~k)
="0!k
2~
Q2
(~k) +1
"2
0!2
kP2
(~k) +i
"0!ki~
="0!k
2~
Q2
(~k) +1
"2
0!2
kP2
(~k)
1
2
=)N(~k) +1
2="0!k
2~Q2
(~k) +1
2~"0!kP2
(~k)
~!k
N(~k) +1
2
="0!2
k
2Q2
(~k) +1
2"0P2
(~k)
~!k
N(~k) +1
2
=H(~k)
Thus, the Hamiltonian of the system becomes,
H=X
~k~!k
N(~k) +1
2
(1.27)
Photon states
The algebra developed above can be applied to a physical situation in which the
number of photons with given momentum and polarization is increased or decreased.
We interpret an eigenvector of N(~k) as the state vector for a state with a denite
number of photons in state ( ~k;).
To represent a situation in which there are many types of photons with dierent
sets of (~k;), we consider the direct product of eigenvectors as follows:
jn1;k1;n2;k2;:::;ni;ki;:::i=jn1;k1ijn2;k2i:::jni;kii:::
This state vector corresponds to the physical situation in which there are n1;k1
photons present in the state ( ~k1;1),n2;k2photons in the state ( ~k2;2) and so on.
The number n;kis called the occupation number for the state ( ~k;):
14

The state represented by j0i=j01;k1ij02;k2i:::j0i;kii:::has the property that
ifa(~k) is applied to it, then we obtain a null vector for any ( ~k;):Hence the
eigenvalue of N(~k) =ay
(~k)a(~k) is zero for any a(~k). So, the state written above
is called the vacuum state.
a(~k)jn1;k1;n2;k2;:::;ni;ki;:::i=pni;kijn1;k1;n2;k2;:::;ni;ki1;:::i
ay
(~k)jn1;k1;n2;k2;:::;ni;ki;:::i=p
ni;ki+ 1jn1;k1;n2;k2;:::;ni;ki+ 1;:::i
jn1;k1;n2;k2;:::;ni;ki;:::i=Y
ki;
ay
i(~ki)ni;ki
pni;kij0i
ay
i(~ki) has the property of creating an additional photon in the state ( ~ki;i)
leaving the occupation numbers of states, other than ( ~ki;i);unchanged. For this
reason,ay
(~k) is called the creation operator for a photon in the state ( ~k;):Similarly,
a(~k) can be interpreted as the annihilation operator for a photon in the state ( ~k;):
In contrast, N(~k);being diagonal, does not change the occupation number of
photons. It simply gives as its eigenvalue the number of photons in the state ( ~k;):
Field operators
The vector potential in equation 1.14 can be expressed by using the pair of
complex circular polarization vectors dened as,
~ e1(~k) =1p
2
~ 1(~k)~ 2(~k)
(1.28)
The phases are chosen to agree with those for the spherical harmonics with l= 1
andm=1.
These unit vectors satisfy the following relations
~ e
0(~k)
~ e(~k) =0
~ e
0(~k)~ e(~k) =i^k0
i^k~ e(~k) =~ e(~k)
The expressions corresponding to the eld operators are written in terms of the
creation and annihilation operators.
15

~A(~ r;t) =1
21p
VX
~k;
~ e(~k)A(~k)ei(!kt~k
~ r)+~ e
(~k)A
(~k)ei(!kt~k
~ r)
=1
2X
~k;
~ e(~k)C(~k)ei~k
~ r
p
V+~ e
(~k)Cy
(~k)ei~k
~ r
p
V
=1
2X
~k;r
2~
"0!k
~ e(~k)a(~k)ei~k
~ r
p
V+~ e
(~k)ay
(~k)ei~k
~ r
p
V!
=)~A(~ r;t) =X
~k;r
~
2"0!k
~ e(~k)a(~k)ei~k
~ r
p
V+~ e
(~k)ay
(~k)ei~k
~ r
p
V!
(1.29)
~E(~ r;t) =@~A(~ r;t)
@t
=1
2X
~k;r
2~
"0!k
i!k~ e(~k)a(~k)ei~k
~ r
p
V+i!k~ e
(~k)ay
(~k)ei~k
~ r
p
V!
=)~E(~ r;t) =iX
~k;r
~!k
2"0
~ e(~k)a(~k)ei~k
~ r
p
V~ e
(~k)ay
(~k)ei~k
~ r
p
V!
(1.30)
~B(~ r;t) =~r~A(~ r;t)
=)~B(~ r;t) =iX
~k;r
~
2"0!k"
~k~ e(~k)
a(~k)ei~k
~ r
p
V
~k~ e
(~k)
ay
(~k)ei~k
~ r
p
V#
(1.31)
The operator corresponding to the linear momentum of the electromagnetic eld
is dened as,
~P="0
2Z
d~ r
~E~B~B~E
(1.32)
First,R
d~ r(~E~B) is calculated.
16

Z
d~ r(~E~B) =Z
d~ rX
~k;~k0;;0~
2"0r!k
!k0
~ e(~k)a(~k)ei~k
~ r
p
V~ e
(~k)ay
(~k)ei~k
~ r
p
V!




~k0~ e0(~k0)
a0(~k0)ei~k0
~ r
p
V
~k0~ e
0(~k0)
ay
0(~k0)ei~k0
~ r
p
V!
=Z
d~ rX
~k;~k0;;0~
2"0r!k
!k0
~ e(~k)
~k0~ e0(~k0)
a(~k)a0(~k0)ei(~k+~k0)
~ r
p
V
~ e(~k)
~k0~ e
0(~k0)
a(~k)ay
0(~k0)ei(~k~k0)
~ r
p
V
~ e
(~k)
~k0~ e0(~k0)
ay
(~k)a0(~k0)ei(~k~k0)
~ r
p
V+
+~ e
(~k)
~k0~ e
0(~k0)
ay
(~k)ay
0(~k0)ei(~k+~k0)
~ r
p
V!
=X
~k;;0~
2"0r!k
!k~ e(~k)
~k~ e0(~k)
a(~k)a0(~k)+
+X
~k;;0~
2"0r!k
!k~ e(~k)
~k~ e
0(~k)
a(~k)ay
0(~k)+
+X
~k;;0~
2"0r!k
!k~ e
(~k)
~k~ e0(~k)
ay
(~k)a0(~k)+
+X
~k;;0~
2"0r!k
!k~ e
(~k)
~k~ e
0(~k)
ay
(~k)ay
0(~k)
By using the fact that
!k=!k
and the results of the triple vector products,
~ e(~k)
~k~ e0(~k)
=~ e(~k)
~k~ e(~k)
=~k
~ e(~k)
~ e(~k)
~ e(~k)
~ e(~k)
~k
=~k
~ e(~k)
~k~ e
0(~k)
=~k
~ e(~k)
~ e
0(~k)
~ e0(~k)
~ e(~k)
~k
=~k0
~ e
(~k)
~k~ e0(~k)
=~k
~ e
(~k)
~ e0(~k)
~ e0(~k)
~ e
(~k)
~k
=~k0
~ e
(~k)
~k~ e
0(~k)
=~ e
(~k)
~k~ e
(~k)
=~k
~ e
(~k)
~ e
(~k)
~ e
(~k)
~ e
(~k)
~k
=~k
it is obtained that,
17

Z
d~ r(~E~B) =X
~k;;0~
2"0~k
a(~k)a0(~k) +ay
(~k)ay
0(~k)
+
+X
~k;~
2"0~k
a(~k)ay
(~k) +ay
(~k)a(~k)
Analogously, it is obtained,
Z
d~ r(~B~E) =X
~k;;0~
2"0~k
a(~k)a0(~k) +ay
(~k)ay
0(~k)

X
~k;~
2"0~k
a(~k)ay
(~k) +ay
(~k)a(~k)
Back in equation 1.32 the above results are replaced and it is obtained that,
~P="0
2X
~k;~
"0~k
a(~k)ay
(~k) +ay
(~k)a(~k)
Starting from the commutation relation between the annihilation and creation
operators 1.26, the rst term of the sum can be expressed as a(~k)ay
(~k) = 1 +
ay
(~k)a(~k).
=)~P=~
2X
~k;~k
1 +ay
(~k)a(~k) +ay
(~k)a(~k)
=~
2 X
~k;~k+X
~k;2~kay
(~k)a(~k)!
The sum over ~kdisappears due to symmetry reason.
As a result, the nal form of the linear momentum operator is,
~P=~X
~k;~kN(~k) (1.33)
The photon carries not only energy and linear momentum, but also an intrinsic
angular momentum.
~J=~S+~L (1.34)
where,
18

~S="0
2Z
d~ r
~E~A~A~E
(1.35)
~L="0
2Z
d~ r3X
i=1h
Ei
~ r~rAi
+
~ r~rAi
Eii
(1.36)
First, the termR
d~ r~E~Ais calculated.
Z
d~ r~E~A=Z
d~ riX
~k;~k0;;0~
2"0r!k
!k0
~ e(~k)a(~k)ei~k
~ r
p
V~ e
(~k)ay
(~k)ei~k
~ r
p
V!


~ e0(~k0)a0(~k0)ei~k0
~ r
p
V+~ e
0(~k0)ay
0(~k0)ei~k0
~ r
p
V!
=iZ
d~ rX
~k;~k0;;0~
2"0r!k
!k0
~ e(~k)~ e0(~k0)a(~k)a0(~k0)ei(~k+~k0)
~ r
p
V+
+~ e(~k)~ e
0(~k0)a(~k)ay
0(~k0)ei(~k~k0)
~ r
p
V+
+~ e
(~k)~ e0(~k0)ay
(~k)a0(~k0)ei(~k~k0)
~ r
p
V+
+~ e
(~k)~ e
0(~k0)ay
(~k)ay
0(~k0)ei(~k+~k0)
~ r
p
V!
=i~
2"0 X
~k;;0r!k
!k~ e(~k)~ e0(~k)a(~k)a0(~k)+
+X
~k;;0r!k
!k~ e(~k)~ e
0(~k)a(~k)ay
0(~k)
X
~k;;0r!k
!k~ e
(~k)~ e0(~k)ay
(~k)a0(~k)
X
~k;;0r!k
!k~ e
(~k)~ e
0(~k)ay
(~k)ay
0(~k)!
=i~
2"0
X
~k;;0i^k0a(~k)ay
0(~k)X
~k;;0i^k0ay
(~k)a0(~k)!
=ii~
2"0X
~k;^k
a(~k)ay
+ay
(~k)a(~k)
=)Z
d~ r~E~A=~
2"0X
~k;^k
a(~k)ay
+ay
(~k)a(~k)
19

Analogously,
=)Z
d~ r~A~E=~
2"0X
~k;^k
a(~k)ay
+ay
(~k)a(~k)
Substituting the above results in equation 1.35 and using the commutation rela-
tion between the annihilation and creation operators 1.26, it is obtained that,
~S="0
22~
2"0X
~k;^k
a(~k)ay
+ay
(~k)a(~k)
=~
2X
~k;^k
1 +ay
(~k)a(~k) +ay
(~k)a(~k)
=~
2X
~k;^k+~
22X
~k;^kay
(~k)a(~k)
The rst sum vanishes due to symmetry reasons.
As a consequance, the nal form of the spin operator is
~S=~X
~k;^kN(~k) (1.37)
20

Chapter 2
Time Dependent Perturbation
Theory
We consider those phenomena that are described by Hamiltonians which can be
split into two parts, a time independent part H0and a time dependent part V(t),
describing a time dependent external force, which is small compared to H0.
H(t) =H0+V(t)
H0describes the unperturbed system and is assumed to have exact solutions
that are known. We consider a system which, when unperturbed, is described by
a time independent Hamiltonian H0whose solutions – the eigenvalues Enand the
eigenstatesj ni- are known.
H0j ni=Enj ni
and whose most general state vectors are given by stationary states
j n(t)i=ei
~H0tj ni=ei
~Entj ni
In the time interval 0 t, we subject the system to a time dependent
perturbation V(t), that is small compared to H0
V(t) =(
V(t);0t
0; t< 0;t>(2.1)
During the time interval 0 t, the Hamiltonian of the system is H(t) =
H0+V(t) and the corresponding Schrodinger equation is

H0+V(t)
j (t)i=i~dj (t)i
dt
V(t) characterizes the interaction of the system with the external source of per-
turbation.
When the system interacts with V(t), it either emits or absorbs energy. This
process causes the system to undergo transitions from one unperturbed eigenstate
to another.
21

The question that arises is the following: if the system is initially in an unper-
turbed eigenstate j iiofH0, what is the probability that the system will be found
at a later time in another unperturbed eigenstate j fi.
2.1 Interaction picture
It is more convinient to solve the Schrodinger equation in the interaction picture.
The interaction picture, also called the Dirac picture, is useful to describe quantum
phenomena with Hamiltonians that depend explicitly on time. In this picture, both
state vectors and operators evolve in time. The time dependence of the state is
due only to the interaction, while the time dependence of the operator is due to its
explicit time dependence plus the dependence on H0.
i~dj (t)iI
dt=VI(t)j (t)i (2.2)
where,
j (t)iI=ei
~H0tj (t)i (2.3)
VI(t) =ei
~H0tV(t)ei
~H0t(2.4)
The time evolution equation is,
j (t)i=U(t;ti)j (ti)i (2.5)
Replacing back in equation 2.3 we obtain,
j (t)iI=ei
~H0tU(t;ti)j (ti)i
=ei
~H0tU(t;ti)ei
~H0tj (ti)iI
=UI(t;ti)j (ti)iI
=) j (t)iI=UI(t;ti)j (ti)iI (2.6)
Back in equation 2.2,
i~d
dt
UI(t;ti)j (ti)iI
=VI(t)UI(t;ti)j (ti)iI
i~j (ti)iIdUI(t;ti)
dt=VI(t)UI(t;ti)j (ti)iI
=)i~dUI(t;ti)
dt=VI(t)UI(t;ti) (2.7)
As initial condition, we impose UI(ti;ti) =I.
22

Solutions are given by the integral equation,
UI(t;ti) = 1i
~Zt
tiVI(t0)UI(t0;ti)dt0(2.8)
Time dependent perturbation theory provides approximate solution to the in-
tegral equation. This consists in assuming that VIis small and then proceeding
iteratively.
First order approximation
U(0)
I(t0;ti) = 1
Replacing in equation 2.8 we obtain,
U(1)
I(t;ti) = 1i
~Zt
tiVI(t0)dt0(2.9)
Second order approximation
Back in equation 2.8 we get,
U(2)
I(t;ti) = 1i
~Zt
tiVI(t0)U(1)
I(t0;ti)dt0
= 1i
~Zt
tiVI(t0)
1i
~Zt1
tiVI(t2)dt2
dt0
= 1i
~Zt
tiVI(t0)dt0+
i
~2Zt
tiVI(t1)dt1Zt1
tiVI(t2)dt2(2.10)
A repetition of this iterative process yields:
U(n)
I(t;ti) = 1i
~Zt
tiVI(t0)dt0+
i
~2Zt
tiVI(t1)dt1Zt1
tiVI(t2)dt2+:::+
+
i
~nZt
tiVI(t1)dt1Zt1
tiVI(t2)dt2Zt2
tiVI(t3)dt3:::Ztn1
tiVI(tn)dtn
(2.11)
where,t>t 1>t2>t3>:::>t n1>tn>t0.
This is the Dyson series which allows the calculation of the state vector up to
the desired order in the perturbation.
23

2.2 Transition probability
The transition probability corresponding to a transition from an initial unperturbed
statej iito an unperturbed state j fiis obtained from:
Pi!f(t) =jh fjUI(t;ti)j iij2
=jh fj iii
~Zt
0ei!fit0h fjV(t0)j iidt0+
+
i
~2X
nZt
0ei!fnt1h fjV(t1)j nidt1Zt
0ei!nit2h njV(t2)j iidt2+:::j2
(2.12)
where we have used the fact that,
h fjVI(t0)j ii=h fjei
~H0t0V(t0)ei
~H0t0j ii=h fjV(t0)j iiei!fit0(2.13)
!fi=EfEi
~=1
~
h fjH0j fih ijH0j ii
represents the transition frequency
between the initial and nal levels i and f.
Pi!f(t) =jh fjU(t;t0)j iij2(2.14)
We consider the expansion
U(t;t0) =U(0)(t;t0) +infX
n=1U(n)(t;t0) (2.15)
=) h fjU(t;t0)j ii=h fjU(0)(t;t0)j ii+h fjinfX
n=1U(n)(t;t0)j ii (2.16)
First order contribution to the transition amplitude
h fjU(1)(t;t0)j ii=i
~Zt
t0h fjVI(t0)j iidt0(2.17)
Replacing eq. (2 :5VI(t)) in the formula above, we obtain
h fjU(1)(t;t0)j ii=i
~Zt
t0h fjei
~H0t0V(t0)ei
~H0t0j iidt0
=i
~Zt
t0ei
~Eft0ei
~Eit0h fjV(t0)j iidt0
=i
~Zt
t0eiEfEi
~t0h fjV(t0)j iidt0
=i
~Zt
t0ei!fit0Vfi(t0)dt0(2.18)
24

So, the rst order transition probability for j ii!j fiwithi6=fis
Pi!f(t) =i
~Zt
t0h fjV(t0)j iiei!fit0dt02
(2.19)
Terms higher than the rst order become rapidly intrasctable. For most problems
of atomic and nuclear physics, the rst order is usually sucient.
Cases when dealing with the interaction of atoms and radiation:
1. Constant perturbation
2. Harmonic perturbation
25

Bibliography
26