Light - Atom Interaction

PHYS261 autumn 2016

Version 24.11.2016 final

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 Topic 3 Topic 4
Overview Discussion Eigenmodes for coupled vibrations.
Time dep. QM- two-well problem Algebraic Method for Harm. Osc.
One level in continuum Quant. th. of extended systems - fields
Comparing hamiltonians and expansions Electromagnetic fields
Time-Dependent Schrödinger Equation The Quant Th. of Elmag. Field
Perturbation theory for TDSE Charged Particles In an Elmag. Field
Dirac delta-function The Hamiltonian of Interaction
Fermi Golden Rule −
Constant rate and exponent. decay Emission of Radiation by Excited Atom
Line width from exponential decay

Density of States
Algebraic Method for Harm. Osc. Matrix element reduction - Table
Eigenmodes for coupled vibrations. Dipole Approximation
Quant. th. of extended systems - fields Detailed Evaluation of Emission rate
Golden Rule Simulator part Final W = 1/τ result
T.-Dep. Schrödinger Eq. in Simulator Stimulated emission
Atomic Units

2
 − (2)

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Overview

1. Time dependent quantum mechanics:

two isolated states, contrasted to
one isolated level
energetically embedded in (quasi-) continuum of states

2. derivation and understanding of the transition rate (probability change per

time unit)

3. Time dependent perturbation theory

4. Fermi Golden rule derivation

5. The delta function (often mentioned as energy conservation; that is not

precise).

3
 6. In contrast to the delta function, the Lorentzian shape; finite line width
(also known as Breit-Wigner formula or shape)

7. Electromagnetic field is an extended (actually continuous) system, there-

fore we must learn how to understeand eigenmodes of large system. Cou-
pled oscillators transformed to a system of independent, de-coupled eigen-
modes.

8. Any harmonic oscillator can be described by so called creation and anihi-

lation operators. Harmonic oscillator via the algebraic method.

9. Follows quantization of the radiation field; photons

10. Evaluation of the transmissin rate for the emission proces

11. Spontaneous and Stimulated Emission

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4
1 Time dependent Q.M. - Two-well problem

| c (t)| 2
1

time

Figure 1: Top left: Well with 1 bound state; Bottom left: Two wells. System
(particle) placed in a state which is not an eigenstate Right column: Eigenstates
and time oscillations of population

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5
The two eigenstates in each of the isolated wells are ϕ1 and ϕ2 .
The static eigenstates of the particle confined to both of the wells are approxi-
mated by
1
ψ+ = √ (ϕ1 + ϕ2 )
2
1
ψ− = √ (ϕ1 − ϕ2 )
2

Clearly,

1
ϕ1 = √ (ψ+ + ψ− )
2
1
ϕ2 = √ (ψ+ − ψ− )
2
The states ψ+ , ψ− are eigenstates of the Hamiltonian, but
ϕ1 , ϕ2 are not, they are eigenstates for each isolated well.
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6
If at t = 0 the system is brought into the state ϕ1 ,
1
Ψ(t = 0) = ϕ1 = √ (ψ+ + ψ− )
2
then at any other later time
1  
Ψ(t) = √ ψ+ eiE+ t/h̄ + ψ− eiE− t/h̄
2
This can be rewritten schematically as
 
Ψ(t) = C(t) ψ+ + ψ− eiωt

so that we can see immediately that for

π
tn = n
ω
the Ψ(tn ) will be a multiple of ϕ1 , i.e. concentrated to the left,
for even n, and a multiple of ϕ2 , i.e. to the right, for odd n.
At other times there would be a changing distribution between
the two regions.
List of topics 7
 | c (t)| 2
1

time

| c (t)| 2
1

time

Figure 2: Above: 2 isolated states in two equal potential wells ; starting in left
well - oscillations. Below: Many states in two potential wells; starting in
left well - decay Go to list of topics

8
| c (t) | 2
1

time

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9
Comparison of basis sets - and hamiltonians Go to list of topics
Eigenstates of H0 , the set of |ϕi i, any |Φi can be expanded
X
H = H0 + H1 H0 |ϕi i = εi |ϕi i |Φi = ci |ϕi i
i
X X
(H0 + H1 )|ψα i = Eα |ψα i |ψα i = Siα |ϕi i |ϕi i = Sαi |ψα i
i α
X X
|Ψi = aα |ψα i |Ψi = ci |ϕi i
α i
Looking for solution of Schrödinger
∂ X
i h̄ | Ψ(t) i = H | Ψ(t) i |Ψ(t = t0 )i = |ϕ0 i |Ψ(t)i = ci (t) |ϕi i
∂t i
leads to X
i h̄ ċk (t) = Hki ci (t) ci (t = t0 ) = δi0 (3)
i
Or expand in |ψα i of H = H0 + H1 - including their exp(−iEα (t − t0 )/h̄)
Sα0 e−iEα (t−t0 )/h̄ | ψα i
X X
| Ψ(t0 ) i = |ϕ0 i = Sα0 |ψα i | Ψ(t) i =
α α
remembering
that i h̄ ∂/∂t ( exp(−iEα (t − t0 )/h̄)|ψα i ) = H ( exp(−iEα (t − t0 )/h̄)|ψα i)

10
2 Time-Dependent Schrödinger Equation
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The time-dependent Schrödinger equation
∂
i h̄ | ψ(t) i = H(t) | ψ(t) i (4)
∂t
is very often solved via a transformation to the matrix formulation. The matrix
formulation arises from expansion of the unknown wavefunction | ψ(t) i in a set
of basis functions | φi i , much in analogy with Fourier series or expansions using
orthogonal polynomials
X
| ψ(t) i = αi (t) | φi i (5)

The unknown quantities to be found are the expansion coefficients, which form
a vector. In this formulation, the time-dependent Schrödinger equation (eq. (4)
above) is replaced by a set of coupled differential equations.

11
The system of coupled equations is conveniently expressed by matrix notation
α1 H11 (t) H12 (t) ... H1n (t) α1
    

d  α2   H21 (t) H22 (t) ... H2n (t)  α2 

i h̄ = 
    
. ... ... ... ... .
   
dt     
αn Hn1 (t) Hn2 (t) ... Hnn (t) αn
where
Hij (t) = h φi | H(t) | φj i
This can also be written as
α̇1 H11 (t) H12 (t) ... H1n (t) α1
    
 α̇2   H21 (t) H22 (t) ... H2n (t)  α2 
i h̄  
=   
. ... ... ... ... .
    
    
α̇n Hn1 (t) Hn2 (t) ... Hnn (t) αn
which is a short-hand notation for
X
i h̄ α̇k = Hki (t) αi
i

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12
3 Perturbation theory for the Time-dependent
Schrödinger Equation
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To apply the perturbation theory, we must identify the small perturbation term.
Usually the Hamiltonian H(t), differs only slightly from well known Hamiltonian
H 0 so that the perturbation H 0 (t)

H(t) = H 0 + H 0 (t) (6)

If the perturbation H 0 (t) were not present, and chosing then as | φk i the eigen-
solutions of
H 0 | φk i = Ek | φk i (7)
the matrix for H(t) → H 0 would be diagonal with the eigenenergies Ek on the
diagonal. The solutions for the the αk (t) would be then very simply

αk (t) = e−iEk t/h̄ (8)

and for usual stationary states only one of the αk (t) would be different from zero.

13
Since the probability Pa of the system being in a state a is
Pa (t) = |αa (t)|2
it means that the system with probability 1 is in that state a, and with 0 in any
other state, i.e.
|αk (t = 0)| = δka (9)
If the perturbation really is small, the |αa (t)| remains ≈ 1 for all times t, while
all others remain close to zero.
We will thus use the so called perturbation theory, an approximate method to
solve the system of equations
X
i h̄ α̇k = Hki (t) αi
i

while keeping
|αa (t)| ≈ 1 −→ αa (t) = 1
and ”preserving” Go to list of topics

|αk (t = 0)| = δka −→ |αk (t > 0)| ≈ 0 for k 6= a

14
To simplify the notation we make a substitution αk (t) → ck (t) exp(−i Eh̄k t). In-
serting this into the last equation
X
i h̄ α̇k = Hki (t) αi
i

we get (using d/dt(c exp(−iEt)) = ċ exp(−iEt) + −iE c exp(−iEt) )

Ek X
0 Ei X Ei
(i h̄ ċk + Ek ck ) exp(−i t) = Hki ci exp(−i t) + Hki (t) ci exp(−i t)
h̄ i h̄ i h̄

0 Ek − Ea
With Hki = Ek δki ωka = (10)
h̄
X
0 Ek − Ei
i h̄ ċk + Ek ck = Ek ck + Hki (t) ci exp(i t)
i h̄
0 0 0
ċ1 H11 (t) H12 (t)eiω12 t ... H1n (t)eiω13 t c1
    
0 iω21 t 0 0
 ċ2   H21 (t)e H22 (t) ... H2n (t)eiω2n t  c2 
i h̄  
= 
  
. ... ... ... ... .
   
    
0 iωn1 t 0 0
ċn Hn1 (t)e Hn2 (t)eiωn2 t ... Hnn (t) cn
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15
Taking the perturbation theory assumptions
|αa (t)| ≈ 1 |αk6=a (t > 0)| ≈ 0 −→ |αk (t = 0)| = δka
the above matrix system of coupled equations decouple - we get independent
equations for each ck (t)
d 0
ck (t) = Hka
i h̄ (t)eiωka t (11)
dt
where we made a substitution αk (t) → ck (t) exp(−i Eh̄k t).
We say that the original assumption about the coefficients is a ’zero-th order
approximation, thus the superscript (0),
(0)
ck = δka (δ(k−a) ) (12)
while the above equation is the first order, thus we rewrite it
(1) 1 0 (iωba t)
ċb = Hba e (13)
ih̄
A detailed description of perturbation theory pictures it as an iterative process.
Each order is obtained by applying the preceeding order of the approximation.
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16
Equation of the form ċ = f (t) is easily solved by integration
(1) 1 Z t 0 (iωba t0 ) 0 0 0
cb (t) = H e dt Hba (t) ≡ Hba (14)
ih̄ tb ba
So the transition probability for going from a state ’a’ to a state ’b’ is defined
like Go to list of topics

(1)
Pba (t) = |cb (t)|2 (15)

Now we evaluate the integral in eq.(14), which is elementary for a constant

potential: Go to list of topics

(1) 0 1 Z t iωba t
cb (t) = Hba e dt ; t0 = 0
ih̄ t0
0 1 1
= Hba (eiωba t − 1)
ih̄ iωba
0
Hba
= − (eiωba t − 1)
h̄ωba
(16)

17
and rearrange (eiωt − 1) (in the following: ωba −→ ω )

ωt ωt ωt
h i
e(iωt) − 1 = e(i 2 ) e(i 2 ) − e(−i 2 )
ωt ωt
= 2ie(i 2 ) sin
2

Inserting this back into eq.(14)

(1) 1 Z t 0 iωt0 0
cb (t) = H e dt (17)
ih̄ t0

(1) 1 0 2 sin2 ωt
|cb (t)|2 = |H | F (t, ω) F (t, ω) = 4 · 2
(18)
h̄2 ba ω2
where the phase factor reduces to its absolute value one.
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18
4 Dirac delta-function
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It is easy to find that the ’F’ is a function of t and ω, and for each time ’t’ it is
equal to

sin2 ωt
2
F (t, ω) = 4 · 2
ω
t2 sin2 ωt
= 4 · · ω2 t2 2
4 4
sin2 x ωt
= t2 · ; x=
x2 2
(19)

It can be seen that the function F (ω, t) for larger and larger t approaches the
shape of the Dirac delta-function, (see the drawing). (Animated on the www)

19
 1

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0
−10 −8 −6 −4 −2 0 2 4 6 8 10

Plot of the function

sin2 x
x2

20
 t=1 t=2 t=5
1 4 25

3.5
0.8 20
3
4 sin2(ω t/2) / ω2 4 sin2(ω t/2) / ω2 4 sin2(ω t/2) / ω2
0.6 2.5 15
F(ω,t)

F(ω,t)

F(ω,t)
2
0.4 1.5 10

1
0.2 5
0.5

0 0 0
−10 −5 0 5 10 −10 −5 0 5 10 −10 −5 0 5 10
ω ω ω

t=20 t=60 t=99

400
3500
9000
350
3000 8000
300
4 sin2(ω t/2) / ω2 2500 4 sin2(ω t/2) / ω2 7000 4 sin2(ω t/2) / ω2
250 6000
2000
F(ω,t)

F(ω,t)

F(ω,t)
200 5000
1500 4000
150
3000
1000
100
2000
50 500
1000
0 0 0
−10 −5 0 5 10 −10 −5 0 5 10 −10 −5 0 5 10
ω ω ω

Approaching the δ-function with increasing t: Go to list of topics

sin2 ωt
2
F (ω, t) = 4 ·
ω2
21
Trying to integrate the function ’F’ over ω shows this

2
Z ∞
sin2 x Z ∞
sin2 x ωt
t 2
dω = 2t 2
· d( )
−∞ x −∞ x 2
Z ∞ 2
sin x
= 2t dx
−∞ x2
= 2πt
(20)

that for large t it really behaves as

sin2 ωt 2
F (t, ω) = −→ 2πtδ(ω) (21)
( ω2 )2
The summation over all the states used for the expansion can be assumed to be
replaced by the integration over ω or the energy E = h̄ω. This however needs to
determine the factor ρ(E) which accounts for the density of states. Since the ’F’
approaches delta-function, we only need the density of states close to the final
state b:

22
 1 Z Eb +η 0 2
P̄ba (t) = 2 |Hba | F (t, ω)ρ(E)dE (22)
h̄ Eb −η
and also integration over dω gives

Z Z
δ(ω)dω = δ(E)dE
dE
δ(ω) = δ(E) ·
dω
= h̄δ(E − Eb )
(23)

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23
5 Fermi Golden Rule
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The final result for the probability of populating the group of states close to the
final state | b > is

1 Z Eb +η 0 2
Pba (t) = 2πt|Hba | δ(E − Eb )ρ(E)dE
h̄ Eb −η
2π
= t| < b|H 0 |a > |2 ρ(Eb ) (24)
h̄
The derivation of the above formula has been based on the assumption of a small
perturbation. It shows that the probability of transition to the state b or states
close to b increases linearly with time. Thus the rate of probability change per
time is given by
dPba 2π
= Wba = | < b|H 0 |a > |2 ρ(Eb ) (25)
dt h̄
This result is known as Golden Rule formula or Fermi Golden Rule.

24
6 Constant rate and exponential decay
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Fermi golden rule gives a constant rate

dPf
=w (26)
dt
or if we consider the probability decrease to find the system in the original state,
dPi
= −w (27)
dt
If this, instead of a definition, is taken as a differential equation
Pi = P0 − wt (28)
can quickly become negative.However, one can quite easily realize that the correct
differential equation must be

dPi
= −wPi (29)
dt
25
since the loss of probability must be proportional to ’how much is left’, i.e. the
Pi itself.It can also be guessed from the differential equations of Q.M. (leading
dP
to the dtf ) , since they contained the amplitude, which we replaced by 1.
The last differential equation leads to the well known exponential decay, since
its solution is

Pi = e−wt (30)
since Pi (t = 0) = 1

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26
7 Line width from exponential decay
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In the equation (11) the expansion coefficient of the initial state α0 (t) → 1 for
all times leads to the delta function for energy (frequency). In order to take into
account the flow of probability from the initial state 0, the relation
αk (t) = δka
when working with the time dependent problem must be changed to
w
|αk (t)| = δka exp(− t)
2
where w is the constant rate factor obtained in the perturbation theory.
If we take into account the result of previous section 6, we have |c0 (t)| →
exp(−wt/2). Inserting this into equation (11), the integrals can still be per-
formed, but the expressions which were found to converge to the δ-function in
frequency (energy) for time going to infinity are not obtained. The term
Z
0 iωba t
Hba e dt

27
leading to
1 iωba t 0
(e − 1) −Hba
ωba
which has been shown to lead to δ-function like behaviour when integrated over
ω is now replaced by
ei(ω−ωba )t e−wt/2 − 1
i(ω − ωba ) − w2
at large times t the factor e−wt/2 −→ 0 and this leads to
2
1 1
w =
i(ω − ωba ) − 2
(ω − ωba )2 + 14 w2
The rate, which has the dimension of frequency becomes the energy width
Γ = h̄w
when multiplied by h̄. the so called Lorentz shape of the line,
1

0.9

Γ=1 E0=15
0.8
 2
0.7

0.6
Γ
0.5
2
0.4

0.3 I(E) ≈ I0  2 (31)

(E − E0 )2 + Γ
0.2

0.1

0
0 5 10 15 20 25 30
2

28
Thus a more realistic treatment than the perturbation theory with constant decay
rate leads to the natural width of the energy spectrum of the ejeced photons as
observed. This behaviour has been illustrated in our Golden Rule Simulator
program Go to list of topics
1 1

0.9 0.9

0.8
Γ=1 E =15
0 0.8

0.7 0.7

0.6 0.6

0.5 0.5

0.4 0.4

0.3 0.3

0.2 0.2

0.1 0.1
Γ=1 E0=15

0 0
0 5 10 15 20 25 30 14 14.5 15 15.5 16

 2
Γ
2
I(E) ≈ I0  2
(E − E0 )2 + Γ
2

29
8 Generally on quantum treatment of extended
systems - fields.
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The system to be considered consists of an atom and the electromagnetic field.
The field has in principle infinitely many degrees of freedom, i.e. the values of
field variables in each point of space.
The electromagnetic field can be treated by wave equations, it is thus a sort of
a harmonic system. The energy of a finite harmonic system can always be
transformed to a sum of independent harmonic oscillators, each of them is in
fact an eigenmode.
Each harmonic oscillator can be then described using the equations for harmonic
oscillator, with the algebraic method and number of quanta states - The creation
and annihilation operators - click 10.
The normal modes or eigenmodes for a general harmonic system are discussed
in the section 9.

30
The prescription

1. identify the eigenmodes

2. quantize each of the modes as independent harmonic oscillator

3. use quanta of eigenmodes

4. express the general coordinates of the system using the eigenmode coor-
dinates (inverse transformations to those discussed in section 9. Each co-
ordinate χi will then be replaced by its combination of the creation and
annihilation operators as discussed in section 10.

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31
9 Normal Coordinates for coupled harmonic
vibrations.
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Transformation to normal coordinates can be described as follows:

Take the total energy, x represents the vector of all coordinates, xT represents
the transposed vctor, M is the mass matrix, V is the matrix which gives the
potential energy, including the couplings.

Note that the mass matrix is written in a very general form, most often this
matrix would be diagonal.

1 1
E = ẋT Mẋ + xT Vx
2 2

First we transform the kinetic energy to a ”spherical” form

32
 1 T 1 1
ẋ Mẋ = η̇ T ST MSη̇ = η̇ T η̇
2 2 2
This transformation does not conserve the lengths. The N-dimensional ellipsoid
is transformed to an N-dim. sphere.

ST MS = 1

If Mij = mj δij , it is quite easy to find Sij = mj −1/2 δij , Transformation S simplifies
the kinetic energy, but the potential remains complicated. Therefore we use one
more, from η to χ,

1 T 1 1
x Vx = η T ST VSη = χT U−1 ST VSUχ
2 2 2
Transformation η = Uχ is a simple rotation obtained by diagonalizing the matrix
of potential energy: Go to list of topics

 
U−1 ST VSU = Ωj2 δij = Wij
ij

33
Therefore we took U−1 = UT . We can see that the ” length ” is conserved.
1 T 1 1
η̇ η̇ = χ̇T U−1 Uχ̇ = χ̇T χ̇
2 2 2
1 1 1 1
E = ẋT Mẋ + xT Vx = χ̇T χ̇ + χT Wχ
2 2 2 2
This expression however is a sum over the new mutually independent degrees of
freedom, since W is diagonal.
N
1 T 1 T 1 2 1 2 2
X  
χ̇ χ̇ + χ Wχ = χ̇i + Ωi χi
2 2 i=1 2 2

Summary: for any type of coupled oscillations normal modes or eigenmodes

can be found by linear transformations (diagonalization)

N N N 
1 1 X 1 2 1 2 2

mi ẋ2i +
X X
Vij xi xj −→ χ̇i + Ωi χi
i=1 2 2 i,j=1 i=1 2 2

Go to list of topics

34
Example Go to list of topics
A string of balls of mass m connected by springs of equal spring constant is de-
scribed by displacements ui . We can label the displacements ui = u(xi ) where xi
is the equilibrium position of the i-th ball. In the limit of infinitisemal small balls
and short springs this would then lead to a ”wave equations” for a continuous
elastic string, with continuously observed displacement u(x).
The total energy is:
1X 1 X 1 1
m (u̇(xi ))2 + k (u(xi ) − u(xi + 1))2 + ku(x1 )2 + ku(xN )2
2 i 2 i 2 2

The last two terms represent the fixed end springs. The matrix M is simply a
diagonal matrix with m as all diagonal elements. The matrix V is a band matrix
(below).
The transformation from u(xi ) to χi can also be reversed. The reversed transfor-
mation in this case simply consists of eigenvectors of the matrix V, so that each
of the eigenmodes is simply given by a function of time

χk (t) = χk0 eiΩk t

35
where χk0 is the amplitude, and the actual displacements for the k-th mode can
be written as
uk (xi ) = Sik χk (t) = Sik χk0 eiΩk t
where Sik is the i-th component of the k-th eigenmode with the frequency Ωk
The matrix V has the form
 
+2 −1 0 0 0 ... 0

 −1 +2 −1 0 0 ... 0 

0 −1 +2 −1 0 ...
 

 0 


 0 0 −1 +2 −1 ... 0 

0 ... ... ... ... ... ...
 
 
 

 0 ... ... ... +2 −1 0 

0 ... ... ... −1 +2 −1
 
 
0 ... ... ... 0 −1 +2

The eigenvectors - giving the eigenmodes - are eigenvectors of this band matrix.
The modes are ”standing waves”, in the string limit. For infinite system - travel-
ing waves. leads to Fourier expansions for fields ) see (link) A(r) expressed
in eigenmodes Go to list of topics

36
 0.4

0.3
0.4

0.3 0.2

0.2
0.1

0.1
0

−0.1

−0.1

−0.2
−0.2

−0.3
−0.3

−0.4 −0.4
0 2 4 6 8 10 12 14 16 18 20 0 2 4 6 8 10 12 14 16 18 20

0.2 0.15

0.15
0.1

0.1
0.05

0.05
0

−0.05
−0.05

−0.1
−0.1

−0.15
−0.15

−0.2 −0.2
0 10 20 30 40 50 60 70 80 90 100 0 10 20 30 40 50 60 70 80 90 100

0.15

0.1

0.05

−0.05

−0.1

−0.15

−0.2
0 10 20 30 40 50 60 70 80 90 100

Figure 3: Components of eigenmodes at positions xi . For N = 20 first and

second node, fo N = 100 nodes 1,2, and 6. Go to list of topics

37
10 Algebraic Method for Harmonic Oscillator.
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In this section we show how one can work with the harmonic oscillator introduc-
ing so called ladder operators which move from state to state, the states being
separated by the same ’quantum of energy’.
The classical hamiltonian can be transformed:
1 2 mω 2 2 h̄ω  2 
p + q transforms to P + Q2
2m 2 2
s
1 mω
r
P = p Q= q
h̄ m ω h̄
With this the commutator
[ q, p ] = ih̄
becomes
[ Q, P ] = i
This simplifies the equation and brings it into a form where the energy is ex-
pressed in h̄ω.

38
The main transformation, however, is to go over to linear combinations of P and
Q

1 i 1 i
a = √ Q+ √ P and a+ = √ Q − √ P
2 2 2 2
By a very simple algebra we find that for their commutator
h i
a, a+ = 1

and the energy is transferred to Go to list of topics

h̄ω  2  h̄ω  + 
P + Q2 → a a + a+ a
2 2
Using the commutator a a+ − a+ a = 1 we finally get
h̄ω h̄ω
H = h̄ω a+ a + = h̄ω N +
2 2
where we already include that

N = a+ a

39
will be a number operator.
Why number? Let us play with [ a, a+ ] = 1 or alternatively a a + − a+ a = 1
and the operator N . We quickly derive that
h i
N, a+ = a+ and [ N, a ] = −a
This is done simply by writing out Go to list of topics
h i h i
N, a+ = N a+ −a+ N = (a+ a) a+ −a+ (a+ a) = a+ (a a+ −a+ a) = a+ a, a+ = a+
h̄ω
Since H = h̄ω N + 2
we also have
h i
H, a+ = h̄ω a+ and [ H, a ] = − h̄ω a
These are the last equations which bring the physical interpretation.
If there is a Q.M. state ψ(q) → |ψi such that
H |ψi = E |ψi
then
H (a |ψi ) = a H |ψi − h̄ω a | ψi = a E |ψi − h̄ω a | ψi = (E − h̄ω ) (a |ψi )

40
that means that (a |ψi ) is also an eigenstate. It has energy lower by h̄ω . This
we can continue again and again, getting eigenstates for

E − h̄ω , E − 2 h̄ω , E − 3 h̄ω , E − 4 h̄ω , E − 5h̄ω .......

Since this cannot go on infinitely, we get finally a state such that

a |ψ0 i = 0

We quickly verify that this state has Go to list of topics

h̄ω
E0 =
2
further that the same thing is possible with a+ only with opposite sign, so that
we get energies

E0 , E0 + h̄ω , E0 + 2 h̄ω , E0 + 3 h̄ω , E0 + 4 h̄ω , E0 + 5h̄ω .......

Each of the eigenstates is an eigenstate of both H and N with obvious number of

quanta. Therefore we call a+ and a creation and annihilation operators. They
make states with one more ore one less quantum h̄ω

41
We can complete this treatment by a complete solution for the wavefunction:
1 ∂
a |ψ0 i = 0 a = √ (Q + i P ) P = −i
2 ∂Q

Thus, writing |ψ0 i −→ ψ0 (Q)

!
∂
a |ψ0 i = 0 −→ Q+ ψ0 (Q) = 0
∂Q

The solution of 1. order differential eq. to the left is easy

2 /2
ψ0 (Q) = C0 e−Q

and using the expression for a+ and the |ψn i Go to list of topics
! ! !
∂ −Q2 /2 ∂ ∂ 2 /2
ψ1 (Q) = C1 Q− e ψ2 (Q) = C2 Q− Q− e−Q
∂Q ∂Q ∂Q

and n times for general |ψn i. cf. recursion relations for Hermite polynomials.

42
11 Electromagnetic fields
Go to list of topics
In atomic physics we prefer to work in Gaussian units, where the strengths of
the fields have the same physical dimension. In vacuum D = E and H = B
The classical electromagnetic field is described by electric and magnetic field
vectors E and B , which satisfy Maxwell’s equations:

∇ · E = 4πρ (32)
∇·B=0 (33)
1 ∂B
∇×E=− (34)
c ∂t
!
1 ∂E
∇×B= + 4πJ (35)
c ∂t

43
The electric field E and magnetic field B can be generated from scalar and vector
potentials φ and A by

1∂
E(r, t) = −∇φ(r, t) − A(r, t) (36)
c ∂t

B(r, t) = ∇ × A(r, t) (37)

The potentials are not unique, observable field strengths E(r, t) and B(r, t) re-
main the same when the potentials are changed by

A(r, t) → A(r, t) + ∇λ(r, t)

∂λ(r, t)
φ(r, t) → φ(r, t) −
∂t
where λ(r, t) is any scalar field. Go to list of topics
This property is called gauge invariance. It allows us to choose A so that

∇·A=0 (38)

44
When A satisfies this condition, we are using the ”Coulomb Gauge”. From
Maxwell’s equations (without sources) we can show that A satisfies the wave
equation
1 ∂ 2A
∇2 A − 2 2 = 0 (39)
c ∂t
Also φ, B and E satisfy wave equations, but for radiation problems in empty
space (vacuum) we have the potential φ = 0 since there are no charges there.
The energy stored or contained in the electromagnetic field is given by the formula
1 Z 3
Hf ie`d = d r (E · E + B · B) (40)
8π
Eigenmode in an ”infinite box” is a plane wave solution of equations (38) and
(39) with angular frequency ω (i.e. the frequency ν = ω/2π) is given by the
vector potential A Go to list of topics

A(ω; r, t) = 2A0 (ω) cos(k · r − ωt + δω )

= A0 (ω) [exp[i(k · r − ωt + δω )] + c.c.] (41)

45
12 The Quantum Theory of Electromagnetic
Field
Go to list of topics
The energy stored or contained in the electromagnetic field is given by the formula
1 Z 3
Hf ie`d = d r (E · E + B · B) (42)
8π
Gaussian units, the electric and magnetic field strengths have the same physical
dimension.
The energy of extended system can be written as a sum of energies in eigenmodes.
The eigenmodes in ”infinite box” are the plane waves.
The sum over plane waves gives the form of Fourier series - for the operator of
the vector potential A(r)
s
2πh̄c2 
−ik·r

ekλ akλ eik·r + a+
X
A(r) = kλ e (43)
kλ V ωk

46
The sum includes all the possible values of the propagation vector k and also the
two possible polarizations λ = 1, 2.In a finite box the boundary conditions lead
to quantization of wave vectors k as discussed in ”Density of states” section.
Periodicity on the box walls gives allowed values (nx , ny , nz ) 2π
L
where ni are
integers.
The dimensional factor s
2πh̄c2
V ωk
is determined by the necessity that Go to list of topics
Z
(E 2 + B 2 )
Z
hΦ| Hrad |Φi = hΦ| H dV |Φi = hΦ| dV |Φi (44)
V V 8π
must give the total energy of the electromagnetic field inside the box, when
written as sum over quantized eigenmodes
h̄ωk a+
X X
hΦ| Hrad |Φi = hΦ| Hkλ |Φi = hΦ| kλ akλ |Φi (45)
kλ kλ

This is done using

ωk
B(r) = i k × A(r) E(r) = i A(r) ωk = kc
c
47
 The operators akλ , a+
kλ and Nkλ are ”annihilation”, is called ”creation” and
number operators for photons in each eigenmode.
One ignores the term 12 kλ h̄ωk that refers to ”zero point energy” (infinite num-
P

ber of modes - Casimir effect).

The state of the field can be written as a direct product of the vector states for
each of the oscillators Go to list of topics

|Φi → |· · · nkλ · · · nk0 λ0 · · ·i = |· · ·i · · · |nkλ i · · · |nk0 λ0 i · · · (46)

For such state vectors:

√
akλ |· · · nkλ · · · nk0 λ0 · · ·i = nkλ |· · · nkλ − 1 · · · nk0 λ0 · · ·i (47)
√
a+
kλ |· · · nkλ · · · nk0 λ0 · · ·i = nkλ + 1 |· · · nkλ + 1 · · · nk0 λ0 · · ·i (48)
Nkλ |· · · nkλ · · · nk0 λ0 · · ·i = nkλ |· · · nkλ · · · nk0 λ0 · · ·i (49)
Hkλ |· · · nkλ · · · nk0 λ0 · · ·i = h̄ωk nkλ |· · · nkλ · · · nk0 λ0 · · ·i (50)
where
nkλ = 0, 1, 2, 3, . . . , ∞

48
13 Density of States
Go to list of topics
We consider the electromagnetic field enclosed in a box with volume V . For a
finite volume there is a discrete number of modes satisfying the imposed boundary
conditions. Therefore the sum over final states is an ordinary sum. As this
volume approaches infinite size, the summation over k, will be approaching an
integral.
The density of states factor is found from performing such a limiting process,
using following relations.
The summation will be replaced by an integral
X Z
−→ ρ(k) dk (51)
k

ρ(k) is the density of states in the K-space.

The allowed discrete values of k are obtained by combinations of components
2π 2π 2π
kx(nx ) = nx ky(ny ) = ny kz(nz ) = nz (52)
L L L
49
where the numbers nx ,ny ,nz are positive and negative integers. It means that
each of the allowed vectors k occupies a small volume of the K-space
3
2π (2π)3

∆kx ∆ky ∆kz = = (53)
L V
The density of states in the K-space is thus a constant (i.e. one per the above
small k-space volume), and the above relation can be written as
X V Z
−→ dk (54)
k (2π)3

Since the derivation of the golden rule assumes integration over frequencies, or
energies, we shall transform this integral over momenta (i.e. wave numbers k) to
integral over energy, Go to list of topics

V Z Z
dk −→ ρ(E)dE
(2π)3

Z Z " #
V Z V Z Z
V dk
dk = dΩk k 2 dk = k2 dΩk dE (55)
(2π)3 (2π)3 Ωk Ωk (2π)3 dE

50
so that the ρ(E) can be identified as

V 2 dk
Z
k dΩk (56)
(2π)3 dE Ωk

If the processes depend on the direction of the wave vector, which is true in
photon emission case, we must keep the angular information inside of the density
of states, and perform the angular integration afterwards.
We must now evaluate the above density of states in terms of energy only, using
ω
k= E = h̄kc
c

Setting these relations of k and E, we obtain the expression for ρ(E)

V E2 V 1 ω2
ρ(E) = · dΩk = · · dΩk (57)
(2π)3 (h̄c)3 (2π)3 h̄ c3

Go to list of topics

51
 L
L 2π
L

L
L
2π
L

For larger L we get finer division of the k-space, i.e. they get denser - ∆k = 2π/L
Thus the density of states is larger in the k-space, i.e. ρ(k) ∝ L3 List of topics

52
 ω
Note: We got for photons - where k= c E(k) = h̄kc

V E2 V 1 ω2
ρ(E) = · dΩ k = · · dΩk
(2π)3 (h̄c)3 (2π)3 h̄ c3
For electrons: The k-dependence of energy for electrons leads to a different density of
states for electrons
3
h̄2 k 2 V 2m

2 1
E(k) = , ρ(E) = 2 E2
2m 2π h̄2
while for the photons we have obtained entirely different density of states

ρ(ω) ∝ ω 2 or ρ() ∝ E 2

In both cases the dependence on the V = L3 must be cancelled by the normalization

of the states / modes
c.f. in our case List of topics
s
2πh̄c2 
−ik·r

ekλ akλ eik·r + a+
X
A(r) = kλ e
kλ
V ωk

so that the results of calculations are independent of the size of the box L.

53
14 Charged Particle In Electromagnetic Field
In classical mechanics Newton equations with the electromagnetic Lorentz force
1 ∂A 1
 
mr̈ = q −∇Φ + − + v × [∇ × A]
c ∂t c
can be derived from a surprisingly simple Lagrange function
1 q
L(r, ṙ, t) = m v2 − qΦ(r, t) + v · A(r, t)
2 c

From this Lagrange function follows the Hamiltonian for a particle of charge q and
mass m
1 q
H= (p − A)2 + qφ (58)
2m c
where p is the generalized momentum of the particle. For the electron in a hydrogenic
2
atom q = −e. The electrostatic potential is φ = − Zer between the electron and the
nucleus. The radiation field which perturbs the atom is described in terms of a vector
potential A alone List of topics

54
The time dependent Schrödinger equation for a hydrogenic atom in an electro-
magnetic field then reads

Ze2
" #
∂ 1 e
ih̄ ψ(r, t) = (−ih̄∇ + A)2 − ψ(r, t) (59)
∂t 2m c r
where we have written p = −ih̄∇ . Because of the gauge condition ∇ · A = 0,
we have

∇ · (Aψ) = A · (∇ψ) + (∇ · A) ψ = A · (∇ψ) (60)

| {z }
0

so that ∇ and A effectively commute. Then

h̄2 2 Ze2 ih̄e e2

" #
∂
ih̄ ψ(r, t) = − ∇ − − A·∇+ A2 ψ(r, t) (61)
∂t 2m r mc 2mc2

In ”weak field case” the term with A2 is small compared

with the linear term A .
Go to list of topics

55
15 The Hamiltonian of Interaction
Go to list of topics
Now, we study the interaction between radiation and atomic system. In this case
the total Hamiltonian of the system is written as

H = Hatom + Hrad + HI
= H 0 + HI (62)

Hatom is the Hamiltonian of the atomic system and Hrad is the Hamiltonian of the
free radiation field. HI shows the Hamiltonian of the interaction effect between
two previous systems. By replacing HI , we replace p by p + e/cA in Hatom we
can construct HI as
e e2 e
HI = p·A+ 2
A2 HI → p·A (63)
mc 2mc mc
As we see the first term of the eq.(63), is proportional to A that contributes in
the transitions involving the emission or absorption of a single photon and the

56
next term that’s proportional to A2 contributes in the transitions involving the
emission or absorption of two photons. We can also see that the first term; A;
contains just one creation or annihilation operator, but A2 contains the terms
as : akλ akλ , a+ + + +
kλ akλ , akλ akλ , akλ akλ which correspond to absorption of two
photons; emission of two photons. Here we study the first term. The probability
for a transition involving two photons contains e4 while this probability for one
photon is proportional to e2 . This extra power of e2 will lead to one more power
of α ' 1/137.
The eigenfunctions of the unperturbed Hamiltonian H◦ are direct products of
the atomic and radiation wave functions

ψatom ψrad = |ψa ; n1 , n2 , . . . , ni , . . .i (64)

where ψa is the wave function of the atomic Hamiltonian, ni ≡ nki λi is the wave
function describing the mode ki λi in the radiation field.

Go to list of topics

57
16 Emission of Radiation By an Excited Atom:
Go to list of topics
Here we study the emission of a single photon by an excited atom, and we use the
time-dependent perturbation theory to find the transition probability for atom.If
we consider |ii as the initial state and |f i as the final state, then the probability
per unit time for a transition by the emission of a single photon from the initial
to final state is given by
2π
Wi→f (k, λ) = |hf |HI |ii|2 ρ(Ef ) (65)
h̄
where h̄k denotes the momentum and λ denotes the polarization, and HI is the
perturbing Hamiltonian. This equation is known as Fermi’s Golden Rule.
The perturbing Hamiltonian for this process is
e
HI = − p·A (66)
mc

58
by referring to the eq.(43), we have
s
e X 2πh̄c2 
−ik·r

HI = − p · ekλ akλ eik·r + a+
kλ e (67)
mc kλ V ωk

The initial and final states |ii; |f i;, i.e. the unperturbed wavefunctions before
and after the emission of one photon is

|ii = |ai |· · · nkλ · · ·i (68)

|f i = |bi |· · · nkλ + 1 · · ·i (69)

The energy difference of these two states is

Ef − Ei = (Eb + nkλ h̄ωk + h̄ωk ) − (Ea + nkλ h̄ωk ) = Eb − Ea + h̄ωk (70)

For spontaneous emission, no photon is present in the initial state: nk → 0 and

(nk + 1) → 1

Go to list of topics

59
17 The matrix element reduction
Go to list of topics
Inserting the final and initial states and the Hamiltonian into the matrix element,
we obtain
s
e 2πh̄c2 D E√
hf |HI | ii = − b p · ekλ e−ik·r a nkλ + 1 (71)
mc V ωk

For spontaneous emission, no photon is present in the initial state: nk → 0 and

(nk + 1) → 1
By considering the eq.(71) we can transform the Golden Rule formula eq.(65) for
spontaneous emission (note that ρ(Ef ) implies angular integration )

2πh̄c2
2 !
2π e
 D E2
Wi→f (k, λ) = ρ(Ef ) b p · ekλ e−ik·r a (72)
h̄ mc V ωk

60
 e p .A
Ψfin HI Ψinit HI mc

Ψfin e
p. A Ψinit Ψinit 0photon ϕEXC
mc
Ψfin ϕGS hω
e
ϕGS h ω m c p .A 0photon ϕEXC ω ωk λ HI

2
2πhc i k.r +
A Σ ek λ
ωkV
e a
p .A Σ p .ekλ
2
2 π h c −i k. r +
kλ kλ
kλ
ωV e ak λ
2
e 2 π h c −i k. r
ϕGS hω mc Σ p.ek λ ωV e a+k λ 0photon ϕEXC

2
e 2πhc
ϕGS p .ek λe−i k r ϕEXC hω ak+λ 0photon
.
mc ωV
Go to Final W = 1/τ result Go to list of topics

61
18 Dipole Approximation
Go to list of topics
An important simplification is obtained by applying the so called dipole ap-
proximation, considering the case kr  1. This means that the wavelength of
the emitted photon is much larger than atomic dimension. The electric dipole
approximation is obtained by replacing, to a good approximation,

eik·r ≈ 1
i.e.
hb pe−ik·r ai → hb |p| ai
This approximation is limited by these requirements:
λ  1Å and h̄ω  10 keV.

Here we also derive the relation between matrix element of momentum and co-
ordinate using the following commutation relations

62
relation between matrix element of momentum and coordinate derived using the com-
mutation relations ( starting from [x, px ] = ih̄ )
h i
x, p2x = x p2x − px x px + px x px − p2x x = [x, px ] px + px [x, px ] = 2ih̄px (74)

ih̄
p
[r, Hatom ] = (75)
m
we change the matrix element of the momentum to a matrix element of the special
coordinate of the atomic electron
m
hb | p | ai = hb |[r, Hatom ]| ai
ih̄
im
= (Ea − Eb ) hb | r | ai
h̄
= i m ωab hb | r | ai (76)

In summary,
hb p e−ik·r ai −→ i m ωab hb | r | ai

Final W = 1/τ result Go to list of topics

63
19 Detailed Evaluation of Emission rate
Go to list of topics
In applying the density of states ρ(E), the summation - i.e. integration - over all
available final (photon) states specified by wavenumber vector k and two independent
polarisation vectors ek,λ for each k must be performed
V dk V 1 ω2
Z X Z X
k2 dΩk −→ · · dΩk
(2π)3 dE Ωk
λ(k)
(2π)3 h̄ c3 Ωk
λ(k)

If e.g. angular distribution of the radiation is of interest, or even combined with

polarisation analysis, this integration will not be performed. The total summation is
essential for the total transition rate, related to the lifetime, which we consider here.
2 !
2π e 2πh̄c2

Wi→f = ρ(Ef ) |hϕb |p · ekλ | ϕa i|2
h̄ mc V ωk
2 !
2π e 2πh̄c2 V 1 ω2
 Z
|hϕb |p| ϕa i · ekλ |2
X
Wi→f = 3
· · 3k dΩk
h̄ mc V ωk (2π) h̄ c Ωk
λ(k)

For given hϕb | and |ϕa i, we have a vector (three numbers)

( hϕb |px |ϕa i, hϕb |py |ϕa i, hϕb |pz |ϕa i ) = P0 hϕb |p · ekλ | ϕa i → P0 · ekλ

64
z’
k
ek λ z’

p’
p’
θ
θ

o ek 2
90
θ
x’ φ
x’ y’

ek 1
ek 1
k =( sin θcos φ ,sinθ sin φ ,cosθ ) k
k =( sinθ , 0, cosθ ) k ek 1 =(cos θcosφ ,cos θ sin φ , −sin θ )
ek 1 =(cosθ , 0, −sin θ ) ek 2 =(− sin φ , cosφ , 0 ) hϕb |p · ekλ | ϕa i → P0 · ekλ

Polarization vectors ekλ . Go to list of topics

The z 0 -axis is arbitrary, here in the direction of the vector P0 = hϕb |p| ϕa i
Since we are summing over the two polarisations, the polarisation vectors can be chosen for each
k independently as indicated. Then ek2 is always perpendicular to P0 and only P0 ·ek1 = Pz0 sinθ
contributes. Pz0 = hϕb |px |ϕa i2 + hϕb |py |ϕa i2 + hϕb |pz |ϕa i2
p

65
We insert P0 · ek1 = Pz0 sinθ and P0 · ek2 = 0 into
2 !
2π e 2πh̄c2 V 1 ω2
 Z
|hϕb |p| ϕa i · ekλ |2
X
Wa→b = · · dΩk
h̄ mc V ωk (2π)3 h̄ c3 Ωk
λ(k)
q
with Pz0 = hϕb |px |ϕa i2 + hϕb |py |ϕa i2 + hϕb |pz |ϕa i2 we get |Pz0 sinθ|2 = |hb |p| ai|2 sin2 θ

1 e 2 ωk
  Z
Wa→b = = dΩk |hb | p | ai|2 sin2 θ (77)
τ a→b 2πm2 h̄c3 Ωk

with |hb|p|ai|2 = hb|p|ai∗ · hb|p|ai and the integration variables is dΩk = sin θ dθ dϕ
Note that in the above evaluations the volume V in ρ[Ef ] cancels the volume in the
field dimension - normalization factor
s
2πh̄c2 V 1 ωk2
Z X
ρ(E) = · · dΩk
V ωk (2π)3 h̄ c3 Ωk
λ(k)

sin2 θ sin θ dθ = 8π/3

R R
Now we can perform the integration dϕ
The last equation eq.(77) then becomes Go to list of topics
1 4 e 2 ωk
 
Wa→b = = |hb| p |ai|2 (78)
τ a→b 3 m2 h̄c3

66
We now discuss the lifetime of spontaneous emission (states a, b, thus ωk → ωab )
1 4ωab e2
 
Wsp. em. −→ = |hb| p |ai|2 (79)
τ a→b 3h̄m2 c3
Further we use for the matrix element of momentum eq. (76) on this page
hb | p | ai = imωab hb | r | ai and thus Bohr radius fine structure const.
 
1 4e2 ωab
3 h̄2 e2 1
Wsp. em. = = |hb | r | ai|2 a◦ =
me2
α=
h̄c
≈
137.04
τ a→b 3h̄c3
! 3 !3 2 e2
4 e2 h̄ωab a◦ 1 e2 r ε◦ = = 27.21 eV = 2 Ryd
 
= a◦ a2◦ b a a◦
3 a◦ h̄ e2 a3◦ h̄c a◦
3 2 h̄ = 0.6582 10−15 eV s
4 ε◦ h̄ωab r
  
= α3 b a
3 h̄ ε◦ a◦
For one-electron atoms (or ions) with atomic number Z, the lifetime of the 2p −→ 1s
transition is given by
1
 
= 0.6 × 109 Z 4 sec−1 (80)
τ 2p−→1s
D E
h̄ωab
since α3 ≈ 0.4 × 10−6 , ε◦ ∝ Z 2, and b r
a◦ a ∝ Z −1
Go to list of topics

67
20 Stimulated emission
Go to list of topics
For the case of stimulated emission, that
√ is dependent on the number of photons in
the field (intensity), we must keep the nkλ + 1 term of the nkλ initial photon state
and start from (however, see the note at the end)
2 !
2π e 2πh̄c2
 D E2
Wi→f (k, λ) = (nkλ + 1) b p · ekλ e−ik·r a ρ(Ef ) (81)
h̄ mc V ωk
D E √ D E
with nkλ + 1|a+
kλ |nkλ = nkλ + 1 replacing 1kλ |a+
kλ |0kλ = 1.

The emission rate then becomes

1
 
Wst. em. = (82)
τ a→b
3 2
4 ε◦ h̄ωab r
  
3
= α (nω + 1) b a (83)
3 h̄ ε◦ a◦
where nω represents the number of photons in the field, with the energy of Eω = h̄ω.

68
The last equation can be considered as a sum of two terms; one is proportional to the
number of photons in the field nkλ , or nω , so it’s radiation field intensity-dependent,
and describes the stimulated emission; the other one (expressed by number one in the
paranthesis) is independent of the field intensity.
The term from ( + 1 ) is the said to account for the spontaneous emission contribution.
Note however, that the total integrations over the emission angle and polarisation per-
formed in the previous case, would not be possible to carry out. A more careful
treatment of the density of states would be necessary in detailed applica-
tions.

Go to list of topics

69
21 Fermi Golden Rule Simulation
———- Go to list of topics ———- Time dependent wavefunction ———- Model
realization ———- The Hamiltonian of the model ———-

The physical problem: solve the time-dependent Schrödinger equation with hamil-
tonian as in the derivations of the Fermi Golden Rule.
continuum of textbook derivation −→ quasicontinuum here, i.e. a set of many
equidistantly closely spaced levels.
The idea is to demonstrate how the line width is proportional to the density of states
and the strength of the coupling.
Only the couplings from the discrete states to continuum and back are nonzero, have
a constant value.

70
22 Realization of the model
———- Start of the simulator part ———- Go to list of topics ———-

Figure 4: Schematic representation of the Energy levels

The figure 4 shows the energy levels: the single discrete level and the two quasicontinua.
Note that each has a different density of levels.

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———- Go to list of topics ———- Start of the simulator part ———-
The Hamilton operator matrix - see figure 6

Figure 5: Schematic representation of the population probabilities, drawn at the

energy levels

72
23 The time-dependent wavefunction
———- Start of the simulator part ———- Go to list of topics ———-
The time-dependent wavefunction Ψ(t) is expanded in terms of the model states ϕi (t)
as X
Ψ(t) = ci (t) ϕi (t)
i

with the initial condition

c1 (0) = 1, c2 (0) = 0, c3 (0) = 0, · · · · ··

and inserted in the Schrödinger equation

∂
ih̄ Ψ(t) = H Ψ(t)
∂t
In the usual way this results in a set of coupled equations. The populations of the
states, i.e. the absolute value squared of the expansion coefficients ci (t) is shown in
fig. 5.

73
 E0 V1 V1 V1 V1 V1 ... V1

V1 E 0 0 0 0 0
−n

V1 0 E 0 0 0 0
−n+1
V1 0 E
0 −n+2
.. 0 0
V1 0 0 0 .
..
V1 E0
..
. .
V1 0 0 0 0 0 En

Figure 6: Schematic representation of Hamiltonian Matrix

Go to list of topics

74
 0.8

0.6

0.4

0.2

0
−5 −4 −3 −2 −1 0 1 2 3 4 5
N1=61 V1=0 N2=121 V2=0.035 Sc=15.5877

0.8

0.6

|a2|2
0.4

0.2

0
0 5 10 15 20 25 30
t

Figure 7: Decay in the simulator

———- Start of the simulator part ———- Go to list of topics ———-

75
24 Atomic Units
Unit of length is the Bohr radius:
!
h̄2 h̄2
a0 = = 4π0
me e2 me e2

The first is in atomic units, second in SI-units. This quantity can be remembered by
recalling the virial theorem, i.e. that in absolute value, half of the potential energy is
equal to the kinetic energy. This gives us

1 e2 h̄2
=
2 a0 2me a0 2
and if we accept this relation, we have the above value of a0 .
The so called fine structure constant
e2
α=
h̄c
expresses in general the weakness of electromagnetic interaction. List of topics

76
 Some Constants and Quantities
v0 = αc = 2.187106 m s− 1 Bohr velocity
a0 = 0.529177 10−10 m Bohr radius
h̄ = 0.6582 10−15 eV s Planck’s constant
kB = 0.8625 10−4 eV ˚K−1 Boltzmann constant

R = NA kB
NA = 6.0222 1023 Avogadro’s number

µB = 0.579 10−4 eV (Tesla)−1 Bohr magneton

Plank’s formula

3
h̄ωba 1
ρ(ωba ) =
π 2 c3 eh̄ω/kT − 1
List of topics

77
Overview of the topics
Time dependent QM- two-well problem Time-Dependent Schrödinger Equation
Perturbation theory for TDSE Dirac delta-function
Fermi Golden Rule Constant rate and exponential decay
Line width from exponential decay Quantum theory of extended systems - fields
Eigenmodes for coupled vibrations. Algebraic Method for Harmonic Oscillator.
Electromagnetic fields The Quantum Theory of Electromag. Field
Density of States Charged Particles In an Electromag. Field
The Hamiltonian of Interaction Emission of Radiation by an Excited Atom
The matrix element reduction Dipole Approximation
Detailed Evaluation of Emission rate Stimulated emission
Golden Rule Simulator part Time-Dependent Schrödinger
Comparing hamiltonians and expansions Equation in Simulator
One level in continuum
Final W = 1/τ result

78