NUCLEAR REACTIONS

INSIDE THE
WATER MOLECULE

by

J. DICKS

NUCLEAR REACTIONS
INSIDE THE
WATER MOLECULE
by

JESSE DICKS
submitted in part fulfillment of the requirements
for the degree of

MASTER OF SCIENCE
in the subject

PHYSICS
at the

UNIVERSITY OF SOUTH AFRICA

SUPERVISOR : DR S A RAKITYANSKY
JUNE 2005

Declaration

I declare that NUCLEAR REACTIONS INSIDE THE WATER MOLECULE is

my own work and that all sources that I have used or quoted have been indicated
and acknowledged by means of complete references.

Signature Date

ii

Acknowledgements

I would like to thank the following persons :

Dr. S.A.Rakityansky for suggesting this problem and acting as supervisor. Without
his guidance and inspiration this work would not have materialized.

Prof. S.A.Sofianos, Prof. M.Braun, and all the other members of the Unisa Physics
Department for their encouragement and support.

My parents Sarel and Joyce, my grandparents George, Raaitjie, Johannes and Sannie, my aunt Leoline, my friend Japie, my uncle Johannes, my aunt Aletta, and my
sister Marlise, for their patience, understanding and motivation.

Last but not least the glory goes to the Lord for making this universe such a wonderful and interesting place to live in.

iii

Summary

A scheme, analogous to the linear combination of atomic orbitals (LCAO), is used

to calculate rates of reactions for the fusion of nuclei confined in molecules. As an
example, the possibility of nuclear fusion in rotationally excited H2 O molecules of
angular momentum 1 is estimated for the p + p + 16 O 18 Ne (4.522, 1 ) nuclear
transition. Due to a practically exact agreement of the energy of the Ne resonance
and of the p + p + 16 O threshold, the possibility of an enhanced transition probability is investigated.

Keywords: LCAO; Reaction Rate; Nuclear Fusion; H2 O Molecule;

Nuclear Transition; Resonance; Enhanced Transition Probability.

iv

18

Ne Nucleus;

Dedication
To my parents

Contents

1 Introduction

1.1

Energy Producing Nuclear Reactions . . . . . . . . . . . . . . . . . .

1.2

Fusion Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.3

Molecular Fusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1.4

Nuclear Fusion in Water . . . . . . . . . . . . . . . . . . . . . . . . .

2 Theory

2.1

Formal Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.2

LCAO Approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 Method of Solution

15

3.1

The H2 O(1 ) Wavefunction . . . . . . . . . . . . . . . . . . . . . . . 17

3.2

The

3.3

The Proton-Proton Potential . . . . . . . . . . . . . . . . . . . . . . . 29

18

Ne (1 ) Wavefunction . . . . . . . . . . . . . . . . . . . . . . . 21

vi

vii
3.4

The Proton-Oxygen Potential . . . . . . . . . . . . . . . . . . . . . . 30

3.5

Overlap Integral, Matrix Elements, and Energy . . . . . . . . . . . . 34

3.5.1

Overlap Integral I

3.5.2

V11S Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

3.5.3

V12S Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.5.4

V21S Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.5.5

e
Element . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
H22

3.5.6

E1 and E2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4 Results

. . . . . . . . . . . . . . . . . . . . . . . . 34

40

4.1

Constants and Independent Quantities . . . . . . . . . . . . . . . . . 41

4.2

Calculated Results for Dependent quantities . . . . . . . . . . . . . . 42

4.3

4.2.1

E2 = 0.009 to 0.007 MeV . . . . . . . . . . . . . . . . . . 43

4.2.2

E2 = 0.006 to 0.004 MeV . . . . . . . . . . . . . . . . . . 44

4.2.3

E2 = 0.003 to 0.001 MeV . . . . . . . . . . . . . . . . . . 45

4.2.4

E2 = 0.000 to 0.002 MeV . . . . . . . . . . . . . . . . . . . . . 46

4.2.5

E2 = 0.003 to 0.005 MeV . . . . . . . . . . . . . . . . . . . . . 47

Reaction Rate and E2 . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5 Conclusion

50

List of Figures
3.1

Molecular structure of H2 O . . . . . . . . . . . . . . . . . . . . . . . 16

3.2

Rotational state of H2 O(1 ) . . . . . . . . . . . . . . . . . . . . . . . 18

3.3

16

3.4

Diagram of ~a = ~y +

~
x
2

. . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.5

Diagram of ~b = ~y

~
x
2

. . . . . . . . . . . . . . . . . . . . . . . . . . . 27

3.6

The Proton-Proton Potential . . . . . . . . . . . . . . . . . . . . . . . 30

3.7

Nuclear-force part of Proton-Oxygen Potential . . . . . . . . . . . . . 33

3.8

Energy-level diagram of the

4.1

Energy versus Reaction Rate . . . . . . . . . . . . . . . . . . . . . . . 48

4.2

Energy versus Log ( Reaction Rate ) . . . . . . . . . . . . . . . . . . 49

O + p + p cluster structure of

18

18

Ne . . . . . . . . . . . . . . . . . . 22

Ne nucleus . . . . . . . . . . . . . . . . 39

viii

Chapter 1
Introduction

1.1

Energy Producing Nuclear Reactions

Nuclei can produce energy via two different types of reactions, namely, fission and
fusion reactions. Fission is possible for heavy nuclei and fusion for very light nuclei. The energetic preference for nuclei to either undergo fission or fusion can be
understood immediatly from the dependence of the binding energy B of a nucleus
on the number A of nucleons constituting it. Any physical system tends to move
into a state with minimal possible energy by getting rid of excess energy. For light
nuclei this is achieved via fusion to form intermediate nuclei, but for heavy nuclei the
energy is minimized when a nucleus splits into fragments of smaller masses. These
fragments are more tightly bound and therefore occupy lower energy levels.
Nuclear fission, which was discovered in 1939 by Hahn and Strassman, and consequently used first, is a decay process in which an unstable nucleus, such as uranium
or plutonium, splits into two fragments of comparable mass. Fission results primar1

2
ily from the competition between the nuclear and Coulomb forces in heavy nuclei.
The total nuclear binding energy increases roughly in proportion to A, while the
Coulomb repulsion energy of the protons increases faster, like Z 2 . The two fragments which are both positively charged repel each other by an electric force and
move apart at a high velocity, distributing their kinetic energy in the surrounding
material.
Fission can occur spontaneously as a natural decay process, or it can be induced
through the absorbtion of a relatively low-energy particle, such as a neutron or
photon which then produces an excited state or a compound-nucleur state that is
high enough in energy to surmount or more easily penetrate the Coulomb barrier.
The applicability of fission for the obtainment of large total energy releases was
evident soon after its discovery, due to the fact that every neutron-induced fission
event produces, in addition to the two heavy fragments, several neutrons which can
themselves induce new fission events. This is known as a chain reaction of fissions,
which can occur very rapidly and without control as in a fission explosion, or slowly
and under careful control as in a fission reactor.
In nuclear fusion two or more small light nuclei, such as hydrogen or helium isotopes,
come together or fuse, to form a larger nucleus (below A = 56 ). This occurs when
two light nuclei approach each other to a distance of a few fm (1 fm = 1013 cm),
where they then experience a strong attraction which overpowers their Coulomb
repulsion. To approach each other within such a distance they need enough kinetic
energy to overcome the electrical repulsion of their positive charges. When they
fuse, they get rid of the excess energy by ejecting a neutron or photon to form a
stable nucleus.
As an energy source, fusion has several obvious advantages over fission. The light
nuclei are plentiful and easy to obtain, and the end products of fusion are usually
light, stable nuclei rather than heavy radioactive ones. Also, for a given mass of fuel,
a fusion reaction yields several times more energy than a fission reaction. Indeed, a

3
change of the binding energy (for each nucleon) is much more significant for a fusion
reaction than for a fission reaction. Fusion, is therefore, a much more powerful
source of energy. Fusion reactions were also responsible for the synthesis of the
initial amount of light elements at primordial times when the universe was created.
Furthermore, the synthesis of nuclei continues inside the stars where the fusion
reactions produce all the energy which reaches us in the form of light.

1.2

Fusion Reactions

As soon as people realised that fusion could be a very powerful and practically
inexhaustible source of energy, attempts were made to generate fusion in earth laboratories as well. Before discussing how we can achieve thermonuclear fusion on
earth, lets consider the sun. The basic process in the sun (and most other stars) is
the fusion of hydrogen into helium. Hence, the stars are natural laboratories where
fusion reactions occur permanently. Here on earth however, we have only succeeded
so far in making uncontrollable fusion in the form of the hydrogen bomb. As previously mentioned, the difficulty with fusion is that nuclei are positively charged
and repel each other by the Coulomb force. As a result, at ordinary temperatures
achievable in our laboratories, they keep away from each other. However, for a
fusion event to occur, the nuclei must be close enough to feel the strong attraction.
The nuclear forces acting between two nuclei are characterized by the potential
energy, or, simply, potential. The form of the strong interaction potential can be
deduced from nucleon-nucleon scattering experiments where at short distances the
nuclear forces cause a very strong attraction. And, at large distances electric charges
causes Coulomb repulsion. The hump in between the strong attraction and the
Coulomb repulsion is called the Coulomb barrier and when nuclei overcome this
barrier and fall down into the potential well a fusion event occurs. The height of
the Coulomb barrier is approximately 1 MeV which corresponds to a temperature

4
of 10 billion degrees. Of course, the nuclei need not be above the barrier as they
can penetrate through the barrier due to the tunnel effect of quantum mechanics.
Thanks to this effect, it is not necessary to have a temperature of 10 billion degrees
for the reaction to occur. In the sun, for example, fusion occurs at 15 million degrees.
Of course, the higher the temperature is, the thinner the barrier is and the easier a
particle can penetrate through it.
A fusion weapon such as a hydrogen bomb includes a fission explosive such as an
ordinary plutonium bomb as an initiator. The radiation from the fission explosion is
responsible for heating (to solar temperatures) and compressing the thermonuclear
fuel. This approach was followed by many people in an attempt to make controllable fusion, but so far they have only succeeded in producing a high temperature
plasma, for a relatively short time. Most probably this straightforward approach to
controllable fusion is doomed and other possibilities need to be considered. Instead
of lifting the particle against the barrier (which means increasing the temperature),
it seems to be more promising to attempt to make the barrier itself thinner or to
keep the particle close to the barrier for such a long time that even a low penetration
probability would work.
This can be done by putting the nuclei we want to fuse, inside a molecule, where they
can stay close to each other for a long time and, in addition, the Coulomb barrier
becomes thinner because of electron screening. In this way fusion may proceed even
at room temperature. This idea of cold fusion was originally (in 1947) discussed by
F. C. Frank [1] and (in 1948) put forward by A. D. Sakharov [2].

1.3

Molecular Fusion

As an example of molecular fusion, we can first look at the cold fusion achieved in
many laboratories via the formation of muonic molecules (see a review in Ref.[3])
. The muon is an elementary particle which has the same characteristics as an

5
electron. The muon however is 200 times heavier than the electron and hence in a
muonic atom of hydrogen, that is, a bound state of a proton and a muon, the size
of such an atom would be 200 times smaller than that of an ordinary atom. Thus
if you make a muonic molecule it will also be 200 times smaller than an ordinary
molecule, the Coulomb barrier will be 200 times thinner, and the nuclei 200 times
closer to each other.
The muon modifies the nucleus-nucleus potential in such a way that a second minimum appears. To fuse, the system needs to jump from the shallow well to the deep
well through the barrier. The muon however is not a stable particle with a lifetime
of only 106 s. The quantum mechanical wave function oscillates with a frequency
proportional to the energy of the system and since the typical binding energy of a
muonic molecule is 300 eV this frequency is 1017 s1 . Thus, the particle hits the

barrier with this frequency and during 1 microsecond it makes 1011 attempts to jump
through it. With a penetration probability of 107 nuclei can penetrate through

the barrier 10000 times in 1 microsecond and fusion can happen much faster than
the decay rate of the muon.
Unfortunately, the fusion in muonic molecules cannot solve the problem of energy
production due to its negative efficiency. In other words, it takes more energy to
make muonic cold fusion than it produces. Since muons do not exist naturally like
protons or electrons, they have to be produced in accelerators which takes a lot of
energy. When the reaction rates for different muonic molecules was calculated, it was
also noticed (see Ref.[4],[5],[6],[7],[8]) that for certain systems there is a possibility of
resonant nuclear transitions which can significantly amplify the fusion probability.
Studies of these resonant transitions in muonic molecules has led to the conclusion
that under certain conditions nuclear fusion can happen even in ordinary (electronic)
molecules [7]. In paricular, an ordinary water molecule, if excited to a certain
rotational state, can undergo nuclear fusion with the collapse of the two protons
and the nucleus

16

O.

6
When a new nucleus is formed through the molecular fusion of a molecule, and if
the nuclear and molecular levels are close to each other, the probability to penetrate
through the barrier for various values of the energy gap between the two levels, drastically increases (by several orders of magnitude) when the difference between the
levels decreases. In addition to this, the amplification of the penetration probability
becomes even more significant when the width of the compound nucleus resonance
is smaller. Therefore, the smaller the energy gap and , the faster the nuclei fuse.
Thus, if one finds a nuclear system for which the corresponding compound nucleus
has a narrow level very close to the threshold, then a molecule constructed from
these fragments may undergo nuclear fusion. A possibilty for such a system is water
(see Ref.[7],[9]).

1.4

Nuclear Fusion in Water

Among other levels, the spectrum of the

18

Ne nucleus has an excited state at

4.522 MeV with angular momentum 1 and negative parity. The surprising fact
about it is that the threshold for the break-up into two protons and oxygen is at the
same level, i.e. the energy gap between this excited state and the break-up threshold
is zero (at least within the accuracy of the measurements). The bound system of
two protons and oxygen is the molecule H2 O, the water molecule. Thus, to make
cold fusion, no muons are needed and water should burn via the fusion reaction
p + p + 16 O 18 Ne
Water in its normal state does not undergo this fusion reaction since the molecular
and nuclear states can mix with each other only if they have the same quantum
numbers: the same angular momentum and the same parity. The state of the
water molecule with angular momentum 1 and negative parity corresponds to a
state in which the molecule is rotating around the axis through the oxygen nucleus.
Therefore, to undergo the fusion process a molecule of water must rotate. Also, in

7
ordinary water there are no individual molecules as water consists of small groups
of molecules, or so-called clusters, in which several dozens of molecules are tightly
bound to each other, and this prevents the rotation of individual molecules. A
molecule can be liberated from a cluster if we warm water up to 1000 degrees or
higher.
This dissertation is organized as follows. In chapter 2, the scheme, analogous to the
linear combination of atomic orbitals, is proposed and discussed, to calculate the
rates of reactions for a general fusion reaction of nuclei confined in a molecule. In
chapter 3 approximate wave functions for the H2 O molecule and the

18

Ne nucleus

are determined and analyzed, the theory from the previous chapter is applied to the
specific case for the burning of water, and the method in general for the numeric
calculation of the reaction rates is discussed. The results obtained are presented and
discussed in chapter 4. The relevant conclusions drawn are presented in chapter 5.

Chapter 2
Theory

2.1

Formal Theory

Consider a water molecule comprising an

16

O nucleus and two protons with an

angular momentum of 1 and negative parity. The transition to

18

Ne with the same

angular momentum and parity must occur via the fusion reaction,

H2 O(1 ) 18 Ne (1 ) + 10e .

(2.1)

The Hamiltonian of the whole system can be represented in two equivalent ways,

H = H1 + V S ,

(2.2)

H = H2 + H e ,

(2.3)

or,

9
where H1 and H2 are the Hamiltonians of the initial and final molecules, in the
absence of the strong interaction V S , and the total Hamiltonian of the electron
subsystem (10 electrons) H e respectively. H1 describes the molecular state and H2
describes the compound nucleus

18

Ne. The Hamiltonians for the initial and final

molecules are given by,

H1 = H0 + VpC1 O + VpC2 O + VpC1 p2 + H e ,

(2.4)

H2 = H0 + VpC1 O + VpC2 O + VpC1 p2 + V S ,

(2.5)

and,

where H0 is the kinetic energy operator of the nuclear subsystem, VpC1 O and VpC2 O is
the Coulombic potential between the two protons and the oxygen nucleus, and VpC1 p2
is the Coulombic potential between the two protons. The strong interaction is given
by,

V S = VpS1 O + VpS2 O + VpS1 p2 ,

(2.6)

where VpS1 O and VpS2 O is the strong interaction between the two protons and the
oxygen nucleus, and VpS1 p2 is the strong interaction between the two protons. Let 1
and 2 be the corresponding eigenfunctions of H1 and H2 ,

Hi i = E i i ,

i = 1, 2 .

(2.7)

The functions 1 and 2 describe the initial and final systems. Each of the Hamiltonians Hi corresponds to only one of the two possible states. The other one occurs
only if we take into account the perturbation, i.e., the Hamiltonian of the electron
subsystem H e or the potential V S . A well known method to treat similar problems
is the linear combination of atomic orbitals (LCAO) [10], which is used in the theory

10
of diatomic molecules to calculate wave functions of electrons having two centers of
attraction.

2.2

LCAO Approach

The LCAO approach has been discussed and used for several other reactions (see
for instance Ref.[11]). Following the LCAO approach, we will look for a solution of
the full Schrodinger equation,

H = E ,

(2.8)

in the form of a linear combination of the unperturbed states,

= C 1 1 + C 2 2 .

(2.9)

Substituting (2.9) into (2.8) and projecting onto 1 and 2 we derive the system of
linear equations for C1 and C2 :

C1 (H11 E) + C2 (H12 EI) = 0 ,

(2.10)

C1 (H21 EI ) + C2 (H22 E) = 0 ,

(2.11)

where,
Hij = hi |H|j i ,
I = h1 |2 i ,
hi |i i = 1,

i, j = 1, 2 .

The homogeneous system Eqs.(2.10) and (2.11) has solutions if and only if its determinant is zero, i.e.,

11

(H11 E)(H22 E) (H21 EI )(H12 EI) = 0 .

(2.12)

This is a quadratic equation determining the two energy levels E () that emerge as
a result of mutual influence of the nuclear and molecular states. From Eq.(2.12), we
obtain,

H11 H22 E(H11 + H22 ) + E 2 H21 H12 + EIH21 + EI H12 E 2 |I|2 = 0 ,

or,
(1 |I|2 )E 2 + (IH21 + I H12 H11 H22 )E + H11 H22 H21 H12 = 0 .
Thus, the energy values of the two perturbed levels is then given by,

()

1
H11 + H22 2Re(H21 I)
=
2(1 |I|2 )

[H11 + H22 2Re (H21 I)]2

2

4(1 |I| )(H11 H22 |H12 | )

o1
2

(2.13)

Using (2.7), the matrix elements of the Hamiltonian can be reduced to matrix elements of the potentials,

H11 = E1 + V11S ,

(2.14)

H12 = E1 I + V12S ,

(2.15)

H21 = E1 I + V21S ,

(2.16)

e
H22 = E2 + H22
,

(2.17)

e
where H22
is just the energy (negative) of 10 electrons in the

energy of its total ionization taken as a negative value.

18

Ne atom, i.e. the

12
The overlap integral I = h1 |2 i is extremely small. This fact suggests a way of
simplifying Eq.(2.13). It can be shown (see Ref.[8]) that,

E (+) = E1 + 1 ,

(2.18)

E () = E2 + 2 ,

(2.19)

where 1 and 2 are small corrections to the molecular(E1 ) and nuclear(E2 ) levels,
namely,

h
e
1 = V11S + gH11 + (1 + g) ( 1 + s 1)(E1 E2 + V11S H22
)
i

Re (IH21 ) ,

(2.20)

and,
h
e
e
)
2 = H22
+ gH22 (1 + g) ( 1 + s 1)(E1 E2 + V11S H22
i

+Re (IH21 ) ,

(2.21)

where g and s are defined as follows,

and,

1
= 1 + g , g = |I|2 + |I|4 + |I|6 + . . . ,
1 |I|2

(2.22)

4
s =
|I|2 H11 H22 + |H21 |2
e 2
(E1 E2 + V11S H22
)

(H11 + H22 )Re (IH21 ) + [Im (IH21 )]2 .

(2.23)

Note that in numerical calculations,

g |I|2 , because I 1 ,

(2.24)

13
and,

1
1
1 + s 1 1 + s 1 = s , because s 1 .
2
2

(2.25)

From the above expressions for 1 and 2 it is seen that this approach differs from
e
the simple perturbation where 1 V11S and 2 H22
. Thus, the wave function is

written as,

(+)

(+)

(+) = C1 1 + C2 2 ,
(+)

where C1

(+)

and C2

(2.26)

are solutions of Eqs.(2.10) and (2.11) corresponding to the

energy E = E (+) . The probability P for the transition (2.1) is obtained by the
projection,

P = |h2 | (+) i|2 .

P can be interpreted as the transition probability through the potential barrier. The
total wave function (+) exp(iE (+) t/h) oscillates at the barrier with frequency
v = |E (+) |/2h. Hence, in order to obtain the reaction rate , we multiply the
probability P by v. Then solving Eqs.(2.10) and (2.11), we finally get,

=
where

|f I + 1|2
|E (+) |
,
2h |f |2 + f I + f I + 1

(2.27)

|f I + 1|2
,
|f |2 + f I + f I + 1

(2.28)

E (+) I H12
.
H11 E (+)

(2.29)

P =
and

f=

In numerical calculations, the numerator and denominator of Eq.(2.29) may become

zero because they are differences of very close numbers. To avoid this, we use
Eqs.(2.18), (2.14) and (2.15), to get,

14

(E1 + 1 )I (E1 I + V12S )

E1 + V11S E1 1
E1 I + 1 I E1 I + V12S I
=
V11S 1
1 I V12S
.
=
V11S 1

f =

(2.30)

Thus, once we have determined f using Eq.(2.30), we can determine the reaction
rate using Eq.(2.27).

Chapter 3
Method of Solution

To obtain the reaction rate from Eqs.(2.27) and (2.30), we need to know the values
e
of I, V11S , V12S , V21S , and H22
. A vectorial representation of the H2 O molecule can be

seen in Fig.(3.1). According to its definition, the overlap integral is defined as,

I = h1 |2 i
=

d~x d~y 1 (~x, ~y ) 2 (~x, ~y ) .

(3.1)

Inspecting the vectorial representation of the H2 O molecule as seen in Fig.(3.1), we

can now compose the V11S matrix element,

V11S = h1 |V S |1 i
= h1 |VpS1 O + VpS2 O + VppS |1 i
Z
h
i
~x
~x
S
S
=
d~x d~y VpO
(~y ) + VpO
(~y + ) + VppS (~x) |1 (~x, ~y )|2
2
2
15

16

~x

p1 v

- vp2

~y

16

Figure 3.1: Molecular structure of H2 O

=

i
h
~x
S
(~y + ) + VppS (~x) |1 (~x, ~y )|2 .
d~x d~y 2VpO
2

(3.2)

Similarly, the V12S matrix element is defined as,

V12S = h1 |V S |2 i
= h1 |VpS1 O + VpS2 O + VppS |2 i
Z
h
i
~x
~x
S
S
=
d~x d~y VpO
(~y ) + VpO
(~y + ) + VppS (~x) 1 (~x, ~y )2 (~x, ~y )
2
2
Z
i
h
~
x
S
(3.3)
(~y + ) + VppS (~x) 1 (~x, ~y )2 (~x, ~y ) .
=
d~x d~y 2VpO
2
S
S
S
Notice that in Eqs.(3.2) and (3.3), we replace [VpO
(~y ~x2 )+VpO
(~y + ~x2 )] with 2VpO
(~y +
~
x
)
2

because 1 and 2 are both antisymmetric with respect to the proton interchange,

i.e. with respect to a ~x ~x transformation. Next, to determine V21S you simply

need to find the complex conjugate of V12S , hence,

V21S = (V12S ) .

(3.4)

17
e
Finally consider H22
. Now H e (~x, ~y ) parametrically depends on ~x and ~y . When

~x = ~y = 0, H e (0, 0) is the atomic Hamiltonian of 10 electrons,

e
H22
= hn |H e |n i

hn |H e (0, 0)|n i = Ee ,

(3.5)

where Ee is the total energy (negative) of the electron shell of the Neon atom. To
be able to calculate I, V11S , and V12S , we need the wave functions 1 and 2 which we
can now define more clearly as,

1 Wavefunction of H2 O(1 ) w ,
2 Wavefunction of

18

N e(1 ) n ,

(3.6)
(3.7)

S
as well as the proton-oxygen potential (VpO
), and the proton-proton potential (VppS ).

3.1

The H2O(1) Wavefunction

The determination and subsequent analysis of the H2 O(1 ) wavefunction is based

on the following ansatz for the wave function of the water molecule (see Ref.[12]),

wJM (~x, ~y ) =

1 F5/2 (0 , ) JM
e Yl (
x, y) .
Nw
5/2

(3.8)

Here we use, instead of the Jacobi variables {~x, ~y } = {x, x , x , y, y , y }, the set

of hyperspherical variables {, , x , x , y , y }, with = x2 + y 2 being the hyperradius; and = arctan y/x the hyperangle. For the five angles, the notation
= {, x , x , y , y } will be used, Fv denotes the regular solutions of the hyper-

radial Schrodinger equation (the regular Coulomb wave function), and YlJM (
x, y)

18
~
are the eigenfunctions of the total angular momentum operator J~ = (~l + ~) + S.
Also, Nw is the normalization factor, and

|mol | represents the momentum

corresponding to the binding energy mol of the H2 O molecule; 0 = V0 /2 is a kind

of Sommerfeld parameter, where V0 is obtained by averaging V() with the angular

part of wJM (~x, ~y ). The ansatz (3.8) takes correctly into account the Coulomb repulsion between the particles at small distance, as well as the geometric size of the
water molecule.

p1 u

@
@

@u

16

u p2

Figure 3.2: Rotational state of H2 O(1 )

Now for this particular problem we are interested in the rotation state having the
quantum numbers 1 , in other words the state with J = 1 and negative parity.
Since the protons are identical fermions, the wave function must be antisymmetric
with respect to their permutations. If we assume that the rotation is associated
with the coordinate ~x, i.e. is around the axis passing through the oxygen nucleus,
see Fig.(3.2), then l = 1, = 0, and the total spin of the pp-pair is 1 (to make the
wavefunction antisymmetric under the p p permutations). YlJM (
x, y) in eq.(3.8)
is given by,

YlJM (
x, y) =

lz z LLz

JM
z
Yllz (
x)Yz (
CLL
,
y )1 CllLL
z 1
z z

19
so that,
1
1M
z
CLL
Y1lz (
x) 1 C1lLLz 00
z 1
4
lz LLz
X
1
,
=
Y1lz (
x) C1l1M
z 1
4
lz

Y101M (
x, y) =

where 1 is the spin function. Hence, for the three possible values of M namely
M = +1 or 0 or 1 you then have,

11 11
10 11
+ Y10 (
x) C1011
Y1011 (
x, y) = Y11 (
x) C1110
4
4

1
)11 ,
=
Y11 (
x)10 Y10 (~x
8

(3.9)

10 10
11 10
11 10
+ Y10 (
x) C1010
+ Y11 (
x) C1110
Y1010 (
x, y) = Y11 (
x) C1111
4
4
4

1
(3.10)
Y11 (
x)11 Y11 (
x)11 ,
=
8
11 11
10 11
x, y) = Y10 (
x) C1011
Y1011 (
+ Y11 (
x) C1110
4
4

1
=
Y10 (
x)11 Y11 (
x)10 ,
8

(3.11)

where we have made use of some of the following Clebsch-Gordan coefficients,

1
11
10
11
C1011
= C1111
= C1110
= ,
2

(3.12)

1
11
10
11
= ,
= C1011
= C1111
C1110
2

(3.13)

10
C1010
= 0.

(3.14)

Thus, the final wave function for the water molecule is given by,

20

w1M (~x, ~y ) =

1 F5/2 (0 , ) 1M
e Y10 (
x, y) ,
Nw
5/2

(3.15)

x, y) is given by Eqs.(3.9), (3.10), and (3.11), depending on the value of

where Y101M (
M.
The binding energy of the water molecule can be found in the literature (see Ref.[13]),

mol = 232.59 kcal mol1

232590 cal
1J
=
1 mol
0.239 cal

1 eV
973179.9163 J
=
1 mol
1.602 1019 J

!
!
6.074 1024 eV
1 mol
=
1 mol
6.02 1023 molecules
10.087 eV per molecule .

(3.16)

Then is given by,

|mol |2

16
)mp
|mol |2( 33

48.547 105 fm1 .

As a simple estimate we take 0 to be given by,

2 e2
h
2 2
e2 c
=
h2 c

0 =

(3.17)

21
c
h
34.654 .
=

(3.18)

The normalization factor Nw can be determined by solving the following,

d~x d~y w1M (~x, ~y ) w1M (~x, ~y ) = M 0 M .

(3.19)

Since this is independent of the choice of M , let M 0 = M = 1 and then,

Nw2 =

d~x d~y

|F5/2 (0 , )|2 2 11
e
|Y10 (
x, y)|2
5

Z
x2 y 2 e2
2
d
x d
y |Y1011 (
x, y)|2
|F
(
,
)|
5/2 0
5
0
0

Z Z
q
2 2 2 x2 +y 2
xy e
|F5/2 (0 , x2 + y 2 )|2 .
=
dx dy
2
2
5/2
(x + y )

3.2

The

18

dx

dy

(3.20)

Ne(1 ) Wavefunction

We assume that the nucleus 18 Ne has the cluster structure 16 O + p + p (see Fig.(3.3)).
This assumption is very reasonable because
to show that in the ground state of

18

16

O is a closed shell nucleus. It is easy

Ne, both protons occupy the 2s state (in the

harmonic oscillator model). Indeed,

|16 Oi = |(1s)4 (1p)12 i ,

and,
|18 Nei = |(1s)4 (1p)12 (2s)2 i .
Therefore, after inspection of the Jacobi vectors of this system as seen in Fig.(3.3),
we can write the ground state wave function as follows,

22

~x

p1 v

- vp2

~y

16

Figure 3.3:

16

O + p + p cluster structure of

18

Ne

~x
~x
n (~x, ~y ) = 2s (~y + )2s (~y )0 ,
2
2

(3.21)

where,
q

8/3
1
2s (~q) =
4 1/4 r03/2

|~q|2 3
|~q|2
exp

r02
2
2r02
!

(3.22)

0 is the spin function of two protons, describing the state with total spin zero. The
spatial part of eq.(3.21) is obviously symmetric with respect to the proton permutations, while 0 is antisymmetric. Therefore eq.(3.21) need not be antisymmetrized.
To form the excited state

18

Ne(1 ), we can put one of the protons in the state

2p. The excited state has negative parity which means that the spatial part of the
corresponding wave function must be antisymmetric and, therefore, the spin part
symmetric, and it has a total angular momentum of J = 1, hence,

n1M (~x, ~y ) =

o
n
~x
~x
,
A 2s (~y + )2p (~y , lz ) 1 C1l1M
z 1
2
2
lz

(3.23)

where A, lz , , M , are the antisymmetrizer, the third component of the angular mo-

23
mentum, the pp-spin, and the total angular momentum of the nucleus respectively.
The normalized eigenfunction 2p of the harmonic oscillator is,

20/3

2 |~q|2
|~q|2
2p (~q, lz ) =
|~
q
|
1

exp

Y1lz (~q) .
5/2
5 r02
2r02
1/4 r0

(3.24)

The parameter r0 , characterizing the oscillator, is related to its frequency 0 as

follows,
v
u
u
r0 = t

,
mp 0

where mp is the proton mass. The frequency 0 can be found from ( see ref.[14] ),

h
0

40 M eV
,
A1/3

where A = 18 is the number of nucleons. Hence you now have,

r0 =

v
u
2 2
u h
t c

c2 h
0
1.64893 fm .

(3.25)

The antisymmetrizer A in eq.(3.23) consists of the following two terms,

A = P(1, 2) P(2, 1) ,
where P(i, j) is the permutation operator that places the protons in the order i, j.
In other words,

24

o
n
~x
~x
~x
~x
A 2s (~y + )2p (~y , lz ) = 2s (~y + )2p (~y , lz )
2
2
2
2
~x
~x
2s (~y )2p (~y + , lz ) ,
2
2

because permutation of the protons is equivalent to replacing ~r ~r (see the Jacobi

coordinates). We then have,

n1M (~x, ~y ) =

1 Xh
~x
~x
2s (~y + )2p (~y , lz )
N n lz
2
2

i
~x
~x
,
2s (~y )2p (~y + , lz ) 1 C1l1M
z 1
2
2

(3.26)

where Nn is the normalization factor. Inspecting the harmonic functions 2s and

2p it is clear that 2s (~q) depends only on |~q|, whereas, 2p (~q) depends on both |~q|
as well as the angles q and q . Hence we can rewrite the function 2p as,

2p (~q, lz ) = 2p (|~q|)Y1lz (q , q ) .
Now performing some algebraic manipulations on the vectors in order to simplify
the numerical calculations that will be executed on the wave functions, let ~a = ~y + ~x2
(see Fig.(3.4)), then,

~a = ~y +

~x
,
2

~a2 = y 2 + xy cos +

|~a| =

x2
,
4

y 2 + xy cos +

x2
.
4

25
Next, let ~b = ~y

~
x
2

(see Fig.(3.5)), then,

~b = ~y ~x ,
2
2
~b2 = y 2 xy cos + x ,
4

|~b| =
Note that

~
x
2

= ~a ~y

x2
4

y 2 xy cos +

x2
.
4

= a2 2ay cos a + y 2 , and hence,

cos a

y 2 + a2
=
2ay
=

x2
4

2y + x cos
q

2 y 2 + xy cos +

x2
4

a = .

And since

~
x
2

= ~y ~b

x2
4

= y 2 2by cos b + b2 , you have,

cos b

y 2 + b2
=
2by
=

x2
4

2y x cos

2 y 2 xy cos +

x2
4

b = + .

Now we have the final form for the wave function and it is given by,

26

a =
~a
~y

~x/2

Figure 3.4: Diagram of ~a = ~y +

n1M (~x, ~y ) = Nn

Xh
lz

~
x
2

2s (|~a|)2p (|~b|)Y1lz (b , b )

2s (|~b|)2p (|~a|)Y1lz (a , a ) 1 C1l1M

.
z 1
i

(3.27)

Also, the normalization factor Nn can be determined by solving the following,

d~x d~y n1M (~x, ~y ) n1M (~x, ~y ) = M 0 M .

(3.28)

This value of Nn does not depend on M . Therefore, in the above integral we let
M 0 = M = 1 which leads to,

Z
2
X
1

2
11
~
(
,

)
d~
x
d~
y

(|~
a
|)
(|
b|)Y
)
(C
=
b
2s
2p
1lz b
1lz 1
Nn2
lz

27

b = +

~y
~b
2

~x/2
Figure 3.5: Diagram of ~b = ~y

~
x
2

+2s (|~b|)2p (|~a|)Y1lz (a , a )

2s (|~a|)2p (|~b|)Y1lz (b , b )2s (|~b|)2p (|~a|)Y1lz (a , a )

2s (|~a|)2p (|~b|)Y1lz (b , b )2s (|~b|)2p (|~a|)Y1lz (a , a ) .

Replacing ~x by ~x in the second and fourth terms, we can simplify the last formula
to get,

Z
2
X
1

11
2
~
=
2
(C
)
d~
x
d~
y

(|~
a
|)
(|
b|)Y
(
,

)
2s
2p
1lz b
b
1lz 1
Nn2
lz

2s (|~a|)2p (|~b|)Y1lz (b , b )2s (|~b|)2p (|~a|)Y1lz (a , a )

We choose the k-axis of the ~x-coordinate system to coincide with ~y . Now the integrand does not depend on the orientation of ~y . Also, note that lz = M thus
= 1 lz and hence the previous equation becomes,

28

Z2
Z Z
Z
1
X
1
11
2
(C1lz 1,1lz ) dx dy d d x2 y 2 sin
= 8
Nn2
lz =0
0

2
2s (|~a|)2p (|~b|)Y1lz (b , b )

2s (|~a|)2p (|~b|)Y1lz (b , b )2s (|~b|)2p (|~a|)Y1lz (a , a )

Next, we again make use of the Clebsch-Gordan coefficients as given in Eqs.(3.12),

(3.13), and (3.14), as well as the following spherical harmonics,

1
Y11 (, ) =
2

3
sin ei ,
2

(3.29)

1
Y10 (, ) =
2

3
cos ,

(3.30)

1
Y1,1 (, ) =
2

3
sin ei .
2

(3.31)

The previous equation then becomes,

Z
Z Z
Z2
1
= 4 dx dy d d x2 y 2 sin
2
Nn
0

2 3
3

2
2
~
2s (|~a|)2p (|b|)
cos b +
sin b
4
8

2s (|~a|)2p (|~b|)2s (|~b|)2p (|~a|)

= 3

dx

dy

3
3
cos a cos b
sin a sin b
4
8

d x2 y 2 sin

2
2s (|~a|)2p (|~b|) (2 cos2 b + sin2 b )

2s (|~a|)2s (|~b|)2p (|~a|)2p (|~b|)(2 cos a cos b sin a sin b ) .

29
Finally, to further simplify the numerical calculations, let z = cos . Then,

Z
Z
Z1

1
2 2
~b|)2 (1 + cos2 b )
=
3
dx
dy

(|~
a
|)
(|
dz
x
y
2s
2p
Nn2
0

2s (|~a|)2s (|~b|)2p (|~a|)2p (|~b|)(2 cos a cos b sin a sin b ) , (3.32)

where,
|~a| =
cos a =

y 2 + xyz +

x2
,
4

2y + xz
q

2 y 2 + xyz +

x2
4

(3.33)

a = ,

|~b| =
cos b =

(3.35)

y 2 xyz +

2y xz

x2
,
4

2 y 2 xyz +

x2
4

(3.36)

b = + .

3.3

(3.34)

(3.37)

(3.38)

The Proton-Proton Potential

The proton-proton potentials (VppS ) in the integral expressions (3.49)-(3.51) is constructed according to isotriplet VN1 N Malfliet-Tjon N N potentials due to the three
possible states of M [15] :

VN1 N = A1 f1 (r) B 1 f2 (r) ,

30
where (A1 , B 1 ) = (1438.72 MeVfm, 626.875 MeVfm), f1 (r) = exp(3.11r)/r, and
f2 (r) = exp(1.55r)/r. Thus, the proton-proton potential is given by,

VppS (r) = 1438.72

exp(1.55r)
exp(3.11r)
626.875
.
r
r

(3.39)

A graphical representation of this potential can be seen in Fig.(3.6).

100

V (MeV)

50

-50

-100

0.5

1.5

2.5
r(fm)

3.5

4.5

Figure 3.6: The Proton-Proton Potential

3.4

The Proton-Oxygen Potential

S
For the proton-oxygen potentials (VpO
) in the integral expressions (3.49)-(3.51) we

will be using a single-particle potential for spherical nuclei ( see Ref.[16] ). The
single-particle potential is composed of the nuclear-force part and the Coulomb part.
S
Since we are determining VpO
we will only concern ourselves with the nuclear-force

part. The nuclear-force part has a central component and a spin-orbit component

31
which is negligible for our approximations. The central component is an extension
of the Woods-Saxon potential and has the following form,
#

"

Vdp
V0
.
1+
Vcen (r) =
a
/k
v
{1 + exp[(r Rv )/av ]}
1 + exp[(r Rv )/av ]
We assume the five potential parameters V0 , Vdp , Rv , av , and k vary smoothly with
Z and N , and express them in a power series in A1/3 and (N Z)/A. To embody
the charge symmetry of nuclear forces in these expressions Ics is defined as,

Ics =

N Z
for the neutron
A

Since we are working with

16

&

Ics =

Z N
for the proton .
A

O we then have Ics = 0 and the simplified expressions

for the five potential parameters are given by (from [16]),

V0 = v 0

"

1
Z2
1 v3 1/3
CexV 4/3 ,
A + v1
A

Vdp = vdp0

"

(3.41)

1
,
+ d1

(3.42)

1
1 vdp3 1/3
,
A + vdp1

Rv = r m + r w c w + d 0 d 3

A1/3

(3.40)

1
,
av = a0 exp a3 1/3
A + a1

(3.43)

1
kv = k0 exp k3 1/3
.
A + k1

(3.44)

Also, the quantity rm in eq.(3.42) is determined as follows,

32

rm =

rm0 A1/3 (1 + CexR Z 2 /A4/3 )

.
1 + 4CexR

(3.45)

And the quantity cw in eq.(3.42) is determined as follows,

cw =

1
.
A2/3

(3.46)

Thus, the proton-oxygen potential is given by,

S
VpO
(r)

"

Vdp
V0
,
=
1
+
{1 + exp[(r Rv )/av ]}av /k
1 + exp[(r Rv )/av ]

(3.47)

where the five potential parameters are given by (3.40)-(3.44), rm is given by (3.45),
cw is given by (3.46), and where,

v0 = 64.981 MeV ,
v3 = 1.523 ,
v1 = 2 ,
CexV

= 0.006 ,

vdp0 = 0.467 ,
vdp3 = 5.4 ,
vdp1 = 0.5 ,
rw = 1 fm ,
d0 = 1.5088 fm ,
d3 = 11.247 fm ,
d1 = 1.5 ,
a0 = 1.4449 fm ,

33
a3 = 1.55 ,
a1 = 0.5 ,
k0 = 1.508 fm ,
k3 = 1.1763 ,
k1 = 2 ,
CexR = 0.005 .

A graphical representation of the nuclear-force part of the single-particle potential

for spherical nuclei, can be seen in Fig(3.7).

0
-5

V (MeV)

-10
-15
-20
-25
-30
-35

0.5

1.5

2.5
r(fm)

3.5

4.5

Figure 3.7: Nuclear-force part of Proton-Oxygen Potential

34

3.5

Overlap Integral, Matrix Elements, and Energy

We are now able to calculate the integrals defined by Eqs.(3.1), (3.2), and (3.3).
These integrals do not depend on M . We therefore assume that M = 0.

3.5.1

Overlap Integral I

Using Eq.(3.1) the overlap integral is given by,

I =

d~x d~y w (~x, ~y ) n (~x, ~y )

1
d~x d~y
Nw

Nn

Xh
lz

F5/2 (0 , )
e
Y
(
x
)

Y
(
x
)
11
11
11
11
5/2

2s (|~a|)2p (|~b|)Y1lz (b , b )

2s (|~b|)2p (|~a|)Y1lz (a , a ) 1 C1l10z 1

i

(0 , x2 + y 2 ) x2 +y2
F5/2
Nn Z
e
d~x d~y
=
(x2 + y 2 )5/4
8Nw

10

10
2s (|~a|)2p (|~b|) Y11 (b , b )Y11 (
x)C1111
Y11 (b , b )Y11
(
x)C1111

10

10
2s (|~b|)2p (|~a|) Y11 (a , a )Y11 (
x)C1111
Y11 (a , a )Y11
(
x)C1111

Using the properties of spherical harmonics it can be seen that the first and second
terms of the previous equation are equal since,

Y11 (~x) = Y11 (~x) ,

Y11 (~x) = Y11 (~x) .

35
Therefore, the previous equation now becomes,

(0 , x2 + y 2 ) x2 +y2
F5/2
2Nn
I =
e
d~x d~y
(x2 + y 2 )5/4
2 8Nw
h
i

2s (|~a|)2p (|~b|) Y11 (b , b )Y11 (

x) + Y11 (b , b )Y11
(
x) .
We again choose the k-axis of the ~x-coordinate system to coincide with ~y so that
the integrand does not depend on the orientation of ~y . The only place where such
dependence could exist now is in the spherical angles b and b of the vector ~b = ~y ~x2 .
However, these angles can be expressed in terms of spherical angles of vector ~x.
Using Eqs.(3.36)-(3.38), as well as the spherical harmonics from Eqs.(3.29) and
(3.31), together with the following,

Z2

d =

Z2

ei d = 2 ,

Z2

ei(+) ei d =

Z2

ei d = 2 ,

i(+) i

the overlap integral then becomes,

Z
Z
Z
4Nn 1 3
(2) dx dy d x2 y 2 sin
I =
2 Nw 2 2
0
0
0
2

2
q
h
i
F5/2 (0 , x + y ) x2 +y2
2 .
~

(|~
a
|)
(|
b|)
2
sin

e
1

cos

2s
2p
b
(x2 + y 2 )5/4

Finally, making use of the substitution z = cos as well as Eqs.(3.33) and (3.36)
the overlap integral is given by,

I =

3 Nn
Nw

dx

dy

Z1

dz

x2 y 2 (1 z 2 )(1 cos2 b )
(x2

y 2 )5/4

x2 +y 2

36

F5/2

0 ,

x2

y2

x2
x2
2s y 2 + xyz + 2p y 2 xyz + ,
4
4

(3.48)

where cos b is given by Eq.(3.37).

3.5.2

V11S Element

We now calculate the V11S matrix element by using Eq.(3.2),

V11S

h
i
~x
S
d~x d~y 2VpO
(~y + ) + VppS (~x) |w (~x, ~y )|2
2

1 Z
d~x d~y
=
|Nw |2
=

2
F ( , )
h
i
~x
5/2 0

10
S
S

e
Y
(
x
,
y

)
)
+
V
(~
x
)
2V
(~
y
+
10
pp
pO

5/2
2

Z
i
|F5/2 (0 , )|2 2 h
1
2
2
e
d~
x
d~
y
|Y
(
x
)|
+
|Y
(
x
)|
11
11
8|Nw |2
5
i
h
~x
S
2VpO
(~y + ) + VppS (~x)
2

Z Z
|F5/2 (0 , x2 + y 2 )|2 2x2 +y2
2 3 4 2 Z
2 2
=
e
dx dy d x y sin
8 8 |Nw |2
(x2 + y 2 )5/2
0

sin

S
2VpO
(|~a|)

VppS (|~x|)

Making use of the substitution z = cos as well as Eq.(3.33) this then becomes,

V11S

2
2
Z
Z1
3 Z
x2 y 2 (1 z 2 )e2 x +y
=
dx
dy
dz
4|Nw |2
(x2 + y 2 )5/2
0

2
2

F5/2 0 , x + y

37

x2
S
2VpO
y 2 + xyz + + VppS (x) .
4

3.5.3

(3.49)

V12S Element

Using Eq.(3.3) the V12S matrix element is given by,

V12S

h
i
~x
S
d~x d~y 2VpO
(~y + ) + VppS (~x) 1 (~x, ~y )2 (~x, ~y ) .
2

Inspecting this equation, it is clear that V12S is expressed by the same integral as I
S
(see Eq.(3.48)), with the additional factor of [2VpO
(~y + ~x2 ) + VppS (~x)] added to the

integrand. Therefore,

V12S

Z
Z1
x2 y 2 (1 z 2 )(1 cos2 b ) x2 +y2
3 Nn Z
=
e
dx dy dz
Nw
(x2 + y 2 )5/4
0

F5/2

0 , x2 + y 2

x2
x2
2s y 2 + xyz + 2p y 2 xyz +
4
4

h
i
~x
S
2VpO
(~y + ) + VppS (~x) ,
2

(3.50)

where cos b is given by Eq.(3.37).

3.5.4

V21S Element

This element is given by Eq.(3.4),

V21S = (V12S ) .

(0 , ) = F5/2 (0 , ) since
However, after inspecting Eq.(3.50), it is clear that F5/2

0 , , and are all real numbers. Also Nw = Nw , hence the equation for V12S contains

38
no complex numbers. Thus, you finally have,
V21S = V12S .

3.5.5

(3.51)

e
Element
H22

From Eq.(3.5), this element is given by,

e
H22
Ee ,

where Ee is the total energy (negative) of the electron shell of the Neon atom. From
the literature (see Ref.[17]) we then have,
e
H22
962 eV .

3.5.6

(3.52)

E1 and E2

E1 is the energy of the molecular level and as a result is equal to the binding energy
of the water molecule (see Eq.(3.16)). Hence,

E1 = mol 10.087 eV .
From Fig.(3.8), which displays the energy-level diagram of the

(3.53)
18

Ne-nucleus (see

Ref.[18] ) we find the energy of the nuclear subsystem EN S = 4.522 MeV. We take
the origin of the energy scale at EN S , and, since there is an uncertainty of 7 keV,
we then have the possible values of the nuclear level E2 namely,
[(4.520 0.007) 4.522] MeV E2 [(4.520 + 0.007) 4.522] MeV ,
0.009 MeV E2 0.005 MeV .

(3.54)

39

= 9 6 keV

16

4.520 7 (1 )

4.522

17

O + 2p

F+p

3.922
1.89 (2+ )

G.s. (0+ )
Figure 3.8: Energy-level diagram of the

18

Ne nucleus

Chapter 4
Results

We now present all the results we obtained through the numerical evaluation of
the formulas presented in the previous two chapters. We firstly determined the
normalization factors Nw and Nn by solving Eqs.(3.20) and (3.32) numerically. Next,
we chose several different values for E2 , over the range of possible values as given
by Eq.(3.54), and for each of these values we then proceeded with the following
calculations:
(i) Determining the overlap integral and matrix elements using Eqs.(3.48), (3.49),
(3.50), (3.51), (3.52).
(ii) Determining the matrix elements of the Hamiltonian by using Eqs.(2.14), (2.15),
(2.16), (2.17).
(iii) Determining g, s, 1 , and 2 using Eqs.(2.22), (2.23), (2.20), (2.21).
(iv) Determining the energy E (+) and E () by using Eqs.(2.18) and (2.19).

40

41
(v) Determining the factor f and the probability using Eqs.(2.30) and (2.28).
(vi) Determining the reaction rate using Eq.(2.27).

4.1

Constants and Independent Quantities

The following are all the constants used throughout the evaluation of the formulas,
as well as the numerically determined values, of the quantities that did not depend
on the value of E2 :

= 48.547 105 fm1

0 = 34.654

r0 = 1.6489 fm

Nw = 1.33322890 1032
Nn = 7.9599343 101
V11S = 5.35704941 1041 MeV
V12S = 9.82139868 1023 MeV
V21S = 9.82139868 1023 MeV

42
e
H22
= 9.62 104 MeV

H11 = 1.00870002 105 MeV

H12 = 9.82165014 1023 MeV
H21 = 9.82165014 1023 MeV
g = 6.21461306 1044

4.2

Calculated Results for Dependent quantities

We now present all the calculated results obtained for a few different values of E 2
in a tabular format.

43

4.2.1

E2 = 0.009 to 0.007 MeV

E2 (MeV)

-0.009

-0.008

-0.007

H22 (103 MeV)

9.962

8.962

7.962

s (1040 )

3.797

4.705

5.979

1 (1041 MeV)

-5.166

-5.144

-5.117

2 (104 MeV)

-9.620

-9.620

-9.620

E (+) (105 MeV)

-1.009

-1.009

-1.009

E () (103 MeV)

-9.962

-8.962

-7.962

f (1019 )

-5.131

-4.610

-4.090

P (1040 )

3.701

4.598

5.858

(1024 s1 )

0.903

1.122

1.429

Discussion

Theoretically, 2 and E (+) does not stay constant. However,

the numerically determined values for different values of E2 stays the same. This is
due to round-off errors the computer makes when adding or subtracting a number
from another which are several orders of magnitude smaller.

44

4.2.2

E2 = 0.006 to 0.004 MeV

E2 (MeV)

-0.006

-0.005

-0.004

H22 (103 MeV)

6.962

5.962

4.962

s (1040 )

7.843

10.73

15.54

1 (1041 MeV)

-5.082

-5.035

-4.970

2 (104 MeV)

-9.620

-9.620

-9.620

E (+) (105 MeV)

-1.009

-1.009

-1.009

E () (103 MeV)

-6.962

-5.962

-4.962

f (1019 )

-3.571

-3.053

-2.537

P (1040 )

7.704

10.57

15.34

(1024 s1 )

1.879

2.577

3.742

Discussion

As can be seen from both the first table as well as this one: the

more the value of the energy increases, the greater the probability for a transition
to occur, and hence the higher the reaction rate.

45

4.2.3

E2 = 0.003 to 0.001 MeV

E2 (MeV)

-0.003

-0.002

-0.001

H22 (103 MeV)

3.962

2.962

1.962

s (1040 )

24.46

43.95

100.8

1 (1041 MeV)

-4.871

-4.706

-4.371

2 (104 MeV)

-9.620

-9.620

-9.620

E (+) (105 MeV)

-1.009

-1.009

-1.009

E () (103 MeV)

-3.962

-2.962

-1.962

f (1019 )

-2.022

-1.508

-0.996

P (1040 )

24.21

43.62

100.3

(1024 s1 )

5.906

10.64

24.46

Discussion

As we now move from E2 = 0.003 MeV to E2 = 0.001 MeV,

the probability for a transition to occur, and hence the reaction rate, is still increasing, but at a much more rapid rate.

46

4.2.4

E2 = 0.000 to 0.002 MeV

E2 (MeV)

0.000

0.001

0.002

H22 (103 MeV)

0.962

0.038

1.038

s (1040 )

424.8

166880

352.2

1 (1041 MeV)

-3.333

-45.48

-7.200

2 (104 MeV)

-9.620

-9.620

-9.620

E (+) (105 MeV)

-1.009

-1.009

-1.009

E () (103 MeV)

-0.962

0.038

1.038

f (1019 )

-0.485

0.024

0.533

P (1040 )

423.8

166900

353.1

(1024 s1 )

103.4

40700

86.13

Discussion

This is what we have been looking for. Recalling the previous

tables, there has been an increase in the transition probability and reaction rates,
at first gradual and then more rapid, until it hits a peak at E2 = 0.001 MeV, where
P and are orders of magnitude greater than at the other values of E2 .

47

4.2.5

E2 = 0.003 to 0.005 MeV

E2 (MeV)

0.003

0.004

0.005

H22 (103 MeV)

2.038

3.038

4.038

s (1040 )

92.46

41.85

23.79

1 (1041 MeV)

-6.301

-5.992

-5.836

2 (104 MeV)

-9.620

-9.620

-9.620

E (+) (105 MeV)

-1.009

-1.009

-1.009

E () (103 MeV)

2.038

3.038

4.038

f (1019 )

1.040

1.546

2.050

P (1040 )

92.94

42.17

24.03

(1024 s1 )

22.67

10.29

5.862

Discussion

After having reached a peak at E2 = 0.001 MeV, the transition

probability and reaction rate starts to decrease for any further increases in the energy
E2 .

48

4.3

Reaction Rate and E2

We now present the reaction rate and the different values of E2 in graphical format
for a clear indication of the peak reached at E2 = 0.001 MeV.

45000
40000

(1024 s1 )

35000
30000
25000
20000
15000
10000
5000
0
-0.01 -0.008 -0.006 -0.004 -0.002
E2 (MeV)

0.002 0.004 0.006

Figure 4.1: Energy versus Reaction Rate

Plotting the energy versus the logarithmic values of the reaction rates produces the
following figure.

49

100000

log[(1024 )]

10000
1000
100
10
1
0.1
-0.01 -0.008 -0.006 -0.004 -0.002
0
E2 (MeV)

0.002 0.004 0.006

Figure 4.2: Energy versus Log ( Reaction Rate )

Chapter 5
Conclusion

Our conclusions can be summarized as follows. We employed a scheme analogous to

the LCAO-motivated approach to calculate the transition probability and reaction
rate for the nuclear fusion inside the H2 O molecule according to the nuclear transition p + p + 16 O 18 Ne (4.522, 1 ). The wave function for the H2 O molecule we
employed, was based on the ansatz for the wave function of the water molecule from
an article (see Ref.([12])). From this formula we constructed the specific H 2 O(1 )
wave function which was relevant for this transition. The wave function of the
18

Ne (1 ) nuclear subsystem used in the calculations were constructed from anti-

symmetrized products of the normalized eigenfunctions 2s and 2p of the harmonic

oscillator.
Using these wave functions we derived formulas from which we could determine the
overlap integral and matrix elements, needed to calculate the transition probability
and reaction rate. Since there is an uncertainty (of a few keV) in the known value
of the energy of the nuclear subsystem, we performed our calculations for several
of the possible values of this energy. From our results we concluded that there is a
50

51
huge peak in the transition probability and reaction rate where E2 = 0.001 MeV.
This indicates an optimal energy (of the nuclear subsystem) for the transition to
take place,

E = (4.522 + 0.001) MeV = 4.523 MeV

At this point the reaction rate is in fact five orders of magnitude greater than at
E2 = 0.009 MeV and four orders of magnitude greater than at E2 = 0.005 MeV.
However, the actual value of E is not yet known at this time (its only known that
it is somewhere in the interval), and even at this peak the reaction rate is only
4.070 1020 transitions per second. Thus, it is clear that this is not an effective
energy producing nuclear reaction. Firstly, water in its natural state will not contain
H2 O molecules that are rotating (such as the H2 O(1 ) molecule), since the molecules
are bound in clusters. Secondly, as we concluded from our results, the reaction rate
is very low.
It is interesting to note that experimental searches for traces of nuclear reactions in
condensed and vaporous phases of water, with the use of low-background annihilation spectrometry, has also been carried out (see Ref.[19]). From these experiments,
the value of 71018 was estimated for the lower limit of a half-life, for the decay of
water molecules, through the studied process of molecular-nuclear transitions. This
value was related to the specific condition the water was under in the experiment
(phase of state, temperature, pressure, density of vapor, etc.). For the value of the
half-life of water in condensed state with respect to the H2 O

18

Ne decay, the

lower limit was estimated as 4 1021 years. The total statistics was insufficient

for a decisive conclusion.

It is important to note that many approximations and estimates have been used
throughout the numerical calculations performed in order to determine the transition
probabilities and reaction rates. More accurate calculations, demands more refined
wave functions of the nuclei, and especially the H2 O(1 ) molecule.

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