PHYSICS 430
NUCLEAR PHYSICS
SYNOPSIS AND STUDY GUIDE FOR THE FERMI GAS MODEL
In studying radioactive decay, nuclear reactions, and models of the nucleus, it is often useful to
invoke the concept of the "phase space" and, from this, an estimate of the density of states
available. The concept of phase space has applications in nuclear physics, particle physics, and
statistical mechanics.
In this discussion we will be describing the state of a system of identical particles, all of which are
fermions (spin = 1/2) and, which therefore obey Fermi-Dirac statistics, which implies that they
obey the Pauli exclusion principle. This means that in a prescribed volume containing the system
of identical particles, no two identical fermions in that volume can simultaneously occupy exactly
the same quantum state. (This is not true of particles called bosons (spin = 0,1,...) which obey
Bose-Einstein statistics which simultaneously occupy the same quantum state in large numbers.)
PHASE SPACE
Let us suppose that the fermions in this system are essentially independent of one another, but
that all are subject to the same potential V. [Examples: (a) In radioactive decay, the final state
particles are weakly interacting, behaving as nearly independent particles. (b) In a nuclear
reaction, the same situation often applies to the particles in the final state. (c) In the extreme
independent particle model of the nucleus, the nucleons can be considered essentially
independent, confined to a volume in which they all move in a common potential.]
For these essentially independent particles, the Heisenberg uncertainty principle imposes limits
on the simultaneous specification of a particle's spatial coordinates and linear momentum
components. Neglecting particle spin for the moment, a complete specification of the state of the
particle will require three spatial coordinates and three linear momentum coordinates. But, in a
spatial interval dx, there will be a minimum uncertainty to the simultaneous specification of the xcomponent of linear momentum which is dpx = h/dx due to the Heisenberg uncertainty
principle. Thus, if a particle occupies a location x in an interval dx, and has a momentum p x , the
next particle in dx must have a momentum which is
h
px + dpx px + dx .
(1a)
Correspondingly, particles whose momentum lies within dpx cannot have their x-location nearer
than
dx · dpx = h.
(1b)
A similar limitation applies to the simultaneous specification for these values along the y and z axes.
These can be viewed as limitations on the proximity of two fermions in this six dimensional space
such that only one particle can occupy a cell in this phase space volume which is computed as
dVps = dx dpx · dy dpy · d z dpz
(2a)
dVps = dV · dp3
(2b)
dVps  h3 .
(2c)
That is, the smallest volume of phase space which can be occupied by at most one uniquely
identified particle is h3. Therefore, if one has an arbitrary phase space volume dVps , the largest
number of unique quantum states within that volume can be found as,
dN =
dVps
h3
,
(3a)
or, writing dp3 in a spherical polar representation we obtain an estimate of the number of
possible states in dV for all particles with momentum between p and p + dp as
dN =
dV · 4  p2 dp
.
(2)3
(3b)
Since this equation was obtained using only the Heisenberg uncertainty relation, its application
for fermions is completely general.
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FERMI GAS MODEL FOR THE NUCLEUS
One model for the nucleons inside the nucleus is called the Fermi gas model wherein the nucleons
are considered essentially individually independent and moving in a common potential. In this
model then, the nucleons are individually (neutrons separately from protons) constrained to have
no more than one particle per phase space cell. Some useful concepts about nucleon energies
(motion) inside the nucleus can be formulated in this model and estimates can be taken which are
in reasonable agreement with measurements.
We begin by approximating a volume for these particles which is cubic and defined by sides of
length L. (If this volume is to be representative of a nucleus, we have every reason to believe that
the nucleus is probably much more spherical than cubic in shape, but this model is a bit easier to
deal with and the results are not unreasonable.) We wish to write the Schrödinger equation for
any one of these fermions and to do so, we postulate that the potential in which the fermion
moves can be characterized by a rectangular well of infinite sides such that
V(x) = 0 for 0 < x < L ; V(x) =  elsewhere.
V(y) = 0 for 0 < y < L ; V(y) =  elsewhere.
(4)
V(z) = 0 for 0 < z < L ; V(z) =  elsewhere.
Here, we have assumed that the potential can be written simply as V = V(x) + V(y) + V(z). Now,
if we substitute this potential into the Schrödinger equation, we have
2
2m 2 + V  = E ,
where
 = (x,y,z) .
(5)
If we let
(x,y,z) = u(x) · u(y) · u(z) , and,
(6a)
E = E x + Ey + E z ,
(6b)
we can show that the three dimensional Schroedinger equation (5) separates into three eigenvalue
equations, one for each of the functions u(x), u(y), and u(z), with corresponding eigenvalue Ex,
Ey, and, Ez. We can also show that, on requiring the functions u(x), u(y), and u(x) each to vanish
at the boundaries 0 and L, we obtain solutions:
u(x) =
2
L
[
sin
nx x
L
], and,
(7)
Ex =
()2
n 2,
2mL2 x
(8)
and similarly for y and z. Thus, the total wave function (6a) is the product of the individual
solutions and the total energy (6b) is the sum of the individual eigenvalues giving
E =
()2
(n 2 + ny2 + nz2 ) .
2mL2 x
(9)
As an exercise, you should try to write out the quantum numbers for the 25 lowest energy
neutron states. Include also the spin giving two fermion states for each unique combination of
three
(nx, ny ,nz) quantum numbers. Note that there is a significant amount of degeneracy, i.e., unique
quantum states with the same total energy eigenvalue. You can also show in tabular form which
of these 25 states are degenerate and note the energy eigenvalue for these degenerate states. See
if you can derive a simple formula which will tell the number of degenerate states for any given
energy state.
The eigenstate for this particle is thus completely and uniquely determined by specifying the
three quantum numbers (nx, ny ,nz), i.e., we could write a state as
n , n ,n (x,y,z).
x y z
(10)
Using the relationship for the non-relativistic kinetic energy, E = p2/2m (remember that V = 0
everywhere in the volume), we can use equation (8) to obtain an expression for the linear
momentum along the x-axis as

px = L nx .
(11)
Note that the components of the linear momentum are quantized. The total linear momentum
has a magnitude

p = L
(12)
(nx2 + ny2 + nz2) .
(As an aside, equation (11) is precisely what you would expect if you envisioned the stationary
states in the volume as corresponding to standing waves whose wavelengths were related to the
deBroglie wavelengths along the separate axes as
h
x = p
x
=
2 
(/L) nx
2L
 x = n
x
.
(13)
Thus, the lowest x-axis momentum state corresponds to the longest de Broglie wavelength which
will fit into the infinite potential well, subject to the condition that the wavefunction vanish
identically at both boundaries. Taking nx = 1 we get 2L which is the longest de Broglie
wavelength.)
It is evident from equation (12) that each triplet coordinate,
(nx, ny ,nz)
(14)
identifies a unique momentum eigenvalue. We can envision the momentum eigenvalue as the
length of the momentum vector from equation (12). The components along the axes are
quantized and give one vector per cell whose center is at

(px, py ,pz) = L (nx, ny ,nz)
(15)
and whose momentum cell size is
dp3 =
(2 )3
L3
.
(16)
Now, let us assume that this system (nucleus) is in a state of extreme degeneracy such that all the
low lying energy states are filled up to some maximum value, calling the maximum value EF, the
Fermi energy level. In equivalent terms, this means that all momentum states whose momentum is
less than or equal to the maximum momentum p (called the Fermi momentum) are filled. Thus,
F
the total number of unique momentum states (N), whose momentum is less than or equal to the Fermi
momentum, is found by taking a volume equal to 1/8 the volume of a sphere, whose radius is p ,
F
divided by the momentum cell size, i.e.,
3
1 4  pF
N=8 3
h3
L3
(17)
(The factor of 1/8 comes from the fact that only the one octant of the entire spherical volume can
have physical values of momentum since the quantum numbers nx, ny ,nz must be positive.)
If we include the fermion spin, there are two allowed fermions per momentum eigenstate, and
substituting V = L3 , we have
N=
2
4
p 3 V.
3
F
(2) 3
(18)
This is to be compared with equation (6-6.1) in Segre's book.
Substituting for the Fermi momentum p =
F
[
2mEF
V
N =  3 
2 2
(19)
]
2mEF , we have
3/2
From equation (19) we can obtain the Fermi energy,
()2
EF = 2m
[3N
V ]
2/3
,
(20)
and from equation (18) we obtain the Fermi momentum,
p
F
=  (32)
1/3
[NV ]
1/3
= h kF .
(21)
Recall that we protons and neutrons are not identical particles and we can calculate the Fermi
energy (momentum) for them separately. In the equations above, N is the number of states filled,
which for protons is Z and for neutrons is (A - Z) as seen in Segre, equations (6-6.2) and (6-6.3)
which agree with (21). The volume V can be estimated assuming the nucleus to be a sphere with
radius r = Ro A1/3 .