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Atomic shell model
Nuclear shell model
each nucleon, separately occupies its own energy level,
with each nucleon-type (p,n)
filling energy levels separately
Each energy level, n exists with an angular quantum number:
s p d f g …
ℓ = 0, 1, 2, 3, 4, …
each with a different
angular wave function
Yℓ,m(,)
Because of spin (2 possible orientations of each nucleon)
any energy level n,ℓ accommodates
2ℓ+1 protons
2ℓ+1 neutrons
s: ℓ = 0
p: ℓ = 1
d: ℓ = 2
f: ℓ = 3
These are different particles –
They don’t
mutually exclude each other!
coupled with spin:
s1/2
p1/2
p3/2
d3/2
d5/2
f 5/2
f 7/2
J=0+½
The total (magnitude)
of J is limited to
-|ℓ-s|…|ℓ+s|
in unit steps!
As an example, there is a
1f5/2 energy level that can hold
5
2( ) ????
1  6 protons with different mj values
2
and another that can hold 6 neutrons.
Consider:
27
13
Al
(ground state)
2
4
2
Protons: (1s1/2) (1p3/2) (1p1/2) (1d5/2)
2
4
2
5
Neutrons: (1s1/2) (1p3/2) (1p1/2) (1d5/2)
6
1 un-paired proton with j=5/2
so aluminum: I=5/2
Consider:
9
Be
(ground state)
2
2
2
3
p: (1s1/2) (1p3/2) (1p1/2) (1d5/2)
n: (1s1/2) (1p3/2) (1p1/2) (1d5/2)
1 un-paired neutron with j=3/2 so: I=3/2
Consider:
9
B
(ground state)
2
3
2
2
p: (1s1/2) (1p3/2) (1p1/2) (1d5/2)
n: (1s1/2) (1p3/2) (1p1/2) (1d5/2)
1 un-paired proton with j=3/2
so: I=3/2
Magnetic Dipole Moment
B
classically
energy Emag = -B
=
1
c
current  area
For a point particle q, velocity v, circular orbital radius r
v
I q
2r
frequency!
1
 = c current  area
 
q
1 qvr
 v 
2
( mvr )
=
q
  r = c
2mc
2
 2r 
L
q 


L but maybe too classical!
2mc
 
We still expect   J quantum mechanically
and write
e  g measures the

g
J
(expected)
2mc
deviation from
1
= c
quantum mechanically
J is quantized!
J  j
e
 g
j
2mc

 g 0 j
0 = “magneton”
For atomic physics (involving electrons)
“Bohr magneton”
e
14
B 
 0.5788 10 MeV / G
2me c
5
 5.7885 10 eV / T
For nuclear physics
“nuclear magneton”
e
N 
 3.1525 1018 MeV / G
2m p c
 3.15255 10 8 eV / T
For the neutron: g= 3.826083 0.0000018  0 !
With this abbreviation:



   (r , t )    (r ) (t )
  n,l,ml,s,ms,j,…
d
i    H  
dt
 d

i  (r )  (t )   (t )E n  (r )
dt
we represent the exact solutions to
some time-independent potential
(constant Hamiltonian)
d
i  (t )  E n (t )
dt
d (t ) E n

 (t )
dt
i
 (t )  exp( iEnt / )
  e

  (r )
 iE n t / 
Where you note we used the fact that
Simply a complex
phase that
oscillates in time.


H  (r )  E n  (r )
With En the eigenvalues, and  the eigenfunctions
which form a COMPLETE ORTHONORMAL SET.
 n...  m...
3
*

  dr  n ( r )  m ( r )  ?m n
As long as H is just constant, these are considered stationary states.

Imagine, initially, a particle was in one of the stationary states
of
H (the simple Hamiltonian with exact solutions).
which is NOT an exact solution of some NEW Hamiltonian
H + Hint
…no reason to expect it to be “stationary” now!
Obviously there is the probability of a transition to some OTHER state
(decay to a lower energy state, scattering to a new momentum state,
or even transmutation into some new particle)!
Maclaurin (or "power") Series
1
2
3
1 x  x  x 
1 x
1
2
3
5
7
x 3x
5x
1 


2
8
16
1 x
3
x
x
x
sin x  x 



3! 5! 7!
2
4
6
x
x
x
cos x  1 



2! 4! 6!
Axn
Functions that form a
COMPLETE
ORTHONORMAL SET
for x measured in
radians (not degrees!)
p( p  1) 2 p( p  1)( p  2) 3
ln( 1  x)  1  px 
x 
x 
2!
3!
p
Fourier Series
Acos(nx) + Bsin(nx)
Functions that form a
COMPLETE
ORTHONORMAL SET
Imagine a free particle (with an exact eigensolution) of initial energy E
Dt ) passes near the
source of some potential and is scattered into a final state of energy Em.
At time to=0 momentarily ( for a short total time,
H  H + Hint
And now we need to try and solve

i   H  H int  
t
Because the |n> comprise a
“complete” orthogonal “basis set”
   anEn  n e
iEnt / 
n
|an(t)|2  probability
of finding the particle
in state n at time t.

i   H  H int  
t
i  an  n e
 iE nt / 
  an E n  n e
n
 iE n t / 

n
 an H  H int n
e
 iE n t / 
n
cancel
i an  n e
n
iEnt / 
  an Hint  n e
n
iEnt / 
i an  n e
iEnt / 
n
  an Hint  n e
iEnt / 
n
Initially for t < to = 0 (when Hint=0)
a (t )  1
an (t )  0
Vint
once t > to = 0 (when Vint ,t small)
~
to = 0
a (t )  1
an (t )  1
but not 0
Watch how we can pluck out a specific coefficient by:
 mm i an  n
n
e
 iEn t / 
 
mm  an Hint  n e
n
 iEn t / 
 m i an  n e
 iEn t / 
n
 m
 an Hint  n
e
 iEn t / 
n
m n  ?
iam e
 iEmt / 
 
  an  m Hint  n e
 iEn t / 
n
with the only substantial contribution coming from n=?
am  (i)
1
m | H int |  e
i ( En E )t / 
We wanted to solve:

i   H  H int  
t
using, because the |n> comprise a “complete” orthogonal “basis set”
   anEn  n e
n
iEnt / 
|an(t)|2  probability
of finding the particle
in state n at time t.
We found:
am  (i)
1
m | H int |  e
i ( En E )t / 
am  (i)
1
m | H int |  e
i ( En E )t / 
If we simplify the action (as we do impulse in momentum problems)
to an average, effective potential V(t) during its action from (t0,t)
≈
Then, for a (short) time T following to:
1
am ( T ) 
m Veff 
i

m Veff 
Em  E
factor out
T
0
1  e
dt e
i ( Em  E ) t / 
i ( Em  E )T / 

So the probability of finding this system in
a different state m after a time T
2
am (T ) 
2
m Veff 
Em  E 
2
 4 m Veff 
 
( Em  E )T 

21  cos


 
2
sin ( Em  E )T / 2 
2
Em  E 
2

E f  E0  t 

2
 4 sin

2


1
Em  E 
2
DE=2h/Dt
The probability of a transition to a particular final state |Ef t>
P4
 E f | V | Ei 
E
2
 Ei 
sin
2
f
2
E
f
2
The total transition probability:
Ptotal   4
N
 E N | V | Ei 
2
EN  Ei 
2
 Ei  t
sin
2
EN  Ei  t
2
If < EN|V|Ei > ~ constant over the narrowly allowed DE
Ptotal  4  EN | V | Ei 
2

N
sin
2
EN  Ei t
2
2
EN  Ei 
for scattering, the final state
particles are free, & actually
in the continuum
 Ei t
sin
2
2
Ptotal  4  EN | V | Ei  
2
N
EN  Ei 
2
E
n=
N
n=3

Ptotal  4  E( N) | V | Ei   dN
2
sin
2
E( N)  E t
2
2
E( N)  Ei 
i
n=2
n=1

Ptotal  4  E( N) | V | Ei   dN
2
sin
2
E( N)  E t
i
2
2
E( N)  Ei 
With the change of variables:
t dE
dx 
dN
2 dN
x  E ( N )  Ei  t / 2

2
dN
2

sin
x


Ptotal  4  EN | V | Ei  
dx

2
dE
t


2x / t


2

2
 dN  t sin x
Ptotal  4  E N | V | Ei  
dx

2

 dE  2  x
2

Notice the total transition probability 
t
2 t
2  dN 
Ptotal 
 E N | V | Ei  


 dE 
and the transition rate
2
2  dN 
W  Ptotal / t 
 EN | V | Ei  


 dE 
Does the density
of states vary
through the
continuum?
n=
dN/dE
n=3
E
n=2
n=1
Classically, for free particles
E = ½ mv2 = ½ m(vx2 + vy2 + vz2 )
vz
vy
Notice for any fixed E, m this defines
spherical surface of velocity points
all which give the same kinetic energy.
vx
The number of “states” accessible by that energy are within the
infinitesimal volume (a shell a thickness dv on that sphere).
dV = 4v2dv
Classically, for free particles
E = ½ mv2 = ½ m(vx2 + vy2 + vz2 )
dE  mv dv
dE
dE
dv 

mv
2mE
We just argued the number of accessible states (the “density of states”)
is proportional to
4v2dv
 2 E  dE 
dN  C 4v dv  C 4 


 m  2mE 
 4C
 1/ 2
dN
dN   3/ 2 2  E dE
 E1/2
dE
m

2
The Maxwell-Boltzmann distribution describes not only how rms velocity
increases with T but the spread about  in the distribution as well
T  absolute zero
With increasing energy (temperature)
a wider range of velocities become probable.
There are MORE combinations of molecules
that can display this total E.
That’s exactly what “density of states” describes.
The absolute number is immaterial.
It’s the slope with changing E,
dN/dE that’s needed.
For a particle confined to a cubic box of volume L3
the wave number vector
p  
is quantum mechanically constrained
2
i 
ni
L
2

p
 2 2 
2
2
2
E

 
n

n

n
x
y
z
2
2m 2m
mL
2
2 2

The continuum of a free particle’s momentum is realized by the limit L
For a LARGE VOLUME these energy levels may be spaced very closely
with many states corresponding to a small range in wave number
L
dnx 
d x
2
The total number of states within a momentum range 
N

dnx dn y dnz 
3
L

3
2 


L3
3
d
3
2 
4V
 d d 
3
2 
2

 d
2
The total number of final states per unit volume
N
4V
2

 d
3
V V 2 
or
dN
4 2
4p dp

 d 
3
3
V
2 
2 
2
From E=p2/2m

dE=p dp/m
dN
4p

dE
3
V
2  v
2

dE = v dp
number of final states per unit volume whose
energies lie within the range from E to E+dE
1 dN
4p
E 

V dE 2 3 v
2
Giving (FINALLY) the transition rate of
2
2
W
 E N | V | Ei   E

Since transition rate = FLUX × cross-section
or

 rate/FLUX
4p f
2
2

 E N | V | Ei 

2 3 v f
2
/nvi
Particle density
of incoming beam
but here we’re talking
about one-on-one collisions
If final states of target and projectile have non-zero spins
c
a
b
 EN | V | Ei  (2 sc  1)( 2 sd  1) 2
pf
4 2

vi v f
2

d
Total cross sections
for incident proton,
antiproton, positive
and negative pions,
and positive and
negative kaons on
proton and neutron
targets.
Breakdown of Rutherford
Scattering formula
When an incident  particle
gets close enough to the target
Pb nucleus so that they interact
through the nuclear force (in
addition to the Coulomb force
that acts when they are further
apart) the Rutherford formula
no longer holds.
The point at which this
breakdown occurs gives a
measure of the size
of the nucleus.
R.M.Eisberg and C.E.Porter
Rev. Mod. Phys, 33, 190 (1961)
Total cross sections for + p  + p and + p   oD++
The total +p cross section
Electron scattering off nuclei
12C
40Ca
16O
Left: total cross sections for K+ on
protons (top)
deuterons (bottom)
Above: K-p total cross sections