Transcript 投影片 1

Chapter 5
Intrinsic Properties of a Nucleus
◎ Total Angular momentum and Nuclear spin
● Parity
◎ The Electric field outside an arbitrary charge distribution
● Nuclear Electromagnetic moments
5-1 Total Angular momentum and Nuclear spin
For nuclei:
The nucleus is an isolated system and so often acts like a single
entity with has a well defined total angular momentum.
It is common practice to represent this total angular momentum
of a nucleus by the symbol I and to call it nuclear spin.
[Associated with each nuclear spin is a nuclear magnetic moment
which produces magnetic interactions with its environment.]
For electrons in atoms:
For electrons in atoms we make a clear distinction between electron
spin and electron orbital angular momentum and then combine
them to give the total angular momentum.
Angular momentum
For nuclei:
total angular momentum of a nucleus = I ― "nuclear spin"
This is the VECTOR sum of:
the intrinsic spins of the individual nucleons (S)
HALF-INTEGER
the intrinsic orbital angular momentum of the individual nucleons (L)
INTEGER
Angular momentum
For nucleons:
Every nucleon has quantum numbers: L, S, J
L - from orbital angular momentum quantum number ― INTEGER
S - from spin quantum number ― HALF INTEGER
J - total angular momentum quantum number of single nucleon ― HALF INTEGER
Because the value of the J quantum number is
always half-integral, if there is an even number of
component angular moments J, then I will be
integral
If there is an odd number, then I will be halfintegral. In nuclear systems it is J that is good
quantum number.
for odd A nuclei I is half integral
for even A nuclei I is integral.
All nuclei with even Z and even N have zero total nuclear angular momentum, I = 0.
Use I for total nuclear spin; use J for nucleon spin
(1) A single valence nucleon may determine the spin so: I = J.
(2) Two valence nucleons may determine the spin so:
I = J1 + J 2 there can be several values of I.
(3) A valence odd particle and the remaining core contribute:
I = Jparticle + Jcore
Total angular momentum of the nucleus - nuclear spin - is the vector sum of each
nucleon = Total angular momentum quantum number of NUCLEUS = I
I 2   2 I ( I  1)
I Z  m (m  I ,......, I )
(1) The nucleus behaves like a single entity with
angular momentum I
(2) In ordinary magnetic fields I splits into 2I +1
Zeeman effect
(3) In strong fields individual nucleon states split
into 2J +1
Notation and concepts
“sharp”
“principal”
“diffuse”
“fundamental”
s
p
d
f
L=0
L=1
L=2
L=3
for example: for a nucleon, n = 2, L= 1 the state is designated 2p
The vector coupling of L and S suggests J = L +1/2 or L - 1/2
Thus for L = 1 (p) we have p3/2 and p1/2
we can count higher states e.g. 2p3/2 .. 3p3/2
5-2 Parity
If we want to describe a nuclear state (for a nucleus) completely
we need to identify its parity. Strong nuclear interactions will not
alter the “parity”. The parity is conserved under strong nuclear
interactions. Every nuclear eigenstate has its own “parity”.
Real mathematical functions can be categorized into three types:
1. If f(-x) = f(x) then f(x) is called an “even function”,
or a function of even parity.
2. If g(-x) = -g(x) then g(x) is called an “odd function”,
or a function of even parity.
3. There are functions of mixed parities.
Ex. h(x) = x2 + x
parity cover
An even function
An odd function
It is quite often that functions we are dealing with are of mixed parities.
In terms of quantum mechanical wave functions we may get from the
Schrödinger’s equation solutions of even and odd parity separately.
1-dimensional case
 ( x)   (  x)
A wave function of even parity
 ( x)   ( x)
A wave function of odd parity
(1)
In the 3-dimensional case
 (r )  (r )
 (r)   (r)
r  r
 ( x, y, z)   ( x, y, z)
even parity
 ( x, y, z )   ( x, y, z ) odd parity
parity transformation
(2)
z
θ
(π-θ)
r
(π+ φ)
y
φ
x
In fact the parity
transformation is to
change r into – r.
r
In the spherical coordinates it is to change
   
  
and leave the radial coordinate r unchanged.
that is
(r ,  ,  )  (r ,    ,    )
Example:
In describing the hydrogen atom we solve the Schrödinger equation and come up the solution
which is written as
ψnlm (r ) = ψnlm (r , θ , φ) = Rnl (r )Ylm (θ , φ)
(3)
where n, l, and m are quantum numbers. Rnl(r) is the radial part of the wave function and
Ylm(θ,φ) is the angular part of the wave function. Ylm(θ,φ) is generally referred as the
spherical harmonics
In this case
 nlm (r )   nlm (r,    ,    )  Rnl (r )Ylm (   ,    )
r
θ
Ylm ( , )  (1)l Ylm ( , )
therefore if l = even then
(π+ φ)
y
φ
x
-r
(4)
ψ nlm (r ) is a function of even parity.
if l = odd then
ψ nlm (r ) is a function of odd parity.
The parity transformation changes a right-handed coordinate system
into a left-handed one or vice versa.
Two applications of the parity transformation restores the coordinate
system to its original state.
It is a reasonable presupposition that nature should not care whether its coordinate
system is right-handed or left-handed, but surprisingly, that turns out not to be so.
In a famous experiment by C. S. Wu, the non-conservation of parity in beta decay was
demonstrated.
Parity
(1) Central potentials only depend upon the magnitude of |r| and so are
invariant with respect to the parity operation, ie V(r) = V(-r).
(2) Measurable effects of such potentials should also be invariant with
respect to the parity operation.
(3) Observable quantities depend upon the square of the modulus of the
wavefunction and so individual nuclear wavefunctions will be either
even or odd but not a mixture of the two. (symmetric, antisymmetric)
(4) That is to say, all nuclear states have a definite parity and
conventionally this is used together with the total angular momentum
to label the states.
e.g. 0+ (I = 0, even parity) ; 2- (I = 2, odd parity)
(5) The definite parity of states means that the distribution of electric
charge in the nucleus is even.
Parity π
Every nuclear state also has a parity that is the product of the
parity of each nucleon π = π1π2π3π4…πA = “+” or “-”
Every nuclear state also has a parity: π
This is denoted: Iπ i.e. 0+, 2-, (1/2)-, (5/2)+.......
•
•
•
not so far possible to know individual nucleon parity
we have an overall parity that is a measured quantity
no theoretical relationship between I and π
 (r )  (r )
 (r)   (r)
 ( x, y, z)   ( x, y, z)
even parity
 ( x, y, z )   ( x, y, z ) odd parity
(5)
5-3 The Electric Filed outside an arbitrary charge
distribution
Much of what we know about nuclear structure comes from studying not the strong nuclear
interaction of nuclei with their surroundings, but instead the much weaker electromagnetic
interaction.
The strong nuclear interaction establishes the distribution and motion of nucleons in the
nucleus, and we probe that distribution with the electromagnetic interaction.
In doing so, we can use electromagnetic fields that have less effect on the motion of nucleons
than strong force of the nuclear environment; thus our measurements do not seriously distort the
object we are trying to measure.
In the absence of a magnetic field,
spins are randomly oriented.
Exposed to an external magnetic field, each spin
or magnetic moment can assume two different
orientations, denoted “parallel” (spin up) and
“anti-parallel” (spin down) respectively.
Let us consider an arbitrary charge distribution of density ρ(x’, y’, z’) occupying a volume τ’
and extending to a maximum distance r’max from the origin of coordinates O. We select O
either within the volume or close to it. Such a distribution is illustrated in the figure.
r”
r’
The electric potential V at some
point P(x,y,z) such that r > r’max is
r
ρ( x' , y' , z ' )dτ '
V =∫
τ'
4πε0r"
(6)
where r” is the distance between the point of observation P and the position P’(x’,y’,z’) of
the element of charge ρ(x’, y’, z’)dτ’ :
r"  [(x  x' )2  ( y  y' )2  ( z  z' )2 ]1/ 2
(7)
Since r” is a function of x’, y’, z’ we may expand (1/r”) as a
Taylor series near the origin:
1 1  

  1
   x'
 y'
 z'   
r" r  x'
y '
z '  0  r" 
2

1 

  1
x
'

y
'

z
'
     ,


2!  x'
y '
z '  0  r" 
(8)
where the subscript 0 indicate that the derivatives are evaluated at the origin.
 1
1 r" x  x'



 
2
x'  r" 
r" x'
r"3
and
   1 
x
l




 x' r"  r 3 r 2
   0
(9)
where l = x/r is the cosine of the angle between the vector r and the x-axis.
We treat the terms concerning y’ and z’ accordingly and define m = y/r and n = z/r
Here l, m, and n are called “directional cosines” and satisfy the following requirement.
l 2 + m2 + n 2 = 1
(10)
The result of the calculation is that at the point P
1 d '
1
d '
  2 (lx' m y' nz' )
 ' r 4
' r
4 0
0
V 
1
[3m ny' z '3nlz' x'3lm x' y '
 ' r3
1
1
1
d '
 (3l 2  1) x'2  (3m2  1) y '2  (3n 2  1)]
 
2
2
2
4 0

 V1  V2  V3    
V1 
V2 
1
40
1
4 0

'

d '
'
V1 is called the monopole term and is zero only if the total net charge
is zero. The electrical field derived from V1 decreases as 1/r2.
r
(lx'm y' nz' )
(11)
d '
r2
V2 is called the dipole term which varies as 1/r2. The electrical
field derived from V2 decreases as 1/r3.
If we define an electrical dipole moment p
p   r 'd '   ( x' iˆ  y' ˆj  z ' kˆ) d '
'
'
V2 
then the dipole term V2 can be written as
r1 is the unit vector along r in the direction of P.
p  rˆ1
4 0
(12)
rˆ1  liˆ  mˆj  nkˆ
V3 is called the quadrupole term which varies as 1/r3. The electrical field derived from V3
decreases as 1/r4.
In general there are totally six components for a quadrupole moment. If the charge
distribution is of cylindrical symmetry, as it is in all the nuclides, then three components
become vanished and it is convenient to define a single quantity q, often called the
quadrupole moment of the charge distribution:
q   (3 z '2  r '2 ) d '
'
(13)
5-4 Nuclear Electromagnetic moments
1. The electrical monopole moment of a nucleus is simply the total
charge Ze.
2. Because of the nearly perfect cylindrical symmetry of all nuclei,
spherical or non-spherical, the electric dipole moments are too small
to be detected by the existing techniques.
3. However, nuclei with prolate- or
oblate-like shape, non-spherical, can
have measurable electric quadrupole
moments. A nucleus of large measured
quadrupole moment q is geometrically
non-spherical.
Some measured electrical quadrupole moment q.
17
8
O
-0.026 barns
39
19
K
0.11 barns
209
83
115
49
Bi
In
165
67
Ho
175
71
Lu
-0.35 barns
181
73
1.16 barns
197
79
2.82 barns
5.68 barns
Ta
4.20 barns
Au
0.59 barns
Nuclear Magnetic Dipole Moment
l
μ
A circular loop carrying current I and enclosing area A has a magnetic
dipole moment of magnitude ∣μ∣= iA. Here we want to consider a simple
classical model of an orbiting point charge circling around a center O
chosen to be the origin of the coordinate system. The orbiting point
charge with charge e and mass m is circling around the origin O with a
radius r and speed v.
In this case the magnitude of the magnetic dipole moment is
O
e
r
 
v
e
evr
e
r 2 

l
(2r / v)
2
2m
(14)
where l is the classical angular momentum mvr.
In quantum mechanics, we operationally define the observable magnetic
moment to correspond to the direction of greatest component of l; thus we
can take equation (14) directly into the quantum regime by replacing l with
the expectation value relative to the axis where it has maximum projection,
which is which is ml (h/2π) with ml = + l. Thus
(15)
e

l
2m
where now l is the angular momentum
quantum number of the orbit.
Magnetic dipole moment of a nucleon due to its orbital motion

e
l
2m
(15)
l
O
e
is called a magneton.
2m
e
v
For atomic motion we use the electron mass and obtain the Bohr magneton
μ
r
B  5.7884105 eV/T
In nuclear system it is the proton mass that we should use to calculate the
magneton and we have the nuclear magneton:
N  3.1525108 eV/T
Note that μN << μB owing to the difference in the masses; thus under most
circumstances atomic magnetism has much larger effects than nuclear magnetism.
For a proton of orbital angular momentum l its magnetic
dipole moment due to the orbital motion is
  gllN
(16)
where gl is the g factor associated with the orbital angular momentum l.
  gllN
(16)
where gl is the g factor associated with the orbital angular momentum l.
For protons gl = 1; because neutrons have no electric charge, we can use
equation (16) to describe the orbital motion of neutrons if we put gl = 0.
Magnetic dipole moment of a nucleon due to its spin
Due to proton spin its magnetic dipole moment can be described in a similar way:
  gs sN
(17)
where s = 1/2 for proton, neutrons and electrons and the quantity gs
is known as the spin g factor and is calculated by solving a
relativistic quantum mechanical equation.
For a spin-1/2 point particle such as the electron, the Dirac equation gives gs =2,
and the measured value from the electron is gs = 2.002319304386 which is quite
consistent with the theoretical predicted value. More accurate value can be
obtained by calculating higher order corrections from quantum electrodynamics.
  gs sN
(17)
For free nucleons the experimental values are far from the
expected value for point particles:
proton:
gs  5.5856912 0.0000022
neutron: gs  3.8260837 0.0000018
It is evident that :
(18)
(1) The proton value is far from the expected value of
2 for a point particle.
(2) The uncharged neutron has a nonzero magnetic
moment!
This is the evidence that the nucleons are not elementary point particles like
the electron, but have an internal structure; the internal structure of the
nucleons must be due to charged particles in motion, whose resulting
currents give the observed spin magnetic moments.
proton:
neutron:
gs  5.5856912 0.0000022
gs  3.8260837 0.0000018
(18)
It is interesting to note that gs for the proton is greater than its
expected value by about 3.6, while gs for the neutron is less than
its expected value (0) by roughly the same amount.
Magnetic dipole moments of proton and neutron are vector
sums of quarks which constitute two different kinds of nucleons.
In nuclei, the pairing force favors the coupling of
nucleons so that their orbital angular momentum and
spin angular momentum each add to zero. Thus the
paired nucleons do not contribute to the magnetic
moment, and we need only consider a few valence
nucleons. No nucleus has been observed with a
magnetic dipole moment larger than about 6 μN.
~The End ~