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16.451 Lecture 19:
The deuteron
d
Basic properties:
1
13/11/2003
2
1H
mass: mc2 = 1876.124 MeV
binding energy:
B 
m
i
 M  m p  mn  md  2.2245731 MeV
i
RMS radius:
(measured via -ray energy in n + p  d + )
1.963  0.004 fm
quantum numbers:
(from electron scattering)
J  , T  1 , 0
magnetic moment:
electric quadrupole moment:
(lectures 13, 14)
 = + 0.8573 N
Q = + 0.002859  0.00030 bn
( the deuteron is not spherical! ....)
Important because:
• deuterium is the lightest nucleus and the only bound N-N state
• testing ground for state-of-the art models of the N-N interaction.
Electron scattering measurements:
2
Because the deuteron has spin 1,
there are 3 form factors to describe
elastic scattering: the “charge” (Gc),
“electric quadrupole” (Gq) and “magnetic
(GM) form factors. (JLab data)
Interpretation of quantum numbers:


S  Sn 


J  S 
J  , T  1 , 0
3

Sp


L  1
  () () (1) L    L  0, 2, 4 .....
• Of the possible quantum numbers, L = 0 has the lowest energy, so we expect
the ground state to be L = 0, S = 1 (the deuteron has no excited states!)
• The nonzero electric quadrupole moment suggests an admixture of L = 2
(more later!)
introduce Spectroscopic Notation:
with naming convention:
2 S 1
LJ
L = 0 is an S-state, L = 1 is a P-state, L = 2: D-state, etc...
 the deuteron configuration is primarily 3 S
1
4
Isospin and the N-N system: (recall lecture 13)

T 


1
1

2
2

T  0, 1
The total wavefunction for two identical Fermions has to be antisymmetric w.r.to
particle exchange:
total   space  spin   isospin
Central force problem:
 space (r, , )  f (r ) YLM ( , )
with symmetry (-1)L given by the spherical harmonic functions
Spin and Isospin configurations:
S = 0 and T = 0 are antisymmetric ;
S = 1 and T = 1 are symmetric
3
S1
state can only
be T = 0 !
Understanding “L” -- the 2-body problem and orbital angular momentum:
Classical Mechanics:

m2

r2
x

r1

v1
m1

v2
CM
• set origin at the center of mass, and let the CM be fixed in space
 particles orbit the CM with angular speed 
• Kinetic energy for each particle and angular momentum L = I:
Ki 
1
2
mi vi 2 
1
2
mi ri 2 2  L2i / 2I i
• Total angular momentum: L = L1 + L2 with I = I1 + I2
• Total kinetic energy:
K = L2/2I
5
6
Equivalent one-body problem:

v

r
x

CM

r 
Particles orbit the center of mass
because of an attractive potential V(r)
with no external force, particle kinematics
are related in the CM system and the 2-body
problem is equivalent to a 1-body problem:
 
r1  r2 , r  r1  r2 ,
Moment of inertia can be written:
Total energy:
E 
L2
2 r
2
 
m1
1  (m1 / m2 )
I  m1 r12  m2 r22   r 2
 V (r )
Note: L = total angular momentum of the 2-particle system.
7
Quantum mechanics version:


H  (r )  E  (r )
H = total energy operator, E = eigenvalue)
 2 2
H 
  V (r ) ;
2
Angular derivatives define
an orbital angular
momentum operator:
 ( r , ,  ) 
R(r )
Ylm ( ,  )
r
 2
ˆ 2    
  2


2


 cot

2
r 2
2 
ˆ 2

 2
r r
r
 2 

sin 2   2 
1
ˆ 2 Ylm ( , )   (  1) Ylm ( , )
wave function in the relative coordinate has
this generic form, where R(r) depends on V(r):
  2  (  1)

 2 d 2R
 
 V ( r )  R ( r )   R( r )
2
2
2 d r
 2 r

Analog of L2/2r2 in the classical problem!
lowest energy state has minimum total L!
8
Deuteron Magnetic Moment: d = 0.857 N
In general, the magnetic moment is a quantum-mechanical vector; it must be aligned
along the “natural symmetry axis” of the system, given by the total angular momentum:

 ~ J


But we don’t know the direction of J , only its “length” and z-component as expectation
values:
J 2  J ( J  1) ;
Jz
 mJ  ( J ....  J )
 
in a magnetic field, the energy depends on mJ via E     B   g J mJ B  N
Strategy:
we will define the magnetic moment by its maximal projection on the
z-axis, defined by the direction of the magnetic field, with mJ = J

    zˆ
mJ  J
 gJ J N
Use expectation values of operators to calculate the result.....
9
Calculation of  :

z
    zˆ

J
mJ



 gJ J N
mJ  J
Subtle point: we have to make two successive
projections to evaluate the magnetic moment
according to our definition, and the spin and
orbital contributions enter with different weights.

1. Project onto the direction of J:
 ˆ
  J   cos 
2. Project onto the z-axis with mJ = J:
cos  
  gJ J N

  J

1
( J  1)
mJ

|J|
 
J
J ( J  1)
J
J ( J  1)
Next, we need to figure

out the operator for 
10
spin and orbital contributions:
We already know the intrinsic magnetic moments of the proton and neutron, so these
must correspond to the spin contributions to the magnetic moment operator:
 p   2.79  N  g s, p S  N 
g s, p   5.58
 n   1.91  N  g s,n S  N 
g s,n   3.83
For the orbital part, there is a contribution from the proton only, corresponding
to a circulating current loop (semiclassical sketch, but the result is correct)


r

  I  r ˆ  g   N ,

I
p
2
For the deuteron, we want to use the magnetic moment operator:

 
g
s, p




S p  g s,n S n  L p  N
g  1
11
Details:
1.

Lp 
because mn  mp
2.
L = 0 in the “S-state” (3S1 ) but we will consider also a contribution from
the “D-state” (3D1) as an exercise
3.
The proton and neutron couple to S = 1, and the deuteron has J = 1
 
 
J 
1
2
Trick: use


L
1
2
 

1
2
1
2
g
s, p


S p  g s,n S n 
1
2

L



J
N



S n  S p  S and write the operator as:
( g s, p

 g s,n ) S

1
2


( g s, p  g s,n ) (S p  S n ) 
But the proton and neutron spins are aligned, and
so the second term has to give zero!
1
2
 
 
S p  J  Sn  J


L N
12
continued....
So, effectively we can write for the deuteron:
 



( g s, p  g s,n ) S  L
1
4


J
N
Trick for expectation values:



J  L  S;
J
2
J 2  J ( J  1)
 
 
(L  S )  (L  S ) 

 
 
 
S J  S L  S S 
 
 
 
LJ  LL  LS
 ( 3S1 ) 
 ( 3D1 ) 
1
2

1
2
1
2
L S
2
2
 
 2L  S
J 2  L2  S 2  S 2

1
2
J 2  L2  S 2
J 2  L2  S 2
( g s , p  g s , n )  N   p  n  0.880  N
1
4
3

 ( g s , p  g s , n )  N  0.310  N
13
Comparison to experiment:
 d  0.857  N
 ( 3S1 )  0.880  N
This is intermediate between the
S-state and D-state values:
 ( 3D1 )  0.310  N
Suppose the wave function of the deuteron is a linear combination of S and D states:
d
 a
3
S1  b
3
D1
with
a2  b2  1
Then we can adjust the coefficients to explain the magnetic moment:
 d  (1  b 2 )  (3S1 )  b 2  (3D1 )
b2 = 0.04, or a 4% D-state admixture accounts for the magnetic moment !