Transcript Slide 1

16.451 Lecture 19:
The deuteron
d
Basic properties:
1
13/11/2003
2
1H
mass: mc2 = 1876.124 MeV
binding energy:
B 
m  M
i
i
RMS radius:
(measured via -ray energy in n + p  d + )
1.963  0.004 fm
quantum numbers:
 m p  mn  md  2.2245731 MeV
(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  2 mi vi 2 
1
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.
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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 

sin2   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!
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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.....
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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 
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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
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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



Sn  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 a 2  b 2  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 !