PHY431 Lecture 4
1
2/27/2000
That pion exchange between nucleons is able to describe only the longer distance behavior of the
force, and not the shortdistance repulsion, nor the transition region to attraction, is highlighted in the
article by Philip Anderson:
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PHY431 Lecture 4
2
2/27/2000
6.1. The Deuteron
The deuteron is the only bound nucleonnucleon state: the bound neutronproton state. Its binding en
ergy is 2.23 MeV, and it has a total angular momentum
j
=1, and the deuteron parity is even, compactly
denoted as
J
P
=1
+
.
The deuteron consists of a neutron and proton circling their common center of mass. The spatial vari
ables describing the system are thus the distance of separation, and the polar and azimuthal angles that
this distance vector makes with the coordinate axes. The neutronproton system therefore has an orbital
angular momentum (with respect to the center of mass), characterized by the orbital quantum number
l
.
In addition, the nucleons each have spin
s
=
√
[
s
(
s
+1)]
ħ
. From the rules of addition of spins, the total
spin
S
of a system of two nucleons
a
and
b
has a quantum number
S
either
S
=0 (a singlet state), or
S
=1
(a triplet of states):
1
1
1
1
1
1
1
1
1
1
1
1
1
1,1
,
,
,
1,0
,
,
,
,
,
1, 1
,
,
2
2
2
2
2
2
2
2
2
2
2
2
2
1
1
1
1
1
1
1
1
1
0,0
,
,
,
,
(Singlet)
2
2
2
2
2
2
2
2
2
a
b
a
b
a
b
a
b
a
b
a
b
=
+
+
=
+
−
+
−
+
−
=
−
−
=
+
−
−
+
−
(6.1)
Because the Pauli principle does not allow two identical nucleons to coexist, the nucleons in a neutron
neutron or protonproton
L
=0,1,2,... state should have antiparallel spins and
S
=0. Experimentally, no
nn or pp bound states are observed, and thus an antiparallel spin state (
S
=0) for neutronproton state
should also not bind. We are led to conclude that the deuteron consists of a
parallelspin
S
=1 neutron
proton system.
The total angular momentum
J
of the deuteron is the vector sum of the orbital angular momentum
L
and the total spin
S
:
J
=
L+S
. Thus, with
J
=1 and
S
=1, the possible values for
L
are
L
=0,1,2: denoted by
3
S
1
,
3
P
1
,
3
D
1
respectively, in the usual spectroscopic notation:
2
S
+1
L
J
. Because the deuteron parity is
even, the orbital angular momentum quantum number also must be even:
P
=(1)
L
, thus
L
=0, 2; with
notation
3
S
1
,
3
D
1
respectively. The predicted magnetic moment of the deuteron in case
L
=0 comes
purely from the nucleons’ moments:
µ
d
predicted
=
µ
p
+
µ
n
= (2.7925
µ
N
) + (1.9128
µ
N
) = 0.8797
µ
N
, close
to the experimental value:
µ
d
expt
=0.85742
µ
N
(the Nuclear magneton
µ
N
≡
e
ħ
/2
m
p
).
The existence of a nonzero quadrupole moment (0.286 fm
2
) indicates that the
L
=2 state also contrib
utes a bit: the deuteron ground state is an admixture of
3
S
1
(96%) and
3
D
1
(4%). The classical quadru
pole moment is defined as:
2
2
2
2
2
(3
)
( )
cos
(3cos
1)
( )
Q
dV
z
r
e
e r drd
d
r
ρ
θ
ϕ
θ
ρ
=
−
=
−
∫
∫
r
r
(6.2)
which vanishes if
ρ
(
r
)=
ρ
(
r
), i.e. spherically symmetric as in a
L
=0 state,
because
2
(3
1)
0
x
dx
−
=
∫
. The
existence of an Dstate admixture then implies that the strong nucleon potential cannot be completely
spherically symmetric and is not a pure central force: a “tensor interaction” must be present! Such an
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 Spring '01
 Rijssenbeek
 Physics, Radioactive Decay, Angular Momentum, Force, Neutron, Light, Nuclear physics, Atomic nucleus

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