PHY431 Lecture 4
2
2/27/2000
6.1. The Deuteron
The deuteron is the only bound nucleon-nucleon state: the bound neutron-proton 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 neutron-proton 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):
11 11
1
1
1
1
1
1,1
,
,
,
1, 0
,
,
,
,
,
1, 1
,
,
22 22
2
2
2
2
2
11111
1111
0,0
,
,
,
,
(Singlet)
2
ab
a
ba
b
=+
+
=
+
−
+
−
+
−
=−
−
−
−
+
−
(6.1)
Because the Pauli principle does not allow two identical nucleons to coexist, the nucleons in a neutron-
neutron or proton-proton
L
=0,1,2,.
.. state should have anti-parallel spins and
S
=0. Experimentally, no
nn or pp bound states are observed, and thus an anti-parallel spin state (
S
=0) for neutron-proton state
should also not bind. We are led to conclude that the deuteron consists of a
parallel-spin
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
=
p
+
n
= (2.7925
µ
N
) + (-1.9128
µ
N
) = 0.8797
µ
N
, close
to the experimental value:
d
expt
=0.85742
µ
N
(the Nuclear magneton
µ
N
≡
e
ħ
/2
m
p
).
The existence of a non-zero 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:
22
2
2
2
(3
)
( )
cos
(3cos
1) ( )
Q
dV
z
r e
e r drd
d
r
ρ
θϕ
θ
=−=
−
∫∫
rr
(6.2)
which vanishes if
(
r
)=
(
r
), i.e. spherically symmetric as in a
L
=0 state,
because
2
(3
1)
0
xd
x
−=
∫
. The
existence of an D-state 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