3Skyrmions
III Classical and Quantum Solitons
3.6 Rigid body quantization
We now make the rigid body approximation. We allow the Skyrmion to translate,
rotate and isorotate rigidly, but do not allow it to deform. This is a low-
energy approximation, and we have reduced the infinite dimensional space of
configurations to a finite-dimensional space. The group acting is
(translation) × (rotation) × (isorotation).
The translation part is trivial. The Skyrmion just gets the ability to move, and
thus gains momentum. So we are going to ignore it. The remaining group acting
is
SO
(3)
× SO
(3). This is a bit subtle. We have
π
1
(
SO
(3)) =
Z
2
, so we would
expect
π
1
(
SO
(3)
× SO
(3)) = (
Z
2
)
2
. However, in the full theory, we only have a
single
Z
2
. So we need to identify a loop in the first
SO
(3) with a loop in the
second SO(3).
Our wavefunction is thus a function on some cover of
SO
(3)
× SO
(3). While
SO
(3)
×SO
(3) is the symmetry group of the full theory, for a particular classical
Skyrmion solution, the orbit is smaller than the whole group. This is because
the Skyrmion often is invariant under some subgroup of
SO
(3)
× SO
(3). For
example, the
B
= 3 solution has a tetrahedral symmetry. Then we require our
wavefunction to be invariant under this group up to a sign.
For a single
SO
(3), we have a rigid-body wavefunction
|J L
3
J
3
i
, where
J
is the spin,
L
3
is the third component of spin relative to body axes, and
J
3
is
the third component of spin relative to space axes. Since we have two copies of
SO(3), our wavefunction can be represented by
|J L
3
J
3
i ⊗ |I K
3
I
3
i.
I
3
is the isospin we are physically familiar with. The values of
J
3
and
I
3
are not
constrained, i.e. they take all the standard (2
J
+ 1)(2
I
+ 1) values. Thus, we are
going to suppress these labels.
However, the symmetry of the Skyrmion places constraints on the body
projections, and not all values of J and I are allowed.
Example.
For the
B
= 1 hedgehog, there is a lot of symmetry. Given an axis
ˆ
n
and an angle
α
, we know that classically, if we rotate and isorotate by the same
axis and angle, then the wavefunction is unchanged. So we must have
e
iα
ˆ
n·L
e
iα
ˆ
n·K
|Ψi = |Ψi.
It follows, by considering α small, and all
ˆ
n, that
(L + K) |Ψi = 0.
So the “grand spin”
L
+
K
must vanish. So
L·L
=
K·K
. Recall that
L·L
=
J·J
and K · K = I · I. So it follows that we must have J = I.
Since 1 is an odd number, for any axis
ˆ
n, we must also have
e
i2π
ˆ
n·L
|Ψi = −|Ψi.
So I and J must be half-integer.
Thus, the allowed states are
J = I =
n
2
for some n ∈ 1 + 2Z. If we work out the formula for the energy in terms of the
spin, we see that it increases with
J
. It turns out the system is highly unstable
if n ≥ 5, and so
1
2
and
3
2
are the only physically allowed values.
The
J
=
I
=
1
2
states corresponds to
p
and
n
, with spin
1
2
. The
J
=
I
=
3
2
correspond to the ∆
++
, ∆
+
, ∆
0
and ∆
−
baryons, with spin
3
2
.
Example.
If
B
= 2, then we have a toroidal symmetry. This still has one
continuous SO(2) symmetry. Our first constraint becomes
e
iαL
3
e
i2αJ
3
|Ψi = |Ψi.
Note that we have 2
α
instead of
α
. If we look at our previous picture, we see
that when we rotate the space around by 2π, the pion field rotates by 4π.
There is another discrete symmetry, given by turning the torus upside down.
This gives
e
iπL
1
e
iπK
1
|Ψi = −|Ψi.
The sign is not obvious from what we’ve said so far, but it is correct. Since we
have an even baryon number, the allowed states have integer spin and isospin.
States with isospin 0 are most interesting, and have lowest energy. Then the
K operators act trivially, and we have
e
iαL
3
|Ψi = |Ψi, e
iπL
1
|Ψi = −|Ψi.
We have reduced the problem to one involving only body-fixed spin projection,
because the first equation tells us
L
3
|Ψi = 0.
Thus the allowed states are |J, L
3
= 0i.
The second constraint requires
J
to be odd. So the lowest energy states with
zero isospin are
|1, 0i
and
|3, 0i
. In particular, there are no spin 0 states. The
state
|1, 0i
represents the deuteron. This is a spin 1, isospin 0 bound state of
p
and n. This is a success.
The
|3, 0i
states have too high energy to be stable, but there is some evidence
for a spin 3 dibaryon resonance that decays into two ∆’s.
There is also a 2-nucleon resonance with
I
= 1 and
J
= 0, but this is not a
bound state. This is also allowed by the Skyrme model.
Example.
For
B
= 4, we have cubic symmetry. The symmetry group is
rather complicated, and imposes various constraints on our theory. But these
constraints always involve + signs on the right-hand side. The lowest allowed
state is |0, 0i ⊗ |0, 0i, which agrees with the α-particle. The next I = 0 state is
|4, 4i +
r
14
5
|4, 0i + |4, −4i
!
⊗ |0, 0i
involving a combination of
L
3
values. This is an excited state with spin 4, and
would have rather high energy. Unfortunately, we haven’t seen this experimentally.
It is, however, a success of the Skyrme model that there are no
J
= 1
,
2
,
3 states
with isospin 0.