3Skyrmions
III Classical and Quantum Solitons
3.2 Skyrmion solutions
The simplest Skyrmion solution has baryon number
B
= 1. We will continue to
set m = 0.
B = 1 hedgehog Skyrmion
Consider the spherically symmetric function
U(x) = cos f(r)1 + i sin f(r)
ˆ
x · τ .
This is manifestly in
SU
(2), because
cos
2
f
+
sin
2
f
= 1. This is known as a
hedgehog, because the unit pion field is
ˆ
x
, which points radially outwards. We
need some boundary conditions. We need
U → 1
at
∞
. On the other hand, we
will see that we need U → −1 at the origin to get baryon number 1. So f → π
as r → 0, and f → 0 as r → ∞. So it looks roughly like this:
r
f
π
After some hard work, we find that the energy is given by
E = 4π
Z
∞
0
f
02
+
2 sin
2
f
r
2
(1 + f
02
) +
sin
4
f
r
4
r
2
dr.
From this, we can obtain a second-order ordinary differential equation in
f
,
which is not simple. Solutions have to be found numerically. This is a sad truth
about Skyrmions. Even in the simplest
B
= 1
, m
= 0 case, we don’t have an
analytic expression for what f looks like. Numerically, the energy is given by
E = 1.232 × 12π
2
.
To compute the baryon number of this solution, we plug our solution into the
formula, and obtain
B = −
1
2π
2
Z
∞
0
sin
2
f
r
2
df
dr
· 4πr
2
dr.
We can interpret
df
dr
as the radial contribution to
B
, while there are two factors
of
sin f
r
coming from the angular contribution due to the i sin f(r)
ˆ
x · τ term.
But this integral is just an exact differential. It simplifies to
B =
1
π
Z
π
0
2 sin
2
f df.
Note that we have lost a sign, because of the change of limits. We can integrate
this directly, and just get
B =
1
π
f −
1
2
sin 2f
π
0
= 1,
as promised.
Intuitively, we see that in this integral,
f
goes from 0 to
π
, and we can think
of this as the field U wrapping around the sphere S
3
once.
More hedgehogs
We can consider a more general hedgehog with the same ansatz, but with the
boundary conditions
f(0) = nπ, f(∞) = 0.
In this case, the same computations gives us
B
=
n
. So in principle, this gives a
Skyrmion of any baryon number. The solutions do exist. However, it turns out
they have extremely high energy, and are nowhere near minimizing the energy.
In fact, the energy increases much faster than
n
itself, because the Skyrmion
“onion” structure highly distorts each
B
= 1 Skyrmion. Unsurprisingly, these
solutions are unstable.
This is not what we want in hadronic physics, where we expect the energy to
scale approximately linearly with
n
. In fact, since baryons attract, we expect
the solution for B = n to have less energy than n times the B = 1 energy.
We can easily get energies approximately
n
times the
B
= 1 energy simply
by having very separated baryons, and then since they attract, when they move
towards each other, we get even lower energies.
A better strategy — rational map approximation
So far, we have been looking at solutions that depend very simply on angle. This
means, all the “winding” happens in the radial direction. In fact, it is a better
idea to wind more in the angular direction.
In the case of the B = 1 hedgehog, the field looks roughly like this:
∞
σ = 1
σ = −1
In our
B >
1 spherically symmetric hedgehogs, we wrapped radially around
the sphere many times, and it turns out that was not a good idea.
Better is to use a similar radial configuration as the
B
= 1 hedgehog, but
introduce more angular twists. We can think of the above
B
= 1 solution as
follows — we slice up our domain
R
3
(or rather,
S
3
since we include the point
at infinity) into 2-spheres of constant radius, say
S
3
=
[
r∈[0,∞]
S
2
r
.
On the other hand, we can also slice up the
S
3
in the codomain into constant
σ
levels, which are also 2-spheres:
S
3
=
[
σ
S
2
σ
.
Then the function
f
(
r
) we had tells us we should map the 2-sphere
S
2
r
into the
two sphere
S
2
cos f(r)
. What we did, essentially, was that we chose to map
S
2
r
to
S
2
cos f(r)
via the “identity map”. This gave a spherically symmetric hedgehog
solution.
But we don’t have to do this! Pick any function
R
:
S
2
→ S
2
. Then we
can construct the map Σ
cos f
R
that sends
S
2
r
to
S
2
cos f(r)
via the map
R
. For
simplicity, we will use the same
R
for all
r
. If we did this, then we obtain a
non-trivial map Σ
cos f
R : S
3
→ S
3
.
Since
R
itself is a map from a sphere to a sphere (but one dimension lower),
R
also has a degree. It turns out this degree is the same as the degree of the
induced map Σ
cos f
R
! So to produce higher baryon number hedgehogs, we simply
have to find maps R : S
2
→ S
2
of higher degree.
Fortunately, this is easier than maps between 3-spheres, because a 2-sphere is
a Riemann surface. We can then use complex coordinates to work on 2-spheres.
By complex analysis, any complex holomorphic map between 2-spheres is given
by a rational map.
Pick any rational function
R
k
(
z
) of degree
k
. This is a map
S
2
→ S
2
. We
use coordinates
r, z
, where
r ∈ R
+
and
z ∈ C
∞
=
C ∪ {∞}
∼
=
S
2
. Then we can
look at generalized hedgehogs
U(x) = cos f(r)1 + i sin f(r)
ˆ
n
R
k
(z)
· τ ,
with
f
(0) =
π
,
f
(
∞
) = 0, and
ˆ
n
R
k
is the normalized pion field
ˆ
π
, given by the
unit vector obtained from
R
k
(
z
) if we view
S
2
as a subset of
R
3
in the usual
way. Explicitly,
ˆ
n
R
=
1
1 + |R|
2
(
¯
R + R, i(
¯
R − R), 1 − |R|
2
).
This construction is in some sense a separation of variables, where we separate
the radial and angular dependence of the field.
Note that even if we find a minimum among this class of fields, it is not a
true minimum energy Skyrmion. However, it gets quite close, and is much better
than our previous attempt.
There is quite a lot of freedom in this construction, since we are free to pick
f
(
r
), as well as the rational function
R
k
(
z
). The geometric degree
k
of
R
k
(
z
) we
care about is the same as the algebraic degree of R
k
(z). Precisely, if we write
R
k
(z) =
p
k
(z)
q
k
(z)
,
where
p
k
and
q
k
are coprime, then the algebraic degree of
R
k
is the maximum of
the degrees of
p
k
and
q
k
as polynomials. Since there are finitely many coefficients
for
p
k
and
q
k
, this is a finite-dimensional problem, which is much easier than
solving for arbitrary functions. We will talk more about the degree later.
Numerically, we find that minimal energy fields are obtained with
R
1
(z) = z R
2
(z) = z
2
R
3
(z) =
√
3iz
2
− 1
z
3
−
√
3iz
R
4
(z) =
z
4
+ 2
√
3iz
2
+ 1
z
4
− 2
√
3iz
2
+ 1
.
The true minimal energy Skyrmions have also been found numerically, and are
very similar to the optimal rational map fields. In fact, the search for the true
minima often starts from a rational map field.
Constant energy density surfaces of Skyrmions up to baryon number 8
(for m = 0), by R. A. Battye and P. M. Sutcliffe
We observe that
– for n = 1, we recover the hedgehog solution.
– for n = 2, our solution has an axial symmetry.
–
for
n
= 3, the solution might seem rather strange, but it is in fact the
unique solution in degree 3 with tetrahedral symmetry.
– for n = 4, the solution has cubic symmetry.
In each case, these are the unique rational maps with such high symmetry. It
turns out, even though these are not the exact Skyrmion solutions, the exact
solutions enjoy the same symmetries.
The function f can be found numerically, and depends on B.
Geometrically, what we are doing is that we are viewing
S
3
as the suspension
Σ
S
2
, and our construction of Σ
cos f
R
from
R
is just the suspension of maps.
The fact that degree is preserved by suspension can be viewed as an example of
the fact that homology is stable.
More on rational maps
Why does the algebraic degree of
R
k
agree with the geometric degree? One way
of characterizing the geometric degree is by counting pre-images. Consider a
generic point c in the target 2-sphere, and consider the equation
R
k
(z) =
p
k
(z)
q
k
(z)
= c
for a generic c. We can rearrange this to say
p
k
− cq
k
= 0.
For a generic
c
, the
z
k
terms do not cancel, so this is a polynomial equation of
degree
k
. Also, generically, this equation doesn’t have repeated roots, so has
exactly
k
solutions. So the number of points in the pre-image of a generic
c
is
k
.
Because
R
k
is a holomorphic map, it is automatically orientation preserving
(and in fact conformal). So each of these
k
points contributes +1 to the degree.
In the pictures above, we saw that the Skyrmions have some “hollow polyhe-
dral” structures. How can we understand these?
The holes turn out to be zeroes of the baryon density, and are where energy
density is small. At the center of the holes, the angular derivatives of the Skyrme
field U are zero, but the radial derivative is not.
We can find these holes precisely in the rational map approximation. This
allows us to find the symmetry of the system. They occur where the derivative
dR
k
dz
= 0. Since R =
p
q
, we can rewrite this requirement as
W (z) = p
0
(z)q(z) − q
0
(z)p(z) = 0.
W (z) is known as the Wronskian.
A quick algebraic manipulation shows that
W
has degree at most 2
k −
2,
and generically, it is indeed 2
k −
2. This degree is the number of holes in the
Skyrmion.
We can look at our examples and look at the pictures to see this is indeed
the case.
Example. For
R
4
(z) =
z
4
+ 2
√
3iz
2
+ 1
z
4
− 2
√
3iz
2
+ 1
,
the Wronskian is
W (z) = (4z
3
+ 4
√
3iz)(z
4
− 2
√
3iz
2
+ 1) − (z
4
+ 2
√
3iz
2
+ 1)(4z
3
− 4
√
3iz).
The highest degree
z
7
terms cancel. But there isn’t any
z
6
term anywhere either.
Thus W turns out to be a degree 5 polynomial. It is given by
W (z) = −8
√
3i(z
5
− z).
We can easily list the roots — they are z = 0, 1, i, −1, −i.
Generically, we expect there to be 6 roots. It turns out the Wronksian has
a zero at
∞
as well. To see this more rigorously, we can rotate the Riemann
sphere a bit by a M¨obius map, and then see there are 6 finite roots. Looking
back at our previous picture, there are indeed 6 holes when B = 4.
It turns out although the rational map approximation is a good way to find
solutions for
B
up to
≈
20, they are hollow with
U
=
−
1 at the center. This is
not a good model for larger nuclei, especially when we introduce non-zero pion
mass.
For m ≈ 1, there are better, less hollow Skyrmions when B ≥ 8.