3Skyrmions

III Classical and Quantum Solitons



3.1 Skyrme field and its topology
Before we begin talking about the Skyrme field, we first discuss the symmetry
group this theory enjoys. Before symmetry breaking, our theory has a symmetry
group
SU(2) × SU(2)
(1, 1)}
=
SU(2) × SU(2)
Z
2
.
This might look like a rather odd symmetry group to work with. We can
begin by understanding the
SU
(2)
× SU
(2) part of the symmetry group. This
group acts naturally on SU(2) again, by
(A, B) · U = AUB
1
.
However, we notice that the pair (
1, 1
)
SU
(2)
×SU
(2) acts trivially. So the
true symmetry group is the quotient by the subgroup generated by this element.
One can check that after this quotienting, the action is faithful.
In the Skyrme model, the field will be valued in
SU
(2). It is convenient to
introduce coordinates for our Skyrme field. As usual, we let
τ
be the Pauli
matrices, and 1 be the unit matrix. Then we can write the Skyrme field as
U(x, t) = σ(x, t)1 + iπ(x, t) · τ .
However, the values of
σ
and
π
cannot be arbitrary. For
U
to actually lie in
SU(2), we need the coefficients to satisfy
σ, π
i
R, σ
2
+ π ·π = 1.
This is a non-linear constraint, and defines what known as a non-linear
σ
-model.
From this constraint, we see that geometrically, we can identify
SU
(2) with
S
3
. We can also see this directly, by writing
SU(2) =

α β
¯
β ¯α
M
2
(C) : |α|
2
+ |β|
2
= 1
,
and this gives a natural embedding of
SU
(2) into
C
2
=
R
4
as the unit sphere,
by sending the matrix to (α, β).
One can check that the action we wrote down acts by isometries of the
induced metric on S
3
. Thus, we obtain an inclusion
SU(2) × SU(2)
Z
2
SO(4),
which happens to be a surjection.
Our theory will undergo spontaneous symmetry breaking, and the canonical
choice of vacuum will be
U
=
1
. Equivalently, this is when
σ
= 1 and
π
=
0
.
We see that the stabilizer of 1 is given by the diagonal
∆ : SU(2)
SU(2) × SU(2)
Z
2
,
since A1B
1
= 1 if and only if A = B.
Geometrically, if we view
SU(2)×SU(2)
Z
2
=
SO
(4), then it is obvious that the
stabilizer of
1
is the copy of
SO
(3)
SO
(4) that fixes the
1
axis. Indeed, the
image of the diagonal is SU(2)/1}
=
SO(3).
Note that in our theory, for any choice of
π
, there are at most two possible
choices of
σ
. Thus, despite there being four variables, there are only three degrees
of freedom. Geometrically, this is saying that
SU
(2)
=
S
3
is a three-dimensional
manifold.
This has some physical significance. We are using the
π
fields to model pions.
We have seen and observed pions a lot. We know they exist. However, as far as
we can tell, there is no
σ
meson”, and this can be explained by the fact that
σ
isn’t a genuine degree of freedom in our theory.
Let’s now try to build a Lagrangian for our field
U
. We will want to introduce
derivative terms. From a mathematical point of view, the quantity
µ
U
isn’t
a very nice thing to work with. It is a quantity that lives in the tangent space
T
U
SU(2), and in particular, the space it lives in depends on the value of U .
What we want to do is to pull this back to
T
1
SU
(2) =
su
(2). To do so, we
multiply by U
1
. We write
R
µ
= (
µ
U)U
1
,
which is known as the right current . For practical, computational purposes, it is
convenient to note that
U
1
= σ1 iπ · τ .
Using the (+
, , ,
) metric signature, we can now write the Skyrme Lagrangian
density as
L =
F
2
π
16
Tr (R
µ
R
µ
) +
1
32e
2
Tr ([R
µ
, R
ν
][R
µ
, R
ν
])
1
8
F
2
π
m
2
π
Tr(1 U).
The three terms are referred to as the
σ
-model term, Skyrme term and pion
mass term respectively.
The first term is the obvious kinetic term. The second term might seem a
bit mysterious, but we must have it (or some variant of it). By Derrick’s scaling
argument, we cannot have solitons if we just have the first term. We need a
higher multiple of the derivative term to make solitons feasible.
There are really only two possible terms with four derivatives. The alternative
is to have the square of the first term. However, Skyrme rejected that object,
because that Lagrangian will have four time derivatives. From a classical point
of view, this is nasty, because to specify the initial configuration, not only do we
need the initial field condition, but also its first three derivatives. This doesn’t
happen in our theory, because the commutator vanishes when
µ
=
ν
. The pieces
of the Skyrme term are thus at most quadratic in time derivatives.
Now note that the first two terms have an exact chiral symmetry, i.e. they
are invariant under the
SO
(4) action previously described. In the absence of
the final term, this symmetry would be spontaneously broken by a choice of
vacuum. As described before, there is a conventional choice
U
=
1
. After this
spontaneous symmetry breaking, we are left with an isospin
SU
(2) symmetry.
This isospin symmetry rotates the π fields among themselves.
The role of the extra term is that now the vacuum has to be the identity
matrix. The symmetry is now explicitly broken. This might not be immediately
obvious, but this is because the pion mass term is linear in
σ
and is minimized
when
σ
= 1. Note that this theory is still invariant under the isospin
SU
(2)
symmetry. Since the isospin symmetry is not broken, all pions have the same
mass. In reality, the pion masses are very close, but not exactly equal, and we
can attribute the difference in mass as due to electromagnetic effects. In terms
of the π fields we defined, the physical pions are given by
π
±
= π
1
±
2
, π
0
= π
3
.
It is convenient to draw the target space SU(2) as
σ = 1
σ = 1
Potential
energy
σ = 0
|π| = 1
S
3
=
SU(2)
σ
=
1 is the anti-vacuum. We will see that in the core of the Skyrmion,
σ
will
take value σ = 1.
In the Skyrme Lagrangian, we have three free parameters. This is rather few
for an effective field theory, but we can reduce the number further by picking
appropriate coefficients. We introduce an energy unit
F
π
4e
and length unit
2
eF
π
.
Setting these to 1, there is one parameter left, which is the dimensionless pion
mass. In these units, we have
L =
Z
1
2
Tr(R
µ
R
µ
) +
1
16
Tr([R
µ
, R
ν
][R
µ
, R
ν
]) m
2
Tr(1 U)
d
3
x.
In this notation, we have
m =
2m
π
eF
π
.
In general, we will think of m as being “small”.
Let’s see what happens if we in fact put
m
= 0. In this case, the lack of mass
term means we no longer have the boundary condition that
U
1 at infinity.
Hence, we need to manually impose this condition.
Deriving the Euler–Lagrange equations is slightly messy, since we have to
vary
U
while staying inside the group
SU
(2). Thus, we vary
U
multiplicatively,
U 7→ U (1 + εV )
for some
V su
(2). We then have to figure out how
R
varies, do some
differentiation, and then the Euler–Lagrange equations turn out to be
µ
R
µ
+
1
4
[R
ν
, [R
ν
, R
µ
]]
= 0.
For static fields, the energy is given by
E =
Z
1
2
Tr(R
i
R
i
)
1
16
Tr([R
i
, R
j
][R
i
, R
j
])
d
3
x E
2
+ E
4
.
where we sum
i
and
j
from 1 to 3. This is a sum of two terms the first is
quadratic in derivatives while the second is quartic.
Note that the trace functional on
su
(2) is negative definite. So the energy is
actually positive.
We can again run Derrick’s theorem.
Theorem
(Derrick’s theorem)
.
We have
E
2
=
E
4
for any finite-energy static
solution for m = 0 Skyrmions.
Proof.
Suppose
U
(
x
) minimizes
E
=
E
2
+
E
4
. We rescale this solution, and
define
˜
U(x) = U(λx).
Since U is a solution, the energy is stationary with respect to λ at λ = 1.
We can take the derivative of this and obtain
i
˜
U(x) = λ
˜
U(λx).
Therefore we find
˜
R
i
(x) = λR
i
(λx),
and therefore
˜
E
2
=
1
2
Z
Tr(
˜
R
i
˜
R
i
) d
3
x
=
1
2
λ
2
Z
Tr(R
i
R
i
)(λx) d
3
x
=
1
2
1
λ
Z
Tr(R
i
R
i
)(λx) d
3
(λx)
=
1
λ
E
2
.
Similarly,
˜
E
4
= λE
4
.
So we find that
˜
E =
1
λ
E
2
+ λE
4
.
But this function has to have a minimum at
λ
= 1. So the derivative with
respect to λ must vanish at 1, requiring
0 =
d
˜
E
dλ
=
1
λ
2
E
2
+ E
4
= 0
at λ = 1. Thus we must have E
4
= E
2
.
We see that we must have a four-derivative term in order to stabilize the
soliton. If we have the mass term as well, then the argument is slightly more
complicated, and we get some more complicated relation between the energies.
Baryon number
Recall that our field is a function
U
:
R
3
SU
(2). Since we have a boundary
condition
U
=
1
at infinity, we can imagine compactifying
R
3
into
S
3
, where
the point at infinity is sent to 1.
On the other hand, we know that
SU
(2) is isomorphic to
S
3
. Geometrically, we
think of the space and
SU
(2) as “different”
S
3
. We should think of
SU
(2) as the
“unit sphere”, and write it as
S
3
1
. However, we can think of the compactification
of
R
3
as a sphere with “infinite radius”, so we denote it as
S
3
. Of course,
topologically, they are the same.
So the field is a map
U : S
3
S
3
1
.
This map has a degree. There are many ways we can think about the degree.
For example, we can think of this as the homotopy class of this map, which is
an element of π
3
(S
3
)
=
Z. Equivalently, we can think of it as the map induced
on the homology or cohomology of S
3
.
While
U
evolves with time, because the degree is a discrete quantity, it has
to be independent of time (alternatively, the degree of the map is homotopy
invariant).
There is an explicit integral formula for the degree. We will not derive this,
but it is
B =
1
24π
2
Z
ε
ijk
Tr(R
i
R
j
R
k
) d
3
x.
The factor of 2
π
2
comes from the volume of the three sphere, and there is also
a factor of 6 coming from how we anti-symmetrize. We identify
B
with the
conserved, physical baryon number.
If we were to derive this, then we would have to pull back a normalized
volume form from
S
3
1
and then integrate over all space. In this formula, we chose
to use the
SO
(4)-invariant volume form, but in general, we can pull back any
normalized volume form.
Locally, near σ = 1, this volume form is actually
1
2π
2
dπ
1
dπ
2
dπ
3
.
Faddeev–Bogomolny bound
There is a nice result analogous to the Bogomolny energy bound we saw for
kinks and vortices, known as the Faddeev–Bogomolny bound . We can write the
static energy as
E =
Z
1
2
Tr
R
i
1
4
ε
ijk
[R
j
, R
k
]
2
!
d
3
x ± 12π
2
B.
This bound is true for both sign choices. However, to get the strongest result,
we should choose the sign such that ±12π
2
B > 0. Then we find
E 12π
2
|B|.
By symmetry, it suffices to consider the case B > 0, which is what we will do.
The Bogomolny equation for B > 0 should be
R
i
1
4
ε
ijk
[R
j
, R
k
] = 0.
However, it turns out this equation has no non-vacuum solution.
Roughly, the argument goes as follows by careful inspection, for this to
vanish, whenever
R
i
is non-zero, the three vectors
R
1
, R
2
, R
3
must form an
orthonormal frame in
su
(2). So
U
must be an isometry. But this isn’t possible,
because the spheres have “different radii”.
Therefore, true Skyrmions with B > 0 satisfy
E > 12π
2
B.
We get a lower bound, but the actual energy is always greater than this lower
bound. It is quite interesting to look at the energies of true solutions numerically,
and their energy is indeed bigger.