III Classical and Quantum Solitons

3Skyrmions

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)

This might look like a rather odd symmetry group to work with. We can

begin by understanding the

(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

(2). It is convenient to

introduce coordinates for our Skyrme field. As usual, we let

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

and

cannot be arbitrary. For

to actually lie in

SU(2), we need the coefficients to satisfy

σ, π

∈ R, σ

+ π ·π = 1.

This is a non-linear constraint, and defines what known as a non-linear

-model.

From this constraint, we see that geometrically, we can identify

(2) with

. We can also see this directly, by writing

SU(2) =



α β

−

β ¯α



∈ M

+ |β|

= 1



and this gives a natural embedding of

(2) into

∼

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

. Thus, we obtain an inclusion

SU(2) × SU(2)

→ SO(4),

which happens to be a surjection.

Our theory will undergo spontaneous symmetry breaking, and the canonical

choice of vacuum will be

. Equivalently, this is when

= 1 and

We see that the stabilizer of 1 is given by the diagonal

∆ : SU(2) →

SU(2) × SU(2)

since A1B

−1

= 1 if and only if A = B.

Geometrically, if we view

SU(2)×SU(2)

∼

(4), then it is obvious that the

stabilizer of

is the copy of

(3)

⊆ SO

(4) that fixes the

axis. Indeed, the

image of the diagonal ∆ is SU(2)/{±1}

∼

SO(3).

Note that in our theory, for any choice of

, there are at most two possible

choices of

. Thus, despite there being four variables, there are only three degrees

of freedom. Geometrically, this is saying that

(2)

∼

is a three-dimensional

manifold.

This has some physical significance. We are using the

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 “

meson”, and this can be explained by the fact that

isn’t a genuine degree of freedom in our theory.

Let’s now try to build a Lagrangian for our field

. We will want to introduce

derivative terms. From a mathematical point of view, the quantity

∂

isn’t

a very nice thing to work with. It is a quantity that lives in the tangent space

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

(2) =

(2). To do so, we

multiply by U

−1

. We write

= (∂

U)U

−1

which is known as the right current . For practical, computational purposes, it is

convenient to note that

−1

= σ1 − iπ · τ .

Using the (+

, −, −, −

) metric signature, we can now write the Skyrme Lagrangian

density as

L = −

Tr (R

) +

32e

Tr ([R

, R

][R

, R

]) −

Tr(1 − U).

The three terms are referred to as the

-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

. 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

(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

. After this

spontaneous symmetry breaking, we are left with an isospin

(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

and is minimized

when

= 1. Note that this theory is still invariant under the isospin

(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

± iπ

, π

= π

It is convenient to draw the target space SU(2) as

σ = 1

σ = −1

Potential

energy

σ = 0

|π| = 1

∼

SU(2)

−

1 is the anti-vacuum. We will see that in the core of the Skyrmion,

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

and length unit

Setting these to 1, there is one parameter left, which is the dimensionless pion

mass. In these units, we have

L =



−

Tr(R

) +

Tr([R

, R

][R

, R

]) − m

Tr(1 − U)



In this notation, we have

m =

In general, we will think of m as being “small”.

Let’s see what happens if we in fact put

= 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

while staying inside the group

(2). Thus, we vary

multiplicatively,

U 7→ U (1 + εV )

for some

V ∈ su

(2). We then have to figure out how

varies, do some

differentiation, and then the Euler–Lagrange equations turn out to be

∂



, [R

, R

]]



= 0.

For static fields, the energy is given by

E =



−

Tr(R

) −

Tr([R

, R

][R

, R

])



x ≡ E

+ E

where we sum

and

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

(2) is negative definite. So the energy is

actually positive.

We can again run Derrick’s theorem.

Theorem

(Derrick’s theorem)

We have

for any finite-energy static

solution for m = 0 Skyrmions.

Proof.

Suppose

(

) minimizes

. 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

∂

U(x) = λ

U(λx).

Therefore we find

(x) = λR

(λx),

and therefore

= −

Tr(

) d

= −

Tr(R

)(λx) d

= −

Tr(R

)(λx) d

(λx)

Similarly,

= λE

So we find that

E =

+ λE

But this function has to have a minimum at

= 1. So the derivative with

respect to λ must vanish at 1, requiring

0 =

dλ

= −

+ E

= 0

at λ = 1. Thus we must have E

= E

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

→ SU

(2). Since we have a boundary

condition

at infinity, we can imagine compactifying

into

, where

the point at infinity is sent to 1.

On the other hand, we know that

(2) is isomorphic to

. Geometrically, we

think of the space and

(2) as “different”

. We should think of

(2) as the

“unit sphere”, and write it as

. However, we can think of the compactification

as a sphere with “infinite radius”, so we denote it as

∞

. Of course,

topologically, they are the same.

So the field is a map

U : S

∞

→ S

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

)

∼

Z. Equivalently, we can think of it as the map induced

on the homology or cohomology of S

While

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 = −

24π

ijk

Tr(R

) d

The factor of 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

with the

conserved, physical baryon number.

If we were to derive this, then we would have to pull back a normalized

volume form from

and then integrate over all space. In this formula, we chose

to use the

(4)-invariant volume form, but in general, we can pull back any

normalized volume form.

Locally, near σ = 1, this volume form is actually

2π

dπ

∧ dπ

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 =

−



∓

ijk

, R

]



x ± 12π

This bound is true for both sign choices. However, to get the strongest result,

we should choose the sign such that ±12π

B > 0. Then we find

E ≥ 12π

|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

−

ijk

, R

] = 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

is non-zero, the three vectors

, R

must form an

orthonormal frame in

(2). So

must be an isometry. But this isn’t possible,

because the spheres have “different radii”.

Therefore, true Skyrmions with B > 0 satisfy

E > 12π

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.