III Classical and Quantum Solitons (Full)

Part III — Classical and Quantum Solitons

Based on lectures by N. S. Manton and D. Stuart

Notes taken by Dexter Chua

Easter 2017

Solitons are solutions of classical field equations with particle-like properties. They

are localised in space, have finite energy and are stable against decay into radiation.

The stability usually has a topological explanation. After quantisation, they give rise

to new particle states in the underlying quantum field theory that are not seen in

perturbation theory. We will focus mainly on kink solitons in one space dimension, on

gauge theory vortices in two dimensions, and on Skyrmions in three dimensions.

Pre-requisites

This course assumes you have taken Quantum Field Theory and Symmetries, Fields

and Particles. The small amount of topology that is needed will be developed during

the course.

Reference

N. Manton and P. Sutcliffe, Topological Solitons, CUP, 2004

Contents

0 Introduction

1 φ

kinks

1.1 Kink solutions

1.2 Dynamic kink

1.3 Soliton interactions

1.4 Quantization of kink motion

1.5 Sine-Gordon kinks

2 Vortices

2.1 Topological background

2.2 Global U (1) Ginzburg–Landau vortices

2.3 Abelian Higgs/Gauged Ginzburg–Landau vortices

2.4 Bogomolny/self-dual vortices and Taubes’ theorem

2.5 Physics of vortices

2.6 Jackiw–Pi vortices

3 Skyrmions

3.1 Skyrme field and its topology

3.2 Skyrmion solutions

3.3 Other Skyrmion structures

3.4 Asymptotic field and forces for B = 1 hedgehogs

3.5 Fermionic quantization of the B = 1 hedgehog

3.6 Rigid body quantization

0 Introduction

Given a classical field theory, if we want to “quantize” it, then we find the

vacuum of the theory, and then do perturbation theory around this vacuum. If

there are multiple vacua, then what we did was that we arbitrarily picked a

vacuum, and then expanded around that vacuum.

However, these field theories with multiple vacua often contain soliton so-

lutions. These are localized, smooth solutions of the classical field equations,

and they “connect multiple vacuums”. To quantize these solitons solutions, we

fix such a soliton, and use it as the “background”. We then do perturbation

theory around these solutions, but this is rather tricky to do. Thus, in a lot of

the course, we will just look at the classical part of the theory.

Recall that when quantizing our field theories in perturbation theory, we

obtain particles in the quantum theory, despite the classical theory being com-

pletely about fields. It turns out solitons also behave like particles, and they are

a new type of particles. These are non-perturbative phenomena. If we want to

do the quantum field theory properly, we have to include these solitons in the

quantum field theory. In general this is hard, and so we are not going to develop

this a lot.

What does it mean to say that solitons are like particles? In relativistic field

theories, we find these solitons have a classical energy. We define the “mass”

of the soliton to be the energy in the “rest frame”. Since this is relativistic, we

can do a Lorentz boost, and we obtain a moving soliton. Then we obtain an

energy-momentum relation of the form

− P · P = M

This is a Lorentz-invariant property of the soliton. Together with the fact that

the soliton is localized, this is a justification for thinking of them as particles.

These particles differ from the particles of perturbative quantum fields, as

they have rather different properties. Interesting solitons have a topological

character different from the classical vacuum. Thus, at least naively, they cannot

be thought of perturbatively.

There are also non-relativistic solitons, but they usually don’t have interpre-

tations as particles. These appear, for example, as defects in solids. We will not

be interested in these much.

What kinds of theories have solitons? To obtain solitons, we definitely need

a non-linear field structure and/or non-linear equations. Thus, free field theories

with quadratic Lagrangians such as Maxwell theory do not have solitons. We

need interaction terms.

Note that in QFT, we dealt with interactions using the interaction picture.

We split the Hamiltonian into a “free field” part, which we solve exactly, and

the “interaction” part. However, to quantize solitons, we need to solve the full

interacting field equations exactly.

Having interactions is not enough for solitons to appear. To obtain solitons,

we also need some non-trivial vacuum topology. In other words, we need more

than one vacuum. This usually comes from symmetry breaking, and often gauge

symmetries are involved.

In this course, we will focus on three types of solitons.

–

In one (space) dimension, we have kinks. We will spend 4 lectures on this.

– In two dimensions, we have vortices. We will spend 6 lectures on this.

–

In three dimensions, there are monopoles and Skyrmions. We will only

study Skyrmions, and will spend 6 lectures on these.

These examples are all relativistic. Non-relativistic solitons include domain walls,

which occur in ferromagnets, and two-dimensional “baby” Skyrmions, which are

seen in exotic magnets, but we will not study these.

In general, solitons appear in all sorts of different actual, physical scenarios

such as in condensed matter physics, optical fibers, superconductors and exotic

magnets. “Cosmic strings” have also been studied. Since we are mathematicians,

we probably will not put much focus on these actual applications. However, we

can talk a bit more about Skyrmions.

Skyrmions are solitons in an effective field theory of interacting pions, which

are thought to be the most important hadrons because they are the lightest.

This happens in spite of the lack of a gauge symmetry. While pions have no

baryon number, the associated solitons have a topological charge identified with

baryon number. This baryon number is conserved for topological reasons.

Note that in QCD, baryon number is conserved because the quark number

is conserved. Experiments tried extremely hard to find proton decay, which

would be a process that involves baryon number change, but we cannot find

such examples. We have very high experimental certainty that baryon number is

conserved. And if baryon number is topological, then this is a very good reason

for the conservation of baryon number.

Skyrmions give a model of low-energy interactions of baryons. This leads

to an (approximate) theory of nucleons (proton and neutron) and larger nuclei,

which are bound states of any number of protons and neutrons.

For these ideas to work out well, we need to eventually do quantization.

For example, Skyrmions by themselves do not have spin. We need to quantize

the theory before spins come out. Also, Skyrmions cannot distinguish between

protons and neutrons. These differences only come up after we quantize.

1 φ

kinks

1.1 Kink solutions

In this section, we are going to study

kinks. These occur in 1 + 1 dimensions,

and involve a single scalar field

(

x, t

). In higher dimensions, we often need

many fields to obtain solitons, but in the case of 1 dimension, we can get away

with a single field.

In general, the Lagrangian density of such a scalar field theory is of the form

L =

∂

φ∂

φ − U(φ)

for some potential

(

) polynomial in

. Note that in 1 + 1 dimensions, any

such theory is renormalizable. Here we will choose the Minkowski metric to be

µν



1 0

0 −1



with µ, ν = 0, 1. Then the Lagrangian is given by

L =

∞

−∞

L dx =

∞

−∞



∂

φ∂

φ − U(φ)



dx,

and the action is

S[φ] =

L dt =

L dx dt.

There is a non-linearity in the field equations due to a potential

(

) with

multiple vacua. We need multiple vacua to obtain a soliton. The kink stability

comes from the topology. It is very simple here, and just comes from counting

the discrete, distinct vacua.

As usual, we will write

φ =

∂φ

∂t

, φ

∂φ

∂x

Often it is convenient to (non-relativistically) split the Lagrangian as

L = T − V,

where

T =

dx, V =



+ U(φ)



dx.

In higher dimensions, we separate out ∂

φ into

φ and ∇φ.

The classical field equation comes from the condition that

[

] is stationary

under variations of

. By a standard manipulation, the field equation turns out

to be

∂

φ +

dφ

= 0.

This is an example of a Klein–Gordon type of field equation, but is non-linear if

U is not quadratic. It is known as the non-linear Klein–Gordon equation.

We are interested in a soliton that is a static solution. For a static field, the

time derivatives can be dropped, and this equation becomes

∂x

dφ

Of course, the important part is the choice of U! In φ

theory, we choose

U(φ) =

(1 − φ

)

This is mathematically the simplest version, because we set all coupling constants

to 1.

The importance of this U is that it has two minima:

U(φ)

−1 1

The two classical vacua are

φ(x) ≡ 1, φ(x) ≡ −1.

This is, of course, not the only possible choice. We can, for example, include

some parameters and set

U(φ) = λ(m

− φ

)

If we are more adventurous, we can talk about a φ

theory with

U(φ) = λφ

− φ

)

In this case, we have 3 minima, instead of 2. Even braver people can choose

U(φ) = 1 − cos φ.

This has infinitely many minima. The field equation involves a

sin φ

term,

and hence this theory is called the sine-Gordon theory (a pun on the name

Klein–Gordon, of course).

The sine-Gordon theory is a special case. While it seems like the most

complicated potential so far, it is actually integrable. This implies we can find

explicit exact solutions involving multiple, interacting solitons in a rather easy

way. However, integrable systems is a topic for another course, namely IID

Integrable Systems.

For now, we will focus on our simplistic

theory. As mentioned, there are

two vacuum field configurations, both of zero energy. We will in general use

the term “field configuration” to refer to fields at a given time that are not

necessarily solutions to the classical field equation, but in this case, the vacua

are indeed solutions.

If we wanted to quantize this

theory, then we have to pick one of the vacua

and do perturbation theory around it. This is known as spontaneous symmetry

breaking. Of course, by symmetry, we obtain the same quantum theory regardless

of which vacuum we expand around.

However, as we mentioned, when we want to study solitons, we have to

involve both vacua. We want to consider solutions that “connect” these two

vacua. In other words, we are looking for solutions that look like

This is known as a kink solution.

To actually find such a solution, we need the full field equation, given by

= −2(1 − φ

)φ.

Instead of solving this directly, we will find the kink solutions by considering the

energy, since this method generalizes better.

We will work with a general potential

with minimum value 0. From

Noether’s theorem, we obtain a conserved energy

E =



+ U(φ)



dx.

For a static field, we drop the

term. Then this is just the

appearing in the

Lagrangian. By definition, the field equation tells us the field is a stationary

point of this energy. To find the kink solution, we will in fact find a minimum

of the energy.

Of course, the global minimum is attained when we have a vacuum field, in

which case

= 0. However, this is the global minimum only if we don’t impose

any boundary conditions. In our case, the kinks satisfy the boundary conditions

“

(

∞

) = 1”, “

(

−∞

) =

−

1” (interpreted in terms of limits, of course). The kinks

will minimize energy subject to these boundary conditions.

These boundary conditions are important, because they are “topological”.

Eventually, we will want to understand the dynamics of solitons, so we will want

to consider fields that evolve with time. From physical considerations, for any

fixed

, the field

(

x, t

) must satisfy

(

x, t

)

→ vacuum

x → ±∞

, or else the

field will have infinite energy. However, the vacuum of our potential

is discrete.

Thus, if

is to evolve continuously with time, the boundary conditions must

not evolve with time! At least, this is what we expect classically. Who knows

what weird tunnelling can happen in quantum field theory.

So from now on, we fix some boundary conditions

(

∞

) and

(

−∞

), and

focus on fields that satisfy these boundary conditions. The trick is to write the

potential in the form

U(φ) =



dW (φ)

dφ



is non-negative, then we can always find

in principle — we take the

square root and then integrate it. However, in practice, this is useful only if we

can find a simple form for

. Let’s assume we’ve done that. Then we can write

E =



dφ





∓

dφ



dx ±

dφ



∓

dφ



dx ±



∓

dφ



dx ± (W (φ(∞)) − W (φ(−∞))).

The second term depends purely on the boundary conditions, which we have

fixed. Thus, we can minimize energy if we can make the first term vanish! Note

that when completing the square, the choice of the signs is arbitrary. However, if

we want to set the first term to be 0, the second term had better be non-negative,

since the energy itself is non-negative! Hence, we will pick the sign such that

the second term is ≥ 0, and then the energy is minimized when

= ±

dφ

In this case, we have

E = ±(W (∞) − W (−∞)).

These are known as the Bogomolny equation and the Bogomolny energy bound.

Note that if we picked the other sign, then we cannot solve the differential

equation φ

= ±

dφ

, because we know the energy must be non-negative.

For the φ

kink, we have

dφ

= 1 − φ

So we pick

W = φ −

So when

1, we have

. We need to choose the + sign, and then we

know the energy (and hence mass) of the kink is

E ≡ M =

We now solve for φ. The equation we have is

= 1 − φ

Rearranging gives

1 − φ

dφ = dx,

which we can integrate to give

φ(x) = tanh(x − a).

This

is an arbitrary constant of integration, labelling the intersection of the

graph of φ with the x-axis. We think of this as the “location” of the kink.

Note that there is not a unique solution, which is not unexpected by trans-

lation invariance. Instead, the solutions are labeled by a parameter

. This is

known as a modulus of the solution. In general, there can be multiple moduli,

and the space of all possible values of the moduli of static solitons is known as

the moduli space. In the case of a kink, the moduli space is just R.

Is this solution stable? We obtained this kink solution by minimizing the

energy within this topological class of solutions (i.e. among all solutions with

the prescribed boundary conditions). Since a field cannot change the boundary

conditions during evolution, it follows that the kink must be stable.

Are there other soliton solutions to the field equations? The solutions are

determined by the boundary conditions. Thus, we can classify all soliton solutions

by counting all possible combinations of the boundary conditions. We have, of

course, two vacuum solutions

φ ≡

1 and

φ ≡ −

1. There is also an anti-kink

solution obtained by inverting the kink:

φ(x) = −tanh(x − b).

This also has energy

1.2 Dynamic kink

We now want to look at kinks that move. Given what we have done so far, this

is trivial. Our theory is Lorentz invariant, so we simply apply a Lorentz boost.

Then we obtain a field

φ(x, t) = tanh γ(x − vt),

where, as usual

γ = (1 − v

)

−1/2

But this isn’t all. Notice that for small

, we can approximate the solution

simply by

φ(x, t) = tanh(x − vt).

This looks like a kink solution with a modulus that varies with time slowly. This

is known as the adiabatic point of view.

More generally, let’s consider a “moving kink” field

φ(x, t) = tanh(x − a(t))

for some function

(

). In general, this is not a solution to the field equation,

but if ˙a is small, then it is “approximately a solution”.

We can now explicitly compute that

φ = −

Let’s consider fields of this type, and look at the Lagrangian of the field theory.

The kinetic term is given by

T =

dx =





dx =





To derive this result, we had to perform the integral

, and if we do

that horrible integral, we will find a value that happens to be equal to

Of course, this is not a coincidence. We can derive this result from Lorentz

invariance to see that the result of integration is manifestly M.

The remaining part of the Lagrangian is less interesting. Since it does not

involve taking time derivatives, the time variation of

is not seen by it, and we

simply have a constant

V =

Then the original field Lagrangian becomes a particle Lagrangian

L =

M ˙a

−

Note that when we first formulated the field theory, the action principle

required us to find a field that extremizes the action among all fields. However,

what we are doing now is to restrict to the set of kink solutions only, and then

when we solve the variational problem arising from this Lagrangian, we are

extremizing the action among fields of the form

tanh

(

x −a

(

)). We can think of

this as motion in a “valley” in the field configuration space. In general, these

solutions will not also extremize the action among all fields. However, as we

said, it will do so “approximately” if ˙a is small.

We can obtain an effective equation of motion

M¨a = 0,

which is an equation of motion for the variable a(t) in the moduli space.

Of course, the solution is just given by

a(t) = vt + const,

where

is an arbitrary constant, which we interpret as the velocity. In this

formulation, we do not have any restrictions on

, because we took the “non-

relativistic approximation”. This approximation breaks down when v is large.

There is a geometric interpretation to this. We can view the equation of

motion

M¨a

= 0 as the geodesic equation in the moduli space

, and we can think

of the coefficient

as specifying a Riemannian metric on the moduli space. In

this case, the metric is (a scalar multiple of) the usual Euclidean metric (da)

This seems like a complicated way of describing such a simple system, but

this picture generalizes to higher-dimensional systems and allows us to analyze

multi-soliton dynamics, in particular, the dynamics of vortices and monopoles.

We might ask ourselves if there are multi-kinks in our theory. There aren’t in

the

theory, because we saw that the solutions are classified by the boundary

conditions, and we have already enumerated all the possible boundary conditions.

In more complicated theories like sine-Gordon theory, multiple kinks are possible.

However, while we cannot have two kinks in

theory, we can have a kink

followed by an anti-kink, or more of these pairs. This actually lies in the “vacuum

sector” of the theory, but it still looks like it’s made up of kinks and anti-kinks,

and it is interesting to study these.

1.3 Soliton interactions

We now want to study interactions between kinks and anti-kinks, and see how

they cause each other to move. So far, we were able to label the position of the

particle by its “center”

, and thus we can sensibly talk about how this center

moves. However, this center is well-defined only in the very special case of a

pure kink or anti-kink, where we can use symmetry to identify the center. If

there is some perturbation, or if we have a kink and an anti-kink, it is less clear

what should be considered the center.

Fortunately, we can still talk about the momentum of the field, even if

we don’t have a well-defined center. Indeed, since our theory has translation

invariance, Noether’s theorem gives us a conserved charge which is interpreted

as the momentum.

Recall that for a single scalar field in 1 +1 dimensions, the Lagrangian density

can be written in the form

L =

∂

φ∂

φ − U(φ).

Applying Noether’s theorem, to the translation symmetry, we obtain the energy-

momentum tensor

∂L

∂(∂

φ)

∂

φ − δ

L = ∂

φ∂

φ − δ

Fixing a time and integrating over all space, we obtain the conserved energy and

conserved momentum. These are

E =

∞

−∞

dx =

∞

−∞



+ U(φ)



dx,

P = −

∞

−∞

dx = −

∞

−∞

φφ

dx.

We now focus on our moving kink in the adiabatic approximation of the

theory. Then the field is given by

φ = tanh(x − a(t)).

Doing another horrible integral, we find that the momentum is just

P = M ˙a.

This is just as we would expect for a particle with mass M!

Now suppose what we have is instead a kink-antikink configuration

−a

Here we have to make the crucial assumption that our kinks are well-separated.

Matters get a lot worse when they get close to each other, and it is difficult

to learn anything about them analytically. However, by making appropriate

approximations, we can understand well-separated kink-antikink configurations.

When the kink and anti-kink are far away, we first pick a point

lying

in-between the kink and the anti-kink:

−a

The choice of

is arbitrary, but we should choose it so that it is far away

from both kinks. We will later see that, at least to first order, the result of our

computations does not depend on which

we choose. We will declare that the

parts to the left of

belongs to the kink, and the parts to the right of

belong

to the anti-kink. Then by integrating the energy-momentum tensor in these two

regions, we can obtain the momentum of the kink and the anti-kink separately.

We will focus on the kink only. Its momentum is given by

P = −

−∞

dx = −

−∞

φφ

dx.

Since T

is conserved, we know ∂

= 0. So we find

∂

∂t

∂

∂x

= 0.

By Newton’s second law, the force

on the kink is given by the rate of change

of the momentum:

F =

= −

−∞

∂

∂t

−∞

∂

∂x

= T





−

+ U



Note that there is no contribution at the

−∞

end because it is vacuum and

vanishes.

But we want to actually work out what this is. To do so, we need to be more

precise about what our initial configuration is. In this theory, we can obtain it

just by adding a kink to an anti-kink. The obvious guess is that it should be

φ(x)

= tanh(x + a) − tanh(x − a),

but this has the wrong boundary condition. It vanishes on both the left and the

right. So we actually want to subtract 1, and obtain

φ(x) = tanh(x + a) − tanh(x − a) − 1 ≡ φ

+ φ

− 1.

Note that since our equation of motion is not linear, this is in general not a

genuine solution! However, it is approximately a solution, because the kink

and anti-kink are well-separated. However, there is no hope that this will be

anywhere near a solution when the kink and anti-kink are close together!

Before we move on to compute

and

explicitly and plugging numbers

in, we first make some simplifications and approximations. First, we restrict

our attention to fields that are initially at rest. So we have

= 0 at

= 0. Of

course, the force will cause the kinks to move, but we shall, for now, ignore what

happens when they start moving.

That gets rid of one term. Next, we notice that we only care about the

expression when evaluated at

. Here we have

−

≈

0. So we can try to

expand the expression to first order in φ

− 1 (and hence φ

), and this gives

F =



−

+ U(φ

) − φ

+ (φ

− 1)

dφ

(φ

)



We have a zeroth order term

−

(

). We claim that this must vanish.

One way to see this is that this term corresponds to the force when there is no

anti-kink φ

. Since the kink does not exert a force on itself, this must vanish!

Analytically, we can deduce this from the Bogomolny equation, which says

for any kink solution φ, we have

dφ

It then follows that



dφ



= U(φ).

Alternatively, we can just compute it directly! In any case, convince yourself

that it indeed vanishes in your favorite way, and then move on.

Finally, we note that the field equations tell us

dφ

(φ

) = φ

So we can write the force as

F =



− φ

+ (φ

− 1)φ



That’s about all the simplifications we can make without getting our hands dirty.

We might think we should plug in the

tanh

terms and compute, but that is too

dirty. Instead, we use asymptotic expressions of kinks and anti-kinks far from

their centers. Using the definition of tanh, we have

= tanh(x + a) =

1 − e

−2(x+a)

1 + e

−2(x+a)

≈ 1 − 2e

−2(x+a)

This is valid for

x  −a

, i.e. to the right of the kink. The constant factor of 2 in

front of the exponential is called the amplitude of the tail. We will later see that

the 2 appearing in the exponent has the interpretation of the mass of the field

For φ

, take the approximation that x  a. Then

− 1 = −tanh(x − a) − 1 ≈ −2e

2(x−a)

We assume that our

satisfies both of these conditions. These are obviously

easy to differentiate once or twice. Doing this, we obtain

−φ

= (−4e

−2(x+a)

)(−4e

2(x−a)

) = 16e

−4a

Note that this is independent of

. In the formula, the

will turn into a

, and

we see that this part of the force is independent of

. Similarly, the other term is

(φ

− 1)φ

= (−2e

2(x−a)

)(−8e

−2(x+a)

) = 16e

−4a

Therefore we find

F = 32e

−4a

and as promised, this is independent of the precise position of the cutoff

chose.

We can write this in a slightly more physical form. Our initial configuration

was symmetric around the

-axis, but in reality, only the separation matters.

We write the separation of the pair as s = 2a. Then we have

F = 32e

−2s

What is the interpretation of the factor of 2? Recall that our potential was given

U(φ) =

(1 − φ

)

We can do perturbation theory around one of the vacua, say

= 1. Thus, we

set φ = 1 + η, and then expanding gives us

U(η) ≈

(−2η)

where m = 2. This is the same “2” that goes into the exponent in the force.

What about the constant factor of 32? Recall that when we expanded the

kink solution, we saw that the amplitude

of the tail was

= 2. It turns out if

we re-did our theory and put back the different possible parameters, we will find

that the force is given by

F = 2m

−ms

This is an interesting and important phenomenon. The mass

was the per-

turbative mass of the field. It is something we obtain by perturbation theory.

However, the same mass appears in the force between the solitons, which are

non-perturbative phenomenon!

This is perhaps not too surprising. After all, when we tried to understand

the soliton interactions, we took the approximation that

and

are close to

1 at b. Thus, we are in some sense perturbing around the vacuum φ ≡ 1.

We can interpret the force between the kink and anti-kink diagrammatically.

From the quantum field theory point of view, we can think of this force as

being due to meson exchange, and we can try to invent a Feynman diagram

calculus that involves solitons and mesons. This is a bit controversial, but at

least heuristically, we can introduce new propagators representing solitons, using

double lines, and draw the interaction as

So what happens to this soliton? The force we derived was positive. So the

kink is made to move to the right. By symmetry, we will expect the anti-kink to

move towards the left. They will collide!

What happens when they collide? All our analysis so far assumed the kinks

were well-separated, so everything breaks down. We can only understand this

phenomenon numerically. After doing some numerical simulations, we see that

there are two regimes:

–

If the kinks are moving slowly, then they will annihilate into meson radia-

tion.

–

If the kinks are moving very quickly, then they often bounce off each other.

1.4 Quantization of kink motion

We now briefly talk about how to quantize kinks. The most naive way of doing

so is pretty straightforward. We use the moduli space approximation, and then

we have a very simple kink Lagrangian.

L =

M ˙a

This is just a free particle moving in

with mass

. This

is known as the

collective coordinate of the kink. Quantizing a free particle is very straightforward.

It is just IB Quantum Mechanics. For completeness, we will briefly outline this

procedure.

We first put the system in Hamiltonian form. The conjugate momentum to

a is given by

P = M ˙a.

Then the Hamiltonian is given by

H = P ˙a − L =

Then to quantize, we replace

by the operator

−i~

∂

∂a

. In this case, the quantum

Hamiltonian is given by

H = −

∂

∂a

A wavefunction is a function of

and

, and this is just ordinary QM for a single

particle.

As usual, the stationary states are given by

ψ(a) = e

iκa

and the momentum and energy (eigenvalues) are

P = ~κ, H = E =

Is this actually “correct”? Morally speaking, we really should quantize the

complete 1 + 1 dimensional field theory. What would this look like?

In normal quantum field theory, we consider perturbations around a vacuum

solution, say

φ ≡

1, and we obtain mesons. Here if we want to quantize the

kink solution, we should consider field oscillations around the kink. Then the

solution contains both a kink and a meson. These mesons give rise to quantum

corrections to the kink mass M .

Should we be worried about these quantum corrections? Unsurprisingly, it

turns out these quantum corrections are of the order of the meson mass. So we

should not be worried when the meson mass is small.

Meson-kink scattering can also be studied in the full quantum theory. To

first approximation, since the kink is heavy, mesons are reflected or transmitted

with some probabilities, while the momentum of the kink is unchanged. But

when we work to higher orders, then of course the kink will move as a result.

This is all rather complicated.

For more details, see Rajaraman’s Solitons and Instantons, or Weinberg’s

Classical Solutions in Quantum Field Theory.

The thing that is really hard to understand in the quantum field theory

is kink-antikink pair production. This happens in meson collisions when the

mesons are very fast, and the theory is highly relativistic. What we have done

so far is perturbative and makes the non-relativistic approximation to get the

adiabatic picture. It is very difficult to understand the highly relativistic regime.

1.5 Sine-Gordon kinks

We end the section by briefly talking about kinks in a different theory, namely

the sine-Gordon theory. In this theory, kinks are often known as solitons instead.

The sine-Gordon theory is given by the potential

U(φ) = 1 − cos φ.

Again, we suppress coupling constants, but it is possible to add them back.

The potential looks like

U(φ)

2π 4π2π4π

Now there are infinitely many distinct vacua. In this case, we find we need to

pick W such that

dφ

= 2 sin

φ.

Static sine-Gordon kinks

To find the static kinks in the sine-Gordon theory, we again look at the Bogomolny

equation. We have to solve

dφ

= 2 sin

φ.

This can be solved. This involves integrating a

csc

, and ultimately gives us a

solution

φ(x) = 4 tan

−1

x−a

We can check that this solution interpolates between 0 and 2π.

2π

Unlike the

theory, dynamical multi-kink solutions exist here and can be

derived exactly. One of the earlier ways to do so was via B¨acklund transforms,

but that was very complicated. People later invented better methods, but they

are still not very straightforward. Nevertheless, it can be done. Ultimately, this

is due to the sine-Gordon equation being integrable. For more details, refer to

the IID Integrable Systems course.

Example. There is a two-kink solution

φ(x, t) = 4 tan

−1



v sinh γx

cosh γvt



where, as usual, we have

γ = (1 − v

)

−1/2

For v = 0.01, this looks like

2π

−2π

t = 0

t = ±400

Note that since

(

x, t

) =

(

x, −t

), we see that this solution involves two

solitons at first approaching each other, and then later bouncing off. Thus, the

two kinks repel each other. When we did kinks in

theory, we saw that a kink

and an anti-kink attracted, but here there are two kinks, which is qualitatively

different.

We can again compute the force just like the

theory, but alternatively,

since we have a full, exact solution, we can work it out directly from the solution!

The answers, fortunately, agree. If we do the computations, we find that the

point of closest approach is ∼ 2 log





if v is small.

There are some important comments to make. In the sine-Gordon theory, we

can have very complicated interactions between kinks and anti-kinks, and these

can connect vastly different vacua. However, static solutions must join 2

nπ

and

n ±

for some

, because if we want to join vacua further apart, we will

have more than one kink, and they necessarily interact.

If we have multiple kinks and anti-kinks, then each of these things can have

their own velocity, and we might expect some very complicated interaction

between them, such as annihilation and pair production. But remarkably, the

interactions are not complicated. If we try to do numerical simulations, or use

the exact solutions, we see that we do not have energy loss due to “radiation”.

Instead, the solitons remain very well-structured and retain their identities. This,

again, is due to the theory being integrable.

Topology of the sine-Gordon equation

There are also a lot of interesting things we can talk about without going into

details about what the solutions look like.

The important realization is that our potential is periodic in

. For the

sine-Gordon theory, it is much better to think of this as a field modulo 2

, i.e.

as a function

φ : R → S

In this language, the boundary condition is that

(

) = 0

mod

x → ±∞

Thus, instead of thinking of the kink as joining two vacua, we can think of it as

“winding around the circle” instead.

We can go further. Since the boundary conditions of

are now the same on

two sides, we can join the ends of the domain

together, and we can think of

as a map

φ : S

→ S

instead. This is a compactification of space.

Topologically, such maps are classified by their winding number , or the degree,

which we denote

. This is a topological (homotopy) invariant of a map, and

is preserved under continuous deformations of the field. Thus, it is preserved

under time evolution of the field.

Intuitively, the winding number is just how many times we go around the

circle. There are multiple (equivalent) ways of making this precise.

The first way, which is the naive way, is purely topological. We simply have

to go back to the first picture, where we regard

as a real value. Suppose the

boundary values are

φ(−∞) = 2n

−

π, φ(∞) = 2n

π.

Then we set the winding number to be Q = n

− n

−

Topologically, we are using the fact that

is the universal covering space of

the circle, and thus we are really looking at the induced map on the fundamental

group of the circle.

Example. As we saw, a single kink has Q = 1.

2π

Thus, we can think of the Q as the net soliton number.

But this construction we presented is rather specific to maps from

We want something more general that can be used for more complicated systems.

We can do this in a more “physics” way. We note that there is a topological

current

2π

µν

∂

φ,

where

µν

is the anti-symmetric tensor in 1 + 1 dimensions, chosen so that

= 1.

In components, this is just

2π

(∂

φ, −∂

φ).

This is conserved because of the symmetry of mixed partial derivatives, so that

∂

2π

µν

∂

φ = 0.

As usual, a current induces a conserved charge

Q =

∞

−∞

dx =

2π

∞

−∞

∂

φ dx =

2π

(φ(∞) − φ(−∞)) = n

− n

−

which is the formula we had earlier.

Note that all these properties do not depend on

satisfying any field equa-

tions! It is completely topological.

Finally, there is also a differential geometry way of defining

. We note that

the target space S

has a normalized volume form ω so that

ω = 1.

For example, we can take

ω =

2π

dφ.

Now, given a mapping

R → S

, we can pull back the volume form to obtain

∗

ω =

2π

dφ

dx.

We can then define the degree of the map to be

Q =

∗

ω =

2π

∞

−∞

dφ

dx.

This is exactly the same as the formula we obtained using the current!

Note that even though the volume form is normalized on

and has integral

1, the integral when pulled back is not 1. We can imagine this as saying if we

wind around the circle

times, then after pulling back, we would have pulled

back

“copies” of the volume form, and so the integral will be

times that of

the integral on S

We saw that these three definitions gave the same result, and different

definitions have different benefits. For example, in the last two formulations, it

is not a priori clear that the winding number has to be an integer, while this is

clear in the first formulation.

2 Vortices

We are now going to start studying vortices. These are topological solitons in

two space dimensions. While we mostly studied

kinks last time, what we are

going to do here is more similar to the sine-Gordon theory than the

theory,

as it is largely topological in nature.

A lot of the computations we perform in the section are much cleaner when

presented using the language of differential forms. However, we shall try our

best to provide alternative versions in coordinates for the easily terrified.

2.1 Topological background

Sine-Gordon kinks

We now review what we just did for sine-Gordon kinks, and then try to develop

some analogous ideas in higher dimension. The sine-Gordon equation is given by

∂

∂t

−

∂

∂x

+ sin θ = 0.

We want to choose boundary conditions so that the energy has a chance to

be finite. The first part is, of course, to figure out what the energy is. The

energy-momentum conservation equation given by Noether’s theorem is

∂



+ θ

+ (1 − cos θ)



+ ∂

(−θ

) = ∂

= 0.

The energy we will be considering is thus

E =

dx =



+ θ

+ (1 − cos θ)



dx.

Thus, to obtain finite energy, we will want

(

)

→

for some integers

as x → ±∞. What is the significance of this n

Example. Consider the basic kink

(x) = 4 tan

−1

Picking the standard branch of tan

−1

, this kink looks like

2π

This goes from θ = 0 to θ = 2π.

To better understand this, we can think of

as an angular variable, i.e. we

identify

θ ∼ θ

+ 2

nπ

for any

n ∈ Z

. This is a sensible thing because the energy

density and the equation etc. are unchanged when we shift everything by 2

nπ

Thus, θ is not taking values in R, but in R/2πZ

∼

Thus, for each fixed t, our field θ is a map

θ : R → S

The number

− n

−

equals the number of times

covers the circle

on going from

−∞

= +

∞

. This is the winding number, which is

interpreted as the topological charge.

As we previously discussed, we can express this topological charge as the

integral of some current. We can write

Q =

2π

θ(R)

dθ =

2π

∞

−∞

dθ

dx.

Note that this formula automatically takes into account the orientation. This is

the form that will lead to generalization in higher dimensions.

This function

dθ

appearing in the integral has the interpretation as a topo-

logical charge density. Note that there is a topological conservation law

∂

∂j

∂t

∂j

∂x

= 0,

where

= θ

, j

= −θ

This conservation law is not a consequence of the field equations, but merely a

mathematical identity, namely the commutation of partial derivatives.

Two dimensions

For the sine-Gordon kink, the target space was a circle

. Now, we are concerned

with the unit disk

D = {(x

, x

) : |x|

< 1} ⊆ R

We will then consider fields

Φ : D → D.

In the case of a sine-Gordon kink, we still cared about moving solitons. However,

here we will mostly work with static solutions, and study fields at a fixed time.

Thus, there is no time variable appearing.

Using the canonical isomorphism

∼

, we can think of the target space

as the unit disk in the complex plane, and write the field as

Φ = Φ

+ iΦ

However, we will usually view the D in the domain as a real space instead.

We will impose some boundary conditions. We pick any function

→ R

and consider

g = e

iχ

: S

→ S

= ∂D ⊆ D.

Here

is a genuine function, and has to be single-valued. So

must be single-

valued modulo 2π. We then require

∂D

= g = e

iχ

In particular, Φ must send the boundary to the boundary.

Now the target space

has a canonical choice of measure dΦ

∧

dΦ

. Then

we can expect the new topological charge to be given by

Q =

dΦ

∧ dΦ

det

∂Φ

∂x

∂Φ

∂x

∂Φ

∂x

∂Φ

∂x

∧ dx

Thus, the charge density is given by

∂Φ

∂x

∂Φ

∂x

Crucially, it turns out this charge density is a total derivative, i.e. we have

∂V

∂x

for some function

. It is not immediately obvious this is the case. However, we

can in fact pick

∂Φ

∂x

To see this actually works, we need to use the anti-symmetry of

and observe

that

∂

∂x

= 0.

Equivalently, using the language of differential forms, we view the charge density

as the 2-form

= dΦ

∧ dΦ

= d(Φ

dΦ

) =

d(Φ

dΦ

− Φ

dΦ

By the divergence theorem, we find that

Q =

2π

∂D

dΦ

− Φ

dΦ

We then use that on the boundary,

= cos χ, Φ

= sin χ,

Q =

2π

∂D

(cos

χ + sin

χ) dχ =

2π

∂D

dχ = N.

Thus, the charge is just the winding number of g!

Now notice that our derivation didn’t really depend on our domain being

. It could have been any region bounded by a simple closed curve in

. In

particular, we can take it to be a disk D

of arbitrary radius R.

What we are actually interested in is a field

Φ : R

→ D.

We then impose asymptotic boundary conditions

Φ ∼ g = e

iχ

as |x| → ∞. We can still define the charge or degree by

Q =

lim

R→∞

This is then again the winding number of g.

It is convenient to rewrite this in terms of an inner product.

itself has an

inner product, and under the identification

∼

, the inner product can be

written as

(a, b) =

¯ab + a

Use of this expression allows calculations to be done efficiently if one makes use

of the fact that for real numbers a and b, we have

(a, b) = (ai, bi) = ab, (ai, b) = (a, bi) = 0.

In particular, we can evaluate

(iΦ, dΦ) = (iΦ

− Φ

, dΦ

+ idΦ

) = Φ

dΦ

− Φ

dΦ

This is just (twice) the current V we found earlier! So we can write our charge

Q =

2π

lim

R→∞

|x|=R

(iΦ, dΦ).

We will refer to (

dΦ) as the current, and the corresponding charge density is

d(iΦ, dΦ).

This current is actually a familiar object from quantum mechanics: recall

that for the Schr¨odinger’s equation

∂Φ

∂t

= −

∆Φ + V (x)Φ, ∆ = ∇

the probability

is conserved. The differential form of the probability

conservation law is

∂

(Φ, Φ) +

∇ · (iΦ, ∇Φ) = 0.

What appears in the flux term here is just the topological current!

2.2 Global U (1) Ginzburg–Landau vortices

We now put the theory into use. We are going to study Ginzburg–Landau vortices.

Our previous discussion involved a function taking values in the unit disk

We will not impose such a restriction on our vortices. However, we will later see

that any solution must take values in D.

The potential energy of the Ginzburg–Landau theory is given by

V (Φ) =



(∇Φ, ∇Φ) +

(1 − (Φ, Φ))



where λ > 0 is some constant.

Note that the inner product is invariant under phase rotation, i.e.

iχ

a, e

iχ

b) = (a, b)

for χ ∈ R. So in particular, the potential satisfies

V (e

iχ

Φ) = V (Φ).

Thus, our theory has a global U(1) symmetry.

The Euler–Lagrange equation of this theory says

−∆Φ =

(1 − |Φ|

)Φ.

This is the ungauged Ginzburg–Landau equation.

To justify the fact that our Φ takes values in

, we use the following lemma:

Lemma.

Assume Φ is a smooth solution of the ungauged Ginzburg–Landau

equation in some domain. Then at any interior maximum point

∗

, we

have |Φ(x

∗

)| ≤ 1.

Proof. Consider the function

W (x) = 1 − |Φ(x)|

Then we want to show that

W ≥

0 when

is minimized. We note that if

at a minimum, then the Hessian matrix must have non-negative eigenvalues. So,

taking the trace, we must have ∆

(

∗

)

≥

0. Now we can compute ∆

directly.

We have

∇W = −2(Φ, ∇Φ)

∆W = ∇

= −2(Φ, ∆Φ) − 2(∇Φ, ∇Φ)

= λ|Φ|

W − 2|∇Φ|

Thus, we can rearrange this to say

2|∇Φ|

+ ∆W = λ|Φ|

But clearly 2

|∇

≥

0 everywhere, and we showed that ∆

(

∗

)

≥

0. So we

must have W (x

∗

) ≥ 0.

By itself, this doesn’t force

| ∈

1], since we could imagine

having no

maximum. However, if we prescribe boundary conditions such that

= 1 on

the boundary, then this would indeed imply that

| ≤

1 everywhere. Often, we

can think of Φ as some “complex order parameter”, in which case the condition

|Φ| ≤ 1 is very natural.

The objects we are interested in are vortices.

Definition

(Ginzburg–Landau vortex)

A global Ginzburg–Landau vortex of

charge

N >

0 is a (smooth) solution of the ungauged Ginzburg–Landau equation

of the form

Φ = f

(r)e

iNθ

in polar coordinates (r, θ). Moreover, we require that f

(r) → 1 as r → ∞.

Note that for Φ to be a smooth solution, we must have

(0) = 0. In fact, a

bit more analysis shows that we must have

(

) as

r →

0. Solutions for

do exist, and they look roughly like this:

In the case of

= 1, we can visualize the field Φ as a vector field on

Then it looks like

This is known as a 2-dimensional hedgehog.

For general

, it might be more instructive to look at how the current looks

like. Recall that the current is defined by (

dΦ). We can write this more

explicitly as

(iΦ, dΦ) = (if

iNθ

, (df

iNθ

+ if

N dθe

iNθ

)

= (if

, df

+ if

N dθ).

We note that

and d

are orthogonal, while

and

are parallel.

So the final result is

(iΦ, dΦ) = f

N dθ.

So the current just looks this:

|x| → ∞

, we have

→

1. So the winding number is given as before,

and we can compute the winding number of this system to be

2π

lim

R→∞

(iΦ, dΦ) =

2π

lim

R→∞

N dθ = N.

The winding number of these systems is a discrete quantity, and can make the

vortex stable.

This theory looks good so far. However, it turns out this model has a problem

— the energy is infinite! We can expand out

(

iNθ

), and see it is a sum of a

few non-negative terms. We will focus on the

∂

∂θ

term. We obtain

V (f

iNθ

) ≥



∂Φ

∂θ



r dr dθ

= N

r dr dθ

= 2πN

∞

dr.

Since f

→ 1 as r → ∞, we see that the integral diverges logarithmically.

This is undesirable physically. To understand heuristically why this occurs,

decompose dΦ into two components — a mode parallel to Φ and a mode

perpendicular to Φ. For a vortex solution these correspond to the radial and

angular modes respectively. We will argue that for fluctuations the parallel mode

is massive, while the perpendicular mode is massless. Now given that we just

saw that the energy divergence of the vortex arises from the angular part of the

energy, we see that it is the massless mode that leads to problems. We will see

below that in gauge theories, the Higgs mechanism serves to make all modes

massive, thus allowing for finite energy vortices.

We can see the difference between massless and massive modes very explicitly

in a different setting, corresponding to Yukawa mesons. Consider the equation

−∆u + M

u = f.

Working in three dimensions, the solution is given by

u(x) =

4π

−M|x−y|

|x − y|

f(y) dy.

Thus, the Green’s function is

G(x) =

−M|x|

4π|x|

If the system is massless, i.e.

= 0, then this decays as

|x|

. However, if the

system is massive with

M >

0, then this decays exponentially as

|x| → ∞

. In

the nonlinear setting the exponential decay which is characteristic of massive

fundamental particles can help to ensure decay of the energy density at a rate

fast enough to ensure finite energy of the solution.

So how do we figure out the massive and massless modes? We do not have

a genuine decomposition of Φ itself into “parallel” and “perpendicular” modes,

because what is parallel and what is perpendicular depends on the local value of

Φ.

Thus, to make sense of this, we have to consider small fluctuations around a

fixed configuration Φ. We suppose Φ is a solution to the field equations. Then

δV

δΦ

= 0. Thus, for small variations Φ 7→ Φ + εϕ, we have

V (Φ + εϕ) = V (Φ) + ε



|∇ϕ|

+ λ(ϕ, Φ)

−

2λ

(1 − |Φ|

)|ϕ|



dx + O(ε

Ultimately, we are interested in the asymptotic behaviour

|x| → ∞

, in which

case 1

− |

→

0. Moreover,

| →

1 implies (

ϕ,

Φ) becomes a projection along

the direction of Φ. Then the quadratic part of the potential energy density for

fluctuations becomes approximately

|∇ϕ|

+ λ|ϕ

parallel

for large

. Thus, for

λ >

0, the parallel mode is massive, with corresponding

“Yukawa” mass parameter

√

, while the perpendicular mode is massless.

The presence of the massless mode is liable to produce a soliton with slow

algebraic decay at spatial infinity, and hence infinite total energy. This is all

slightly heuristic, but is a good way to think about the issues. When we study

vortices that are gauged, i.e. coupled to the electromagnetic field, we will see

that the Higgs mechanism renders all components massive, and this problem

does not arise.

2.3 Abelian Higgs/Gauged Ginzburg–Landau vortices

We now consider a theory where the complex scalar field Φ is coupled to a

magnetic field. This is a U(1) gauge theory, with gauge potential given by a

smooth real 1-form

A = A

+ A

The coupling between Φ and

is given by minimal coupling: this is enacted by

introduction of the covariant derivative

DΦ = D

Φ = dΦ − iAΦ =

j=1

Φ dx

To proceed, it is convenient to have a list of identities involving the covariant

derivative.

Proposition.

is a smooth real-valued function, and Φ and Ψ are smooth

complex scalar fields, then

D(fΦ) = (df) Φ + f DΦ,

d(Φ, Ψ) = (DΦ, Ψ) + (Φ, DΨ).

(Here (

) is the real inner product defined above.) In coordinates, these take

the form

(fΦ) = (∂

f) Φ + f D

∂

(Φ, Ψ) = (D

Φ, Ψ) + (Φ, D

Ψ).

The proofs just involve writing all terms out. The first rule is a version of

the Leibniz rule, while the second, called unitarity, is analogous to the fact that

if V, W are smooth vector fields on a Riemannian manifold, then

∂

g(V, W ) = g(∇

V, W ) + g(V, ∇

W )

for the Levi-Civita connection ∇ associated to a Riemannian metric g.

The curvature term is given by the magnetic field.

Definition

(Magnetic field/curvature)

The magnetic field, or curvature is given

B = ∂

− ∂

We can alternatively think of it as the 2-form

F = dA = B dx

∧ dx

The formulation in terms of differential forms is convenient for computations,

because we don’t have to constrain ourselves to working in Cartesian coordinates

— for example, polar coordinates may be more convenient.

Proposition. If Φ is a smooth scalar field, then

− D

)Φ = −iBΦ.

The proof is again a direct computation. Alternatively, we can express this

without coordinates. We can extend D to act on

-forms by letting

act as

A∧

Then this result says

Proposition.

DDΦ = −iF Φ.

Proof.

DDΦ = (d − iA)(dΦ − iAΦ)

= d

Φ − id(AΦ) − iA dΦ − A ∧ A Φ

= −id(AΦ) − iA dΦ

= −idA Φ + iA dΦ − iA dΦ

= −i(dA) Φ

= −iF Φ.

The point of introducing the covariant derivative is that we can turn the global

U(1) invariance into a local one. Previously, we had a global U(1) symmetry,

where our field is unchanged when we replace Φ

7→

iχ

for some constant

χ ∈ R

With the covariant derivative, we can promote this to a gauge symmetry.

Consider the simultaneous gauge transformations

Φ(x) 7→ e

iχ(x)

Φ(x)

A(x) 7→ A(x) + dχ(x).

Then the covariant derivative of Φ transforms as

(d − iA)Φ 7→ (d − i(A + dχ))(Φe

iχ

)

= (dΦ + iΦdχ − i(A + dχ)Φ)e

iχ

= e

iχ

(d − iA)Φ.

Similarly, the magnetic field is also invariant under gauge transformations.

As a consequence, we can write down energy functionals that are invariant

under these gauge transformations. In particular, we have (using the real inner

product defined above)

(DΦ, DΦ) 7→ (e

iχ

DΦ, e

iχ

DΦ) = (DΦ, DΦ).

So we can now write down the gauged Ginzburg–Landau energy

(A, Φ) =



+ |DΦ|

(1 − |Φ|

)



This is then manifestly gauge invariant.

As before, the equations of motion are given by the Euler–Lagrange equations.

Varying Φ, we obtain

−(D

+ D

)Φ −

(1 − |Φ|

)Φ = 0.

This is just like the previous vortex equation in the ungauged case, but since we

have the covariant derivative, this is now coupled to the gauge potential

. The

equations of motion satisfied by A are

∂

B = (iΦ, D

Φ)

−∂

B = (iΦ, D

Φ).

These are similar to one of Maxwell’s equation — the one relating the curl of

the magnetic field to the current.

It is again an exercise to derive these. We refer to the complete system as

the gauged Ginzburg–Landau, or Abelian Higgs equations. In deriving them, it

is helpful to use the previous identities such as

∂

(Φ, Ψ) = (D

Φ, Ψ) + (Φ, D

Ψ).

So we get the integration by parts formula

Φ, Ψ) d

x = −

(Φ, D

Ψ) d

under suitable boundary conditions.

Lemma.

Assume Φ is a smooth solution of the gauged Ginzburg–Landau

equation in some domain. Then at any interior maximum point

∗

, we

have |Φ(x

∗

)| ≤ 1.

Proof. Consider the function

W (x) = 1 − |Φ(x)|

Then we want to show that

W ≥

0 when

is minimized. We note that if

at a minimum, then the Hessian matrix must have non-negative eigenvalues. So,

taking the trace, we must have ∆

(

∗

)

≥

0. Now we can compute ∆

directly.

We have

∂

W = −2(Φ, D

Φ)

∆W = ∂

∂

= −2(Φ, D

Φ) − 2(D

Φ, D

Φ)

= λ|Φ|

W − 2|∇Φ|

Thus, we can rearrange this to say

2|∇Φ|

+ ∆W = λ|Φ|

But clearly 2

|∇

≥

0 everywhere, and we showed that ∆

(

∗

)

≥

0. So we

must have W (x

∗

) ≥ 0.

As before, this suggests we interpret

as an order parameter. This model

was first used to describe superconductors. The matter can either be in a

“normal” phase or a superconducting phase, and

measures how much we are

in the superconducting phase.

Thus, in our model, far away from the vortices, we have

| ≈

1, and so we

are mostly in the superconducting phase. The vortices represent a breakdown of

the superconductivity. At the core of the vortices, we have |Φ| = 0, and we are

left with completely normal matter. Usually, this happens when we have a strong

magnetic field. In general, a magnetic field cannot penetrate the superconductor

(the “Meissner effect”), but if it is strong enough, it will cause such breakdown

in the superconductivity along vortex “tubes”.

Radial vortices

Similar to the ungauged case, for

λ >

0, there exist vortex solutions of the form

Φ = f

(r)e

iNθ

A = Nα

(r) dθ.

The boundary conditions are f

, α

→ 1 as r → ∞ and f

, α

→ 0 as r → 0.

Let’s say a few words about why these are sensible boundary conditions from

the point of view of energy. We want

(1 − |Φ|

)

< ∞,

and this is possible only for

→

1 as

r → ∞

. What is less obvious is that we

also need α

→ 1. We note that we have

Φ =

∂Φ

∂θ

− iA

Φ = (iNf

− iNα

iNθ

We want this to approach 0 as r → ∞. Since f

→ 1, we also need α

→ 1.

The boundary conditions at 0 can be justified as before, so that the functions

are regular at 0.

Topological charge and magnetic flux

Let’s calculate the topological charge. We have (assuming sufficiently rapid

approach to the asymptotic values as r → ∞)

Q =

(Φ) d

= lim

R→∞

2π

|x|=R

(iΦ, dΦ)

= lim

R→∞

2π

(if

iNθ

, iNf

iNθ

) dθ

2π

· N lim

R→∞

2π

dθ

= N.

Previously, we understood

as the “winding number”, and it measures how

“twisted” our field was. However, we shall see shortly that there is an alternative

interpretation of this

. Previously, in the sine-Gordon theory, we could think of

as the number of kinks present. Similarly, here we can think of this

-vortex

as a superposition of

vortices at the origin. In the case of

= 1, we will

see that there are static solutions involving multiple vortices placed at different

points in space.

We can compute the magnetic field and total flux as well. It is convenient to

use the d

definition, as we are not working in Cartesian coordinates. We have

dA = Nα

(r) dr ∧ dθ =

Nα

∧ dx

Thus it follows that

B = ∂

− ∂

Working slightly more generally, we assume given a smooth finite energy con-

figuration

Φ, and suppose in addition that

→

1 and

r|A|

is bounded as

r → ∞, and also that D

Φ = o(r

−1

). Then we find that

(dΦ)

= iA

Φ + o(r

−1

Then if we integrate around a circular contour, since only the angular part

contributes, we obtain

|x|=R

(iΦ, dΦ) =

|x|=R

(iΦ, iA

Φ) dθ + o(1) =

|x|=R

A + o(1).

Note that here we are explicitly viewing

as a differential form so that we can

integrate it. We can then note that

|x|

is the boundary of the disk

|x| ≤ R

So we can apply Stokes’ theorem and obtain

|x|=R

(iΦ, dΦ) =

|x|≤R

dA =

|x|≤R

B d

Now we let R → ∞ to obtain

lim

R→∞

x≤R

B d

x = lim

R→∞

|x|=R

A = lim

R→∞

|x|=R

(iΦ, dΦ) = 2πQ.

Physically, what this tells us then is that there is a relation between the

topological winding number and the magnetic flux. This is a common property

of topological gauge theories. In mathematics, this is already well known —

it is the fact that we can compute characteristic classes of vector bundles by

integrating the curvature, as discovered by Chern.

Behaviour of vortices as r → 0

We saw earlier that for reasons of regularity, it was necessary that

, α

→

as r → 0.

But actually, we must have

∼ r

r →

0. This has as a consequence

that Φ

∼

(

iθ

)

. So the local appearance of the vortex is the zero of an

analytic function with multiplicity N.

To see this, we need to compute that the Euler–Lagrange equations are

−f

(r) −

(r) +

(N − α

)

(1 − f

With the boundary condition that

and

vanish at

= 0, we can approxi-

mate this locally as

−f

−

≈ 0

since we have a

on the left hand side. The approximating equation is

homogeneous, and the solutions are just

= r

±N

So for regularity, we want the one that → 0, so f

∼ r

as r → 0.

2.4 Bogomolny/self-dual vortices and Taubes’ theorem

As mentioned, we can think of the radial vortex solution as a collection of

vortices all superposed at the origin. Is it possible to have separated vortices all

over the plane? Naively, we would expect that the vortices exert forces on each

other, and so we don’t get a static solution. However, it turns out that in the

= 1 case, there do exist static solutions corresponding to vortices at arbitrary

locations on the plane.

This is not obvious, and the proof requires some serious analysis. We will

not do the analysis, which requires use of Sobolev spaces and PDE theory.

However, we will do all the non-hard-analysis part. In particular, we will obtain

Bogomolny bounds as we did in the sine-Gordon case, and reduce the problem

to finding solutions of a single scalar PDE, which can be understood with tools

from calculus of variations and elliptic PDEs.

Recall that for the sine-Gordon kinks, we needed to solve

= sin θ,

with boundary conditions

(

)

→

0 or 2

x → ±∞

. The only solutions we

found were

(x − X)

for any

X ∈ R

. This

is interpreted as the location of the kink. So the moduli

space of solutions is M = R.

We shall get a similar but more interesting description for the

= 1 vortices.

This time, the moduli space will be

, given by

complex parameters

describing the solutions.

Theorem

(Taubes’ theorem)

For

= 1, the space of (gauge equivalence classes

of) solutions of the Euler–Lagrange equations

δV

= 0 with winding number

is M

∼

To be precise, given

N ∈ N

and an unordered set of points

, ··· , Z

}

there exists a smooth solution

(

;

, ··· , Z

) and Φ(

;

, ··· , Z

) which

solves the Euler-Lagrange equations

δV

= 0, and also the so-called Bogomolny

equations

Φ + iD

Φ = 0, B =

(1 − |Φ|

Moreover, Φ has exactly

zeroes

, ··· , Z

counted with multiplicity, where

(using the complex coordinates z = x

+ ix

)

Φ(x; Z

, ··· , Z

) ∼ c

(z − Z

)

z → Z

, where

|{k

is the multiplicity and

is a nonzero

complex number.

This is the unique such solution up to gauge equivalence. Furthermore,

(A( ·, Z

, ··· , Z

), Φ( ·; Z

, ··· , Z

)) = πN (∗)

and

2π

B d

x = N = winding number.

Finally, this gives all finite energy solutions of the gauged Ginzburg–Landau

equations.

Note that it is not immediately clear from our description that the mod-

uli space is

. It looks more like

quotiented out by the action of the

permutation group

. However, the resulting quotient is still isomorphic to

. (However, it is important for various purposes to remember this quotient

structure, and to use holomorphic coordinates which are invariant under the

action of the permutation group — the elementary symmetric polynomials in

, . . . , Z

There is a lot to be said about this theorem. The equation (

∗

) tells us the

energy is just a function of the number of particles, and does not depend on

where they are. This means there is no force between the vortices. In situations

like this, it is said that the Bogomolny bound is saturated. The final statement

suggests that the topology is what is driving the existence of the vortices, as we

have already seen. The reader will find it useful to work out the corresponding

result in the case of negative winding number (in which case the holomorphicity

condition becomes anti-holomorphicity, and the sign of the magnetic field is

reversed in the Bogomolny equations).

Note that the Euler–Lagrange equations themselves are second-order equa-

tions. However, the Bogomolny equations are first order. In general, this is a

signature that suggests that interesting mathematical structures are present.

We’ll discuss three crucial ingredients in this theorem, but we will not

complete the proof, which involves more analysis than is a pre-requisite for this

course. The proof can be found in Chapter 3 of Jaffe and Taubes’s Vortices and

Monopoles.

Holomorphic structure

When there are Bogomolny equations, there is often some complex analysis

lurking behind. We can explicitly write the first Bogomolny equation as

Φ + iD

Φ =

∂Φ

∂x

+ i

∂Φ

∂x

− i(A

+ iA

)Φ = 0.

Recall that in complex analysis, holomorphic functions can be characterized as

complex-valued functions which are continuously differentiable (in the real sense)

and also satisfy the Cauchy–Riemann equations

∂f

∂¯z



∂f

∂x

+ i

∂f

∂x



= 0.

So we think of the first Bogomolny equation as the covariant Cauchy–Riemann

equations. It is possible to convert this into the standard Cauchy–Riemann

equations to deduce the local behaviour at Φ.

To do so, we write

Φ = e

Then

∂f

∂¯z

= e

−ω



∂Φ

∂¯z

−

∂ω

∂¯z



= e

−ω



i(A

+ iA

)

−

∂ω

∂¯z



Φ.

This is equal to 0 if ω satisfies

∂ω

∂¯z

= i

+ iA

So the question is — can we solve this? It turns out we can always solve this

equation, and there is an explicit formula for the solution. In general, if

smooth, then the equation

∂w

∂¯z

= β,

has a smooth solution in the disc {z : |z| < r}, given by

ω(z, ¯z) =

2πi

|w|<r

β(w)

w − z

dw ∧ d ¯w.

A proof can be found in the book Griffiths and Harris on algebraic geometry,

on page 5. So we can write

Φ = e

where

is holomorphic. Since

is never zero, we can apply all our knowledge

of holomorphic functions to

, and deduce that Φ has isolated zeroes, where

Φ ∼ (z − Z

)

for some integer power n

The Bogomolny equations

We’ll now show that (

Φ) satisfies

(

Φ) =

πN ≥

0 iff it satisfies the

Bogomolny equations, i.e.

Φ + iD

Φ = 0, B =

(1 − |Φ|

We first consider the simpler case of the sine-Gordon equation. As in the

kinks, to find soliton solutions, we write the energy as

E =

∞

−∞



+ (1 − cos θ)



∞

−∞



+ 4 sin



∞

−∞



− 2 sin



+ 4θ

sin

∞

−∞



− 2 sin



dx +

∞

−∞

∂

∂x



−4 cos



∞

−∞



− 2 sin



dx +



−4 cos

θ(+∞)

+ 4 cos

θ(−∞)



We then use the kink asymptotic boundary conditions to obtain, say,

∞

) = 2

and

(

−∞

) = 0. So the boundary terms gives 8. Thus, subject to these boundary

conditions, we can write the sine-Gordon energy as

E =

∞

−∞



− 2 sin



dx + 8.

Thus, if we try to minimize the energy, then we know the minimum is at least 8,

and if we could solve the first-order equation

= 2

sin

, then the minimum

would be exactly 8. The solution we found does satisfy this first-order equation.

Moreover, the solutions are all of the form

θ(x) = θ

(x − X), θ

(x) = 4 arctan e

Thus, we have shown that the minimum energy is 8, and the minimizers are all

of this form, parameterized by X ∈ R.

We want to do something similar for the Ginzburg–Landau theory. In order

to make use of the discussion above of the winding number

, we will make

the same standing assumptions as used in that discussion, but it is possible

to generalize the conclusion of the following result to arbitrary finite energy

configurations with an appropriate formulation of the winding number.

Lemma. We have

(A, Φ) =



B −

(1 − |Φ|

)



+ 4|

∂

Φ|

x + πN,

where

∂

Φ =

Φ + iD

Φ).

It is clear that the desired result follows from this.

Proof. We complete the square and obtain

(A, Φ) =



B −

(1 − |Φ|

)



+ B(1 − |Φ|

) + |D

Φ|

+ |D

Φ|

We now dissect the terms one by one. We first use the definition of

∧

dA and integration by parts to obtain

(1 − |Φ|

) dA = −

d(1 − |Φ|

) ∧ A = 2

(Φ, DΦ) ∧ A.

Alternatively, we can explicitly write

B(1 − |Φ|

) d

x =

(∂

− ∂

)(1 − |Φ|

) d

∂

|Φ|

− A

∂

|Φ|

) d

= 2

(Φ, D

Φ) − A

(Φ, D

Φ).

Ultimately, we want to obtain something that looks like

∂

. We can write

this out as

Φ + iD

Φ, D

Φ + iD

Φ) = |D

Φ|

+ |D

Φ|

+ 2(D

Φ, iD

Φ).

We note that

Φ and Φ are always orthogonal, and

is always a real coefficient.

So we can write

Φ, iD

Φ) = (∂

Φ − iA

Φ, i∂

Φ + A

Φ)

= (∂

Φ, i∂

Φ) + A

(Φ, ∂

Φ) − A

(Φ, ∂

Φ).

We now use again the fact that (Φ

, i

Φ) = 0 to replace the usual derivatives with

the covariant derivatives. So we have

Φ, iD

Φ) = (∂

Φ, i∂

Φ) + A

(Φ, D

Φ) − A

(Φ, D

Φ).

This tells us we have



B(1 − |Φ|

) + |D

Φ|

+ |D

Φ|



x =



∂

Φ|

+ 2(∂

Φ, i∂

Φ)



It then remains to show that (∂

Φ, i∂

Φ) = j

(Φ). But we just write

(∂

+ i∂

, −∂

+ i∂

) = −(∂

, ∂

) + (∂

, ∂

)

= −j

(Φ)

= −det



∂



Then we are done.

Corollary.

For any (

Φ) with winding number

, we always have

(

Φ)

≥

πN, and those (A, Φ) that achieve this bound are exactly those that satisfy

∂

Φ = 0, B =

(1 − |Φ|

Reduction to scalar equation

The remaining part of Taubes’ theorem is to prove the existence of solutions to

these equations, and that they are classified by

unordered complex numbers.

This is the main analytic content of the theorem.

To do so, we reduce the two Bogomolny equations into a scalar equation.

Note that we have

Φ + iD

Φ = (∂

Φ + i∂

Φ) − i(A

+ iA

)Φ = 0.

So we can write

+ iA

= −i(∂

+ i∂

) log Φ.

Thus, once we’ve got Φ, we can get A

and A

The next step is to use gauge invariance. Under gauge invariance, we can fix

the phase of Φ to anything we want. We write

Φ = e

(u+iθ)

Then |Φ|

= e

We might think we can get rid of

completely. However, this is not quite

true, since the argument is not well-defined at a zero of Φ, and in general we

cannot get rid of θ by a smooth gauge transformation. But since

Φ ∼ c

(z − Z

)

near Z

, we expect we can make θ look like

θ = 2

j=1

arg(z − Z

We will assume we can indeed do so. Then we have

(∂

u + ∂

θ), A

= −

(∂

u − ∂

θ).

We have now solved for

using the first Bogomolny equation. We then use

this to work out

and obtain a scalar equation for

by the second Bogomolny

equation.

Theorem.

In the above situation, the Bogomolny equation

− |

) is

equivalent to the scalar equation for u

−∆u + (e

− 1) = −4π

j=1

This is known as Taubes’ equation.

Proof. We have

B = ∂

− ∂

= −

∂

u −

∂

u +

(∂

∂

− ∂

∂

)θ

= −

∆u +

(∂

∂

− ∂

∂

)θ.

We might think the second term vanishes identically, but that is not true. Our

has some singularities, and so that expression is not going to vanish at the

singularities. The precise statement is that (

∂

− ∂

∂

)

is a distribution

supported at the points Z

To figure out what it is, we have to integrate:

|z−Z

|≤ε

(∂

∂

− ∂

∂

)θ d

x =

|z−Z

|=ε

∂

θ dx

+ ∂

θ dx

|z−Z

|=ε

dθ = 4πn

where n

is the multiplicity of the zero. Thus, we deduce that

(∂

∂

− ∂

∂

)θ = 2π

But then we are done!

We can think of this

as a non-linear combination of fundamental solutions

to the Laplacian. Near the

functions, the

−

1 term doesn’t contribute much,

and the solution looks like the familiar fundamental solutions to the Laplacian

with logarithmic singularities. However, far away from the singularities,

−

forces u to tend to 0, instead of growing to infinity.

Taubes proved that this equation has a unique solution, which is smooth on

\{Z

}

, with logarithmic singularities at

, and such that

u →

0 as

|z| → ∞

Also, u < 0.

It is an exercise to check that the Bogomolny equations imply the second-order

Euler–Lagrange equations.

For example, differentiating the second Bogomolny equation and using the

first gives

∂

B = −(Φ, D

Φ) = (Φ, iD

Φ).

We can similarly do this for the sine-Gordon theory.

2.5 Physics of vortices

Recall we began with the ungauged Ginzburg–Landau theory, and realized the

solitons didn’t have finite energy. We then added a gauge field, and the problem

went away — we argued the coupling to the gauge field “gave mass” to the trans-

verse component, thus allowing the existence of finite energy soliton solutions. In

the book of Jaffe and Taubes there are results on the exponential decay of gauge

invariant combinations of the fields which are another expression of this effect

— the Higgs mechanism. However, there is a useful and complementary way of

understanding how gauge fields assist in stabilizing finite energy configurations

against collapse, and this doesn’t require any detailed information about the

theory at all — only scaling. We now consider this technique, which is known

either as the Derrick or the Pohozaev argument.

Suppose we work in

space dimensions. Then a general scalar field Φ :

→

has energy functional given by



|∇Φ|

+ U(Φ)



for some

. In the following we consider smooth finite energy configurations

for which the energy is stationary. To be precise, we require that the energy is

stationary with respect to variations induced by rescaling of space (as is made

explicit in the proof); we just refer to these configurations as stationary points.

Theorem

(Derrick’s scaling argument)

Consider a field theory in

-dimensions

with energy functional

E[Φ] =



|∇Φ|

+ U(Φ)



x = T + W,

with

the integral of the gradient term and

the integral of the term involving

– If d = 1, then any stationary point must satisfy

T = W.

– If d = 2, then all stationary points satisfy W = 0.

– If d ≥ 3, then all stationary points have T = W = 0, i.e. Φ is constant.

This forbids the existence of solitons in high dimensions for this type of

energy functional.

Proof.

Suppose Φ were such a stationary point. Then for any variation Φ

of Φ

such that Φ = Φ

, we have

dλ



λ=1

E[Φ

] = 0.

Consider the particular variation given by

(x) = Φ(λx).

Then we have

W [Φ

] =

U(Φ

(x)) d

x = λ

−d

U(Φ(λx)) d

(λx) = λ

−d

W [Φ].

On the other hand, since

contains two derivatives, scaling the derivatives as

well gives us

T [Φ

] = λ

2−d

T [Φ].

Thus, we find

E[Φ

] = λ

2−d

T [Φ] + λ

−d

W [Φ].

Differentiating and setting λ = 1, we see that we need

(2 − d)T [Φ] − dW [Φ] = 0.

Then the results in different dimensions follow.

The

= 2 case is rather interesting. We can still get interesting soliton

theories if we have sufficiently large space of classical vacua {Φ : W (Φ) = 0}.

Example. In d = 2, we can take  = 3 and

W (φ) = (1 − |φ|

)

Then the set

= 0 is given by the unit sphere

⊆ R

. With

constrained

to this 2-sphere, this is a

-model , and there is a large class of such maps

(

) which minimize the energy (for a fixed topology) — in fact they are just

rational functions when stereographic projection is used to introduce complex

coordinates.

Derrick’s scaling argument is not only a mathematical trick. We can also

interpret it physically. Increasing

corresponds to “collapsing” down the field.

Then we see that in

d ≥

3, both the gradient and potential terms favour collapsing

of the field. However, in

= 1, the gradient term wants the field to expand, and

the potential term wants the field to collapse. If these two energies agree, then

these forces perfectly balance, and one can hope that stationary solitons exist.

We will eventually want to work with theories in higher dimensions, and

Derrick’s scaling argument shows that for scalar theories with energy functionals

as above this isn’t going to be successful for three or more dimensions, and

places strong restrictions in two dimensions. There are different ways to get

around Derrick’s argument. In the Skyrme model, which we are going to study

in the next chapter, there are no gauge fields, but instead we will have some

higher powers of derivative terms. In particular, by introducing fourth powers of

derivatives in the energy density, we will have a term that scales as

4−d

, and

this acts to stabilize scalar field solitons in three dimensions.

Now let’s see how gauge theory provides a way around Derrick’s argu-

ment without having to depart from employing only energy densities which

are quadratic in the derivatives (as is highly desirable for quantization). To

understand this, we need to know how gauge fields transform under spatial

rescaling.

One way to figure this out is to insist that the covariant derivative D

must scale as a whole in the same way that ordinary derivatives scale in scalar

field theory. Since

∂

= λ(∂

Φ)(λx),

we would want A

to scale as

)

(x) = λA

(λx).

We can also see this more geometrically. Consider the function

: R

→ R

x 7→ λx.

Then our previous transformations were just given by pulling back along

Since A is a 1-form, it pulls back as

∗

) = λA

(λx) dx

Then since B = dA, the curvature must scale as

(x) = λ

B(λx).

Thus, we can obtain a gauged version of Derrick’s scaling argument.

Since we don’t want to develop gauge theory in higher dimensions, we will

restrict our attention to the Ginzburg–Landau model. Since we already used the

letter λ, we will denote the scaling parameter by µ. We have

, Φ

) =



B(µx)

+ µ

|DΦ(µx)|

(1 − |Φ(µx)|

)



(µx)



(y) + |DΦ(y)|

4µ

(1 − |Φ(y)|

)



Again, the gradient term is scale invariant, but the magnetic field term counter-

acts the potential term. We can find the derivative to be

dµ



µ=1

, Φ

) =



−

(1 − |Φ|

)



Thus, for a soliton, we must have

x =

(1 − |Φ|

)

Such solutions exist, and we see that this is because they are stabilized by the

magnetic field energy in the sense that a collapse of the configuration induced

by rescaling would be resisted by the increase of magnetic energy which such a

collapse would produce. Note that in the case

= 1, this equation is just the

integral form of one of the Bogomolny equations!

Scattering of vortices

Derrick’s scaling argument suggests that vortices can exist if

λ >

0. However, as

we previously discussed, for

λ 6

= 1, there are forces between vortices in general,

and we don’t get static, separated vortices. By doing numerical simulations,

we find that when

λ <

1, the vortices attract. When

λ >

1, the vortices repel.

Thus, when

λ >

1, the symmetric vortices with

N >

1 are unstable, as they

want to break up into multiple single vortices.

We can talk a bit more about the

= 1 case, where we have static multi-

vortices. For example, for

= 2, the solutions are parametrized by pairs of

points in C, up to equivalence

, Z

) ∼ (Z

, Z

We said the moduli space is

, and this is indeed true. However,

and

are

not good coordinates for this space. Instead, good coordinates on the moduli

space

are given by some functions symmetric in

and

. One valid

choice is

Q = Z

+ Z

, P = Z

In general, for vortex number

, we should use the elementary symmetric

polynomials in Z

, ··· , Z

as our coordinates.

Now suppose we set our vortices in motion. For convenience, we fix the

center of mass so that Q(t) = 0. We can then write P as

P = −Z

If we do some simulations, we find that in a head-on collision, after they collide,

the vortices scatter off at 90

◦

. This is the 90

◦

scattering phenomenon, and holds

for other λ as well.

In terms of our coordinates,

is smoothly evolving from a negative to a

positive value, going through 0. This corresponds to

7→ ±iZ

−Z

Note that at the point when they collide, we lose track of which vortex is which.

Similar to the

kinks, we can obtain effective Lagrangians for the dynamics

of these vortices. However, this is much more complicated.

2.6 Jackiw–Pi vortices

So far, we have been thinking about electromagnetism, using the abelian Higgs

model. There is a different system that is useful in condensed matter physics. We

look at vortices in Chern–Simons–Higgs theory. This has a different Lagrangian

which is not Lorentz invariant — instead of having the Maxwell curvature term,

we have the Chern–Simons Lagrangian term. We again work in two space

dimensions, with the Lagrangian density given by

L =

µνλ

νλ

− (iΦ, D

Φ) −

|DΦ|

2κ

|Φ|

where κ is a constant,

νλ

= ∂

− ∂

is the electromagnetic field and, as before,

Φ =

∂Φ

∂t

− iA

Φ.

The first term is the Chern–Simons t erm, while the rest is the Schr¨odinger

Lagrangian density with a |Φ|

potential term.

Varying with respect to Φ, the Euler–Lagrange equation gives the Schr¨odinger

equation

Φ +

|Φ|

Φ = 0.

If we take the variation with respect to A

instead, then we have

κB + |Φ|

= 0.

This is unusual — it looks more like a constraint than an evolution equation,

and is a characteristic feature of Chern–Simons theories.

The other equations give conditions like

∂

= ∂

(iΦ, D

Φ)

∂

= ∂

−

(iΦ, D

Φ).

This is peculiar compared to Maxwell theory — the equations relate the current

directly to the electromagnetic field, rather than its derivative.

For static solutions, we need

∂

(iΦ, D

Φ).

To solve this, we assume the field again satisfies the covariant anti-holomorphicity

condition

Φ = iε

Φ = 0.

Then we can write

∂

= +

(iΦ, D

Φ) = −

∂

|Φ|

and similarly for the derivative in the second coordinate direction. We can then

integrate these to obtain

= −

|Φ|

2κ

We can then look at the other two equations, and see how we can solve those.

For static fields, the Schr¨odinger equation becomes

Φ +

|Φ|

Φ = 0.

Substituting in A

, we obtain

Φ = −

|Φ|

Let’s then see if this makes sense. We need to see whether this can be consistent

with the holomorphicity condition. The answer is yes, if we have the equation

κB + |Φ|

= 0. We calculate

Φ = D

Φ + D

Φ = i(D

− D

)Φ = +BΦ = −

|Φ|

Φ,

exactly what we wanted.

So the conclusion (check as an exercise) is that we can generate vortex

solutions by solving

Φ − iε

Φ = 0

κB + |Φ|

= 0.

As in Taubes’ theorem there is a reduction to a scalar equation, which is in this

case solvable explicitly:

∆ log |Φ| = −

|Φ|

Setting ρ = |Φ|

, this becomes Liouville’s equation

∆ log ρ = −

ρ,

which can in fact be solved in terms of rational functions — see for example

Chapter 5 of the book Solitons in Field Theory and Nonlinear Analysis by Y.

Yang. (As in Taubes’ theorem, there is a corresponding statement providing

solutions via holomorphic rather than anti-holomorphic conditions.)

3 Skyrmions

We now move on to one dimension higher, and study Skyrmions. In recent years,

there has been a lot of interest in what people call “Skyrmions”, but what they

are studying is a 2-dimensional variant of the original idea of Skyrmions. These

occur in certain exotic magnetic systems. But instead, we are going to study the

original Skyrmions as discovered by Skyrme, which have applications to nuclear

physics.

With details to be filled in soon, hadronic physics exhibits (approximate)

spontaneously broken chiral symmetry

SU(2)

×SU(2)

∼

(4), where the

unbroken group is (diagonal)

(3) isospin, and the elements of

(3) are

(g, g) ∈ SU(2) × SU(2).

This symmetry is captured in various effective field theories of pions (which

are the approximate Goldstone bosons) and heavier mesons. It is also a feature

of QCD with very light u and d quarks.

The special feature of Skyrmion theory is that we describe nucleons as solitons

in the effective field theory. Skyrme’s original idea was that nucleons and bigger

nuclei can be modelled by classical approximations to some “condensates” of

pion fields. To explain the conservation of baryon number, the classical field

equations have soliton solutions (Skyrmions) with an integer topological charge.

This topological charge is then identified with what is known, physically, as the

baryon number.

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.

3.2 Skyrmion solutions

The simplest Skyrmion solution has baryon number

= 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)

x · τ .

This is manifestly in

(2), because

cos

sin

= 1. This is known as a

hedgehog, because the unit pion field is

, which points radially outwards. We

need some boundary conditions. We need

U → 1

∞

. 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:

After some hard work, we find that the energy is given by

E = 4π

∞



2 sin

(1 + f

) +

sin



dr.

From this, we can obtain a second-order ordinary differential equation in

which is not simple. Solutions have to be found numerically. This is a sad truth

about Skyrmions. Even in the simplest

= 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

To compute the baryon number of this solution, we plug our solution into the

formula, and obtain

B = −

2π

∞

sin

· 4πr

dr.

We can interpret

as the radial contribution to

, while there are two factors

sin f

coming from the angular contribution due to the i sin f(r)

x · τ term.

But this integral is just an exact differential. It simplifies to

B =

2 sin

f df.

Note that we have lost a sign, because of the change of limits. We can integrate

this directly, and just get

B =



f −

sin 2f



= 1,

as promised.

Intuitively, we see that in this integral,

goes from 0 to

, and we can think

of this as the field U wrapping around the sphere S

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

. 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

itself, because the Skyrmion

“onion” structure highly distorts each

= 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

. 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

times the

= 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

= 1 hedgehog, but

introduce more angular twists. We can think of the above

= 1 solution as

follows — we slice up our domain

(or rather,

since we include the point

at infinity) into 2-spheres of constant radius, say

[

r∈[0,∞]

On the other hand, we can also slice up the

in the codomain into constant

levels, which are also 2-spheres:

[

Then the function

(

) we had tells us we should map the 2-sphere

into the

two sphere

cos f(r)

. What we did, essentially, was that we chose to map

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

→ S

. Then we

can construct the map Σ

cos f

that sends

cos f(r)

via the map

. For

simplicity, we will use the same

for all

. If we did this, then we obtain a

non-trivial map Σ

cos f

R : S

→ S

Since

itself is a map from a sphere to a sphere (but one dimension lower),

also has a degree. It turns out this degree is the same as the degree of the

induced map Σ

cos f

! So to produce higher baryon number hedgehogs, we simply

have to find maps R : S

→ S

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

(

) of degree

. This is a map

→ S

. We

use coordinates

r, z

, where

r ∈ R

and

z ∈ C

∞

C ∪ {∞}

∼

. Then we can

look at generalized hedgehogs

U(x) = cos f(r)1 + i sin f (r)

(z)

· τ ,

with

(0) =

(

∞

) = 0, and

is the normalized pion field

, given by the

unit vector obtained from

(

) if we view

as a subset of

in the usual

way. Explicitly,

1 + |R|

(

R + R, i(

R − R), 1 − |R|

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

(

), as well as the rational function

(

). The geometric degree

(

) we

care about is the same as the algebraic degree of R

(z). Precisely, if we write

(z) =

(z)

where

and

are coprime, then the algebraic degree of

is the maximum of

the degrees of

and

as polynomials. Since there are finitely many coefficients

for

and

, 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

(z) = z R

(z) = z

(z) =

√

3iz

− 1

−

√

3iz

(z) =

+ 2

√

3iz

+ 1

− 2

√

3iz

+ 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

= 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

as the suspension

, and our construction of Σ

cos f

from

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.