notes topol sol

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We are about to study something called topological solitons. Roughly speaking, solitons
are solutions of classical field equations having a particle-like behaviour. Quantisation appears
as corrections. The topology or the topological properties of these solutions are responsible
for their stability.
1 Theories with a scalar field in 1 + 1 dimensions
Lets consider a real field ϕ(t, x) ∈ IR in a Minkowski spacetime with metric diag(1,−1) and
whose dynamics is governed by the action
S =
∫
dt
∫
dx
1
2
∂µϕ ∂
µϕ− U(ϕ) (1.1)
where in the last term, U(ϕ), we put the interactions, and therefore, it’s there where the
non-linearity is introduced.
The lagrangian density reads L = 1
2
ϕ˙2− 1
2
ϕ′2−U(ϕ) and the stress-energy tensor is easily
calculated from T µν = ∂L
∂∂µϕ
∂νϕ − gµνL. The energy density is the T 00 component of it, and
the energy is obtained by integrating it over the entire space:
E =
∫
dx
1
2
ϕ˙2 +
1
2
ϕ′2 + U(ϕ). (1.2)
The potential U(ϕ) must be bounded from below, so we take its minimum to be zero and then
U(ϕ) > 0.
The equation of motion for this field is
ϕ¨− ϕ′′ + dU
dϕ
= 0 (1.3)
and in the static case, i.e., ϕ˙ = 0, one can integrate it by quadrature. First, we multiply
the static equation by ϕ′ getting ϕ′ϕ′′ − dU
dx
= 0. Then, we recognize that the LHS is a total
derivative: d
dx
(
1
2
ϕ′2 − U) = 0. Integration of that gives
1
2
ϕ′2 − U(ϕ) = const.
and the constant of integration is fixed by the boundary conditions, which in turn are chosen
as a finite energy requirement: ϕ′ → 0, U → 0 as |x| → ∞. Thus
1
2
ϕ′2 − U(ϕ) = 0 (1.4)
1
For the non-static case we also impose ϕ˙ → 0. Then, at spatial infinity the field becomes
static and constant: ϕ± = limx→±∞ ϕ(t, x).
Among the finite energy configurations we pick the one that has a minimum energy. This
is the vacuum state. Notice that such boundary conditions (and therefore the finite energy
requirement) cannot be fulfilled for a theory with a single vacuum. This can be easily seen by
using an analogy with a mechanical system. If we regard x in our theory as the time, ϕ(x)
can be seen as the position of a particle, and taking V ≡ −U , the static version of equation
(1.3) is just the Newton’s equation of a particle under the action of the force −dV
dϕ
.
Now, if U has a single minimum at ϕ0, say U(ϕ0) = 0, or equivalently, V has a single
maximum at that point, it’s easy to conclude that we never get to a static configuration (i.e.,
to a constant value of ϕ) after we leave ϕ0, since the kinetic energy
1
2
ϕ′2 is always growing1
(1
2
ϕ′2 = −V ). On the other hand, if the potential has, say, two vacua at ϕ0 and ϕ1, then two
possibilities of motion connecting these points arise: from ϕ0 to ϕ1 and the opposite, in the
case we start at ϕ1. In general, for a potential with n vacua there are 2(n− 1) possibilities of
motion between any two neighbouring vacua as x goes from −∞ to +∞.
V
ϕ0
quinta-feira, 5 de julho de 2012
quinta-feira, 5 de julho de 2012
(a) Single vacuum poten-
tial.
ϕ0 ϕ1
V
quinta-feira, 5 de julho de 2012
(b) Two vacua potential.
Figure 1: A multiple vacua potential is needed to fulfil the finite energy requirement.
1.1 Bogomol’nyi bound
Using the topological properties of the solution we can find an energy bound for them. The
question we must ask is: what’s the lowest possible energy for such a solution connecting the
two vacua ϕ+ and ϕ−?
Consider the expression of the energy density: E = 1
2
ϕ˙2 + 1
2
ϕ′2 + U . Notice that the
1Remember that the constant that appears when we integrate the static equation of motion if fixed to zero,
which in turn is the total energy of the mechanical system.
2
potential part can be rewritten as
1
2
ϕ˙2 + U =
(
1√
2
ϕ′ ±
√
U
)2
∓ ϕ′
√
2U .
The first term at the RHS is obviously positive, and therefore 1
2
ϕ˙2 + U ± ϕ′√2U > 0, which
in turn implies that 1
2
ϕ˙2 + U > ϕ′√2U . Then, writing E = T + V where T ≡ ∫ dx 1
2
ϕ˙2 and
V ≡ ∫ dx 1
2
ϕ˙2 + U , we have
V > ±
∫
dx ϕ′
√
2U = ±
∫ ϕ+
ϕ−
√
2Udϕ.
Now, since T is positive we can write
E >
∣∣∣∣∫ ϕ+
ϕ−
√
2Udϕ
∣∣∣∣ . (1.5)
Remarkably the minimum value of the energy does not depend on how you go from one vacuum
to the other. This is exactly because it is a topological bound: the energy bound depends
basically on on the values of ϕ− and ϕ+, and not on the details of the path you choose linking
them.
Define now what’s called the superpotential
U(ϕ) ≡ 1
2
(
dW
dϕ
)2
.
Then, equation (1.5) becomes
E > |W (ϕ+)−W (ϕ−)| (1.6)
and |W (ϕ+) −W (ϕ−)| is the least potential energy one needs for a transition between the
vacua.
Before we discuss some examples we remark that once the static solution is found, the
moving one is obtained by taking a Lorentz boost.
1.2 ϕ4 kinks
Consider the potential U = λ (m2 − ϕ2)2 with λ and m positive constants. The vacuum state
is obtained by minimizing this function:
dU
dϕ
= 2λ(m2 − ϕ2)(−2λϕ) != 0 ⇒ ϕ0 = ±m & ϕ0 = 0.
3
-2 -1 1 2
2
4
6
8
Figure 2: A plot of the ϕ4 potential with λ = 1 and m = 1. The vacua are at ϕ = 0 and
ϕ = ±m.
Now (1.4) gives x− x0 = ± 1√2λm2
∫
dϕ
1−( ϕm)
2 and defining ψ ≡ ϕm this can be written as
x− x0 = ± 1√
2λm
∫
dψ
1− ψ2 .
Changing the variable again to ψ = tanh z the integration becomes trivial:
x− x0 = ± 1√
2λm
arctanh
( ϕ
m
)
.
The final step is to invert this relation, which gives the kink and anti-kink solutions:
ϕ = ±m tanh
(√
2λm(x− x0)
)
(1.7)
The position of the kink can be understood as the point x0. One can see the kink moving
-10 -5 5 10
-1.0
-0.5
0.5
1.0
(a) A kink (from ϕ− to ϕ+) and an anti-kink.
-10 -5 5 10
-1.0
-0.5
0.5
1.0
(b) The solutions with x0 6= 0. One can see
x0 as the position of the (anti-)kink.
Figure 3: The solutions ϕ = ± tanh (√2(x− x0)) for x0 = 0 and x0 = 2.
with velocity v ∈ (−1, 1) by replacing x − x0 by x−x0−vt√1−v2 , which corresponds to the Lorentz
4
boost:
ϕv = ±m tanh
(√
2λm
(
x− x0 − vt√
1− v2
))
. (1.8)
One can easily compute the energy of this static solution: M =
∫ +∞
−∞ dx
(
1
2
ϕ′2 + U). In fact,
-10 -5 5 10
-1.0
-0.5
0.5
1.0
(a) The kink at some t < 0.
-10 -5 5 10
-1.0
-0.5
0.5
1.0
(b) The kink at some t > 0.
Figure 4: The moving kink ϕ = ± tanh
(√
2( x−vt√
1−v2 )
)
with v = 0.8.
from (1.4) the energy density is simply E = 2U . So, plugging (1.7) here we get easily
E = 2λm4sech4
(√
2λm(x− x0)
)
.
The energy density is a lump. In fact, another way to define the position of the kink is from
the center of the lump. Let us perform the integration of the energy density over space2:
M =
∫
dx 2λm4sech4
(√
2λm(x− x0)
)
=
√
2λm3
∫
dy sech4y
=
√
2λm3
∫
dy
16
(ey + e−y)4
=
√
2λm3
∫
dy
16e4y
(1 + e2y)4
=
√
2λm3
∫
du
8(u− 1)
u4
=
4
3
√
2λm3.
Using the moving solution (1.8) we calculate the energy3 E =
∫
dx 1
2
ϕ˙2 + 1
2
ϕ′2 + U .
E = 1
2
ϕ˙2 +
1
2
ϕ′2 + U = 1
2
2λm4γ2v2sech4z +
1
2
m42λγ2sech4z + λm4
(
1− tanh2 z)2
= λγ2m4sech4z
(
v2 + 1
)
+ λm4sech4z
=
2
1− v2λm
4sech4z
2We used first y =
√
2λm(x− x0) and then u = 1 + e2y.
3Use γ = 1√
1−v2 and z =
√
2λmγ(x− x0 − vt).
5
-10 -5 5 10
-1
1
2
3
4
5
Figure 5: A plot of the kink and its energy density.
Then, we get E = M√
1−v2 . We see that the static energy of the kink can be seen as the rest
mass of a relativistic particle and when it moves, we have the relativistic energy relation.
Now, the vacua (where the potential is minimum) are just limx→±∞ ϕ(t, x) =
±m, so we
have the four possibilities for the values of the field:
(ϕ−, ϕ+) = {(−m,−m), (−m,+m), (+m,−m), (+m,+m)} .
We can divide this set into two categories: the vacuum sectors, which are the first and fourth
possibilities above, and the non-perturbative sectors, which correspond to the transition be-
tween minima.
The boundary conditions hold the kink stable in the sense that one cannot deform it using
perturbations.
Consider the quantity
N =
1
2m
(ϕ+ − ϕ−) . (1.9)
This is clearly time independent since the vacua values are time independent. Then, concerning
the four possibilities above we get N = 0 for the vacuum sectors and N = ±1 for the non-
perturbative sectors. This quantity classifies the sectors and also it says whether we are going
from ϕ− to ϕ+ or the opposite. Notice that we can write it in terms of the derivative of the
field:
N =
∫ +∞
−∞
ϕ′
2m
dx. (1.10)
This number is called the topological charge.
6
Using the superpotential we have
U = 1
2
(
dW
dϕ
)2
= λ
(
m2 − ϕ2)2
and therefore
W =
√
2λϕ
(
m2 − 1
3
ϕ2
)
.
So, we get W (+m) = 2
3
√
2λm3 and W (−m) = −2
3
√
2λm3, and from the Bogomol’nyi bound
the minimum value of the potential is |W (+m) −W (−m)| = 4
3
√
2λm3. Well, we can write
the superpotential as W (ϕ±) = 23
√
2λm2ϕ±, which gives |W (+m)−W (−m)| = 43
√
2λm3|N |,
showing that for a kink or anti-kink whose topological charges are 1 and −1 respectively we
get the energy M = 4
3
√
2λm3 given above as the minimum possible for the transition between
the vacua.
1.3 Sine-Gordon kinks
The lagrangian density of the sine-Gordon theory reads
L = 1
2
∂µϕ∂
µϕ− (1− cosϕ). (1.11)
Clearly the potential has infinitely many vacua: dU
dϕ
= sinϕ
!
= 0 ⇒ ϕ0 = 2pin, with
n ∈ ZZ. One vacuum can be mapped into another with the use of the symmetry transformation
ϕ→ ϕ+ 2pi.
Now, the Bogomol’nyi equation gives
1
2
ϕ′2 = 1− cosϕ ⇒ dx = ± 1√
2
dϕ√
1− cosϕ.
In order to integrate that we use cosϕ = 1 − 2 sin2(ϕ
2
) and define first ψ = ϕ
2
, and then
u = cosψ so that we get
x−x0 = ±
∫
dψ sinψ
1− cos2 ψ = ±
∫ −du
(1− u)(1 + u) = ∓
∫
du
2(1− u)∓
∫
du
2(1 + u)
=
1
2
ln
(
1− u
1 + u
)
.
Finally this can be inverted to give the sine-Gordon kink
ϕ = 4arctan
(
e±(x−x0)
)
. (1.12)
Again, for the moving soliton one takes the Lorentz boost x− x0 → x−x0−vt√1−v2 .
7
The topological charge is now defined as
N =
∫ +∞
−∞
1
2pi
ϕ′ dx (1.13)
Let us assume that we start from ϕ− = 0. Then, the neighbouring vacuum is ϕ+ = 2pi
and the topological charge for the solution (kink) linking them is therefore N = 1. Next
5 10 15 20 25
0.5
1.0
1.5
2.0
(a) U = 1− cosϕ.
-10 -5 5 10
1
2
3
4
5
6
(b) Kink of sine-Gordon.
Figure 6: The kink links two neighbouring vacua of the sine-Gordon potential.
we calculate the Bogomol’nyi bound. Considering ϕ− = 0 we get ϕ+ = 2piN and therefore
E >
∫ ϕ+
ϕ−
√
2U dϕ = ∫ 2piN
0
∣∣sin ϕ
2
∣∣ dϕ = −4|N | cos ϕ
2
∣∣2pi
0
= 8|N |. The energy of the kink, with
topological charge N = 1 is E = 8. As an exercise we check this explicitly. The energy density
(for the static solution) reads E = 1
2
ϕ′2 + 1− cosϕ. From (1.12) we get
tan
ϕ
4
= ex−x0 ⇒ 1
4
ϕ′ sec2
(ϕ
4
)
= ex−x0 ⇒ 1
4
(
1 + tan2
(ϕ
4
))
ϕ′ = ex−x0
Writing 1 + tan2
(
ϕ
4
)
= 1 + e2(x−x0) we get the first term of E :
ϕ′2
2
=
8e2(x−x0)
(1 + e2(x−x0))2
.
For the cosϕ appearing there we have
cosϕ = 1− 2 sin2
(ϕ
2
)
= 1− 2 · 4 sin2
(ϕ
4
)
cos2
ϕ
4
= 1− 8e2(x−x0) cos4
(ϕ
4
)
= 1− 8e
2(x−x0)
(1 + e2(x−x0))2
.
Then, finally,
E = 16e
2(x−x0)
(1 + e2(x−x0))2
= 4sech2 (x− x0) .
8
Notice that it peaks at the point x = x0, which can be regarded as the position of the kink.
Furthermore, this same conclusion can be derived from (1.12): ϕ(x0) = 4 arctan(1) = pi,
which is half-way between the vacua 0 and 2pi.
Integration of this energy density over the whole space gives the energy of the kink4:
M = 16
∫ +∞
−∞ dx
e2(x−x0)
(1+e2(x−x0))
2 = 8
∫
du
u2
= − 8
1+e2(x−x0)
∣∣+∞
−∞ = 8, as expected. Again, for a moving
solution one gets a relativistic-like relation for the mass of a particle.
2 Topology in field theory
Lets reconsider the sine-Gordon model. Take the vector~ϕ = (ϕ1, ϕ2) such that ~ϕ · ~ϕ = 1. Its
components can be written in terms of a single parameter φ as ~ϕ = (sinφ, cosφ). Then, the
lagrangian
L = 1
2
∂µ~ϕ · ∂µ~ϕ− (1− ϕ2) + λ(1− ~ϕ · ~ϕ)
describes the dynamics of φ, the sine-Gordon field. The vacuum state is ~ϕ = (0, 1), i.e., for
ϕ2 = 1 we get φ = 2pin, the vacua we met before. However the field ~ϕ has now the same limit
for x → −∞ and x → −∞: lim|x|→∞ ~ϕ = (0, 1). The points ±∞ are therefore identified:
IR ∪ {∞} ' S1. So, the field ~ϕ that was a map ~ϕ : IR → S1 is now, due to the boundary
conditions, the map ~ϕ : S1 → S1. One could write ~ϕ : S1space → S1field, but we shall not bother
with that.
The topological charge is now the number of times the field ~ϕ winds around the circle
while x goes from −∞ to +∞, i.e., while the compactified space is covered once:
N =
1
2pi
∫
dx �abϕ
′
a ϕb. (2.14)
2.1 Basic notions on homotopy theory
Suppose X and Y two manifolds without boundary. The map Ψ between them is such that
x0 ∈ X is mapped into y0 ∈ Y : Ψ(x0) = y0. If one can continuously deform a map Ψ0 into
another, Ψ1, then these maps are said to be homotopic. If X is a sphere, things become very
interesting.
The set of homotopy classes of Ψ : Sn → Y is denoted by Πn(Y ). Π1(Y ) is called the first
fundamental group. For instance, the map S1 → S1 is in Π1(S1) = ZZ. Its not difficult to see
that Π1(S
n) = 1l for n ≥ 2, since any loop on Sn can be contracted to the point-loop based at
y0. A small list of some homotopy groups is the following:
4Using y = x− x0 and u = 1 + e2y.
9
quinta-feira, 12 de julho de 2012
Figure 7: Any loop on S2 can be contracted to the point-loop 1lp.
Πn(S
n) = ZZ, ∀ n ≥ 1; Πn(Sd) = 1l, ∀ n < d;
Π3(S
2) = ZZ; Πn+1(S
n) = ZZ2, ∀ n ≥ 3; Πn+2(Sn) = ZZ2, ∀ n ≥ 2.
Π0(Y ) is the set of unbased homotopy classes of maps from a single point to Y . It is the
set of distinct connected components of Y . We have that Π0(Y ) = 1l ⇔ Y is connected and
Π1(Y ) = 1l⇔ Y is simply connected.
Sometimes a related quantity to the homotopy class is the topological degree. Take Ψ :
X → Y with dimX = dimY . Let Ω be the volume measured on Y normalised to unit:∫
Y
Ω = 1; For instance, if Y = S2 then Ω = 1
4pi
sin θdθdφ. Locally
Ω = β(~y)dy1 ∧ · · · ∧ dyd
and Ψ = ~y(~x). Then the pull-back of Ψ reads5
Ψ∗(Ω) = β(~y(~x))
∂y1
∂xj
dxj ∧ · · · ∧ ∂y
d
∂xk
dxk = β(~y(~x)) det
(
∂yi
∂xj
)
dx1 ∧ · · · ∧ dxd.
The topological degree is then defined as
degΨ =
∫
X
Ψ∗(Ω). (2.15)
2.1.1 1 dimensional case
The sine-Gordon theory as we say can be described by a field ϕ : IR ∪ {∞} = S1 → S1.
The volume form of the target space S1 is simply dl = Rdϕ (but R = 1!). Integrating it
gives L = 2pi, so, in order to normalize we divide by 2pi: Ω = 1
2pi
dϕ. From here we can
5It is convenient to denote the jacobian by J(~x) = det
(
∂yi
∂xj
)
.
10
recognize that β = 1
2pi
, and dϕ is just the basis of the target space. The map is Ψ = ϕ(x),
just to compare with our previous abstract notation. Then, the pull-back of the volume form
is Ψ∗(Ω) ≡ ϕ∗(Ω) = 1
2pi
∂ϕ
∂x
dx, and the topological degree becomes
degϕ =
∫ 2pi
0
1
2pi
∂ϕ
∂x
dx =
ϕ(2pi)− ϕ(0)
2pi
which is just the topological charge N ∈ Π1(S1) = ZZ.
2π0 2π0
4π
quinta-feira,
12 de julho de 2012
(a) As x goes from −∞ to∞, the target space
is covered twice .
 0
 2
 4
 6
 8
 10
 12
 14
-20 -15 -10 -5 0 5 10 15 20
Va
lue
 o
f t
he
 fie
ld 
Space 
The 2-kink solution
kink
Initial guess
(b) 2-kink solution of sine-Gordon.
Figure 8: The topological charge of the 2-kink solution is N = 2.
2.1.2 2 dimensional case
Now consider the field ~ϕ : S2 → S2; ~ϕ(x1, x2) = (ϕ1, ϕ2, ϕ3). Notice that since the target
space is the sphere S2, the coordinates their satisfy ~ϕ · ~ϕ = 1. Finite energy requires that as
|x| → ∞ the field ~ϕ approaches a constant vector. Here the homotopy group is Π2(S2) = ZZ,
thus each field configuration is characterized by an integer.
The normalized volume of the target space is (the area of the sphere) Ω = 1
4pi
sin θ dθ∧dφ.
Then, in terms of the physical space with coordinates x we get dθ ∧ dφ = ∂θ
∂xi
∂φ
∂xj
dxi ∧ dxj =
∂θ
∂xi
∂φ
∂xj
�ijdx1dx2, so ϕ∗(Ω) = 1
4pi
sin θ (∂1θ∂2φ− ∂2θ∂1φ) dx1dx2. It’s straightforward to check
that writing ~ϕ = (cos θ, sin θ cosφ, sin θ sinφ) one gets ~ϕ·(∂1~ϕ× ∂2~ϕ) = sin θ (∂1θ∂2φ− ∂2θ∂1φ).
Then, the degree of the field is
deg~ϕ =
1
8pi
∫
�ij ~ϕ · (∂i~ϕ× ∂j ~ϕ) dx1dx2
where the extra 1
2
factor comes from ∂1~ϕ× ∂2~ϕ = 12�ij∂i~ϕ× ∂j ~ϕ.
11
3 Derrick’s theorem
Consider time-independent fields. Let E be the energy of a field configuration and let E(λ)
be the energy after a spatial rescaling ~x→ λ~x, λ > 0. If there are no critical points when one
considers the variation due to rescaling then there is no chance to get a soliton-like solution.
In (flat) IRd (where d is the number of spatial dimensions), a theory with a scalar field ϕ
has the energy functional6
E =
∫
ddx ∂iϕ∂iϕ+ U(ϕ) ≡ E2 + E0. (3.16)
A rescaling xi → yi = λxi implies that7
E → E(λ) =
∫
ddy λ−d
(
λ2∂iϕ∂iϕ+ U(ϕ)
)
.
This can be better visualised as
E(λ) = λ2−dE2 + λ−dE0. (3.17)
Stationary points of this energy with respect to the parameter λ are given by
(2− d)λ1−dE2 − dλ−(d+1)E0 = 0.
• For d = 1 we get E(λ) = λE2 + E0λ and stationary points are found if E2 − E0λ2 = 0 is
satisfied, which is possible when λ =
√
E0
E2
. The static solution corresponds to λ = 1,
which gives E2 = E0, the virial theorem.
• From d = 2 we have E(λ) = E2 + E0λ2 , which has no stationary points... unless we take
E0 = 0. In this case the static theory is conformally invariant. The solutions are not
exactly solitons; they are called lumpsm since their size is not fixed.
One can get solitons with E = E4 + E2 + E0. After rescaling
E(λ) = λ4−dE4 + λ2−dE2 + λ−dE0.
• For d = 2 we get E(λ) = λ2E4 + E2 + E0λ2 , and the solutions are the Baby-Skyrmions.
• In d = 3 the energy rescales as E(λ) = λE4 + E2λ + E0λ3 (the E0 term is in fact optional!).
Solutions here are Skyrmions.
6The indices 2 and 0 in E2 and E0 stand for the number of derivatives in each term.
7Here ϕ→ ϕ(λ)(λ~x), but we are using the same old notation!
12
In a gauge theory with a scalar field one gets
E =
∫
F 2 + (Dϕ)2 + U(ϕ) d4x = E4 + E2 + E0
the covariant derivative is D = d + A and A scales like d: A → A
λ
. One finds solitons in
such theories in d = 2 (vortices) and d = 3 (monopoles). The case d = 4 is important for
instantons.
4 (2 + 1) non-linear O(3) σ-model lumps
As we saw before in d = 2 with a scalar field the Derrick’s theorem states that E0 = 0 gives
you a possibility of finding soliton-like solutions. In order to turn this theory less trivial than
just a free field lets take the target space to be S2, so that the lagrangian is now augmented
by a constraint:
L = 1
4
∂µ~ϕ · ∂µ~ϕ+ λ
(
1− |~ϕ|2) . (4.18)
The equations of motion are then
1
2
∂2~ϕ+ 2λ~ϕ = 0 1− ~ϕ · ~ϕ = 0.
In fact, one can solve the constraint as follows. First take the inner product of the equation
by ~ϕ, which gives 1
2
~ϕ · ∂2~ϕ + 2λ~ϕ · ~ϕ = 0. Then, notice that in the last term we can use
~ϕ · ~ϕ = 1. For the first term we also use this constraint to realise that ∂µ (~ϕ · ~ϕ) = 2~ϕ ·∂µ~ϕ = 0
and therefore 0 = ∂µ (~ϕ · ∂µ~ϕ) = ∂µ~ϕ · ∂µ~ϕ+ ~ϕ · ∂2~ϕ. Thus λ = 14∂µ~ϕ · ∂µ~ϕ, and the equation
of motion becomes
∂2~ϕ+ (∂µ~ϕ · ∂µ~ϕ) ~ϕ = 0. (4.19)
The static energy is
E =
1
4
∫
d2x ∂i~ϕ · ∂i~ϕ (4.20)
and in the ground state it is minimized: goes to zero, in this case. This requires the field
to be homogeneous at infinity (∂i~ϕ = 0). By choosing a constant vector (the ground state)
~ϕ(∞) = (0, 0, 1) the O(3) symmetry is broken to O(2). Then, all the points at infinity are
identified, so that the space is compactified as IR2∪{∞} ' S2 and the field becomes the map
~ϕ : S2 → S2, and the relevant homotopy group is Π2(S2) = ZZ, so that each configuration is
characterized by an integer.
Lets recover the expression for the topological charge in a more visual way. Consider a
map of the 2 dimensional region around the point ~x in S2phys to the region around ~ϕ(~x) in
13
dx1
dx2
(d�ϕ)1
(d�ϕ)2
quinta-feira, 12 de julho de 2012
Figure 9: A map of a region around ~x to that around ~ϕ(~x).
S2target. A vector perpendicular to the plane in the target space will be d~Σ = (d~ϕ)1 × (d~ϕ)2.
This can be parallel or anti-parallel to ~ϕ, which is orthogonal to the sphere. The area of the
unit sphere is 4pi, thus if this sphere is covered N times, this number is given by
N =
1
4pi
∫
~ϕ · d~Σ.
The numerator in this expression is basically the surface area swept by the mapping. The
inner product with ~ϕ was considered in order to take into account the plus or minus sign due
to the orientation of the coordinate basis of S2target with respect to that of S
2
phys. With that
we recover the expression for the topological charge
N =
1
8pi
∫
�ij ~ϕ · (∂i~ϕ× ∂j ~ϕ) dx1dx2 (4.21)
4.1 What is topological in the charge?
Lets consider a variation ~ϕ → ~ϕ + δ~ϕ in the above expression for the topological charge and
see what happens. First, rewrite it as
N =
1
8pi
∫
d2x �abc�ijϕ
a∂iϕ
b∂jϕ
c.
Then,
δN =
1
8pi
∫
d2x �abc�ijδϕ
a∂iϕ
b∂jϕ
c +
1
4pi
∫
d2x �abc�ijϕ
aδ∂iϕ
b∂jϕ
c.
Notice that δ~ϕ and ∂i~ϕ are parallel, thus δ~ϕ · (∂i~ϕ× ∂j ~ϕ) �ij vanishes since the quantity inside
the parenthesis is orthogonal to δ~ϕ. So,
δN =
1
4pi
∫
d2x �abc�ij∂i
(
ϕaδϕb∂jϕ
c
)− 1
4pi
∫
d2x �abc�ijδϕ
b∂i (ϕ
a∂jϕ
c)
14
This last term on the RHS also vanishes due to the fact that ∂i~ϕ × ∂j ~ϕ is orthogonal to δ~ϕ
and also because of the quantity �ij∂i∂j ~ϕ = 0.
Finally, we conclude that δN = 0, i.e., the topological charge doesn’t change when we
change the field configuration.
4.2 Bogomol’nyi bound
Consider the vector F ai = ∂iϕ
a ± �abc�ijϕb∂jϕc, and construct the scalar
F ai F
a
i =
(
∂iϕ
a ± �abc�ijϕb∂jϕc
) (
∂iϕ
a ± �ade�ikϕd∂kϕe
)
= ∂iϕ
a∂iϕ
a ± 2�abc�ij∂iϕaϕb∂jϕc + �abc�ij�ade�ikϕb∂jϕcϕd∂kϕe.
The last term is δjk
(
δbdδce − δbeδcd)ϕbϕd∂jϕc∂kϕ = ϕbϕb∂jϕc∂jϕc−ϕbϕc∂jϕc∂jϕb =���>1|~ϕ|2∂j ~ϕ ·
∂j ~ϕ−�����
�: 0
(~ϕ · ∂j ~ϕ)2 = ∂j ~ϕ · ∂j ~ϕ. Then
F ai F
a
i = ∂iϕ
a · ∂iϕa ± 2�abc�ij∂iϕaϕb∂jϕc + ∂iϕa∂iϕa
= 2∂iϕ
a∂iϕ
a ± 2�abc�ij∂iϕaϕb∂jϕc︸ ︷︷ ︸
�bac�ij∂iϕbϕa∂jϕc
= 2∂iϕ
a∂iϕ
a ∓ 2�abc�ijϕa∂iϕb∂jϕc.
Next we note that
∫
F ai F
a
i d
2x > 0, which implies that∫
d2x ∂iϕ
a∂iϕ
a > ±
∫
�abc�ijϕ
a∂iϕ
b∂jϕ
c = 8piN
and from equation (4.20) we get the Bogomol’nyi bound E > 2piN , which defines the minimum
energy for each topological sector (each N).
Notice that the Bogomol’nyi equation
∂iϕ
a ± �abc�ijϕb∂jϕc = 0 (4.22)
is first order while the equation of motion (4.19) is second order. This simplifies the problem
a lot! Lets check that solving the Bogomol’nyi equation is as good as solving the equations of
motion.
Taking
the derivative of (4.22) we have
∂i∂iϕ
a = ∓∂i
(
�abc�ijϕ
b∂jϕ
c
)
= ∓�abc�ij∂iϕb∂jϕc.
15
Now we replace, say, the first derivative term above ∂iϕ
b by ∓�bde�ikϕd∂kϕe, using the Bogo-
mol’nyi equation again, which gives
∂i∂iϕ
a = δjk
(
δcdδae − δceδad)ϕd∂kϕe∂jϕc = (ϕc∂jϕc) ∂jϕa − (∂jϕc∂jϕc)ϕa.
Using now ∂j (ϕ
aϕa) = 2∂jϕ
aϕa = 0 we get
∂i∂iϕ
a + (∂jϕ
c∂jϕ
c)ϕa = 0
which are exactly the equations of motion.
4.3 The N = 1 axially symmetric solution
Lets consider the following ansatz for the field:
~ϕ = (
x1
r
sin f(r),
x2
r
sin f(r), cos f(r)).
It is convenient to introduce the notation nα = x
α
r
, with α = 1, 2, so that ~n is the unit vector
in two dimensions: ϕaϕa = ϕαϕα + ϕ3ϕ3 =����:
1
nαnα sin2 f + cos2 f = 1.
Now we have
∂iϕ
α = ∂in
α sin f + nα cos f∂if and ∂in
α =
∂ix
α
r
− x
α
r2
∂ir.
Since r =
√
xjxj we get ∂ir =
1
2
∂i(xjxj)√
xkxk
=
2δijx
j
2r
= ni, where we used ∂ix
α = δiα. Then
∂iϕ
α = sin f
(
δiα
r
− x
αni
r2
)
+ nα cos f ∂if︸︷︷︸
=f ′ni
=
1
r
(
δiα − ninα
)
sin f + ninαf ′ cos f
the f ′ being ∂f
∂r
. And also
∂iϕ
3 = −f ′ sin f ni.
In the Bogomol’nyi equation (4.22) we take a = 3 which leads to the following:
�3αβ�ijϕ
α∂jϕ
β = �αβ�ijn
α sin f
[
1
r
(
δjβ − njnβ
)
sin f + njnβf ′ cos f
]
= �αβ�iβn
α sin f
sin f
r
− �αβ�ijnα sin f 1
r
njnβ sin f + �αβ�ijn
α sin fnjnβf ′ cos f
=
ni
r
sin2 f ; the other two terms vanish because �αβnαnβ = 0.
16
Then what we have is
−nif ′ sin f ± n
i
r
sin2 f = 0,
i.e.
f ′ =
sin f
r
. (4.23)
Now lets consider a = 1 in (4.22):
1
r
sin fδi1 − 1
r
nin1 sin f + nin1f ′ cos f = �ijnjn2f ′ + �ij
1
r
sin f cos f
(
δj2 − njn2
)
.
For i = 1 this becomes
1
r
sin f − 1
r
sin f
(
n1
)2
+
(
n1
)2
f ′ cos f =
(
n2
)2
f ′ +
1
r
sin f cos f
(
1− (n2)2)
and using the fact that ~n is a unit vector:
sin f
r
(n2)2 + f ′ cos f(n1)2 = (n2)2f ′ +
sin f
r
cos f(n1)2
and for each component we get again equation (4.23). This is also the case for a = 2.
What we need now is to fix suitable boundary conditions for f(r) such that ~ϕ→ (0, 0, 1)
for r →∞. Well, it’s quite direct to see that this is achieved if f(r)→ 2pi as r →∞. On the
other hand, we also need to consider f(0) = 0 so that ϕα = nα sin f can have a well behaviour
for r → 0. In fact, it is easy to check that f(r) = 2arctan( r
r0
) is the solution, with r0 being
an arbitrary number. We write tan
(
f
2
)
= r
r0
and calculate the derivative with respect to r on
both sides, which gives
1
cos2
(
f
2
) 1
2
f ′ =
1
r0
1 2 3 4 5 6
0.2
0.4
0.6
0.8
1.0
1.2
1.4
Figure 10: The behaviour of the function arctan(r) suites the required boundary conditions
expected for f(r).
17
and the cosine function can be replaced by
1
cos2
(
f
2
) = 1 + tan2(f
2
)
=
r20 + r
2
r20
.
Finally one gets
f ′ =
2r0
r20 + r
2
.
On the other hand, sin f = 2 sin f
2
cos f
2
. Using sin f
2
= r√
r20+r
2
and cos f
2
= r0√
r20+r
2
we end up
with
sin f
r
=
2r0
r20 + r
2
which agrees with the result for f ′.
With this solution we can recover the components of the field ~ϕ
ϕα = nα sin f =
2r0x
α
r20 + r
2
ϕ3 =
r20 − r2
r20 + r
2
,
thus
~ϕ =
(
2r0x
1
r20 + r
2
,
2r0x
2
r20 + r
2
,
r20 − r2
r20 + r
2
)
. (4.24)
The static energy density E = 1
4
∂i~ϕ · ∂i~ϕ can be calculated for the solution (4.24) above and
-100
-50
0
50
100
-100
-50
0
50
100
0.000
0.005
0.010
0.015
(a) Sketch of the energy density of a solution
with a given r0.
(b) Sketch of the energy density for twice the
previous value of r0.
Figure 11: Sketch of lumps of energy density with different sizes. Notice that r0 defines de
size of the lump.
the expression is very simple: E = 2r20
(r20+x2+y2)
2 . This can be easily integrated over the whole
18
p�
p
N
S
θ
2
θ
r
1
1
segunda-feira, 23 de julho de 2012
Figure 12: Stereographic projection of a point p on the sphere to a point p′ on the plane.
plane and the result is E = 2pi, exactly what is predicted by the Bogomol’nyi bound8 for
|N | = 1. Moreover one should notice that the energy is independent of r0, the size of the
lump!
5 The fields on the Riemann sphere
The field ~ϕ : IR1,2 → S2 defines points on the sphere ~ϕ · ~ϕ = 1 labelled by p in our scheme. A
solution of the constraint is given by
ϕ1 = sin θ cosψ ϕ2 = sin θ sinψ ϕ3 = cos θ
and the lagrangian of the σ-model becomes
L = 1
4
(
∂µθ∂
µθ + (sin θ)2 ∂µφ∂
µφ
)
(5.25)
Then, one can project these points onto the plane getting the points p′, with polar coordinates
(r, φ) =
(
2 tan
(
θ
2
)
, φ
)
. Now we have ∂r = sec2
(
θ
2
)
∂θ and therefore ∂µθ∂
µθ = cos4
(
θ
2
)
∂µr∂
µr.
Now, multiplying the identity sin2 θ = 4 sin2
(
θ
2
)
cos2
(
θ
2
)
by
cos2( θ2)
cos2( θ2)
and using again the
expression for r in terms of θ we get sin2 θ = r2 cos4
(
θ
2
)
, and the lagrangian becomes
L = 1
4
(
1
sec2( θ2)
)2
(∂r∂r + r2∂φ∂φ). Finally using tan2 θ
2
= sec2 θ
2
− 1 and the expression
8As a tedious exercise one can plug the solution (4.24) into the expression for the topological charge (4.21)
and check that N = 1.
19
for r we end up with
L = 1(
1 + r
2
4
)2 (∂r∂r + r2∂φ∂φ)4 .
Define now R = r
2
eiφ = tan
(
θ
2
)
eiφ, and the lagrangian of the CP1 model reads
L = ∂µR∂
µR
(1 + |R|2)2 . (5.26)
The equations of motion are (
1 + |R|2) ∂2R− 2R∂µR∂µR = 0 (5.27)
and its complex conjugate.
The cartesian coordinates on the plan are (this is in the target space!) u = tan
(
θ
2
)
cosφ and
v = tan
(
θ
2
)
sinφ. It’s not difficult to show that9 tan θ
2
= 1−cos θ
sin θ
, so we write u = 1−cos θ
sin θ
cosφ
and v = 1−cos θ
sin θ
sinφ. Now, u = 1−ϕ
3
sin θ
cosφ = (1− ϕ3) ϕ1
1−(ϕ3)2 =
(1−ϕ3)(1+ϕ3)ϕ1
(1+ϕ3)(1−(ϕ3)2) =
ϕ1
1+ϕ3
, where
we used ϕ1 = sin θ cosφ, (ϕ1)
2
+ (ϕ2)
2
= sin2 θ = 1 − (ϕ3)2. We do an analogous thing for
v = 1−cos θ
sin θ
sinφ, using ϕ2 = sin θ sinφ. We then get v = ϕ
2
1+ϕ3
. Also, ϕ3 = 1−|R|
2
1+|R|2 . Finally, from
R = u+ iv we get R + R = 2ϕ
1
1+ϕ3
and R− R = 2iϕ2
1+ϕ3
. Also, we have 1 + ϕ3 = 2
1+|R|2 and with
these relations one can write
~ϕ =
(
R +R
1 + |R|2 ,−i
R−R
1 + |R|2 ,
1− |R|2
1 + |R|2
)
.
Now consider the derivative of R = ϕ
1+ϕ3
, where ϕ ≡ ϕ1 + iϕ2:
∂iR =
∂iϕ+ ϕ
3∂iϕ− ϕ∂iϕ3
(1 + ϕ3)2
.
Lets calculate ∂iϕ, using the Bogomol’nyi equations (4.22):
∂1ϕ
1 ± (ϕ2∂2ϕ3 − ϕ3∂2ϕ2) = 0 ∂1ϕ2 ± (ϕ3∂2ϕ1 − ϕ1∂2ϕ3) = 0
∂2ϕ
1 ± (−ϕ2∂1ϕ3 + ϕ3∂1ϕ2) = 0 ∂2ϕ2 ± (−ϕ3∂1ϕ1 − ϕ1∂1ϕ2) = 0.
9tan
(
θ
2
)
=
sin θ2
cos θ2
=
sin2 θ2
cos θ2 sin
θ
2
=
2 sin2 θ2
sin θ =
2(1−cos2 θ2 )
sin θ =
1−cos θ
sin θ
20
We can arrange the ϕ1 and ϕ2 appearing above together so that we get ϕ, instead. Then
∂1R =
±i (ϕ∂2ϕ3 − ϕ3∂2ϕ) + ϕ3∂1ϕ− ϕ∂1ϕ3
(1 + ϕ3)2
and
∂2R =
±i (ϕ3∂1ϕ− ϕ∂1ϕ3) + ϕ3∂2ϕ− ϕ∂2ϕ3
(1 + ϕ3)2
.
Notice that
i∂1R = ±∂2R
and in terms of u and v we have
∂1u = ∂2v ∂2u = −∂1v
which are the Cauchy-Riemann equations.
The energy density is calculated using T 00 =
∂L
∂∂0R
∂0R+
∂L
∂∂0R
∂0R−L, with the lagrangian
given in (5.26). Then, the static energy reads
E =
∫
d2x
~∇R · ~∇R
(1 + |R|2)2 .
Now consider the new coordinates z = x1 + ix2 and z = x1 − ix2. Also, lets call ∂ ≡ ∂
∂z
and
∂ = ∂
∂z
. Then, ∂1 = ∂ + ∂
and ∂2 = i
(
∂ − ∂), so we have
~∇R · ~∇R = ∂1R∂1R + ∂2R∂2R = 2
(
∂R∂R + ∂R∂R
)
= 2
(|∂R|2 + |∂R|2)
and
E = 2
∫
d2x
(|∂R|2 + |∂R|2)
(1 + |R|2)2 .
For the volume element:
d2z ≡ dx1 ∧ dx2 = (∂x∂y − ∂x∂y) dz ∧ dz = i
2
dz ∧ dz,
where we used x1(z, z) = z+z
2
and x2(z, z) = z−z
2i
.
The R and R fields are coordinates in the target space. Take the “volume form”
Ω˜ =
dR ∧ dR
(1 + |R|2)2
and lets write it in terms of θ and φ through the relation R = tan
(
θ
2
)
eiφ. We have dR =
21
1
2
sec2
(
θ
2
)
eiφdθ + i tan
(
θ
2
)
eiφdφ and dR is just the complex conjugate. Then dR ∧ dR =(
∂θR∂φR− ∂φR∂θR
)
dθ ∧ dφ = −i sin(
θ
2)
cos3 θ
2
dθ ∧ dφ, and with (1 + |R|2)2 = 1
cos4( θ2)
we get Ω˜ =
− i
2
sin θdθ ∧ dφ. Now, recognizing the normalized volume form of S2 as Ω = 1
4pi
sin θdθ ∧ dφ
we write
Ω =
i
2pi
dR ∧ dR
(1 + |R|2)2 .
The topological charge is then given by the pull-back of Ω above:
N =
i
2pi
∫ (
∂1R∂2R− ∂1R∂2R
) d2x
(1− |R|2)2 =
1
pi
∫ (|∂R|2 − |∂R|2)
(1− |R|2)2 d
2x.
Notice that
E = 2piN + 4
∫ |∂R|2
(1− |R|2)2d
2x
and the last term is a positive contribution to E, thus
E > 2pi|N |
and the equality happens when (N > 0) ∂R = 0.
The boundary condition ~ϕ → (0, 0, 1) for |~x| → ∞ implies that R(z) goes to a constant
value as |z| → ∞ and this is necessary if we want a finite energy (the energy density vanishes
for “distant places” since it has a derivative in R, which will then be a constant there!), and
also forces R(z) to be a rational map10
R(z) =
p(z)
q(z)
=
aN−1zN−1 + · · ·+ a0
bNzN + bN−1zN−1 + · · ·+ b0 =
aN−1zN−1 + · · ·+ a0
zN + bN−1zN−1 + . . . b0
such that the degree of the polynomial q(z) is bigger than that of q(z), and its maximum
value is the topological number N .
The number of parameters in the theory is the dimension of the Moduli space MN , the
parameter space of the static N -soliton solutions, which is 4N . The simplest case isM1, with
dim(M1) = 4. One could in principle take a polynomial such as
R(z) = µz, µ ∈ IR.
10where in the last step we divided by bN and renamed everything.
22
The energy density (which is easy to calculate from the definition!) would then be
2µ2
(1 + µ2|z|2)2
and strictly speaking this is not confined within a finite region. Notice that the integration of
it over the entire space is 2pi, as expected from the Bogomol’nyi bound for N = 1.
A 1-lump configuration is characterized by 3 complex parameters:
R(z) =
αz + β
z + γ
,
2 real parameters give the (2-dimensional) position of the lump, 1 gives its width and the
remaining 3 the orientation in target space. The condition R(∞) = 0 implies α = 0 and we
remain with only 4 real parameters: the orientation is defined by only 1 parameter now.
This is a particular case of the general statement that a degree N rational map has 2N +1
complex parameters and 2N if the boundary conditions are included.
The 1-lump can be written as
R(z) =
λeiα
z − a
where λ ∈ IR+ gives the size of the soliton,a ∈ C its position and α ∈ S1 is the phase. The
point R(z = a) =∞ is as far away from the vacuum as possible; it corresponds to the south-
pole. The point R = 0 is the north-pole, which we choose to be the vacuum. This breaks the
symmetry SO(3)→ SO(2).
REMARK dim(MN) = N × dim(M1), so there is no static force between the solitons.
They have the same energy if they’re far away apart or on top of each other! Another case is
with N = 2:
R(z) =
λ2
(z − a)(z + a) .
For a large a, the two lumps are very separated. The a = 0 case is allowed as well:
R(z) =
λ2
z2
.
In that case the two solitons are on top of each other.
For a study of the scattering of the 2-lump solutions we refer to [1].
23
-10
-5
0
5
10
-10
-5
0
5
10
0.00
0.01
0.02
0.03
0.04
(a) Sketch of the energy density of the 1-lump.
-10
-5
0
5
10
-10
-5
0
5
10
0.0
0.5
1.0
(b) Sketch of the energy density of the 1-lump
with different (smaller) λ.
Figure 13: Sketch of lumps of energy density with different sizes.
6 Solitons in gauge theories
We start by considering a lagrangian density for a gauge field Aµ and a scalar field ϕ
L = 1
2
trFµνF
µν + (Dµϕ) (D
µϕ)− U(ϕ)
where Fµν = ∂µAν − ∂νAµ + [Aµ, Aν ] and Dµϕ = ∂µϕ+ T (Aµ)ϕ.
We select configurations such that F0i = 0 and D0ϕ = 0. The possibility of finding solitonic
configurations in such a theory can be understood by using Derrick’s theorem. The energy is
of the “Ginzburg-Landau type”:
E =
∫
ddx
(
−1
2
trFijFij + (Diϕ) (Diϕ) + U(ϕ)
)
≡ E4 + E2 + E0.
Then we consider the transformation ~x → ~y = λ~x under which the fields ϕ(~x) and Aµ(~x),
solutions of the theory, transform like
ϕλ(~x) = ϕ(λ~x) ~Aλ(~x) = λ ~A(λ~x).
Now, it’s easy to see that
~D
(λ)
~x ϕλ(~x) = λ
~D~yϕ(~y) F
(λ)
ij (~x) = λ
2Fij(~y)
24
-10
-5
0
5
10
-10
-5
0
5
10
0.0
0.5
1.0
(a) Sketch of the energy density of the 2-lump
for a = 0.
-10
-5
0
5
10
-10
-5
0
5
10
0
1
2
3
4
(b) Sketch of the energy density of the 2-lump
with a 6= 0.
Figure 14: Sketch of the energy density for the 2-lump configuration.
so that the energy is rescaled as
E(λ) = λ4−dE4 + λ2−dE2 + λ−dE0,
and the extremality condition on it at λ = 1 gives
(2− d)E4 + (2− d)E2 − dE0 = 0.
Such a condition can be solved for d = 2 and d = 3 as well. The case d = 4 gives what we call
instantons, and are also very important and interesting!
6.1 Vortex
Consider the gauge group to be U(1). We have a scalar field ϕ and a gauge field aµ transforming
as
ϕ(x)→ eiα(x)ϕ(x) aµ(x)→ aµ(x)− i∂µα(x).
We need a Higgs mechanism here so we consider the potential to be
U(|ϕ|) = µ+ ν|ϕ|2 + λ
8
|ϕ|4.
25
For negative ν = −|ν| we fix µ = 2|ν|2
λ
for the minimum of U to be zero and define m2 = 4|ν|
λ
,
and the potential can be written as
U = λ
8
(
m2 − |ϕ|2)2 .
The static energy reads
V =
1
2
∫ (
B2 +Diϕ Diϕ+
λ
4
(|ϕ|2 −m2)2) d2x
where fij = ∂iaj − ∂jai = �ijB and Diϕ = ∂iϕ− iauϕ.
The static equations are obtained by taking the variation of V with respect to the fields.
Lets start by δϕ:
δV =
1
2
∫
d2x
(
B2 + δ
(
Diϕ
)
Diϕ+
λ
4
δ
(
m2 − |ϕ|2)2) ,
then,
DiDiϕ+
λ
2
(
m2 − |ϕ|2)ϕ = 0. (6.28)
The equation for ϕ is just the complex conjugate of this. Now, varying with respect to ai:
δV =
1
2
∫
d2x
(
δB2 DiϕDiϕ+B
2δ
(
DiϕDiϕ
))
and using δB2 = 4�ij∂jBδai we get
�ij∂jB +
iB2
2
(
ϕDiϕ− ϕDiϕ
)
= 0. (6.29)
The energy is minimized for
|ϕ| = m Diϕ = 0 B = 0
The first of these conditions11 implies that in such a state of minimal energy ϕ(~x) = meiχ(~x).
The third equation implies that the connection is a pure gauge: ai(~x) = ∂iα(~x). Using these
results in the second equation we get that ∂i (χ− α) = 0 and therefore χ = α + constant,
which in turn implies ai = ∂iχ.
Now, taking a gauge transformation with gauge group element g = e−iχ(~x), (α(~x) = −χ(~x)),
11In fact, the first equation is in part a consequence of the second: ϕDiϕ+ϕDiϕ = ∂i|ϕ|2 = 0⇒ |ϕ| = const..
Thus, we know that |ϕ| must be a constant, and what fiexes this constant to be m is indeed the requirement
for a minimum value of V .
26
the Higgs field becomes ϕ(~x)→ e−iχ(~x)ϕ(~x) = e−iχ(~x)meiχ(~x) = m. The connection transforms
as ai(~x)→ ∂iχ(~x) + ∂i(−χ(~x)) = 0. Finally, we can consider a second transformation with a
constant parameter α = χ. Then, the Higgs field becomes ϕ(~x) = meiχ while the gauge field
remains ai(~x) = 0.
Now we shall work in polar coordinates: x1 = ρ cos θ, x2 = ρ sin θ. The gauge
field becomes
a = aρdρ+ aθdθ = (a1 cos θ + a2 sin θ) dρ+ (−a1ρ sin θ + a2ρ cos θ) dθ. The area element reads
dx1 ∧ dx2 = ρdρ∧ dθ, and the field strength components are fρθ = ∂xi∂ρ ∂x
j
∂θ
fij = ρB. Then, the
static energy simply reads12
V =
1
2
∫ ∞
0
∫ 2pi
0
(
1
ρ2
f 2ρθ +DρϕDρϕ+
1
ρ2
DθϕDθϕ+
λ
4
(
m2 + |ϕ|2)2) ρdρdθ.
Now, if ϕ(~x) and ai(~x) are finite energy configurations, then one feature we have is that
|ϕ(~x)| → m as |~x| → ∞. Thus, for large ρ, ϕ ∼ meiχ and from Dρϕ = 0 we get aρ = ∂ρχ.
Using a gauge transformation with
g(ρ, θ) = e−i
∫ ρ
0 aρ(ρ
′,θ)dρ′
we can make aρ → 0. Then, in this gauge, since Dρϕ = 0, we have ∂ρϕ→ 0 as |~x| → ∞, thus,
along each radial line the Higgs field has a limiting value:
lim
ρ→∞
ϕ(~x) = ϕ(∞)(θ) = meiχ
∞(θ).
The only freedom we are left with is to multiply the field by a constant phase eiα.
Now the condition fρθ → 0 implies that ∂ρaθ = 0 in the gauge aρ = 0. This means that at
ρ ∼ ∞ the gauge field aθ is a constant in the radial direction: a(∞)θ (θ). With these results we
move to the condition Dθϕ = 0 at ρ ∼ ∞, which becomes ∂θχ∞ − a∞θ = 0.
In the boundary of the two dimensional space we have a circle S1 on which the field ϕ can
depend only on the direction along the plane. Thus, the field defines the map ϕ : S1 → S1,
and therefore Π1(S
1) = ZZ; the degree of the Higgs field at infinity is given by an integer
number N , the integral of the pull-back of the normalized volume form of the target space.
We saw how to compute this already13:
N =
1
2pi
∫
dχ =
1
2pi
∫ 2pi
0
∂θχ
∞ dθ =
χ∞(2pi)− χ∞(0)
2pi
.
12For the gradient in polar coordinates we first note that dϕ = ~∇ϕ · d~S. Then, d~S · d~S = dρ2 + ρ2dθ2, so
d~S = (dρ, ρdθ), and therefore (~∇ϕ)ρ = ∂ρϕ and (~∇ϕ)θ = 1ρ∂θϕ, since dϕ = ∂ρϕdρ+ ∂θϕdθ
13Notice that the coordinate (angle) in the target space S1 is χ now that ϕ = eiχ.
27
Notice that a constant gauge transformation can change χ∞ by a constant factor, and therefore
its value is not really fixed. The topological charge N is gauge independent, as expected.
We found the topological charge through the scalar field. However, one can also find it
using the gauge field.
The Chern-number of the gauge field aµ is given by
C1 =
1
2pi
∫
f12d
2x.
Using Stokes theorem this becomes C1 =
1
2pi
∫
a1dx
1 +a2dx
2 = 1
2pi
∫
aρdρ+aθdθ, and at ρ ∼ ∞
(where it is constant, making the circle S1), C1 =
1
2pi
∫ 2pi
0
aθdθ
∣∣∣
ρ=∞
. On the other hand, in
order to get a finite energy (and a finite C1) we need f12 → 0 as ρ→∞, and a vanishing field
strength implies a pure gauge connection: aθ = ∂θα. So,
C1 =
1
2pi
(α(2pi)− α(0)) .
The gauge transformation eiα must be single-valued, hence α(2pi) = α(0)+2piN˜ , and therefore
C1 = N˜ ∈ ZZ.
From the condition that Dθϕ = 0 at ρ ∼ ∞, we get ∂θϕ−iaθϕ = 0. Using now ϕ(∞) = eiNθ
this relation gives aθ = N .
On the other hand, if we take ϕ(∞) = eiχ(θ) the equation Dθϕ = 0 gives ∂θχ = aθ, and
since aθ = ∂θα, integration of it leads to the conclusion that χ = α and finally that N˜ = N .
Summarizing: the vortex number can be seen as (i) the winding number of the Higgs field
at infinity, as (ii) the total magnetic fluc through the plane divided by 2pi, i.e.,
N =
1
2pi
∫
B d2x,
or as (iii) the number of zeros of the Higgs field.
6.2 Axial vortex solutions
7 Monopoles
28
A A proof that Π1(S
1) = ZZ.
Let h(z) be a meromorphic function with ni zeros of order i:
h(z) =
(z − bi)ni
(z − ai)mi g(z),
where g(z) is an analytic function.
Taking the logarithm of h(z) one gets
lnh(z) = niln(z − bi)−miln(z − ai) + lng(z).
Then, we consider the following quantity
w(z) ≡ d
dz
ln(z) =
ni
z − bi −
mi
z − ai +
d
dz
lng(z),
and its integral
W ≡ 1
2pii
∮
w(z)dz =
ni
2pii
∮
dz
z − bi −
mi
2pii
∮
dz
z − ai +
1
2pii
∮
d
dz
lng(z)dz.
Cauchy’s theorem states that ∮
h(z)dz
z − z0 = 2piih(z0).
Then, one gets
W = ni −mi + 0 = (#of zeros)− (#of poles)
which is clearly an integer number. Moreover
d ln(h1 ◦ h2) = d lnh1 + d lnh2 ⇒ wh1◦h2 = wh1 + wh2
so, it’s not difficult to realise that the winding numbers for a group isomorphic to the integers
under sum.
B Noether’s theorem
Consider an infinitesimal transformation of the fields φa, δφa = ε(x)F a(φ). The lagrangian
varies as
δL = ∂L
∂∂µφa
δ∂µφ
a +
∂L
∂φa
δφa
29
and now we work on this expression. First, δ∂µφ
a = ∂µδφ
a = ∂µ (εF
a) = ∂µεF
a + ε∂µF
a, and
then
δL = ∂L
∂∂µφa
(∂µεF
a + ε∂µF
a) +
∂L
∂φa
εF a = ε
(
∂L
∂∂µφa
∂µF
a +
∂L
∂φa
F a
)
+ ∂µε
(
∂L
∂∂µφa
F a
)
.
Setting ε as a constant then the second term above disappears and
δS =
∫
dDx ε
(
∂L
∂∂µφa
∂µF
a +
∂L
∂φa
F a
)
= 0
if δφa is a symmetry.
The most general solution here is that the integrand equals ∂µχ
µ and χµ
∣∣∣
∂M
= 0, for
instance.
Now, consider ε not a constant. Then
δS =
∫
dDx
(
ε∂µχ
µ + ∂µε
∂L
∂∂µφa
F a
)
.
Integrating the first term by parts we get
δS = εχµ
∣∣∣
∂M
+
∫
dDx ∂µε
(
−χµ + ∂L
∂∂µφa
F a
)
=
∫
d3x ∂µε J
µ
where Jµ = ∂L
∂∂µφa
F a − χµ. Integrating again
δS = εJµ
∣∣∣
∂M
−
∫
dDx ε ∂µJ
µ.
Imposing the symmetry we conclude that ∂µJ
µ = 0.
It might be useful to have also the following notation for what we just did. Consider
the fields to transform as φa → Tabφb so that the infinitesimal transformation reads δφa =
−iθATAabφb, where
[
TA, TB
]
= ifABCTC . Then JAµ = −i ∂L∂∂µφaTAabφb, up to that χµ function
which can be ignored in most cases.
Finally, for the conserved charges we look at the continuity equation:
∂µJ
µ = 0 ⇒ ~Q =
∫
~J0 d
D−1x.
30
B.1 SO(3) ' SU(2)/ZZ2 Noether’s charges
Consider the field ~ϕ, whose dynamics is given by the lagrangian (4.18). This theory is clearly
SO(3) invariant. The transformation of the fields is given by
~ϕ(x)→ e−iθAJA ~ϕ(x)
where the generators are
(
JA
)
ab
= −i�Aab,
J1 =
 0 0 00 0 −i
0 i 0
 J2 =
 0 0 i0 0 0
−i 0 0
 J3 =
 0 −i 0i 0 0
0 0 0
 .
and the algebra reads
[
JA, JB
]
= −i�ABCJC .
The infinitesimal transformation of the fields is then δϕa = −iθAJAϕa, and since it’s a
symmetry, the current from Noether’s theorem is
i = ∂µϕ
a
(−iθAJa)ϕa = −�AabθA∂iϕaϕb.
The above three currents can be written as
~i = ~ϕ× ∂i~ϕ. (B.30)
Each current corresponds to an element of the so(3) algebra. We now ask ourselves how
these currents are mapped into the theory described by (??), i.e., how the SO(3) symmetry
is mapped on the complex plane.
We start by considering the homographic transformation
u′ =
au+ b
cu+ d
ad− bc = 1
where a, b, c, d are complex numbers.
It’s direct to calculate ∂u′ = ∂u
(cu+d)2
, ∂u¯′ = ∂u¯
(c¯u¯+d¯)
2 and (1 + u′u¯′)
2 =
[(cu+d)(c¯u¯+d¯)+(au+b)(a¯u¯+b¯)]
2
(cu+d)2(c¯u¯+d¯)
2 .
Then the lagrangian becomes
2
∂u′∂u¯′
(1 + |u′|2)2 = 2
∂u∂u¯
(cu+ d)
(
c¯u¯+ d¯
)
+ (au+ b)
(
a¯u¯+ b¯
) .
The denominator can be rewritten as (|c|2 + |a|2) |u|2 + (cd¯+ ab¯)u+ (c¯d+ ba¯) u¯+ |b|2 + |d|2
and if we want the lagrangian to remain invariant we need |c|2 + |a|2 = 1, ab¯ + cd¯ = 0 and
31
|b|2 + |d|2 = 1. The matrix
U =
(
a b
c d
)
with the above conditions on the coefficients belongs to SU(2). Such a transformation acts
on a spinor ψ as ψ → Uψ, ψ† → ψ†U †, so that the quantity ψψ† remains invariant. In fact,
this is what defines the SU(2) matrices as U †U = 1l. This equation gives
U =
(
a b
−b¯ a¯
)
and therefore |a|2 + |b|2 = 1.
Consider now the position vector ~r in IR3. Define h = ~σ · ~r, where the
components of ~σ
are the Pauli matrices
σ1 =
(
0 1
1 0
)
σ2 =
(
0 −i
i 0
)
σ3 =
(
1 0
0 −1
)
which together with 1l2×2 for the basis of su(2) algebra. Then
h =
(
z x− iy
x+ iy −z
)
.
Notice that h† = h and trh = 0. It’s direct to check that a transformation h → UhU †
preserves these properties if U ∈ SU(2). Moreover, deth = x2 +y2 +z2 also remains invariant:
deth′ = detUhU † = deth. So, the unitary transformation h → UhU † leaves the quantity
deth = x2 + y2 + z2 invariant, which is equivalent to an SO(3) transformation on ~r. Indeed,
SO(3) has 3 parameters, just like SU(2). The next question is how these parameters are
related.
First, lets write ~r = ~r (ψ1, ψ2), where ψi are the entries of the spinor ψ. This can be done
as follows. Take σih = σi (~σ · ~r), so that we have xi = 12tr (σih). This is easy to see explicitly,
for instance:
σ1h =
(
x+ iy −z
z x− iy
)
then the trace of it is 2x, and half of the trace is exactly x.
In order to identify x, y and z with ψ1 and ψ2 we have to look at ψψ
†. The quantity ψ†ψ
32
remains invariant under an SU(2) transformation, however, this is not the case for
ψψ† =
(
ψ1
ψ2
)(
ψ∗1 ψ
∗
2
)
=
(
|ψ1|2 ψ1ψ∗2
ψ2ψ
∗
1 |ψ2|2
)
which transforms as ψψ† → Uψψ†U †. This happens because ψ and ψ† do not transform in
the same way under U . On the other hand,
ψ =
(
ψ1
ψ2
)
& ζψ∗ =
(
−ψ∗2
ψ∗1
)
with ζ =
(
0 −1
1 0
)
do transform in the same way under
(
a b
−b∗ a∗
)
∈ SU(2):
(
ψ1
ψ2
)
→
(
aψ1 + bψ2
−b∗ψ1 + a∗ψ2
) (
−ψ∗2
ψ∗1
)
→
(
a (−ψ∗2) + bψ∗1
−b∗ (−ψ∗2) + a∗ψ∗1
)
.
The we can identify ψ ∼ ζψ and also ψ† ∼ (ζψ)T . Now,
ζψ =
(
0 −1
1 0
)(
ψ1
ψ2
)
=
(
−ψ2
ψ1
)
then
ψ† =
(
−ψ2 ψ1
)
and
ψψ† =
(
ψ1
ψ2
)(
−ψ2 ψ1
)
=
(
−ψ1ψ2 ψ21
−ψ22 ψ1ψ2
)
≡ −H.
The matrix −H transforms under a similarity transformation UHU †, just like h. Thus we
identify them, which leads to
x− iy = −ψ21 x+ iy = ψ22 z = ψ1ψ2
and finally
~r =
(
ψ22 − ψ21
2
,
ψ21 + ψ
2
2
2i
, ψ1ψ2
)
.
The next step is to write the transformation
ψ′1 = aψ1 + bψ2 ψ
′
2 = −b∗ψ1 + a∗ψ2
33
in terms of x, y and z. In order to do that we use the above results: x′ = ψ
′2
2 −ψ′21
2
, etc. Then
x′ =
1
2
(
a∗2 + a2 − b∗2 − b2)x+ i
2
(
a∗2 − a2 + b∗2 − b2) y − (a∗b∗ + ab) z
y′ =
1
2i
(
a∗2 − a2 + b2 − b∗2)x+ 1
2
(
a2 + a∗2 + b2 + b∗2
)
y +
1
i
(ab− a∗b∗) z
z′ = (ab∗ + ba∗)x+ i (ba∗ − ab∗) y + (|a|2 − |b|2) z.
Now considering b = 0 and a = e−
iα
2 one gets
x′ = cosα x− sinα y y′ = sinα x+ cosα y z′ = z
which is the SO(3) rotation on the plane xy:
U =
(
e−i
α
2 0
0 ei
α
2
)
' Rz(θ) =
 cos θ − sin θ 0sin θ cos θ 0
0 0 1
 .
If we write the fractional linear transformation for this case:
u′ =
au+ b
−b∗u+ a∗ = e
−iαu.
For small α the transformation U becomes
U ≈
(
1− iα
2
0
0 1 + iα
2
)
= 1l− αi
2
σ3
and iterating the series one gets
U = e−i
α
2
σ3 .
Now, for the SO(3) matrix, a small θ leads to
Rz(θ) = 1l− iθJ3,
so
Rz(θ) = e
−iθJ3 .
Now, taking a = cos β
2
and b = − sin β
2
we get
x′ = cos β x+ sin β z y′ = y z′ = − sin β x+ cos β z.
34
Then
U =
(
cos β
2
− sin β
2
sin β
2
cos β
2
)
' Ry(θ) =
 cos β 0 sin β0 1 0
− sin β 0 cos β
 .
The fractional linear transformation reads
u′ =
cos β
2
u− sin β
2
sin β
2
u+ cos β
2
and in particular, for β = pi this gives u′ = − 1
u
. For small β we get
U ≈ 1l− iβ
2
σ2
and iterating the series it becomes
U = e−i
β
2
σ2 .
The SO(3) element reads
Ry(β) = e
−iβJ2 .
Finally setting a = cos γ
2
and b = −i sin γ
2
we have
x′ = x y′ = cos γ y − sin γ z z′ = sin γ y + cos γ z
which is exactly
Rx(γ) = e
−iγJ1
and the SU(2) matrix is
U = e−i
γ
2
σ1 .
The fractional linear transformation in this case reads
u′ =
cos γ
2
u− i sin γ
2
−i sin γ
2
u+ cos γ
2
.
One can check that the inversion u→ 1
u
takes (ϕ1, ϕ2.ϕ3) to (ϕ1,−ϕ2,−ϕ3). This is a discrete
transformation (∈ O(3)) and is not in SO(3). Notice that β → β + 2pi implies R → R and
U → −U , so the elements U and −U correspond to the same element R. We need β → β+4pi
for R→ R and U → U . Then the isomorphism is SO(3) ' SU(2)/± 1.
35
B.1.1 A closer look to the fractional linear transformation
The sphere S2 can be understood as the extended complex plane, which includes the point
u =∞ and the north pole ~ϕ = (0, 0, 1). The notation for this is
Ĉ = C ∪ {∞};
the LHS stands for the Riemann sphere, and the RHS is the complex plane plus points at
infinity.
The most general analytic one-to-one map from Ĉ to Ĉ is the fractional linear transfor-
mation
u′ =
au+ b
cu+ d
, ad− bc 6= 0
with a, b, c and d complex numbers. Notice that limn→∞ u′ = ac , which is a point just like any
other.
Lets see why we need ad− bc 6= 0. If u′(u) is to be one-to-one, we need u′(u1) = u′(u2)⇒
u1 = u2. Then
au1 + b
cu1 + d
=
au2 + b
cu2 + d
.
If cu1 + d = 0 then cu2 + d = 0, and so u1 = u2. If cu1 + d 6= 0 then (au1 + b)(cu2 + d) =
(au2 + b)(cu1 + d) ⇒ (ad − bc)u1 = (ad − bc)u2 and in order to get u1 = u2 6= 0 we need
ad− bc 6= 0.
The fractional linear map is invariant under rescaling. Thus taking
a→ a√
ad− bc, b→
b√
ad− bc, etc.
we get
ad− bc→ ad− bc
ad− bc = 1 ⇒ a
′d′ − b′c′ = 1.
So, we have three independent complex parameters, which means that the fractional linear
transformation maps three punctures on the u-plane into three punctures on the u′-plane.
Lets construct this map. First we want u′ = u′1 if u = u1, then
(factors)(u− u1) = (factors)(u′ − u′1).
Now we also want u′ = u′2 if u = u2, so
(factors)
(u− u1)
u− u2 = (factors)
(u′ − u′1)
u′ − u′2
.
36
Finally, we want u′ = u′3 if u = u3, which gives
(factors)
(u− u1)
u− u2
u3 − u2
u3 − u1 = (factors)
(u′ − u′1)
u′ − u′2
u′3 − u′2
u′3 − u′1
.
Lets consider the upper-half plane H, defined by Im(u) > 0, with a real line and a point at
infinity. The question we address is whether it is possible to build maps Ĉ → Ĉ preserving
H.
First, since the real line must be preserved we have u = u∗, implying u′ = u′∗. Then
ac∗u2 + (ad∗ + bc∗)u+ bd∗ = a∗cu2 + (a∗d+ b∗c)u+ b∗d, and we impose a, b, c, d ∈ IR.
Now, Im(u′) = Im
(
au+b
cu+d
)
, and Im(u′) = u
′−u′∗
2i
= ad−bc|cu+d|2 Im(u). So, as long as ad− bc > 0,
if Im(u) > 0, then Im(u′) > 0.
Two particularly important transformations preserving the upper-half plane are
T : u′ = u+ 1 S : u′ = −1
u
.
For the first we have a = 1, b = 1, c = 0 and d = 1, while for the second a = 0, b = −1, c = 1
and d = 0.
Proposition: The Moebius transformation is generated by the composition of the follow-
ing three transformations:
• Translations u→ u+ k;
• Dilations u→ ku;
• Inversion u→ 1
u
;
For c = 0 its trivial to see: u′ = a
d
u+ b
d
. For c 6= 0, u′ = a
c
− ad−bc
c(cz+d)
.
Now we have to consider the symmetries of the lagrangian (??). It’s direct to see that
dilations and inversions are transformations that leave (??) invariant. However this is not the
case for translations, and in particular for the T transformation. That is because SU(2) is
more restrictive than PSL(2,ZZ) ' SL(2,ZZ)/ZZ2, which is the group containing T and S.
Take the infinitesimal transformation of the field u given by the holomorphic function
δu =
n=∞∑
n=0
anu
n.
It’s non-singular at u = 0, the south-pole, but we also need a nice behavior for the north pole.
37
In order to see that define z = 1
u
, which takes u =∞ to z = 0. Then
δz =
δz
δu
δu = − 1
u2
δu = −
∞∑
n=0
anu
n−2 = −
∞∑
n=0
anz
2−n.
This is only well defined for n = 0, 1, 2, thus the infinitesimal transformation of the field u
(and u∗) reads (using a, b and c instead of a0, a1 and a2)
δu = a+ bu+ cu2.
If this is a symmetry of the lagrangian (??), then the Noether theorem associated to that
reads
Jµ =
2∂µu∗
(1 + |u|2)2 δu+
2∂µu
(1 + |u|2)2 δu
∗ − χµ.
Lets fix the coefficients a, b and c such that ∂µχ
µ = 0. We have ∂µχ
µ = ∂µ
(
2∂µu∗
(1+|u|2)2 δu+
2∂µu
(1+|u|2)2 δu
∗
)
.
∂µχ
µ =
2∂u∗
(1 + |u|2)2 (b∂u+ 2cu∂u) +
2∂u
(1 + |u|2)2 (b
∗∂u∗ + 2c∗u∗∂u∗)
− 4∂u∂u
∗
(1 + |u|2)3u
∗ (a+ bu+ cu2)− 4∂u∂u∗
(1 + |u|2)3u
(
a∗ + b∗u∗ + c∗u∗2
)
∝ 2 (b+ b∗) ∂u∂u∗ − 2 (b+ b∗)uu∗∂u∂u∗ + 4 (c− a∗)u∂u∂u∗ + 4 (c∗ − a)u∗∂u∂u∗.
Setting b+ b∗ = 0 and c = a∗ this quantity vanishes. So, lets take b = i and a = 1 + i. Then
the current is
Jµ =
2
(1 + |u|2)2
{(
∂µu+ u2∂µu∗ + ∂µu∗ + u∗2∂µu
)− i (∂µu+ u2∂µu∗ − ∂µu∗ − u∗2∂µu)}+
+
2i
(1 + |u|2)2 (u∂
µu∗ − u∗∂µu) .
Consider the quantity
µ =
4 (∂µu+ u2∂µu∗)
(1 + |u|2)2 .
Notice that the first two terms of Jµ are simply Reµ = 
µ+µ
2
and Imµ = 
µ−µ
2i
. The last term
is precisely the current (lets call µz ) associated to the rotation around z axis: u→ eiθu ≈ 1+u.
Then, we have finally found the three SU(2) currents of (??) corresponding to the three
SO(3) currents of (4.18):
Jµ = Reµ + Imµ + µz
38
References
[1] Solitons-like scattering in the O(3) sigma model in (2 + 1) dimensions, Nonlinearity, 4,
429 (1991)
39
	Theories with a scalar field in 1+1 dimensions
	Bogomol'nyi bound
	4 kinks
	Sine-Gordon kinks
	Topology in field theory
	Basic notions on homotopy theory
	1 dimensional case
	2 dimensional case
	Derrick's theorem
	(2+1) non-linear O(3) -model lumps
	What is topological in the charge?
	Bogomol'nyi bound
	The N=1 axially symmetric solution
	The fields on the Riemann sphere
	Solitons in gauge theories
	Vortex
	Axial vortex solutions
	A proof that 1(S1)=ZZ.
	Noether's theorem
	SO(3)SU(2)/ZZ2 Noether's charges
	A closer look to the fractional linear transformation