1ϕ4 kinks
III Classical and Quantum Solitons
1.1 Kink solutions
In this section, we are going to study
φ
4
kinks. These occur in 1 + 1 dimensions,
and involve a single scalar field
φ
(
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 =
1
2
∂
µ
φ∂
µ
φ − U(φ)
for some potential
U
(
φ
) polynomial in
φ
. Note that in 1 + 1 dimensions, any
such theory is renormalizable. Here we will choose the Minkowski metric to be
η
µν
=
1 0
0 −1
,
with µ, ν = 0, 1. Then the Lagrangian is given by
L =
Z
∞
−∞
L dx =
Z
∞
−∞
1
2
∂
µ
φ∂
µ
φ − U(φ)
dx,
and the action is
S[φ] =
Z
L dt =
Z
L dx dt.
There is a non-linearity in the field equations due to a potential
U
(
φ
) 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
0
=
∂φ
∂x
.
Often it is convenient to (non-relativistically) split the Lagrangian as
L = T − V,
where
T =
Z
1
2
˙
φ
2
dx, V =
Z
1
2
φ
02
+ U (φ)
dx.
In higher dimensions, we separate out ∂
µ
φ into
˙
φ and ∇φ.
The classical field equation comes from the condition that
S
[
φ
] is stationary
under variations of
φ
. By a standard manipulation, the field equation turns out
to be
∂
µ
∂
µ
φ +
dU
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
d
2
φ
∂x
2
=
dU
dφ
.
Of course, the important part is the choice of U! In φ
4
theory, we choose
U(φ) =
1
2
(1 − φ
2
)
2
.
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:
φ
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
2
− φ
2
)
2
.
If we are more adventurous, we can talk about a φ
6
theory with
U(φ) = λφ
2
(m
2
− φ
2
)
2
.
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
φ
4
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
φ
4
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
x
φ
a
This is known as a kink solution.
To actually find such a solution, we need the full field equation, given by
d
2
φ
dx
2
= −2(1 − φ
2
)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
U
with minimum value 0. From
Noether’s theorem, we obtain a conserved energy
E =
Z
1
2
˙
φ
2
+
1
2
φ
02
+ U (φ)
dx.
For a static field, we drop the
˙
φ
2
term. Then this is just the
V
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
E
= 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
“
φ
(
∞
) = 1”, “
φ
(
−∞
) =
−
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
t
, the field
φ
(
x, t
) must satisfy
φ
(
x, t
)
→ vacuum
as
x → ±∞
, or else the
field will have infinite energy. However, the vacuum of our potential
U
is discrete.
Thus, if
φ
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
φ
(
∞
) and
φ
(
−∞
), and
focus on fields that satisfy these boundary conditions. The trick is to write the
potential in the form
U(φ) =
1
2
dW (φ)
dφ
2
.
If
U
is non-negative, then we can always find
W
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
W
. Let’s assume we’ve done that. Then we can write
E =
1
2
Z
φ
02
+
dW
dφ
2
!
dx
=
1
2
Z
φ
0
∓
dW
dφ
2
dx ±
Z
dW
dφ
dφ
dx
dx
=
1
2
Z
φ
0
∓
dW
dφ
2
dx ±
Z
dW
=
1
2
Z
φ
0
∓
dW
dφ
2
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
φ
0
= ±
dW
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
0
= ±
dW
dφ
, because we know the energy must be non-negative.
For the φ
4
kink, we have
dW
dφ
= 1 − φ
2
.
So we pick
W = φ −
1
3
φ
3
.
So when
φ
=
±
1, we have
W
=
±
2
3
. We need to choose the + sign, and then we
know the energy (and hence mass) of the kink is
E ≡ M =
4
3
.
We now solve for φ. The equation we have is
φ
0
= 1 − φ
2
.
Rearranging gives
1
1 − φ
2
dφ = dx,
which we can integrate to give
φ(x) = tanh(x − a).
This
a
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
a
. 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
4
3
.