1ϕ4 kinks
III Classical and Quantum Solitons
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”
a
, 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 =
1
2
∂
µ
φ∂
µ
φ − U(φ).
Applying Noether’s theorem, to the translation symmetry, we obtain the energy-
momentum tensor
T
µ
ν
=
∂L
∂(∂
µ
φ)
∂
ν
φ − δ
µ
ν
L = ∂
µ
φ∂
ν
φ − δ
µ
ν
L.
Fixing a time and integrating over all space, we obtain the conserved energy and
conserved momentum. These are
E =
Z
∞
−∞
T
0
0
dx =
Z
∞
−∞
1
2
˙
φ
2
+
1
2
φ
02
+ U(φ)
dx,
P = −
Z
∞
−∞
T
0
1
dx = −
Z
∞
−∞
˙
φφ
0
dx.
We now focus on our moving kink in the adiabatic approximation of the
φ
4
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
x
φ
−a
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
b
lying
in-between the kink and the anti-kink:
x
φ
−a
a
b
The choice of
b
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
b
we choose. We will declare that the
parts to the left of
b
belongs to the kink, and the parts to the right of
b
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 = −
Z
b
−∞
T
0
1
dx = −
Z
b
−∞
˙
φφ
0
dx.
Since T
µ
ν
is conserved, we know ∂
µ
T
µ
ν
= 0. So we find
∂
∂t
T
0
1
+
∂
∂x
T
1
1
= 0.
By Newton’s second law, the force
F
on the kink is given by the rate of change
of the momentum:
F =
d
dt
P
= −
Z
b
−∞
∂
∂t
T
0
1
dx
=
Z
b
−∞
∂
∂x
T
1
1
dx
= T
1
1
b
=
−
1
2
˙
φ
2
−
1
2
φ
02
+ U
b
.
Note that there is no contribution at the
−∞
end because it is vacuum and
T
1
1
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
1
+ φ
2
− 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
˙
φ
and
φ
0
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
˙
φ
= 0 at
t
= 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
b
. Here we have
φ
2
−
1
≈
0. So we can try to
expand the expression to first order in φ
2
− 1 (and hence φ
0
2
), and this gives
F =
−
1
2
φ
02
1
+ U(φ
1
) − φ
0
1
φ
0
2
+ (φ
2
− 1)
dU
dφ
(φ
1
)
b
.
We have a zeroth order term
−
1
2
φ
02
1
+
U
(
φ
1
). 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
2
. 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
φ
0
=
dW
dφ
.
It then follows that
1
2
φ
02
=
1
2
dW
dφ
2
= 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
dU
dφ
(φ
1
) = φ
00
1
.
So we can write the force as
F =
− φ
0
1
φ
0
2
+ (φ
2
− 1)φ
00
1
b
.
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
φ
1
= 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
φ
.
For φ
2
, take the approximation that x a. Then
φ
2
− 1 = −tanh(x − a) − 1 ≈ −2e
2(x−a)
.
We assume that our
b
satisfies both of these conditions. These are obviously
easy to differentiate once or twice. Doing this, we obtain
−φ
0
1
φ
0
2
= (−4e
−2(x+a)
)(−4e
2(x−a)
) = 16e
−4a
.
Note that this is independent of
x
. In the formula, the
x
will turn into a
b
, and
we see that this part of the force is independent of
b
. Similarly, the other term is
(φ
2
− 1)φ
00
1
= (−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
b
we
chose.
We can write this in a slightly more physical form. Our initial configuration
was symmetric around the
y
-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
by
U(φ) =
1
2
(1 − φ
2
)
2
.
We can do perturbation theory around one of the vacua, say
φ
= 1. Thus, we
set φ = 1 + η, and then expanding gives us
U(η) ≈
1
2
(−2η)
2
=
1
2
m
2
η
2
,
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
A
of the tail was
A
= 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
2
A
2
e
−ms
.
This is an interesting and important phenomenon. The mass
m
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
φ
1
and
φ
2
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
¯
K
K
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.