3Inverse scattering transform
II Integrable Systems
3.1 Forward scattering problem
Before we talk about the inverse scattering transform, it is helpful to know
what the forward problem is. This is, as you would have obviously guessed,
related to the Schr¨odinger operator we know and love from quantum mechanics.
Throughout this section, L will be the Schr¨odinger operator
L = −
∂
2
∂x
2
+ u(x),
where the “potential”
u
has compact support, i.e.
u
= 0 for
|x|
sufficiently large.
What we actually need is just that
u
decays quickly enough as
|x| → ∞
, but to
make our life easy, we do not figure out the precise conditions to make things
work, and just assume that
u
actually vanishes for large
|x|
. For a fixed
u
, we are
interested in an eigenvalue (or “spectral”) problem, i.e. we want to find solutions
to
Lψ = λψ.
This is the “forward” problem, i.e. given a
u
, we want to find the eigenvalues and
eigenfunctions. The inverse problem is given the collection of all such eigenvalues
and eigenfunctions, some sort of solutions like this, we want to find out what
u
is.
We will divide this into the continuous and discrete cases.
3.1.1 Continuous spectrum
Here we consider solutions to
Lψ
=
k
2
ψ
for real
k
. Since
u
= 0 for
|x|
large, we
must have
ψ
xx
+ k
2
ψ = 0
for large |x|.
So solutions as
|x| → ∞
are linear combinations of
e
±ikx
. We look for specific
solutions for ψ = ϕ(x, k) defined by the condition
ϕ = e
−ikx
as x → −∞.
Then there must be coefficients a = a(k) and b = b(k) such that
φ(x, k) = a(k)e
−ikx
+ b(k)e
ikx
as x → +∞.
We define the quantities
Φ(x, k) =
ϕ(x, k)
a(k)
, R(k) =
b(k)
a(k)
, T (k) =
1
a(k)
.
Here
R
(
k
) is called the reflection coefficient, and
A
(
k
) is the transmission
coefficient. You may have seen these terms from IB Quantum Mechanics. Then
we can write
Φ(x, k) =
(
T (k)e
−ikx
x → −∞
e
−ikx
+ R(k)e
kx
x → +∞
.
We can view the
e
−ikx
term as waves travelling to the left, and
e
ikx
as waves
travelling to the right. Thus in this scenario, we have an incident
e
−ikx
wave
coming from the right, the potential reflects some portion of the wave, namely
R(k)e
ikx
, and transmits the remaining T (k)e
−ikx
. It will be shown on the first
example sheet that in fact |T (k)|
2
+ |R(k)|
2
= 1.
What would happen when we change
k
? Since
k
is the “frequency” of the
wave, which is proportional to the energy we would expect that the larger
k
is,
the more of the wave is transmitted. Thus we might expect that
T
(
k
)
→
1, and
R
(
k
)
→
0. This is indeed true, but we will not prove it. We can think of these
as “boundary conditions” for T and R.
So far, we’ve only been arguing hypothetically about what the solution has
to look like if it existed. However, we do not know if there is a solution at all!
In general, differential equations are bad. They are hard to talk about,
because if we differentiate a function, it generally gets worse. It might cease to
be differentiable, or even continuous. This means differential operators could
take our function out of the relevant function space we are talking about. On
the other hand, integration makes functions look better. The more times we
integrate, the smoother it becomes. So if we want to talk about the existence of
solutions, it is wise to rewrite the differential equation as an integral solution
instead.
We consider the integral equation for f = f(x, k) given by
f(x, k) = f
0
(x, k) +
Z
∞
−∞
G(x − y, k)u(y)f(y, k) dy,
where
f
0
is any solution to (
∂
2
x
+
k
2
)
f
0
= 0, and
G
is the Green’s function for
the differential operator ∂
2
x
+ k
2
, i.e. we have
(∂
2
x
+ k
2
)G = δ(x).
What we want to show is that if we can find an
f
that satisfies this integral
equation, then it also satisfies the eigenvalue equation. We simply compute
(∂
2
x
+ k
2
)f = (∂
2
x
+ k
2
)f
0
+
Z
∞
−∞
(∂
2
x
+ k
2
)G(x − y, k)u(y)f(y, k) dy
= 0 +
Z
∞
−∞
δ(x − y)u(y)f(y, k) dy
= u(x)f(x, k).
In other words, we have
Lf = k
2
f.
So it remains to prove that solutions to the integral equation exists.
We pick f
0
= e
−ikx
and
G(x, k) =
(
0 x < 0
1
k
sin(kx) x ≥ 0
.
Then our integral equation automatically implies
f(x, k) = e
−ikx
as
x → −∞
, because for very small
x
, either
x − y <
0 or
y
is very small, so the
integral always vanishes as u has compact support.
To solve the integral equation, we write this in abstract form
(I − K)f = f
0
,
where I is the identity, and
(Kf)(x) =
Z
∞
−∞
G(x − y, k)u(y)f(y, k) dy.
So we can “invert”
f = (I − K)
−1
f
0
.
We can “guess” a solution to the inverse. If we don’t care about rigour and just
expand this, we get
f = (I + K + K
2
+ ···)f
0
.
It doesn’t matter how unrigorous our derivation was. To see it is a valid solution,
we just have to check that it works! The first question to ask is if this expression
converges. On the second example sheet, we will show that this thing actually
converges. If this holds, then we have
(I − K)f = If
0
+ Kf
0
+ K
2
f
0
+ ··· − (K + K
2
f
0
+ K
3
f
0
+ ···) = f
0
.
So this is a solution!
Of course, this result is purely formal. Usually, there are better ad hoc ways
to solve the equation, as we know from IB Quantum Mechanics.
3.1.2 Discrete spacetime and bound states
We now consider the case
λ
=
−κ
2
<
0, where we wlog
λ >
0. We are going to
seek solutions to
Lψ
κ
= −κ
2
ψ
κ
.
This time, we are going to ask that
kψ
κ
k
2
=
Z
∞
−∞
ψ
κ
(x)
2
dx = 1.
We will wlog ψ
κ
∈ R. We will call these things bound states.
Since u has compact support, any solution
Lϕ = −κ
2
ϕ
must obey
ϕ
xx
− κ
2
φ = 0
for
|x| → ∞
. Then the solutions are linear combinations of
e
±κx
as
|x| → ∞
.
We now fix ϕ
κ
by the boundary condition
ϕ
κ
(x) = e
−κx
as x → +∞
Then as x → −∞, there must exist coefficients α = α(κ), β = β(κ) such that
ϕ
κ
(x) = α(κ)e
κx
+ β(κ)e
−κx
as x → −∞.
Note that for any
κ
, we can solve the equation
Lϕ
=
−κ
2
ϕ
and find a solution
of this form. However, we have the additional condition that
kψ
κ
k
2
= 1, and in
particular is finite. So we must have
β
(
κ
) = 0. It can be shown that the function
β = β(κ) has only finitely many zeroes
χ
1
> χ
2
> ··· > χ
N
> 0.
So we have a finite list of bound-states {ψ
n
}
N
n=1
, written
ψ
n
(x) = c
n
ϕ
χ
n
(x),
where c
n
are normalization constants chosen so that kψ
n
k = 1.
3.1.3 Summary of forward scattering problem
In summary, we had a spectral problem
Lψ = λψ,
where
L = −
∂
2
∂x
2
+ u,
where u has compact support. The goal is to find ψ and λ.
In the continuous spectrum, we have
λ
=
k
2
>
0. Then we can find some
T (k) and R(k) such that
φ(x, k) =
(
T (k)e
−ikx
x → −∞
e
−ikx
+ R(k)e
kx
x → +∞
,
and solutions exist for all k.
In the discrete spectrum, we have
λ
=
−κ
2
<
0. We can construct bound
states {ψ
n
}
N
n=1
such that Lψ
n
= −χ
2
n
ψ
n
with
χ
1
> χ
2
> ··· > χ
N
> 0,
and kψ
n
k = 1.
Bound states are characterized by large, positive x behaviour
ψ
n
(x) = c
n
e
−χ
n
x
as x → +∞,
where {c
n
}
N
n=1
are normalization constants.
Putting all these together, the scattering data for L is
S =
{χ
n
, c
n
}
N
n=1
, R(k), T (k)
.
Example.
Consider the Dirac potential
u
(
x
) =
−
2
αδ
(
x
), where
α >
0. Let’s
try to compute the scattering data.
We do the continuous spectrum first. Since
u
(
x
) = 0 for
x 6
= 0, we must have
Φ(x, k) =
(
T (k)e
−ikx
x < 0
e
−ikx
+ R(k)e
ikx
x > 0
Also, we want Φ(x, k) to be continuous at x = 0. So we must have
T (k) = 1 + R(k).
By integrating
L
Φ =
k
2
Φ over (
−ε, ε
), taking
ε →
0, we find that
∂Φ
∂x
has a jump
discontinuity at x = 0 given by
ik(R − 1) + ikT = −2αT.
We now have two equations and two unknowns, and we can solve to obtain
R(k) =
iα
k −iα
, T (k) =
k
k −iα
.
We can see that we indeed have
|R|
2
+ |T |
2
= 1.
Note that as
k
increases, we find that
R
(
k
)
→
0 and
T
(
k
)
→
1. This makes
sense, since we can think of
k
as the energy of the wave, and the larger the
energy, the more likely we are to pass through.
Now let’s do the discrete part of the spectrum, and we jump through the
same hoops. Since δ(x) = 0 for x 6= 0, we must have
−
∂
2
ψ
n
∂x
2
+ χ
2
n
ψ
n
= 0
for x 6= 0. So we have
ψ
n
(x) = c
n
e
−χ
n
|x|
.
Integrating Lψ
n
= −χ
2
n
ψ
n
over (−ε, ε), we similarly find that
c
n
χ
n
= c
n
α.
So there is just one bound state, with
χ
1
=
α
. We finally find
c
n
by requiring
kψ
1
k = 1. We have
1 =
Z
∞
−∞
ψ
1
(x)
2
dx = c
2
1
Z
∞
−∞
e
−2χ
1
|x|
dx =
c
2
1
α
.
So we have
c
1
=
√
α.
In total, we have the following scattering data:
S =
{α,
√
α},
iα
k −iα
,
k
k −iα
.