MA 342N
Assignment 1
Due 24 February 2016
Id:
342N-s2016-1.m4,v 1.1 2016/02/15 21:25:36 john Exp john
1. Suppose that q, in addition to satisfying the assumptions from lecture,
is an even function. Prove that η(λ) = 0, i.e. that there are no mixed
terms (θϕ or ϕθ terms) in the spectral decomposition.
Hint: Show that
ψ+ (λ, −x) = ψ− (λ, x),
ϕ(λ, −x) = −ϕ(λ, x),
ψ− (λ, −x) = ψ+ (λ, x),
θ(λ, −x) = θ(λ, x).
Solution:
In general, if y satisfies
y ′′ + λy = qy
then ỹ satisfies
ỹ ′′ + λỹ = q̃ỹ
where ỹ(x) = y(−x) and q̃(x) = q(−x). In particular, if q is even then
q̃ = q and ỹ satisfies the same differential equation as y. So ψ+ (λ, −x),
ψ− (λ, −x), ϕ(λ, −x) and θ(λ, −x) all satisfy
y ′′ + λy = qy.
√
√
From ψ+ (λ, x) ≈ ei λx as x → +∞ it follows that ψ+ (λ, −x) ≈ e−i λx
as x → −∞, so ψ+ (λ, −x) = ψ− (λ, x). Similarly, or as a corollary,
ψ− (λ, −x) = ψ+ (λ, x). Also ϕ(λ, −x) and −ϕ(λ, x) satisfy the same
initial value problem and so are, by the fundamental existence and
uniqueness theorem, the same function. Similarly, θ(λ, −x) and θ(λ, x)
are the same.
1
Id:
342N-s2016-1.m4,v 1.1 2016/02/15 21:25:36 john Exp john 2
Substituting −x for x in the equation
α(λ) β(λ)
( ψ− (λ, x) ψ+ (λ, x) ) = ( ϕ(λ, x) θ(λ, x) )
γ(λ) δ(λ)
gives
α(λ) β(λ)
( ψ+ (λ, x) ψ− (λ, x) ) = ( −ϕ(λ, x) θ(λ, x) )
γ(λ) δ(λ)
and hence
−β(λ) −α(λ)
( ψ− (λ, x) ψ+ (λ, x) ) = ( ϕ(λ, x) θ(λ, x) )
.
δ(λ)
γ(λ)
It follows that
α(λ) β(λ)
γ(λ) δ(λ)
and thus
=
−β(λ) −α(λ)
δ(λ)
γ(λ)
1
α(λ)δ(λ)
= .
α(λ)δ(λ) − β(λ)γ(λ)
2
Integrating,
Z
λ+iτ
iτ
and hence
λ
α(µ)δ(µ)
dµ =
α(µ)δ(µ) − β(µ)γ(µ)
2
η(λ) = 0.
2. Consider the set of functions y which satisfy y ′′ + λy = 0 for x 6= 0, are
continuous at x = 0, and whose derivatives are continuous except for a
jump discontinuity of the form
lim y ′ (x) − lim− y ′ (x) = Ay(0).
x→0+
x→0
These are close enough to the situation considered in lecture that we
can still do scattering theory and spectral theory.1
1
If we take the limit of our special solutions to y ′′ + λy = qy with
h if |x| < w/2,
q(x) =
0 if |x| > w/2
with y and y ′ continuous at x = ±w/2 as w → 0 and h → ∞ in such a way that A = wh
is constant then we get solutions to the problem above. Formally this is often written as
y ′′ + λy = qy with q(x) = Aδ(x), where δ is the Dirac delta. Unfortunately that doesn’t
really make any sense, even if the equations are interpreted in the sense of distributions.
The simplest way of treating the problem is to ignore both delta functions and the h → 0
limit and just view the equation as y ′′ + λy = 0 with the jump condition given above.
Id:
342N-s2016-1.m4,v 1.1 2016/02/15 21:25:36 john Exp john 3
Find the scattering matrix for this problem, and verify that it is unitary.
Note: The basic existence and uniqueness theorem still applies here,
despite the strangeness of the equation at x = 0. That’s not hard
to prove, but it’s also not interesting, so don’t bother. You can just
assume there is a unique solution for any choice of initial data at any
point, except that if we are taking data at x = 0 then we need to specify
the average of the limits
1
2
lim+ y ′ (x) + lim− y ′(x)
x→0
x→0
rather than the (probably non-existent) y ′(0).
Hint: The functions θ and ϕ are still useful, but their definitions need
to be modified as described above:
1
2
θ(λ, 0) = 1,
x→0
1
2
cos(sx) −
cos(sx) +
A
2s
A
2s
lim+ θ′ (λ, x) + lim− θ′ (λ, x) = 0,
x→0
′
′
lim ϕ (λ, x) + lim− ϕ (λ, x) = −1.
x→0
√
Solution: With, as usual s = λ, these two solutions are
ϕ(λ, 0) = 0,
θ(λ, x) =
(
x→0+
sin(sx) if x < 0,
sin(sx) if x > 0,
ϕ(λ, x) = −
sin(sx)
.
s
The solutions defined by their behaviour as x → +∞ are given by the
equations
A
= cos(sx) + i sin(sx) = θ(λ, x) +
− is ϕ(λ, x)
2
isx
ψ+ (λ, x) = e
and
−isx
χ+ (λ, x) = e
A
= cos(sx) − i sin(sx) = θ(λ, x) +
+ is ϕ(λ, x)
2
if x > 0. The solutions defined by their behaviour as x → −∞ are
−isx
ψ− (λ, x) = e
A
= cos(sx) − i sin(sx) = θ(λ, x) + − + is ϕ(λ, x)
2
and
isx
χ− (λ, x) = e
A
= cos(sx) + i sin(sx) = θ(λ, x) + − − is ϕ(λ, x)
2
Id:
342N-s2016-1.m4,v 1.1 2016/02/15 21:25:36 john Exp john 4
if x < 0. Forgetting the intermediate steps,
A
− is ϕ(λ, x)
ψ+ (λ, x) = θ(λ, x) +
2
and
χ+ (λ, x) = θ(λ, x) +
A
+ is ϕ(λ, x)
2
if x > 0 and
A
ψ− (λ, x) = θ(λ, x) + − + is ϕ(λ, x)
2
and
A
χ− (λ, x) = θ(λ, x) + − − is ϕ(λ, x)
2
if x < 0. These equations must in fact hold everywhere, as a consequence of the existence and uniqueness theorem.
A little algebra shows that
ψ+ (λ, x) + ψ− (λ, x) = 2θ(λ, x) = χ+ (λ, x) + χ− (λ, x)
ψ+ (λ, x) − ψ− (λ, x) = (A − 2is) ϕ( λ, x)
χ+ (λ, x) − χ− (λ, x) = (A + 2is) ϕ( λ, x)
so
1
1
1
1 A + 2is
(χ+ + χ− ) − (χ+ − χ− ) = (ψ+ + ψ− ) −
(ψ+ − ψ− )
2
2
2
2 A − 2is
−2is
A
ψ− +
ψ+
=
A − 2is
A − 2is
χ− =
and
1
1
1
1 A + 2is
(χ+ + χ− ) + (χ+ − χ− ) = (ψ+ + ψ− ) +
(ψ+ − ψ− )
2
2
2
2 A − 2is
A
−2is
ψ− +
ψ+
=
A − 2is
A − 2is
χ+ =
The scattering matrix is therefore
a(λ) b(λ)
c(λ) d(λ)
1
=
A − 2is
A
−2is
.
−2is
A
Id:
342N-s2016-1.m4,v 1.1 2016/02/15 21:25:36 john Exp john 5
3. Find the spectral representation for the same problem.
Solution: We have
α(λ) β(λ)
( ψ− (λ, x) ψ+ (λ, x) ) = ( ϕ(λ, x) θ(λ, x) )
γ(λ) δ(λ)
where
α=−
A
+ is,
2
β=
A
− is,
2
γ = 1,
δ = 1.
so
αδ − βγ = −A + 2is,
αβ
A
s
= −i ,
αδ − βγ
4
2
1
βγ
=− ,
αδ − βγ
2
and
1
γδ
=
.
αδ − βγ
−A + 2is
Then, integrating,
Z
αβ
A
i
dλ = λ − λ3/2 ,
αδ − βγ
4
3
Z
and
Z
λ
βγ
=− ,
αδ − βγ
2
√ !
√
γδ
A − 2i λ
A
.
= −i λ + log
αδ − βγ
2
A
Integrating from iτ to λ + iτ and taking −1/π times the limit of the
imaginary part as τ → 0+ ,
ξ(λ) =
(
0
λ3/2
3π
if λ ≤ 0,
if λ > 0,
and
η(λ) = 0.
ζ(λ) is more complicated and depends on the sign of A. If A is positive
then
(
0
if λ ≤ 0,
√
ζ(λ) = λ1/2
A
2 λ
+ 2π arctan A if λ ≥ 0 .
π
Id:
342N-s2016-1.m4,v 1.1 2016/02/15 21:25:36 john Exp john 6
If A is negative then
ζ(λ) =





A
2π
0
λ1/2
π
+
A
2π
√
arctan 2 Aλ
if λ < −A2 /4,
if −A2 /4 < λ ≤ 0,
if λ ≥ 0 .
The case A = 0 was of course done in lecture, since that’s just q = 0.
The spectral representation is then
f (x)
1 Z ∞Z ∞
ϕ(λ, x)ϕ(λ, x̃)f (x̃) dx̃ λ1/2 dλ
2π Z0 Z−∞
1 ∞ ∞
4λ1/2
+
θ(λ, x)θ(λ, x̃)f (x̃) dx̃
dλ
2π 0 −∞
4λ + A2
Z ∞
A
ϕ(−A2 /4, x)
−
ϕ(−A2 /4, x̃)f (x̃) dx̃,
2π
−∞
with the last term present only if A < 0.