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§5. Finite Band Potentials 221
it follows that
r
j
where λ is to be replaced by α
2
= Ajϕ
+2Bjϕ1ϕ2+ Cjϕ
1
. We have to show that
j
2
− B
A
jCj
j
2
2
(5.9)
=0. (5.10)
But this follows from the differential equation (5.4). For Γ(x, λ) this equation
takes the form
− 2(q −λ)Γ)Γ − Γ2=4
2(Γ
Inserting the expression (5.7) for Γ and using that ϕ = ϕ
(L − α
)ϕ =0we obtain as coefficients of (λ − αj)−2(dropping the index j)
j
12
4(Aϕ
−4(Aϕ
=4(AC − B
+2Bϕ
ϕ
1
ϕ
+ Cϕ
1
2
+ B(ϕ1ϕ
2
2
)(ϕ1ϕ
22
2
2
)(Aϕ
+2Bϕ1ϕ2+ Cϕ
1
+ ϕ
ϕ2)+Cϕ2ϕ
2
1
− ϕ
ϕ2)2=4(AC −B2).
1
Since the right hand side of (5.11) has only a simple pole at λ = α
indeed (5.10). Hence, if A
and because of (5.8) we have A
=0we can write
j
−1
r
= A
j
ψ
j
(Ajϕ1+ Bjϕ2)2,
j
> 0. Therefore we can take
j
−1/2
= A
Ajϕ1+ Bjϕ2).
j
b
. (5.11)
a
, ϕ2satisfy
1
2
)−
2
)2=
2
it follows
j
If A
=0then also Bj=0and Cj> 0. In this case we can take
j
j
= C
1/2
j
ϕ
2
ψ
proving the proposition.
3. Connection with the mechanical problem
From Proposition 5.1 we have the important representation for the function Γ
related by (5.7) to the Green’s function:
n
2
ψ
(x)
j=1
j
λ − α
. (5.12)
j
Γ(x, λ)=
222 Integrable Hamiltonian Systems and Spectral Theory
Since Γ ∼ λ−1for λ →∞we conclude that
n
2
ψ
=1. (5.13)
j
j=1
Moreover, ψ
the form
The last two equations show that we can interpret x
are solutions of the equation (L − αj)ψj=0which we write in
j
= −αjψj+ qψj. (5.14)
ψ
j
= ψj(t) as the
j
components of a vector x satisfying the differential equation
¨x = −Ax + q(t)x, A = diag(α
where x is restricted to the unit sphere, i. e. ψ
, ..., αn)(5.14)
1
(t) are the components of a
j
solution of the Neumann problem (3.11).
We recall that this mechanical problem possesses the integrals (see (3.7))
=Γ,
=
2
(ψ, ψ)(5.14)
λ
1
Γ
.
2
where
hence
Φ
(ψ,ψ)=(1+Qλ(ψ))Qλ(ψ) − Q
λ
n
n
j=1
2
ψ
j
λ − α
ψjψ
λ − α
j
j
Q
λ
Q
(ψ, ψ)=
λ
(ψ)=
j=1
j
Another differentiation gives with the aid of (5.13)
1
1+Q
(ψ)=
λ
(Γ
− 2(q −λ)Γ).
2
Therefore we find from the differential equation for G, or its equivalent
form (5.11) for Γ,
1
(ψ,ψ)=
Φ
λ
Hence the solution x =(ψ
{2(Γ
4
1,ψ2
− 2(q −λ))Γ − Γ2} =
, ..., ψn) of Proposition (5.1) corresponds to
the solution of the mechanical problem satisfying
(ψ,ψ)=
Φ
λ
b(λ)
a(λ)
b
.
a
.
§5. Finite Band Potentials 223
In other words, the solution ψ lies on the invariant manifold defined by
(ψ,ψ)=0, j =0, 2, ..., n− 1.
Φ
β
j
Theorem 5.2. If G(x, y; λ) is the Green’s function of the potential q and
if G(x, x; λ) satisfies the assumptions (5.2), (5.3) then it can be represented in
the form
G(x, x; λ)=−
"
1
−
2
a(λ)
b(λ)
n
j=1
2
ψ
(x)
j
λ − α
j
(5.15)
where ψ =(ψ
, ..., ψn) is a solution of the Neumann problem (5.14)
1,ψ2
satisfying
Φ
(ψ,ψ)=
λ
b(λ)
a(λ)
.
Moreover, the potential q = q(x) is given by
q(x)=2
n
j=1
αjψ
n−1
2
+
j
k=1
βk−
n
αj. (5.16)
j=1
It suffices to verify (5.16). For this purpose we recall the asymptotic
expansion (4.16) where G
= q/2. Comparing this with the expansion of the
1
right hand side of (5.15) we find (5.16).
4. Solution of the inverse problem
Our problem will be solved if we show the converse to Theorem 5.2, i. e.
Theorem 5.3. If ψ
is any solution of the problem (5.14) with
j
Φ
(ψ,ψ)=
λ
b(λ)
a(λ)
(5.17)
then the function q(x) defined by (5.16) is the potential of an operator (4.1) with
the band spectrum (5.1).
We indicate the proof: With the given ψ
have
(L − α
j)ψj
and the q defined by (5.16) we
j
=0.
224 Integrable Hamiltonian Systems and Spectral Theory
But we need the solutions of
(L − λ)ψ =0 (5.18)
for arbitrary complex λ. For this purpose we define G(x, λ)=G(x, x; λ)
by (5.15). Then the relation (5.17) is equivalent to the differential equation (5.4).
It also clearly satisfies (5.5) and therefore G(x, λ) is the candidate for the Green’s
function on the diagonal. Moreover, we show that
Im G(x, λ) > 0 for Im λ>0. (5.19)
By the maximum principle for harmonic functions it suffices to show
Im G(x, λ) 0 on the real axis. In the gaps one has Im G(x, λ)=0by (5.15)
since both factors are real. In the bands, however, G(x, λ) is purely imaginary
and Im G(x, λ) > 0 for α
that G has no zero in a band. This follows from (5.4) which gives G
<λ<αj+ ε and ε>0 small. It suffices to show
j
= ±1 at
a zero of G, hence G would be real near such a zero, while actually it is purely
imaginary. Hence the zeros of G must lie in the gaps.
In order to determine the spectrum of L we have to find the Green’s function
G(x, y; λ) of L, since we have not yet seen that G(x, λ) is related to G(x, y; λ)
on the diagonal. For this we first show that for Im λ>0,
ϕ(x, λ)=
#
G(x, λ)exp−
x
1
G−1(t, λ) dt
2
0
(5.20)
is a solution of (5.18). Note that the integrand is well defined because of (5.19).
To verify this claim we form
and
ϕ
ϕ
ϕ
ϕ
ϕ
=
ϕ
1
=
2
+
1
G
G
ϕ
ϕ
−1
−
G
2
2
1
=
4G
− 1
G
1
=
2
G
{2GG− G2+1}.
2
Because of (5.4) this agrees with
ϕ
ϕ
1
=
4(q − λ)G2= q − λ,
2
4G
proving our claim.
§5. Finite Band Potentials 225
If we could show that for Im λ =0
x
− 1
lim
x→∞
satisfies Re w(λ) < 0 then the above solution (5.20) belongs to L
can be used for ψ
(x, λ). Similarly we find
+
ψ
(x, λ)=#G(x, λ)exp
−
G−1(t, λ) dt = w(λ)
2x
0
1
2
x
G−1(t, λ) dt
0
2
(0, ∞) and
and therefore
x
G(x, y; λ)=#G(x, λ)G(y, λ)exp−
1
G−1(t, λ) dtfor x>y.
2
y
In particular, we see that G(x, x; λ) agrees with the function given by (5.15).
From the fact that G(x, λ) is real in the gaps and purely imaginary and =0in
the bands one verifies that the spectrum consists of the bands.
To prove the inequality Re w(λ) < 0 we use the fact that the potential q so
constructed is almost periodic, and therefore the solutions ψ
, ψ−decay at an
+
exponential rate for x → +∞ or x →−∞. This follows from (4.15). From
this it follows that the mean value w(λ) of the exponent in (5.20) has a non
vanishing real part. Since Re w(λ) < 0 for λ →−∞we have Re w(λ) < 0 for
Im λ =0. This completes the proof of Theorem 5.3.
5. Finite gap potentials as almost periodic functions
In Section 3 we showed that the solutions of the Neumann problem are
quasi-periodic functions with generally n −1=N basic frequencies. Therefore
also the potentials q defined by (5.16) are quasi-periodic. It is natural to study
the Floquet exponent w(λ) and α(λ)=Imw(λ) as defined in Section 4 for
these potentials.
Theorem 5.4. For the above potentials q defined by (5.16)
dw =
p(λ)
2#−a(λ)b(λ)
dλ =
p(λ) dλ
2#−R(λ)
(5.21)
226 Integrable Hamiltonian Systems and Spectral Theory
is a differential of the third kind, where
n−1
p(λ)=λ
+ ...
is a polynomial of degree n − 1.
This follows immediately from the formula (4.14)
dw
dλ
= −
.
1
2
n
a
−
b
j=1
M(ψ
λ − α
2
)
j
j
= −
.
1
2
p
a
−
a
b
where p is a polynomial of degree n − 1 with highest coefficient
n
j=1
M(ψ
2
)=1
j
(see (5.13)).
Secondly we study α(λ)=Imw(λ) in the bands. We can write G(x, λ)=
= iK(x, λ) with K(x, λ) > 0 in the interior of the band. Hence
w(λ)=M−
(λ)=M(G)=iM(K )
w
1
2G
i
−1
M(K
=
2
)
and for α(λ)=Imw(λ) we obtain
dα
= M(K
2α
dλ
by Schwarz’ inequality. We have equality only if K =const,i.e. G
−1
)M(K) >M(1) = 1
implies q =const. Hence we have
Theorem 5.5. In the bands we have the inequality
=0which
2
)
d(α
> 1.
dλ
This inequality is, of course, not only applicable in the finite band case, but
in all intervals of the real axis where G is purely imaginary, for example, in the
periodic case, where we used it to estimate the band length.
§5. Finite Band Potentials 227
6. The elliptic coordinates on the sphere
At the end of Section 3 we introduced the elliptic coordinates μ
on the
k
sphere; we shall interpret them here for the spectral problem and express the
potential q in terms of them.
These coordinates μ
were introduced as zeros of
k
n
2
ψ
(x)
j=1
j
λ − α
Q
(ψ)=
λ
n−1
(λ − μk)
k=1
=
j
a(λ)
. (5.22)
Because of (5.15) they agree with the zeroes of the Green’s function on the
diagonal G(x, x; λ). They depend on x and are restricted to the gaps. Indeed
we showed that G(x, x; λ) =0in the bands. Moreover, from (5.22) it is clear
that each interval (α
Therefore we conclude that each gap (β
j,αj+1
j,αj+1
) j =1, 2, , ..., n− 1, contains
one zero, or
μj(x) α
β
j
The potential q = q(x) can be expressed in terms of these variables μ
the formula
q(x) − α
n−1
=
k=1
(α
k+1
1
.
j+1
(x) by
j
+ βk− 2μk), (5.23)
which was derived by McKean and Trubowitz [18] in the periodic case, also in
the presence of infinitely many gaps.
−2
For the proof we compare the coefficients of λ
in the expansion of (5.22)
at λ →∞to get
n
j=1
αjψ
n
2
=
j
j=1
αj−
n−1
k=1
μk.
If we insert this into (5.16) we obtain formula (5.23).
Note that the terms on the right hand side of (5.23) lie in the interval
[−α
k+1
+ βk,α
− βk] so that we obtain simple bounds for q(x) − α1:
k+1
n−1
|q(x) − α
1
|
k=1
|α
k+1
− βk|
in terms of the total gap length.
228 Integrable Hamiltonian Systems and Spectral Theory
7. Alternative choice of the branch points
In the above consideration the choice of the α
matters are the branch points λ
, λ1, ..., λ2Nof the Riemann surface
0
#
Z =
−R(λ); R(λ)=
, βjis rather arbitrary. What
j
2N
(λ − λj).
j=0
The factorization
R(λ)=a(λ)b(λ)=
n
j=1
(λ − αj)
n−1
k=1
(λ − βk)
corresponding to (5.6) could be replaced by any other factorization
R(λ)=a
∗
(λ)b∗(λ)=
n
j=1
(λ − α
n−1
∗
)
j
k=1
(λ − β
∗
).
k
Since the Green’s function is independent of this factorization we obtain analogous to (5.15),
"
a
1
−
2
b∗(λ)
)=(q −α
with solutions ψ
G(x, x; λ)=−
∗
(x) of
j
(ψ
∗
j
This gives a large number of quadratic relations between these solutions ψ
∗
(λ)
∗
j
)ψ
n
j=1
∗
.
j
∗2
ψ
j
λ − α
∗
j
; ψ
j
at the various branch points.
This freedom of choice can be used to study those non-degenerate orbits
of the mechanical problem which we neglected so far. Until now we restricted
ourselves to the case where
∗
j
α
1<β1
<α2<... <β
which is equivalent to the condition (see (5.17))
n
j=1
Fj(ψ,ψ)
λ − α
Φ
(ψ,ψ)=
λ
F
(ψ,ψ) > 0(j =1, 2, ..., n).
j
n−1
j
=
<α
b(λ)
a(λ)
n
,
§6. Limit Cases, Bargmann Potentials 229
If we have a different arrangement of the distinct α
∗
to the above case by picking the α
∗
, β
interlacing and making use of the above
j
j
identities.
Fig. 1
, βkwe can always reduce it
j
§ 6. Limit Cases, Bargmann Potentials
1. Schwarz – Christoffel mapping
The Floquet exponent w = w(λ) for the finite band case gives rise to a
conformal mapping of Im λ>0 into a slit domain. From (4.15) it is clear that
the image domain lies in the second quadrant. Moreover, α(λ)=Imw(λ) is
constant in the gaps (see theorem 4.7). Since α(λ)=0for real λ λ
this part of the real axis is mapped on the negative real axis. If one uses that
G(x, x; λ) =0and purely imaginary on the bands one verifies easily that w =
= w(λ) maps the upper half plane into the slit domain shown in the figure. The
n − 1 gaps of finite length go into n − 1 slits, and the bands are mapped onto
parts of the imaginary axis.
0
= α
1
230 Integrable Hamiltonian Systems and Spectral Theory
This mapping is given by the Schwarz –Christoffel formula
w(λ)=
λ
λ
0
p(λ)−
2N
j=0
(λ − λj)
−1/2
dλ
where p(λ) is a polynomial of degree n − 1=N . This is in agreement
with (5.21). The zeros of p(λ) are mapped into the tips of the slits. From the
asymptotic behavior w ∼−
Instead of prescribing λ
slit domain characterized by ω
where iω
corresponds to w(α
j
√
−λ one sees that this mapping is bijective.
, λ1, ..., λ2Nwe could equally well prescribe the
0
, hj> 0, j =1, 2, ..., N = n−1 and λ0= α1,
j
)=w(βj) and hj> 0 to the «height» of the
j+1
j-th slit. By the Riemann mapping theorem there is a unique conformal mapping
of the slit domain characterized by ω
half plane, taking w =0into λ = λ
λ ∼−w
at infinity. Clearly λ
= α1can be normalized to be zero by a translation, so
0
that the spectrum is characterized by the 2N positive numbers λ
=1, 2, ..., N) or equivalently by the 2N positive numbers ω
, hj(j =1, 2, ..., n−1), onto the upper
j
= α1and with the asymptotic behavior
0
2
or w ∼−√−λ
− λ0(j =
j
, hj.
j
2. Basis for the frequency module
We know from the study of the mechanical model that the finite band
potentials are quasi-periodic and the frequency module
(q) is spanned in
general by n − 1=N frequencies.
Theorem 6.1. For a finite band potential q = q(x) the frequency module
is spanned by ω
= w(βj), j =1, 2, ..., n= N −1.
j
The proof turns out to be relatively simple if we use the connection with
the mechanical problem.
Let ω
1
, ω
, ..., ω
2
be a frequency basis of(q), where we assume at
N
first that we are in the generic situation of N rationally independent frequencies.
Then we know from Theorem 4.7 that
N
2iω
=2w(βk)=i
k
ν=1
jνkω
k =1, 2, ..., N (6.1)
ν
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