One-dimensional quantum walks with a general perturbation of the radius 1
M. Ryazanov A. Zamyatin

Maximum Likelihood Estimation and Graph Matching in Errorfully Observed Networks
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We give a complete description of the discrete spectrum of one-dimensional
Hamiltonian with a general perturbation of the radius 1. 









§ INTRODUCTION


We consider the one particle continuous time one-dimensional quantum
walk with an Hamiltonian H acting in the Hilbert space l_2(Z),
where  is the one-dimensional lattice and l_2() is the
space of complex valued square summable sequences f={f_n∈,n∈}.

Let { e_n} _n∈ Z be the standard basis of l_2().
If f∈ l_2() then we have f=∑_n∈ Zf_ne_n.

The Hamiltonian H=H_0+H_1, l_2()→ l_2()
is defined as follows:

    H_0e_n    =-λ(e_n+1-2e_n+e_n-1), n∈
    
    H_1e_0    =-λ_1(e_1-2e_0+e_-1)+μ e_0
    
    H_1e_1    =-λ_1e_0+μ_1e_1
    
    H_1e_-1    =-λ_1e_0+μ_1e_-1
    
    H_1e_n    =0,if |n|>1,

where λ, λ_1, μ, μ_1∈. Without
loss of generality we can assume that λ>0. Our aim is to
describe the discrete spectrum of the Hamiltonian H.

The continuous time quantum walk evolution is defined as 

    f(t)=e^-itHf(0)∈ l_2()

where f(0) is the initial state.

We generalize the result of <cit.>, where authors investigated
the discrete spectrum of the Hamiltonian which is defined by two parameters
μ and μ_1, H=H_0+H_1 : l_2^s()→ l_2^s():

    H_0e_n    =-1/2e_n+1+e_n-1/2e_n-1
    
    H_1e_0    =μ e_0
    
    H_1e_1    =μ_1e_1
    
    H_1e_-1    =μ_1e_-1
    
    H_1e_n    =0, if  |n|>1

where l_2^s()⊂ l_2() is the subspace of symmetric
sequences: f_n=f_-n for all n∈ .



§ MAIN RESULTS




  
*Notation

It will be convenient to introduce new parameters 

    α=λ_1/λ,   δ=μ_1/λ,  σ=2λ_1+μ/2λ

Define the functions of parameters α,δ,σ:

    c_1    =c_1(α,δ,σ)=(δ-1)(σ-1)-(α+1)^2
    
    c_2    =c_2(α,δ,σ)=(δ+1)(σ+1)-(α+1)^2

where (α,δ,σ)∈^3.

Let us introduce the following partition of ^3:

    U_1(+,+)=    {(α,δ,σ)∈^3:c_1>0,c_2>0,δ>1,σ>1}
    
    U_0(+,+)=    {(α,δ,σ)∈^3:c_1⩾0,c_2⩾0,|δ|≤1,|σ|≤1}
    
    U_-1(+,+)=    {(α,δ,σ)∈^3:c_1>0,c_2>0,δ<-1,σ<-1}
    
    U(-,+)=    {(α,δ,σ)∈^3:c_1<0,c_2⩾0}∪
       ∪{(α,δ,σ)∈^3:c_1=0,c_2>0,δ>1,>1}
    
    U(-,-)=    {(α,δ,σ)∈^3:c_1<0,c_2<0}
    
    U(+,-)=    {(α,δ,σ)∈^3:c_1⩾0,c_2<0}∪
       ∪{(α,δ,σ)∈^3:c_1>0,c_2=0,δ<-1,σ<-1}

Figures <ref>–<ref> (see Appendix) show sections of the subsets from partition
(<ref>) by planes α=c for various constants c≠-1.

Finally we put 

    κ=λ(δ+1)^2/δ=(μ_1+λ)^2/μ_1

for μ_1≠0.



  
*Results

It follows from Weyl's theorem <cit.> on essential spectrum
that the essential spectrum of the Hamiltonian H is the same as
for the homogeneous Hamiltonian H_0 and coincides with the segment
[0,4λ].

Denote by s_d(H) the discrete spectrum <cit.> of H.
We give a complete description of s_d(H). We prove that the discrete
spectrum consists of at most three eigenvalues. One of them we find
explicitly. For the other two we define the intervals of the real
axis containing these eigenvalues.

 Let λ>0 and κ be defined
by (<ref>). For α=λ_1/λ≠-1
we have

a) if (α,δ,σ)∈ U_1(+,+),
then s_d(H)={ν_1,ν_2,ν_3}, where ν_3=κ,
ν_1,ν_2∈(4λ,∞) and, moreover, 

    4λ<ν_1<2λ(min{σ,δ^2+1/2δ} +1)≤
      2λ(max{σ,δ^2+1/2δ} +1)<ν_2;


b) if (α,δ,σ)∈ U_0(+,+), then s_d(H)=∅;

c) if (α,δ,σ)∈ U_-1(+,+) then s_d(H)={ν_1,ν_2,ν_3},
where ν_3=κ,ν_1,ν_2∈(-∞,0) and, moreover,

    ν_1<2λ(min{σ,δ^2+1/2δ} +1)≤
      2λ(max{σ,δ^2+1/2δ} +1)<ν_2<0;


d) if (α,δ,σ)∈ U(-,+), then

d1) for |δ|⩽1 s_d(H)={ν_1}, where ν_1∈(4λ,∞),

d2) for |δ|>1 s_d(H)={ν_1,ν_2}, where ν_1∈(4λ,∞),ν_2=κ;

e) if (α,δ,σ)∈ U(-,-), then

e1) for |δ|⩽1 s_d(H)={ν_1,ν_2}, where
ν_1∈(-∞,0), ν_2∈(4λ,∞),

e2) for |δ|>1 s_d(H)={ν_1,ν_2,ν_3}, where
ν_1∈ (-∞,0), ν_2∈ (4λ,∞), ν_3=κ;

f) if (α,δ,σ)∈ U(+,-), then

f1) for |δ|⩽1 s_d(H)={ν_1}, where ν_1∈(-∞,0),
and

f2) for |δ|>1 s_d(H)={ν_1,ν_2}, where ν_1∈(-∞,0),ν_2=κ;

g) geometric multiplicity of all eigenvalues is 1; the eigenvector
f={f_n}∈ l_2() corresponding to the eigenvalue κ
is an odd function (f_n=-f_-n) and the eigenvectors f={f_n}∈ l_2()
corresponding to the eigenvalues ν_1,ν_2 are even functions
(f_n=f_-n).


For α=λ_1 / λ =-1 one can find eigenvalues
in an explicit form.

Let λ>0 and κ be defined
by (<ref>). For α=λ_1 / λ =-1 


  * if |δ|>1, |σ|>1, μ≠κ, then s_d(H)={μ,κ} 

  * if |δ|≤1, |σ|>1, then s_d(H)={μ} 

  * if |δ|>1, |σ|≤1, then s_d(H)={κ} 

  * if |δ|≤1, |σ|≤1, then s_d(H)={∅} 

  * the eigenvalue κ is of the geometrical multiplicity 2 and
has two linearly independent eigenvectors one of which is an odd function
and the other is an even function. 

  * if μ=κ, |δ|>1, |σ|>1, there is only one
eigenvalue of the geometrical multiplicity 2 with the same eigenvectors
as in the previous item. 



Note that for α=λ_1 / λ =-1 there exists
the eigenvalue μ with the eigenvector e_0. By definition
(<ref>), one can obtain

    He_0    =H_0e_0+H_1e_0=
        =-λ(e_1-2e_0+e_-1)-λ_1(e_1-2e_0+e_-1)+μ e_0=μ e_0

So, if μ∈[0,4λ], then there exists the eigenvalue ν=μ
belonging to the essential spectrum of Hamiltonian H.



§ PROOF OF MAIN RESULTS


For f∈ l_2() we have 

    Hf     =∑_n∈ Zλ(-f_n+1+2f_n-f_n-1)e_n+(μ_1f_-1-λ_1f_0)e_-1+
        +((2λ_1+μ)f_0-λ_1f_1-λ_1f_-1))e_0+(μ_1f_1-λ_1f_0)e_1


Let us consider the Hilbert space L_2() of square integrable
functions defined on the unit circle . The elements of L_2()
we will denote by f̂. It is known that any two separable Hilbert
spaces are isomorphic, so we have l_2()≅ L_2().

Consider an isomorphism U : l_2()→ L_2() such
that e_n⟷1/√(2π)e^iφ n.
Then U is an unitary operator and for f=∑_n∈ Zf_ne_n
we have

    Uf=∑_n∈ Zf_n1/√(2π)e^iφ n

Introduce operator Ĥ=UHU^-1 : L_2()→ L_2()
which is unitarily equivalent to H. Using (<ref>) we get

    Ĥf̂=∑_n∈ Zλ(-f_n+1+2f_n-f_n-1)1/√(2π)e^iφ n+(μ_1f_-1-λ_1f_0)1/√(2π)e^-iφ+
    
    +((2λ_1+μ)f_0-λ_1f_1-λ_1f_-1))1/√(2π)+(μ_1f_1-λ_1f_0)1/√(2π)e^iφ=
    
    =∑_n∈ Zλ(-f_n+11/√(2π)e^iφ(n+1)e^-iφ+2f_n1/√(2π)e^iφ n-f_n-11/√(2π)e^iφ(n-1)e^iφ)-
    
    -λ_1/√(2π)f_0(e^-iφ+e^iφ-2)-λ_1/√(2π)(f_-1+f_1)+μ/√(2π)f_0+μ_1/√(2π)(f_-1e^-iφ+f_1e^iφ)=
    
    =-λ(e^-iφ+e^iφ-2)f̂-λ_1/√(2π)(f_-1+f_1+f_0(e^-iφ+e^iφ-2))+
    
    +μ/√(2π)f_0+μ_1/√(2π)(f_-1e^-iφ+f_1e^iφ)

Thus, 

    Ĥf̂=-2λ(cosφ-1)f̂-λ_1/√(2π)(f_-1+f_1+2f_0(cosφ-1))+
    
    +μ/√(2π)f_0+μ_1/√(2π)(f_-1e^-iφ+f_1e^iφ)


Because of unitarily equivalence of operators Ĥ and H,
their point spectra are identical.

If f̂ is an eigenfunction with eigenvalue ν, i.e. Ĥf̂=νf̂,̂
then by (<ref>) we have 

    νf̂=-2λ(cosφ-1)f̂-λ_1/√(2π)(f_-1+f_1+2f_0(cosφ-1))+
    
    +μ/√(2π)f_0+μ_1/√(2π)(f_-1e^-iφ+f_1e^iφ)

and, hence,

    f̂=1/√(2π)-λ_1/2λ(f_-1+f_1+2f_0(cosφ-1))+μ/2λf_0+μ_1/2λ(f_-1e^-iφ+f_1e^iφ)/cosφ+ν/2λ-1=
    
    =1/√(2π)-α/2(f_-1+f_1)+(σ-αcosφ)f_0+δ/2(f_-1e^-iφ+f_1e^iφ)/cosφ+ν/2λ-1

Since 

    f_-1e^-iφ+f_1e^iφ=(f_-1+f_1)cosφ+i(-f_-1+f_1)sinφ

we get 

    f̂=1/√(2π)-α/2(f_-1+f_1)+(σ-αcosφ)f_0/cosφ+ν/2λ-1+
    
    +1/√(2π)δ/2((f_-1+f_1)cosφ+i(-f_-1+f_1)sinφ)/cosφ+ν/2λ-1=
    
    =1/√(2π)f_-1+f_1/2(-α+δcosφ)+(σ-αcosφ)f_0+iδf_1-f_-1/2sinφ/cosφ+ν/2λ-1

Put

    γ=ν/2λ-1

It is evident that γ∈[-1,1]⟺ν∈[0,4λ].
Then 

    f̂=1/√(2π)f_-1+f_1/2(δcosφ-α)+(σ-αcosφ)f_0+iδf_1-f_-1/2sinφ/cosφ+γ

If γ∉[-1,1], then f̂∈ L_2().

Applying the inverse Fourier transform we get 

    f_k   =1/√(2π)∫_-π^πf̂ e^-ikφdφ= 
       =1/2π∫_-π^πf_-1+f_1/2(δcosφ-α)+(σ-αcosφ)f_0+iδf_1-f_-1/2sinφ/cosφ+γe^-ikφdφ

for all k∈. It follows, that the coordinates f_0,f_1,f_-1
of eigenvector f={f_k} satisfy the following homogeneous linear
system:

    f_0=1/2π∫_-π^πf_-1+f_1/2(δcosφ-α)+(σ-αcosφ)f_0+iδf_1-f_-1/2sinφ/cosφ+γdφ


    f_1=1/2π∫_-π^πf_-1+f_1/2(δcosφ-α)+(σ-αcosφ)f_0+iδf_1-f_-1/2sinφ/cosφ+γe^-iφdφ


    f_-1=1/2π∫_-π^πf_-1+f_1/2(δcosφ-α)+(σ-αcosφ)f_0+iδf_1-f_-1/2sinφ/cosφ+γe^iφdφ


Thus, we proved the following lemma:


Real ν∉[0,4λ] is an eigenvalue of the Hamiltonian
H iff linear system (<ref>)–(<ref>) has a nontrivial
solution for γ=ν/2λ-1. The geometric multiplicity
of eigenvalue ν is equal to 3-r, where r is the rank of
linear system (<ref>)–(<ref>).


Now we simplify integrals in equations (<ref>)–(<ref>).
Let us denote 

    f_1^'=f_-1+f_1/2,   f_-1^'=f_1-f_-1/2

We have from equation (<ref>)

    f_0=∫_-π^πf_1^'(δcosφ-α)+(σ-αcosφ)f_0+iδ f_-1^'sinφ/cosφ+γdφ=
    
    =∫_-π^πf_1^'(δcosφ-α)+(σ-αcosφ)f_0/cosφ+γdφ

since 

    ∫_-π^πiδ/2(-f_-1+f_1)sinφ/cosφ+γdφ=0

because of the integrand is an odd function.

Further, from equation (<ref>) we get 

    f_1=∫_-π^πf_1^'(δcosφ-α)+(σ-αcosφ)f_0/cosφ+γ(cosφ-isinφ)dφ+
    
    +∫_-π^πiδ f_-1^'sinφ/cosφ+γ(cosφ-isinφ)dφ

Because of the integrands are odd functions we have 

    ∫_-π^πf_1^'(δcosφ-α)+(σ-αcosφ)f_0/cosφ+γisinφ dφ    =0
    ∫_-π^πsinφ/cosφ+γcosφ dφ    =0


So 

    f_1=∫_-π^πf_1^'(δcosφ-α)+(σ-αcosφ)f_0/cosφ+γcosφ dφ+
    
        ∫_-π^πδ f_-1^'sin^2φ/cosφ+γdφ

Similarly, one can show that 

    f_-1=∫_-π^πf_1^'(δcosφ-α)+(σ-αcosφ)f_0/cosφ+γcosφ dφ
        -∫_-π^πδ f_-1^'sin^2φ/cosφ+γdφ

Hence, system (<ref>)–(<ref>) can be rewritten as follows

    f_0=1/2π∫_-π^πf_1^'(δcosφ-α)+(σ-αcosφ)f_0/cosφ+γdφ


    f_1 =1/2π∫_-π^πf_1^'(δcosφ-α)+(σ-αcosφ)f_0/cosφ+γcosφ dφ
      +1/2π∫_-π^πδ f_-1^'sin^2φ/cosφ+γdφ


    f_-1   =1/2π∫_-π^πf_1^'(δcosφ-α)+(σ-αcosφ)f_0/cosφ+γcosφ dφ
    -1/2π∫_-π^πδ f_-1^'sin^2φ/cosφ+γdφ

Adding equations (<ref>) and (<ref>), subtracting equation
(<ref>) from (<ref>) and dividing the resulting equations
by 2 we come to the linear system with respect to variables f_0,f_1^',f_-1^':

    f_0=1/2π∫_-π^π(σ-αcosφ)f_0+(δcosφ-α)f_1^'/cosφ+γdφ



    f_1^'=1/2π∫_-π^π(σ-αcosφ)cosφ f_0+(δcosφ-α)cosφ f_1^'/cosφ+γdφ



    f_-1^'=f_-1^'1/2π∫_-π^πδsin^2φ/cosφ+γdφ


Finely, we get the system which is equivalent to the original system
(<ref>)–(<ref>):

    f_0(1-1/2π∫_-π^πσ-αcosφ/cosφ+γdφ)+f_1^'/2π∫_-π^πα-δcosφ/cosφ+γdφ=0



    -f_0/2π∫_-π^π(σ-αcosφ)cosφ/cosφ+γdφ+f_1^'(1+1/2π∫_-π^π(α-δcosφ)cosφ/cosφ+γdφ)=0



    f_-1^'(1/2π∫_-π^πδ(1-cos^2φ)/cosφ+γdφ-1)=0

Define the following functions

    g(x)     =    1/2π∫_-π^πdφ/cosφ+x
    
    h(x)     =     -1/2π∫_-π^πcosφ dφ/cosφ+x
    
    v(x)     =    1/2π∫_-π^πcos^2φ dφ/cosφ+x

where |x|>1. By lemma <ref>

    g(x)=1/x√(1-x^-2),


    h(x)=xg(x)-1, v(x)=xh(x),


    g(x)v(x)=h^2(x)+h(x)

Let us express coefficients of linear system (<ref>)–(<ref>)
in terms of functions g,h,v:

    f_0(1-α h(γ)-σ g(γ))+f_1^'(α g(γ)+δ h(γ))=0,


    f_0(σ h(γ)+α v(γ))+f_1^'(1-α h(γ)-δ v(γ))=0


    f_-1^'(δ(g(γ)-v(γ))-1)=0

Coefficients of this system are defined only for |γ|>1. We
need to find such γ that system (<ref>)–(<ref>)
has nontrivial solution. A homogeneous linear system has non trivial
solution iff its determinant equals zero. Denote by D^'(α,δ,σ,γ)
the determinant of this system and let D(α,δ,σ,γ)
be the determinant of system (<ref>)–(<ref>), consisting
of the first two equations. We have 

    D^'(α,δ,σ,γ)=D(α,δ,σ,γ)(δ(g(γ)-v(γ))-1)

So in order to determine eigenvalues we need to find roots of the
equation D^'(α,δ,σ,γ)=0. Simple algebra
gives

    D(α,δ,σ,γ)     =(1-α h(γ)-σ g(γ))(1-α h(γ)-δ v(γ))-
        -(α g(γ)+δ h(γ))(σ h(γ)+α v(γ))=
        =(1-α h(γ))^2-δ v(γ)+α h(γ)δ v(γ)-
        -σ g(γ)(1-δ v(γ))+σ g(γ)α h(γ)-
        -α g(γ)σ h(γ)-α^2g(γ)v(γ)-δσ h^2(γ)-δ h(γ)α v(γ)=
        =(1-α h(γ))^2-δ v(γ)-σ g(γ)(1-δ v(γ))-
        -α^2g(γ)v(γ)-δσ h^2(γ)=
        =(1-α h(γ))^2-δσ h^2(γ)-σ g(γ)-(α^2-σδ)g(γ)v(γ)-δ v(γ)

By (<ref>) we have g(γ)v(γ)=h^2(γ)+h(γ)
and, hence,

    D(α,δ,σ,γ)     =(1-α h(γ))^2-δσ h^2(γ)-σ g(γ)-(α^2-σδ)h^2(γ)-
        -(α^2-σδ)h(γ)-δ v(γ)
        =1-2α h(γ)+α^2h^2(γ)-δσ h^2(γ)-σ g(γ)-α^2h^2(γ)+
        +σδ h^2(γ)-α^2h(γ)+σδ h(γ)-δ v(γ)=
        =1-σ g(γ)-(α^2+2α-σδ)h(γ)-δ v(γ)


Remark that h(γ)≠0 for |γ|>1. Let us divide D(α,δ,σ,γ)
by h(γ). Using (<ref>), we obtain 

    D(α,δ,σ,γ)=h(γ)(1/h(γ)-σg(γ)/h(γ)-α^2-2α+σδ-δγ)

Put 

    l(γ)=γ(1+√(1-γ^-2))=γ+√(γ^2-1)   γ≥1
    γ-√(γ^2-1)   γ≤-1

By (<ref>), we have 

    1/h(γ)   =√(1-γ^-2)/1-√(1-γ^-2)=√(1-γ^-2)-1+1/1-√(1-γ^-2)=
       =-1+1/1-√(1-γ^-2)
    g(γ)/h(γ)   =1/γ(1-√(1-γ^-2))

Since 

    1/1-√(1-γ^-2)    =1+√(1-γ^-2)/(1-√(1-γ^-2))(1+√(1-γ^-2))
        =γ^2(1+√(1-γ^-2))=γ l(γ)

we get 

    1/h(γ)  =-1+γ l(γ),     g(γ)/h(γ)  =l(γ)

and, hence, 

    D(α,δ,σ,γ)   =h(γ)(γ l(γ)-σ l(γ)-δ(γ-σ)-(α+1)^2)=
       =h^-1(γ)((γ-σ)(l(γ)-δ)-(α+1)^2)


Consider the second factor in (<ref>). By (<ref>) we
have 

    g(γ)-v(γ)     =g(γ)-γ h(γ)=g(γ)-γ(γ g(γ)-1)
        =-g(γ)γ^2(1-γ^-2)+γ

Now using (<ref>) we get 

    g(γ)-v(γ)=-γ^2(1-γ^-2)/γ√(1-γ^-2)+γ=γ(1-√(1-γ^-2))=
    
    =γ(1-√(1-γ^-2))(1+√(1-γ^-2))/1+√(1-γ^-2)=
    
    =1/γ(1+√(1-γ^-2))=1/l(γ)

and the second factor in (<ref>) is 

    δ(g(γ)-v(γ))-1=δ/l(γ)-1=-l(γ)+δ/l(γ)

where l(γ) is defined by (<ref>). According to (<ref>)

    D^'(α,δ,σ,γ)=-h(γ)l^-1(γ)((γ-σ)(l(γ)-δ)-(α+1)^2)(l(γ)-δ)

Thus D^'(α,δ,σ,γ)=0 iff or l(γ)-δ=0
or (γ-σ)(l(γ)-δ)-(α+1)^2=0.

Note that |l(γ)|≥1 and the equation l(γ)-δ=0
has a unique root γ_δ=δ^2+1/2δ
iff |δ|≥1 (|γ|=1 iff |δ|=1). Hence, in
case of |δ|>1 we have eigenvalue 

    κ=2λ(δ^2+1/2δ+1)=λ(δ+1)^2/δ=(μ_1+λ)^2/μ_1


Let us find an eigenvector corresponding to this eigenvalue. To satisfy
linear system (<ref>)–(<ref>) one can put f_1^'=f_1+f_-1=0,
f_0=0, f_-1^'=f_1-f_-1=2f_1 and for all |k|>1
by (<ref>) we have

    f_k    =1/2π∫_-π^πδ/2(f_-1e^-iφ+f_1e^iφ)/cosφ+γe^-ikφdφ
        =1/2π∫_-π^πδ/2f_1(-e^-iφ+e^iφ)/cosφ+γe^-ikφdφ=
        =1/2π∫_-π^πδ if_1sinφ/cosφ+γ(cos kφ-isin kφ)dφ=
        =δ f_1/2π∫_-π^πsinφsin kφ/cosφ+γdφ=
        =δ f_1/4π∫_-π^πcos((k-1)φ)-cos((k+1)φ)/cosφ+γdφ


We see that f_k is an odd function of k: f_k=-f_-k.

Let α≠-1. Define the function 

    F(γ)=(γ-σ)(l(γ)-δ)

for |γ|≥1. As γ→±∞ F(γ)→∞.

Consider the equation 

    F(γ)=(α+1)^2

One can see (since l(±1)=±1) 

    c_1=     c_1(α,δ,σ)=(δ-1)(σ-1)-(α+1)^2=F(1)-(α+1)^2
    
    c_2=     c_2(α,δ,σ)=(δ+1)(σ+1)-(α+1)^2=F(-1)-(α+1)^2

The equation l(γ)=δ has a unique root 

    γ_δ=δ^2+1/2δ=1/2(δ+1/δ)

iff |δ|≥1.



a) Let (α,δ,σ)∈ U_1(+,+)={(α,δ,σ)∈ R^3:c_1>0,c_2>0,σ>1,δ>1}

The equation F(γ)=0 has two roots σ and γ_δ
which are greater than 1. Put 

    γ_min=min(σ,γ_δ),γ_max=max(σ,γ_δ)

Function F(γ) is strictly decreasing and positive on the interval
(1,γ_min) and takes all values in the interval (0,F(1));
F(γ) is strictly increasing and positive on the interval (γ_max,∞)
and takes all positive real values. For γ∈(γ_min,γ_max)
we have F(γ)<0. The condition c_1>0 is equivalent to
F(1)>(α+1)^2. So equation F(γ)=(α+1)^2 has
exactly two solutions γ_1∈(1,γ_min),γ_2∈(γ_max,∞).

For γ<-1 there is no solution as F(-1)>F(1)>(α+1)^2
and F(γ) is strictly decreasing on the interval (-∞,-1).



b) Let 

    (α,δ,σ)∈ U_0(+,+)={(α,δ,σ)∈ R^3:c_1⩾0,c_2⩾0,|σ|⩽1,|δ|⩽1}

In this case the function F(γ) is strictly decreasing on the
interval (-∞,-1) and takes all values in the interval (F(-1),∞);
for γ∈(1,∞) F(γ) is strictly increasing and
takes all values in the interval (F(1),∞);

It follows from c_1⩾0,c_2⩾0 that F(-1)≥(α+1)^2
and F(1)≥(α+1)^2. So the equation F(γ)=(α+1)^2
has no solutions satisfying condition |γ|>1.



The proof of c) is similar to the proof of a).



d) Let (α,δ,σ)∈ U(-,+)={(α,δ,σ)∈ R^3:c_1<0,c_2⩾0}∪

∪{(α,δ,σ)∈ R^3:c_1=0,c_2>0,σ>1,δ>1} .
If c_1<0,c_2⩾0 then we have

    F(-1)⩾(α+1)^2>F(1)


The condition c_1<0 implies that equation F(γ)=(α+1)^2
has a unique solution satisfying condition γ>1. Indeed, two
cases are possible. In the first one for γ∈(1,∞) F(γ)
is strictly increasing taking all values in the interval (F(1),∞)
and, since (α+1)^2>F(1) the equation (<ref>) has a
unique solution γ>1. The second case is similar to item a)
except for now the condition (α+1)^2>F(1) holds and, hence,
the equation (<ref>) has no solution belonging to the interval
(1,γ_min), but exactly one solution γ>γ_max.

For γ<-1 there is no solution since F(-1)⩾(α+1)^2
and F(γ) is strictly decreasing for γ<-1.

So there exists a unique solution γ∈(1,+∞) satisfying
(<ref>).

The case {(α,δ,σ)∈ R^3:c_1=0,c_2>0,σ>1,δ>1}
is similar to item a) except for now the condition (α+1)^2=F(1)
holds and, hence, the equation (<ref>) has a unique solution
γ satisfying condition γ>1.



e) Let (α,δ,σ)∈ U(-,-)={(α,δ,σ)∈ R^3:c_1<0,c_2<0}.
As well as for the item d) one can show that conditions c_1<0,c_2<0
imply the existence of exactly two solutions of equation F(γ)=(α+1)^2
one of which is strictly greater than 1 and the other is strictly
less than -1.



The proof of the item f) is similar to the proof of the item e).



To prove the item g) note, that the eigenvector corresponding to κ,
defined by (<ref>), is an odd function.

Let us find an eigenvector corresponding to the eigenvalue ν,
where ν=2λ(γ+1) and γ is a root
of equation (<ref>). Note, that if α≠-1 then
l(γ)≠δ.

It follows from (<ref>) and (<ref>), that f_-1^'=-f_-1+f_1/2=0
and, hence, f_-1+f_1/2=f_1. By (<ref>) we
have 

    f_k=1/2π∫_-π^πf_1(-α+δcosφ)+(σ-αcosφ)f_0/cosφ+γe^-ikφdφ=
    
    =1/2π∫_-π^πf_1(-α+δcosφ)+(σ-αcosφ)f_0/cosφ+γ(cos kφ-isin kφ)dφ

Because of the integrand is an odd function we have

    1/2π∫_-π^πf_1(-α+δcosφ)+(σ-αcosφ)f_0/cosφ+γisin kφ dφ=0

and, hence 

    f_k=1/2π∫_-π^πf_1(-α+δcosφ)+(σ-αcosφ)f_0/cosφ+γcos kφ dφ

It follows that f_k is an even function of k: f_k=f_-k.

This completes the proof of Theorem <ref>.



In order to prove theorem <ref> consider the case of α=-1.
According to (<ref>) we have 

    D^'(α,δ,σ,γ)=-h(γ)l^-1(γ)(γ-σ)(l(γ)-δ)^2

and the equation for eigenvalues will take the simple form 

    (γ-σ)(l(γ)-δ)^2=0

We are looking for roots satisfying the condition |γ|>1. So 


  * if |δ|>1, |σ|>1, then there are two roots γ=σ
and γ=δ^2+1/2δ 

  * if |δ|≤1, |σ|>1, then there is one root γ=σ 

  * if |δ|>1, |σ|≤1, then there is one root γ=δ^2+1/2δ 

  * if |δ|≤1, |σ|≤1, then there are no roots 

Remind that γ=ν/2λ-1 and σ=2λ_1+μ/2λ.
If α=λ_1/λ=-1 then σ=μ/2λ-1
and equality γ=σ is equivalent to ν=μ.

According to (<ref>) equality γ=δ^2+1/2δ
is equivalent to ν=κ. By lemma <ref> this eigenvalue
is of multiplicity 2 since the rank of the system (<ref>)–(<ref>) is equal to 1.

If μ=κ, then there is only one eigenvalue of multiplicity
2 because of the rank of the system (<ref>)–(<ref>)
equals 1.

The theorem is proved.



§ APPENDIX


A



 §.§ Partition of parameter space


Here we give an illustration of partition (<ref>). We consider
sections of the parameter space R^3 by planes α=c
for various constants c.









 §.§ Lemma <ref>


Consider the functions

    g(x)     =1/2π∫_-π^πdφ/cosφ+x,
    
    h(x)     =-1/2π∫_-π^πcosφ dφ/cosφ+x
    
    v(x)     =1/2π∫_-π^πcos^2φ dφ/cosφ+x

where |x|>1.

For |x|>1

    g(x)=1/x√(1-x^-2),


    h(x)=xg(x)-1,v(x)=xh(x),


    g(x)v(x)=h^2(x)+h(x)


Proof
Let us make change of variable z=e^iφ in integral (<ref>).
We get 

    g(x)=1/2π∫_-π^πdφ/cosφ+x=1/2π i∫_Γdz/(z+z^-1/2+x)z=
    
    =1/2π i∫_Γ2dz/z^2+2xz+1=1/2π i∫_Γ2dz/(z-a_1)(z-a_2)

where Γ={z: |z|=1} is the unit circle and a_1=-(x-√(x^2-1)),
a_2=-(x+√(x^2-1)) are roots of square equation
z^2+2xz+1=0. For |x|>1 only one of these roots is strictly
less than 1 by module. If x>1, then |a_1|=|x-√(x^2-1)|<1
and 

    1/2π i∫_Γ2dz/(z-a_1)(z-a_2)=
    
    = Res_a_12/(z-a_1)(z-a_2)=1/√(x^2-1)=1/x√(1-x^-2)

If x<-1, then |a_2|=|x+√(x^2-1)|<1 and 

    1/2π i∫_Γ2dz/(z-a_1)(z-a_2)=
    
    = Res_a_22/(z-a_1)(z-a_2)=1/-√(x^2-1)=1/x√(1-x^-2)

For integral (<ref>) we have

    h(x)=-1/2π∫_-π^πcosφ dφ/cosφ+x=x/2π∫_-π^πdφ/cosφ+x-1/2π∫_-π^πdφ=xg(x)-1.

Further, we get

    v(x)   =1/2π∫_-π^πcos^2φ+2xcosφ+x^2-(2xcosφ+x^2)/cosφ+xdφ=
       =1/2π∫_-π^πcosφ(cosφ+x)-xcosφ/cosφ+xdφ=
       =1/2π(∫_-π^πcosφ dφ-x∫_-π^πcosφ dφ/cosφ+x)=
       =-x/2π∫_-π^πcosφ dφ/cosφ+x=xh(x)


Finally, 

    g(x)v(x)=g(x)xh(x)=(h(x)+1)h(x)=h^2(x)+h(x)

This completes the proof of the lemma. 

1

[1]Lak
  S. Lakaev, E. Ozdemir (2015) The Existence
and Location of Eigenvalues of the One Particle Discrete Schrödinger
Operators, arXiv:1505.03645.

[2]Reed-1
  M. Reed, B. Simon (1977) Methods of modern
mathematical physics, Vol. 1. Functional Analysis, Academic Press.

[3]Reed-4
  M. Reed, B. Simon (1978) Methods of modern
mathematical physics, Vol. 4. Analysis of operators, Academic Press. 


