THE ONE PARTICLE THEORY OF PERIODIC POINT INTERACTIONS

*

**

***

Polymers, Monomolecular layers and crystals

* ** ***

A. Grossman , R.H0egh-Krohn , M. Mebkhout

Centre de Physique Theorique 2, CNRS, Marseille Centre de Physique Theorique, CNRS, Marseille and Matematisk institutt, Universitetet i Oslo

Universit~ d'Aix-Marseille II, Facult~ des Sciences de Luminy and Centre de Physique Theorique 2, CNRS, Marseille

July 1979

Postal Address: CNRS - Luminy - Case 907 Centre de Physique Theorique F - 13288 Marseille Codex 3

ABSTRACT

We solve explicitely and without approximation the problem of a quantum-mechanical particle in R3 subjected to point interactions that are periodic in R3 with periodicity of the type Z,

z

^{2 and}

z3.

In the first case we get a model of a infinite strait polymer, in the second case we get a model of a monomolecular layer and in the third case we get a model of a crystal. In all three cases the unite cell of the Bravais lattice is allowed to contain any finite number of interaction stites (atomes), placed arbitraryly and with arbitrary inter- action strength. In the case one interaction site per unite cell we find explicite formulas for the resonance bands and energy bands and their corresponding wavefunctions. In the two first cases we also find explicitely the scattering matrix.

Introduction

The one-electron theory of solidsis based on the study of a Schrodinger particle in a periodic potential. This theory contains a large body of results that is obtained by perturba- tion methods or by symmetry arguments. However, it has not

been possible up to now to check the perturbation results, which are necessarily only approximate, against an explicitely solv- able model.

A class of non-separable two- and three-dimensional generali- zation of the Kronig-Penney model [1] has been solved in a recent paper by Sutherland [2]. The interactions in [2] are, however carried by lines (in two dimensions) or by planes (in three dimensions) which do not have a direct physical application.

However it has been known for quite a while that there exists a Schrodinger operator with a point interaction in three dimen- sions. These operators and relatives of them have a long histo- ry going back several decades. Their study started with Breit, Thomas, Wigner, and others as a model in nuclear physics for poten- tial with short ran~ interactions [ 3]. They observed that po-

tential scattering converge in the low energy limit to scat- tering from a point interaction. In the late fifties, Huang, Yang, Lee, Luttinger, and Wu studied multiparticle operators with point interactions in low order perturbation theory [4].

Beginning in the early sixties a series of papers by Danilov, Minlos and Faddeev was published concerning three- body operators with two body point interactions

[5].

The

physical motivation was to compute the bound state of tritium.

A survey article by Flamand [6], covers this part very well.

The many centered point interaction in three dimensions was first studied by Albeverio, Fenstad and H0egh-Krohn using methods of non standard analysis [7]o Further work on the many centered situation is to be found in [8] and [9].

As we see the point interactions has a long and venerable history. What we do in this article is to put the point inter- action at work in the important field of solid state physics.

We show that it is possible to use the point interactions to construct realistic models of solid states, models which may be built after specification. For instance in the crystal we may specify the lattice, the number and positions of atoms per lattice unite as well as their relative strength. With this input we construct the corresponding Hamiltonian and give ex- plicit formulas for the resolvent kernel of the reduced

Hamiltonian.

In section 2 we give the general formula for a Hamiltonian with a potential with support on a discreet subset of R2

and R3. We call such Hamiltonians, Hamiltonians given by point interactions. It is worth mentioning that point interactions exist only in dimensions, 1, 2 and

3.

In section 3 we apply the results of section 2 to construct models of polymers, i.e. we consider point interactions in R3 which are periodic with only one period. We compute the resol- vent and scattering matrix explicitely up to the inversion of a n x n matrix where n is the number of atoms per polymer unite. In the case of one atom per polymer unite we get com- pletely explicit formulas for the resonances and energy bands.

In section 4 we consider monomolecular layers, i.e. point

interactions in R3 which is periodic with two independent periods. Again we compute the resolvent and the scattering matrix. The Bragg reflections comes out of the scattering matrix in a very explicit mannero

In section 5 we consider the crystals, i.eo point inter- actions with three independent periods. The resolvent kernel of the reduced Hamiltonian is given explicitely up to the in- version of an nxn matrix where n is the number of atoms per lattice unit. In the case of one atom per lattice unit we give the energy bands and corresponding wavefunctions expli- citely.

In section 6 we consider the case of the grating or the linear interferometer. Again we give explicit formulas for the energy bands and its corresponding wave functions. There are also formulas for the reduced resolvent kernel and the corresponding scattering matrix.

2. Point interactions or potentials with discrete support.

Let Y be a discrete subset of R3. We want to consider Hamiltonians formally of the form

- A - L: "- 6 ( x-y) •

yEY y

In the Fourier transform representation

(2.1)

is given formally by the operator

For finite subsets Y such Hamiltonians has been considered in [8]. The result if this discussion is that (2.2) makes

. R:z,

sense 1n - if Ay is chosen suitably infinite- simal. (For a discussion on this point see

[7].)

It is possible to extend these results to discrete subsets in the following way.

Let first Y be a finite subset of R3, then

(2.3) where

(2.4)

Xw(p)

=

1 if

IPI2w

and zero if not, is a well defined self adjoint operator on L2(R3), if we remark that

cw;,f)w;.

We consider ~ to be an approximation to the formal ex- pression

(2.2).

Hence the problem is to choose AY(w) such that Hw converge to some self adjoint operator as w - co • Since

(2.5)

is a bounded operator we have for complex E

Using now that Vw is an operator of finite dimensional range we may compute (2.6) explicitely in the following way.

A

=

(p2-E)-f vW(p2-E)-f then

Al

=

(p 2-E)-t vW(p 2-E)-1^{o o o} vW(p2-E)-f Let

g~ ⁼ Cw~, ^{(p2-E)-1 $w)} _y then

lt(w)l~(w)(2n)-3 ₁ J

^-i(x-y)p

gW e

=

2 dp

xy

IP

<w ^{p -E}

1 i

= A~(w)A;(w) G~(x-y)

Let gw be the nxn matrix with elements w gx,y n

=

IYI .. Then (2.8) takes the form

where _g 1-1 is the 1-1 power of the matrix

Set

(2.7)

(2 .. 8)

(2.9)

(2.10)

x,y E Y'

(2.11) From (2.11) we get that if 1 is not in the spectrum of A then

Hence by (2.6)

(lPU-E)-1

=

(p 2-E)-1 + (p 2-E)-1[

r:

[1-gwJ-1 1ww>< wyw

I

]Cp2-E)-1 (2.13) x,yEY xy x

where [1-g]~ are the elements of the inverse n x n matrix of 1-gw _xy

=

_uxy-^{2 gw} _xy"

Let now

then

hence

cp~(p)

=

(2n)-3/2 X (p)

e~px

w

p -E

where [

J

^-1 stands for the inverse n x n matrix.

Let now Ax(w) be given by

where a is independent of

w.

Then

X

where

-i(x-y)p

G~(x-y) ^{= (2n)-3}

J

^e dp if x-y ~ ⁰

IP!<w P2 -E

and Let

-~ r e-i(x-y)p

=

(2n) ~ j ₂ dp if x-y ~ 0 R3 p -E

and ~E(O)

=

0. Then we see that for complex E (2.16) converge strongly as w

-oe:-

to

( Ha.-E ) -1

= c

p ² -E ^)-1 + L:

[c

a -

4n

^{iVE) 6 - GE ,....}

c

^{x-y )J-1}

x,yEY x n xy

(2.14)

(2.15)

(2.16)

(2.18)

(2.19)

The Fourier transform version of (2.19) is given by

C

-~a -E)- ¹

^Cx,y)

=

u,;cx-y)

+,i<JeY[

(ax-

¥/!fl6x:y- GEcX-Yl J- ¹

u,;Cx-i)GE(y-J) (2.20) where

(2. 21) and

Let now X be a discrete set and a.(x) is function on X which is bounded below

Let Y c X be a finite subset of X and a.y the restriction of a. to Y. Let now H~ be the self adjoint operator, the resolvent of which is given by (2.19). Since

is positive for

w

large enough we see that

is a monotone decreasing function from finite subsets of X into self adjoint operators, i.e.

y y

y cy => H 1 > H 2

1 2 a . - a . (2.23)

This implies that the corresponding resolvents for large nega- tive E is monotone increasing, i.e. for E<E' we have

y y

y1 c y2 => (H 1 - E) -1 < (H 2 - E) -1 •

a. - a.

Here E^l d epen s on d y 2 • We shall see however that it is possi- ble to pick E¹ independent of Y_{2 •}

From (2.21) we have that for E < 0 we have that GE(x-y)

is the kernel of a bounded operator on _{1 2}(x) which tends strongly to zero as E-+ -ro. We use here that X is discrete and the fact that GE(x) tends exponentially to zero in x. Since ax is bounded below we see that

iVE ,...,

(a - -₄ )6 - G (x-y)

X n x:y E (2.25)

is positive as a self adjoint operator on 12(X) for E large negative. Let now E be the largest negative value of E such that (2.25) is still positive, or if there is no such largest negative value we set E

=

0. We observe that if (2.25) is

0

positive as a kernel on 12(X) it is also positive as a kernel on 12(Y) for any subset YcX.

Let now Y be a finite subset of X then

( Y _{H -E} ) -1

₌

C _{p -E} 2 ) -1 _{+ L:} [ C _ax-

_-zrn:-

iV'E) 6x:y- GE x-y ,..., C )

J

y ^-1

a x,yEY

where [

]y

¹ is the inverse kernel in 1 2(Y). It follows from (2.26) that the spectrum of HY _a. is the interval [O,~]

plus atmost n

=

IYI negative eigenvalues E~ ^{where E} are the points in (-oo,O) where (2.25) as a kernel on

bas eigenvalue zero. From this we get that E₀ is a uniform lower bound on the spectrum of for all finite subsete ^X of Y. Especially we get the uniform normbound for E < E₀

for all finite subsets Y c X. From this uniform bound and the mo- notonicity

(2.24)

we get that the strong limit

(HX- E)-¹

=

st lim(Hay- E)-¹

a ycx

(2.28)

!Yj<OO

over the filter of all finite subsets ycx exists. Since the strong limit of resolvents is again a resolvent

(2.29)

is the resolvent of a self adjoint operator H~ which is bounded below.

From the strong convergence

(2.28)

and

(2.26)

it follows that the resolvent (HX- E)-¹ is given by

a

( x

_{H - E} ^)-1 ₌ ^{( 2} _{p - E} ^)-1 _{+ L:} ^r( _{I a - -}^i{'E' _{4 ) 6 -} ^,..., _{GE (x-y} ^)J-1

a x,yEXL x ^TT xy X

(2.29)

where [

Jx

¹ is the inverse as an operator on 12(x). The sum in

(2.29)

is absolutely convergent in the sense that if we integrate with respect to L₂-functions of p and q respecti- vely, then the sum is,absolutely convergent. Hence we have the following theorem.

Theorem 1.1

Let

Y

be a finite subset of

R3

and let

"' 1 eivE'l xl tion defined on Y. Let GE(x) = 4nlxl

a X

for GE(O)

=

0. Let Hw be the self adjoint operator in by

(f,Hf)

be a func- x I 0 and L

2(R3)

^given

If then ~ converges in the strong resolvent

sense, i.e. ^{(Hw -} E) - 1 .-. (Hy-E) - 1 strongly for complex E , where a.

the limit operator is given by

( Y ^)-1

C

2 )-1

[c

^ivE')

^"'C

^)J-1 (2rr)-3ei(px-qy)

H -E

=

p -E - L: a. - 4TI 5 - ~ x-y

a. x,yEY x xy Y (p 2-E)(q2-E)

where [

]y

1 is the inverse as an operator in 12(Y).

Let X be a discrete subset of R3 and a.x be a real func- tion on X whcih is bounded below. Let Y c X be a finite sub- set then Y .-. (HJ- E)-1 is a monoton function for where E₀ is independent of Y. Moreover

exists and define a self adjoint operator H~ bounded below.

The corresponding resolvent is given by

X )-1 2 -1 [

ik) "' (

^)J-1

(H -E

=

(p -E) - L: (a. - "'4iT 5 - GE x-y

a. x,yEX x TT xy X

(2TT)-3ei(px-qy) (p2-E)(q2-E) where [

Ji

¹ is the inverse as an operator in 12

(x).

If we integrate with respect to L2-functions in p and q the series is abso.Lute.Ly conve1·P;tm"'G ⁰

There is an analog theorem in R2 , the proof of which is com- pletely analog to theorem 1.1.

Theorem 1.2

Let Y be a finite subset of R2 and a. be a function

. X

defined on Y. Let GE(x) ^"'

=

(2rr)-^{2 ~} .§__2 ^lxp dp for x I 0 and

R p --E

GE(O) = 0. Let A.x(w)

= c:.rr

^{ln Cw-2}+1) + a.x)-1 for X E y and let (f,nWf)

=

f,p 2f)- (2rr)-3 L: L_(w)l

J

f(p)eiypdpl 2 define

yEY--y

I PI

<w

a self adjoint operator in L2(R2 ). Then ^(Hw- E)-1 .-. (H~- ^E)-1

strongly for complex E where the limit operator is given by

(HY-E)-1 ( 2

)-·1 [c 1~)

·""' ( )J-1 (2rr)-2ei(px-qy) a.

=

P -E - L: a.x- 2 5xy- GE x-y Y 2 2

x,yEY TT (p -E)(q -E).

Let X be a discrete subset of li2 then the analog state- ment as in theorem 1o1 holdso

3o Infinite strait polymers.

L e t A ₁ = an,nE [ Z) , a E R+ b e a d . 1scre e su group o t b f R •

We consider A₁ to be a discrete subgroup of R3 by the injection A₁ c R c; R2 X R. If X is a finite subset of R:S then Y

=

A₁ +X is a discrete subset of R3 invariant under the group A_{1 •} If a is a real function on Y which is invariant under the action of A_{1 ,} i.e. a.A+X

=

ax for A E A₁ and x EX, then Ha."' H~ of theorem 1.1 is invariant under the unitary group A .... UA., A E A_, where (UA.f)(x) = f(x-A). We consider Ha. to be the Hamiltonian for a model of an infinite strait polymer. The points y in Y are then the sites of the atoms in the polymer while a.y are the relative strengths of the interactions at the site y. a.y is actually the scattering length of the atom at the site y.

Since UA Ha. UA -1 = Ha. we have that

Ha.=

I

^Ha.(k)dk

^(3.1)

A1

"'

where A1 is the dual group of A1 and (3o1) is the direct integral over the spectrum of the unitary representation A .... UA, A E A_{1 •} .A₁ is the circle of radius a-¹ or ;\₁

=

_{R/r 1} where r₁

= (::

n,nEZ). By theorem 1.1 we have

We may now utilize the fact that what is inside the square bracket above is translation invariant under A E A₁ to simplify the

expression (3.2). We recall that [ ]-¹ means the inverse

kernel in 12^{(x x 11.}1 )

=

_12(X) ^® 12(A1 ), and the method for simplifica- tion is to use Fourier analysis in 12(A1 ). To compute the Fourier transform of the square bracket in (3.2) we first compute for

- .!l<k<.!!.

e. -a

~(x-y,k) =

(3.3)

where we have identified _[-

:, ^~

by the mapping

k .... eiA.k. For x-y ~ ^11.1 we have by the Poisson summation formula

~ GE(x-y+A.)e-iA.k A.El\.1

= (2n)-3

~ J

eip(x-y)2ei(p3-k)A. dp A.EA1 R3 P -E

ip1 (x1-y1) + p2(x2-y 2) + ( y+k) Cx-:s-Y-=s)

e - -

p~+p~+ (y+k) 2 -E Hence for x- y E 11.1 we have

~(x-y,k)

=

For x- y

=

A. E ^11.1 we have that

while

h-(O,k) = ~ G Cx)e-iA.k

--~ A.EA E

1 A.IO

-iank e

Hence

b:E(O,k)

= - 2 ^~a

ln[e 2iav'F- 2 cos(ak) eia\IT + 1] • (3.6) In terms of hE(x-y,k) (~.2) now takes the form

i(p1x1 + p 2x2 + (y+k)x;z;)

e -

p~+p~+ (y+k) 2 -E

-i(q1y1 + q2y2 + (y l+kl )y3) e

q~ + q~ + ( y ^I + k ^{I )} 2 - E

6(k-k I)

where kl <.ll.

- a

P 3

=

y+k, q 3

=

Y¹ +k', with y,y ¹ E r 1 and -~.::_k, and [ ]-1 is the inverse n x n matrix, n =

I

XI.

Hence we get the reduced Hamiltonian Ha(k) of (~o1) is given by its resolvent kernel on L2 (R2 x r) by the following theorem Theorem 3o1

Let A1

=

(an,n E Z}, a E R+, be a discrete subgroup of R considered as a discrete subgroup of R3 by A1

= [

(O,O,an) ,n E Z} o Let r1

=

{~TT ^{n ,n E Z}} so that the dual group A1

=

R/r1. By

k ... eikA., A. E A1 , we identify .A₁ with the Brillouin zone [- ; , ~] ^o

Let X be a finite subset of R3 and set Y = X+ A1 , and let a _x+A.

=

a _x be a A₁ invariant real function on Y. Then H _a

=

^Hy _a.

of theorem 1.1 is invariant under A_, so that .!! a

Ha. =

I

^{Ha.(k)dk =}

I

^Ha.(k)dk

A1

-a

TT

"'

where we have identified A₁ with

[- n,

a ^{.ll.] •} a Let h_(x-y,k) --.t;

be given by (3.4), G-5) and (3.6). Then the reduced Hamiltonian Ha.(k) is a self adjoint operator on the reduced Hilbert space L2 (R2 xr1 ) with resolvent kernel given by

i(p1x1 + p2x2 + (y+k)x3) -i(~y1 + q2y2 + (y '+k)y~)

e • e -

p~ + p~ + ( y+k) 2 - E q~ + q~ + ( y ^I +k) 2 - E

where [ ]-1 stands for the inverse n x n matrix, n

= I XI ..

with

since H₀₀ is invariant under the group of translation A_{1 ..}

To prove that the wave operators

-itH W + (Ha. ,Hcc )

=

strong lim e a.

t ... +ex:>

e

itH ex:>

exists, it is enough to prove that the wnve operators -itHa.(k) itH:6k) W (H (k) ,H₀₀ (k))

=

strong lime e ·

:!: a. t--+ro

exists for almost all k, since

Since the scattering matrix S(Ha.,H_{00 )} is given by

(3.8)

Cli.10)

we get in the same way that

S(Ha ,Ho:_)

= Js(%

(k) ,H'Xl (k) )dk, J\1

so that the scattering matrix S(Ha(k),Hoc(k)) for the reduced pair Ha(k),H₀₀(k) is actually the reduced scattering matrix and correspondingly for the wave operators.

From the formula for the resolvent kernel of Ha(k) in

theorem 3.1 we see that the kernel (Ha (k)-E)-1 is analytic in a neighborhood of the cut [k2,oo) with smooth boundary values on the cut from above and from below. This implies by standard techniques that the reduced scattering matrix S(Ha(k),H₀₀(k)) exists and is given by the difference of the boundary values from above and the boundary values from below. Hence we have the

following theorem Theorem ':).2

Let H₀₀

=

p 2 be the free Hamiltonian in L2(R3), and let Ha be as given in theorem

3.1.

Then the wave operators

-itH itHx

= strong lim e a e t-o+oo

exists. The corresponding scattering matrix Sa

=

W+ W_ is given * by

[ 'NE

:r

¹

^r: ·vE ]-

¹

- L: [ (ax-4rr)6xy-h:E+io(x-y,k~J -L(a.x+~rr)⁶xy-h:E-io(x-y,k) } x,yEX

i[p1x1 +p2x2+ (y+k)x3] -i[q1y1 + q2y2+ (y'+k)y';)]

• e • e -

2 2 2

where E = p1 + p₂ + p3 , ^p3

=

y + k, q3 = y ¹ + k ¹ with y, y ¹ E r ₁ and k,k¹ E [-

.rr.,

_a ^.II]. _a

~+io(x-y,k)

=

hE(x-y,k) is given by

(3.4),(3.5)

^and

(3.6),

while ~-io(x-y,k) is the analytic continuation around the cut [k2 , oo). Hence for x .. y i _A1

( ) -iAk ( )

and hE-io A,k

=

e hE-io O,k with

( ) 1 [ -2iaVE' ( ) -ia.JF:'

l

hE-io O,k =- 4na ln e - 2 cos ak e + 1 J •

For a qualitative understanding of the reduced scattering matrix S(Ha(k),H₀₀(k)) the resonances are important. We see from the formula of theorem 3.1 that the reduced resolvent kernel

(Ha (k) -E) -'1 is a meromorphic function of

..fE'

on some covering Riemann surface. The structure of this Riemann surface is quite

complex and we see that there is actually a logarithmic branchcut along each of the halfline [(y+k),oo), recall that K1 (x) has a logarithmic singularity at zero. The resonances are the poles of this meromorphic function on its Riemann surface and the eigen- values of Ha(k) are special cases of this poles for which the corresponding wavefunctions are square integrable.

Let now X consist of one point only and we may by transla- tion invariance take X

=

{0}. In this case we have from

theorem 3.1 that

Hence (Ha. (k) -E) -'1 has the same Riemann surface and the same poles as

( _a.- iv'E' ( )

)-'1

4n -hE O,k ^o

Especially the resonances are given by the equation a.-\?

= - 4~a

ln[ e 2iaVE'- 2 cos (ak)eia/E\

'1 J .

which is equivalent to

or

Hence if z₀ (k) is a solution of z2 - (2 cos (ak) + e-⁴rraa.)z + 1 = 0 then

rfF::

_n

= -

i _a ln( z (k)) _o + 2n n _a

are all the solution of

(3.'17).

If z₀ is one solution of (3.18) then z~¹ is the other, hence if z₀(k) is the solu- tion in the upper half plane then

E~(k) + = ~(! i•ln(z₀ (k)) + 2nn) 2 , n=

0,'1, •••

are all the solutions of (3.17).

(3.16)

(3 ⁰ 21)

There are two different cases (i) 2 cos(ak) + e - 4rraa. > 2

then (3.19) has two real solutions z₀ (k) and z₀(k)-1 • Hence (3.21) is real only for n

=

0 and we find only one value

E₀ (k)

= -

~(ln(z

0

^(k)))2•

a

From (3.15) we have that the corresponding eigenfunction is

which is square integrable. The other solutions E~(k) +

=

~(Z:iln(z

0

^(k))+2nn)2,

a are complex resonances.

The other case is

(ii) 12 cos (ak) + e - 4naa.! _:: 2 ,

n

=

1,2, •••

(3.23)

(3.25)

(3.26) in which case (3.19) has a pair of complex conjugate solutions z₀(k) and

z

0(k) with jz₀(k)j =

lz

₀^{(k)l =} 1. So that

ln(z0^(k~ is imaginary and (3.21) are all real solutions.

If ~ ^•

I

^{arg z}0 (k)

I =

J i ln( z₀ (k))

I

^<

I

^kj then again Eo(k) = 12(arg(zo(k)))2

a is an eigenvalue of Ha.(k) if

~~

^{arg z}0 (k)

I

^>

!

^{kl then}

with eigenfunction given by (3.24)

+

E₀(k) is a resonance. E~(k),

n

=

1,2, ••• given by (3.25) are all resonances imbedded in the continuous spectrum [k2 ,oo) of Ha.(k). Hence we have

Theorem 3o3

Let X= (0}, then the essential spectrum of Ha(k) is abso- lutely continuous and is the half line Ck²,~o

In addition Ha(k) has atmost one simple eigenvalue E₀(k) which satis.fiesE₀ (k) < k 2 • Let z₀ be one of the solutions of

Then

z2 - (2 cos(ak) + e-4rraa)z + '1 = 0.

E₀ (k)

= -

'12(ln(z0(k)))2

a

and the corresponding eigenfunction is

If E

0

^{(k)~k2 then E0}(k) is a resonance imbedded in the con- tinuous spectrumo The other resonances are given by

E~(k) +

=

¹2 C:t i ln z (k) ⁺ 2rrn] 2 ,

a o ^{n = '1,2, •• 0}

These resonances are complex if 2 cos(ak) ⁺ e - 4rraa > 2 and they are all real and on the line [k2 , co) if

!

2 cos(ak) + e - 4rraa I~ 2.

We see that 2 cos(ak) + e -⁴naa. > 2 for all k iff e -⁴naa > 4 or - 4rraa > 2 ln 2. Hence iff

all k. Hence we have Theorem 3.4

a<- - - l n .., 2

2rra we are in case (i) for

Let X= (0}, then the spectrum of Ha is absolutely con- tinuous and if

(i')

..,

a.::_ - 2rra ln 2 then sp Ha ^{= [} e ~ ,-:x:')

(ii) a < -

2 ^~a

ln 2 then sp

~ = [

e

~,

e

~]

^{U [ 0,} oo) +

where e~ < 0 and

e :t

= -

12[ ln(+ 1 ⁺

+

^{e -4rraa +}

J

^{c+ 1 +}

^t

e -4naa)2 -1') ]2 ..

o a

4o Mono molecular layers.

Let 11.2 = [n1a1 + n2a2 ; (n1 ,n2) E z2} where a. E R , be a dis-2

l

crete subgroup of R2 • We shall identify 11.2 as a discrete sub- group of n3 by the standard injection of R2 into R3o Let X be a finite subset of n3, then Y

=

11.2 +X is a discrete subset of n3 invariant under the group 11.2 . Let ~ be a real function on Y which is invariant under the action of .11.₂ so that

~A.+x = ~x·

tary group

Then H~ y of theorem 1o1 is invariant under the uni-

Ha

= J

^{Ha (k)dk}

A2

where the dual group

A

₂ ⁼ _R2^/r_{2 and} ^r₂

lattice i.e. r2 =(n1b1 +n2b 2 ,(n1 ,n2)Ez2 }

is the orthogonal

and (a. , b . ) = 2TT6 ..•

l J lJ

Let B2 be the corresponding Brillouin zone i.e.

As in section 3 we start by computing

For x-y ~ A2 1!Ve have

( ) ( ) -iA.k

~ x-y,k = E G x-y+A. e J..E/1.2

eip(x-y). e i[(p1-k1)"-4 + (p2-k2)A.2]

= (2TT)-3 E

J

dp

A.EA 2

~3

p 2 - E

e i[(y1+k1)(x1-y1) + (y2+k2)(x2-y2) + p3(x3-y3)]

2 2 2 . ~3°

p 3 + ( y 1 + k1 ) + ( y 2 + k2) - E .

Hence for x-y ~ A2 we have h:E(x-y,k) =

For _{A E A2} and

Hence

Let now gE(x-y,k)

=

hE(x-y,k) if x-y I 0 and

lim (2n)-3[

i

^{l:: [} (y1+k1 )2 + (y2+k2)2- E]-i- w 2]

w ... co

I

^{A+ k}

I

^{<w 2n}

yEr2- Theorem 4.1

(4.5)

Let A2

=

{n1 a1 + n 2a 2 ; (n1 ,n2 ) ^E z2 } where a1 and a 2 are two independent vectors in R2 •

We consider A2 ^c R2 c; R3 to be a discrete subgroup of R3 by the standard injection of R2 into R3. Let X be a finite

subset of R3, then Y

=

A2 +X is a discrete subset of R3 in- variant under A2 • Let a be a real function on Y which is invariant under the action of A2 so that aA+X

=

ax. Then Ha of theorem 1.1 is invariant under A2 so that

Ha

= J

Ha ( k) dk ,

!..2

where 11.2 = R2

/r

_{2 ,}

r

2 = [n1 b1 + n2b 2 , (n1 ,n2 ) ^E

z

²J (ai, bj) = 2n6ij.

Let gE(x-y,k) be given by (4.6), then the resolvent kernel of

1\t

(k) is

ei[ (y1+k1 )x1 + (y2+k2)x2 + p3x3]

(y1+k1)2+ (y2+k2)2+p~-E

as a kernel on L2^(r 2 x R), wh1re [

J

^-1 1s the 1nverse n ^{. .} x n matrix. We may also compute the wave operators and the corre-

spending scattering matrix as in the previous section, and we get Theorem 4.2

Let H₀₀ = p 2 be the free Hamiltonian in L2 (R3), and let Ha be given in theorem 4o1, then the wave operators W+(Ha,H ) =

- et::'

-itH itH₀₀

strong lim e a e · existso The corresponding scattering t-+CC

matrix sa = w+w_ * is given by

8a(p1,p2,p3; q1,q2,q3) = 6(p1-~)⁶(p2-q2)⁶(p3-q3)

~

ax6xy- gE+io(x-y),k)]-1- [ax6xy- gE-io(x-y,k)]-1}

x,yEX

ei[(y1+k1)~ + (y2+k2)x2+p3x3] e-i[(y1+k .. py1 + (y2+k2)y2+ q3y3]

6(k1-k1)6(k2-k2)6(Cv1+k1 ) 2 +

Cv 2 ^+k 2 ⁾

²

^+p~-

CY1+k1) 2 - Cv2+k2) 2 -

q~)

2 2 2

where E = p 1 + P 2 + P 3 ' P 1 = Y 1 + k1 ' and q2 = y 2 + k2, ( y 1 , y 2) ^E r _{2 and}

p2=y2+k2' (k1 ,k2 ) E B2^o is defined by analytic continuation around the cut

q1 = Y1 + k..;

gE-io(x-y,k) [O,cc].

Let now X = {0}, in this case eigenvalues and resonances are the solution of a = gE(O,k)\ ioe.

( )^{-~ 1 [ ( )2 ( )2}

-t

a= 2 n - lim[2 ^i: y_1+k + v₂+k -E] -4nw].

w .... T 1 y+kl<w yEr2

Since gE(O,k)- -oo as E .... -oo and gE(O,k)- +OO as E ... k² we see that

(4.7)

as a unique solution E₀ (k) in the interval

(-oo,k2 ). E₀(k) is obviously the bottom of the spectrum of Ha(k)o E₀(k) is the only real solution of

(4.7)

and it is an eigenvalue with corresponding eigenfunction in L₂(r2 x R)

The resonances are the complex solutions of

(4o7).

Theorem 4.3

Let X

=

{OJ then the essential spectrum of Ha(k) is abso- lutely continuous and consists of the halfline [k²,oo). In

addition Ha (k) has exactly one simple eigenvalue E₀ (k) < k² with corresponding eigenfunction

5.

Crystals.

Let A = [n₁a₁ + n 2a2 + n?1a3; ⁽ⁿ¹ ,n2 ,n3) ^{E Z-}7j J where a₁ ,a2 ,a3 are there linearly independent vectors in R3. A is then a

discrete subgroup of R3 and let X be a finite subset of R3.

Then ^y = A+ X is a discrete subset of R3 invariant under the lattice group Ao Let a. be a real function on ^y which is in- variant und~r A, i.e. a.x+A. = ax for )..EA and x ^EX. Then Ha.

of theorem 1o1 is invariant under translations in A, i.e.

-1 = H ' that U)..HaUA. a. so

Ha. = JHa(k)dk ^(5.1)

A

A

where Ha.(k) is the reduced Hamiltonian for fixed lattice

momentum k E

A =

^R3/r where r is the orthogonal lattice, i.e.

r

=

[n₁b₁ + n 2b2 + n7ib7i; (n1,n2 ,n3) ^E z3}, with (ai,bj) = 2m5ij.

The projection R3 .... R3/r

= A

is given by k..._ eikA.. It is con-

venient to identify 'f.. with the Brillouin zone B= [s₁b₁+s 2b 2+!!3^b3;

- f < s. < f

J

by the identification k <-> eikA.. From theorem 1.1

~-

we have that

where GE(x-y)

=

4rr(x-y) e ^{1 iVEI x-yl} if x-yrl-0 and zero if not, and 1: ^J-1 is the inverse kernel as operator on l 2^{(X X} A)

=

l 2^(X)

0

^4?.(A). As in the two previous sections we compute

~(x-y,k)

=

If x- y ~A 111re have that

.A.k

~(x-y,k) ^{= L} ~(x-y+A.)e~

A.EA

_3

J

eip(x-y) ei(p-k)A.

=

^L (2rr) 2 ^dp

A.EA p--E

which by the Poisson summation formula gives

For x- y

=

A. E A we have

while

h_(Q,k)

=

L G (A.)e-idk

--.t; A.EA E

A.;io

hence

. ( )-3[

¹

J ^iVE'

=

11m 2rr L ₂ - 4rrw - -4TT

w-x yEr (y+k) - E

IY+kl<w Let now

gE(x-y,k)

=

^~(x-y,k) for x- y ;i 0

( ( )^-~1

gE O,k)

=

^{lim 2rr}

'L

^L

w-co yEr

Then

IY+kl <w

L[

(ax- i.;')5₀A. 5:xy-GE(x-y+>.) JeiA.k = ax5:xy- gE(x-y,k) A.

(5-5)

(5-7)

(5.8)

which implies that (5.2) takes the form

) -1 (H -E =

a

6 (k-k ^{I )} e i[ (y+k)x-( y ^I +k ^I )y]

( ( y+k) 2-E) ( ( y ^I +k ^{I )} 2- E)

(5 .. 9)

where p = y + k and q = y ¹ + k ¹ with y, y ' E r and k and k ¹ in B and [ ]-1 is the inverse n

x

n matrix.. Hence we have proved

Theorem 5o1

Let A

=

(n₁a₁ + n₂a₂ + n₃a_{3 ;} (n₁ ,n₂,n₃₎ E z3}

lattice subgroup of R3,

r =

(n₁ b₁ + n₂b₂ + n3^b3 },

be a discrete ( a . , b . )

=

2n6 . .

l J lJ

the orthogonal lattice and B = (s₁b₁ +s₂b₂ +s3^b3; f<s; ,:£f} be the corresponding Brillouin zone.. Let X be a finite set in R3

and Y

=

A+ X and let be a A invariant real function

on Y.. Then Ha of theorem 1.1 is translation invariant under translations A. E A hence

where we have identified the dual group ping k ... eiA.k.. The reduced Hamiltonian

"'

A with B via the map- Ha(k) is a self adjoint operator on 12(r) given by its resolvent kernel

where gE(x-y,k) is given by

(5.7)

and [ nx n matrix n

= lXI ..

]-1 is the inverse

It is well known that if H is a Hamiltonian which is invari- ant under A then the corresponding reduced Hamiltonian H(k) has discrete spectrum with eigenvalues En(k) [10]. The functions En(k) over the forms

A

= R3/r are called the energy bands of

the Bloch Hamiltonian H. A detailed knowledge of the energy bands En(k) is of considerable interest in the study of the solid state.

Let us therefore consider the case

lXI ⁼

^{1, i.e.}

X=

^{0}.

In this case Y

=

A and a.A.

=

a. for all A. E A. From theorem 5.1 we have that

(Ha. (k)-E) -1

=

C C A.+k)2-E) -1

6vv

^{I - (2rr)} -3 (a-gE Co ,k)) -1 12 • _ __;_1 --::::-- ( V+k) -E ( y ^I +kf-E where

=

lim (2rr)-3[ ~ ¹ ₂ - 4rrw]

w-oo YEr (y+k) -E lv+kl <w

The eigenvalues of ^{H (k)} a. are the poles of (5.10).

(5.10) (5.11)

There are two possibilities for such poles, first the zeros of a.- gE(O,k) and then the zeros of ( y+k) 2 - E, which are the eigenvalues of the free Hamiltonian. Let us first consider the solutions of

a.-f%(0,k) =

o.

(5.12)

Since

we see that E-gE(O,k) an increasing function everywhere on the real line. gE(O,k) has a positive pole of first order at the

points

I

y+kl 2 , y E

r

with a residue equal to the number of points in r which is mapped into lv+k! 2 • Hence there is exactly one solution of (5.12) in each of the bounded intervals In, n

=

1, 2, •••

where

and In are open intervals numbered in increasing order from left to right and I₀ is the unbounded interval to the left ..

Observing that gE(O,k) ... --:c as E .... -oo, we find that there is exactly one solution in the unbounded interval I_{0 ..} All these solutions are obviously first order poles of (Hc:t (k)- E)-1 and are there- fore simple eigenvalues of Ha(k).. The eigenvalue which is the solution of (5 .. 12) in I₀ is obviously the bottom of the spect- rum of Ha (k) and called E₀ (k) ..

The other possible poles of (5 .. 10) are the points E = (y+k) 2 , y ^E

r

which are the eigenvalues of the free Hamiltonian.. It is easily seen that if (y₀+k) 2 is a simple eigenvalue of the free Hamiltonian i .. e .. there is only one y E r such that (y+k) 2

=

(y₀+k) 2 and that is y

=

^y0 , then E

=

(y+k) 2 is not a pole of the resolvent (5.10) because the poles in the first and in the

second term in (5 .. 10) exactly cancel each other. If

however

(y₀+k) 2 has multiplicity m+1, i.e. there are exactly n+1 different

elements y_{0 , ....} ,ym in r such that (y₀+k) 2 - (y1+k) 2

=

(ym+k) 2 , then (5 .. 10) has a pole at E

=

(y₀+k) 2 and the corresponding

eigenspace is m dimensional and is spent by the vectors

1 m

6yyj -m+1 i~O 6yyi' ^j

=

1, ••• ,m (5.15) where (y0 , . . . . ,ym} is the inverse image of jy₀+kl 2 under the map y ...

I

y+kj 2 ..

Let now E₀ (k) _:: E1 (k) ^~ .. .. .. be the eigenvalues of Ha (k) ..

If En(k) is the unique solution of (5 .. 12) in an interval In' then En is simple and from (5 .. 10) we get that the corresponding

eigenfunction is ¢~( y)

= ( (

y+k) 2 - En (k)) - 1 o Hence we have proved Theorem 5o2

Let

X

consist of only one point, i.e.

X= [0}.

Then the corresponding reduced Bloch Hamiltonian Ha(k) is given by its resolvent kernel on 12 (r)

(Ha(k)-E)-¹(y,y')

=

(y+k) 2-E)-1 6yy'- (2rr)-3(a-gE(O,k))-'1 ¹ • _ ___:..'1~- (y+k)2-E (y'+E) 2-E where

=

lim (2rr)-3 [ 2: 1 - 4rrw

J

w-cr yEr ( y+k) 2-E lv+kl<w

Ha(k) has a pure point spectrum and is bounded below. Let

R- I r+k 12

=

U I~ where I~ are open intervals, I₀ unbounded.

n>O

Then there is exactly one simple eigenvalue En(k) in each I~

with corresponding eigenfunction ljl~(y)

=

((y+k) 2 -En(k))-1 • In addition Ha(k) has eigenvalues at the points lv+kl 2 for which the map r .... R+ given by y ....

I

y+k! 2 is not simple. The multiplicity of the eigenvalue lv₀+kl 2 is one less than the multiplicity of the map y .... IY+kl 2 at the point

IY

⁰ +kl ^e The corresponding eigenspace is spend by the vectors

j = 1 , ••• m

where [y_{0 , •••} ym} is the inverse image of lv1+kl 2 under v ....

I

^v+kl2.

We shall consider the eigenvalues En(k) of Ha(k) of theorem 5.2 to be periodic functions over R3 with periods r,

i.e. En (k+y) = En (k) for y E r. Jfrom theorem 5o 2 we see that there is a natural correspondance between the elements y E r and the eigenvalues [En(k)} of Hcx.(k) i.e. y-.EY(k) where Ey(k) is the largest eigenvalue smaller or equal to jy+k! 2 • Let

Y1 I y be such that IY1+k! 2 is a largest element in lr+kl 2 smaller or equal to

I

^{y+k! 2 • If}

I

y 1 +kl 2 <

I

^{y+kl 2 then}

EY(k) E <IY1+k! 2 , lv+kl ²

>

^{and if IY1+kl 2}

⁼

^{lv+kl2 then}

Ey(k)

=

h+kl 2 •

y ... Ey (k) preservs multiplicity and Ey (k)

=

Ey 1 (k) for

y I y 1 if and only if there is a y" E r different from y and y 1 such that h+k! ²

= h

^{1+kl 2}

=

h"+kl 2 • Since Ey(k) are solutions of (5.12) they are different branches of one and the same analytic function of ko There is a unique lowest band E₀ (k) which is smaller than all the points

I

y+kl 2 , y E r. All the other energy bands are connected.

To prove this let y be arbitrary in r such that EY(O) is not the unique smallest eigenvalue E₀ (0), and let Ey1(0) smallest eigenvalue of Hcx.(k) larger or equal to E₀(0).

We want to prove that there is a k E R3 such that Ey (k)

=

Ey,(k). If Ey(O)

=

Ey1(0) we are finished and if not

IY1

2 <Eyi(O) <

h

¹

1

² and EY(o)_::

h!

2 • Since EY(O) >E₀ (0) there is a y" E r such that Ey" (0) is a largest eigenvalue smaller or equal to Ey(O), and

Moreover from the definition of Ey(k) we have that

I

y"+kl 2_: Ey (k)..::

I

y+kl 2 2Ey ^I (k)

~I

^{y I +kl 2 (5.17)}

in the neighborhood of zero where

By a theorem of Euclid there is a point in R3 such that

IY"+kl

=

IY+kl

=

I y ^I +kl '

(5.19)

in fact

(5.19)

defines a unique line in R3, namely the line

through the barycentre of the triangle (y,y',y") and orthogonal to this triangle. This line intersects the neighborhood given by

(5o18)

and on this intersection we get from

(5o17)

that EY (k)

=

Ey ¹ (k). Hence we have proved ^o

Theorem

5o3

Let the assumptions be as in theorem

5.2.

The energy bands En(k) are branches of an analytic function of k, periodic over R3 with period r. There is a unique simple lowest eigenvalue E₀(k) which is the only eigenvalue smaller than lr+kl 2 • More-

over E₀ (k) <En (k) for all n I 0. There is a natural correspon- dance y ... Ey (k), y E

r

between r and sp Ha (k), where Ey (k) is the largest eigenvalue smaller or equal to I y+kl 2 ^o y .... Ey(k)

preserves multiplicity in the sense that the multiplicity of Ey(k) is the number of elements in the inverse imageo Moreover if Y¹

Er

is so that IY¹+kl 2 is a largest element in lr+k! 2 smaller or

equal to IY+kl

2

_{then IY}¹+kl <E (k)< h+kl

2,

and if h ¹+kl

2

_{< jy+kl}

2

- y -

then I y ¹ +k

!

^{2 <} E~ (k) < I y+k 12 _o All the energy surface Ey (k) apart from the lowest E₀ (k) are connected, and Ey(k) is connected with other surfaces along lines given by

I

y"+kl

=

h+kl

= IY

¹ +kl for y,y ¹ and y" three different points in

r.

It

follows from

(5.11)

and

(5o12)

that E₀ (k) takes its

minimum at k = 0 and maximum at k = k₀ = (ib1 , ib2 , tb3) Moreover we see that the minimum E₀ (0) < 0 and the maximum

is negative if and only if

Hence we get the following theorem Theorem 5o4

a<a

0 where

(5.20)

Let the assumptions be as in theorem 5o2o Then the spectrum of Ha is absolutely continuous and

where

and

if a>a

- 0 then sp ^{H =} [E ( 0) , oc )

0:. 0

if a<a₀ then spHa = [E₀ (0), E₀ (k₀)]U [O,cc) with E₀ (k_{0 )} < O,

ao = go(O,ko) (2n)-3 lim [ I: 1

- 4nw]

= w ... oo YEr ( y+k ) 2

0

IY+kl<w ko = t(b1,b2,b3)o

6o The grating (the linear interferometer)o

In this section we consider a potential with support on a set of equally spaced parallel lines in the plane spanned by the second and the third axis in R3o We take the lines to be parallel to the third axis and we want the potential to be trans- lation invariant along the lines and equally strong on each linea This is the potential of a grating which is often used as an

interferometer in spectroscopyo The corresponding Hamiltonian is now translation invariant in the direction of the third axis, and therefore the third component of the momenta is conservedo Hence the problem reduces to a problem in the plane orthogonal to the third axis, and the lines are represented by a string of equally spaced points along the second axis in the plane spanned by the first and the second axiso

Hence the problem of the grating is that of a Hamiltonian in R² with a potential with support on a discrete subgroup of the second axis and the potential being translation invariant under this discrete subgroupo

Let A1 = (na, n ^{E Z},} a E R+ and consider _A1 as a subgroup of R2 which is a subgroup of the second component of R x R = R2 o Let Y = A1 and let a be a real function on Y invariant

under A_{1 , ioeo} aA =ao The corresponding Hamiltonian Ha is given by theorem 1o2 as

(H - E)-1

=

(p2-E)-1- L: Lr (a- lnV'=E)6 ^{I -}

G

().-A ^I )J-1

a A,A¹EA1 ^2~ AA E

(6o1)

where

Now set

Hence

-2 : eip(A-A.')

= (

2n) j 2 dp

~(O,k)

=

p -E

L: G(A)e-ikA AEA1

for

>..I

A¹

=lim [ f L: ((y+kf-E)-f-

4 ^{~ln Cw}

^2+1)]

w-+ce>

I

y+kl <w YEr1

(6.3)

where r_{1 =} [~TT n, n E Z}. Since Ha is translation invariant under A1 we have that

1I

JHa(k)dk

a

H = a = J Ha(k)dk ^(6.4)

...

TT

A a

Where A" ^fl , = R/r₁ is identified with the interval [- 1I, a .II] .. a From (5 .. 1) and (5 .. 3) we then have

We have the following theorem Theorem 5 .. 1

Let A1 = [ na, n E Z}, a E R +

group of the second component in

and consider A1 to be a sub- R = R x R.. 2 Let a be a real

function on _J\1 and invariant under _{J\1 '} iaeo a A.

=

a. .Then the corresponding Hamiltonian H _a

=

JHa(k)dk where Ha(k) is a self adjoint operator on _{L2 (R} ^X r 1 ), r1

=

{~n,nEZ} with resolvent kernel

where

Now on the interval (-~,k²) gE(O,k) is a monotonefunction of E with ra..YJ.ge eg,ual to R. Hence for any a. E R there is a unique solution E₀ (k,a.) of the equation

(6.6) with E₀ (k,a.) < k 2 • It follows from the expression for the re- solvent kernel in theorem 6.1 that E₀ (k,a) is a simple eigen- value of Ha(k) with corresponding eigenfunction

*

₀ ^(p,y;k)

⁼

²

~

p + (y+k) -E₀ (k,a.)

We see that this is the only eigenvalue of Ha.(k). Hence we have

Theorem 6.2

(6.7)

The essential spectrum of Ha~k) is absolutely continuous and equal to [k2 , oc). Ha. (k) has exactly one simple eigenvalue E₀(k,a) and this lies below the continuous spectrum with corre-

sponding eigenfunction

Remark: The scattering matrix may be computed in the same way as in section 3 or 4o

131 _{, I}

L -'

[5]

REFERENCES G.H.Wannier

Elements of Solid State Theory (Cambridge University Press (1959)) B. Sutherland

Exact solutions of some non separable band problems, Phys.Rev. Letters, Vol. 42 page 950

G. Breit

Phys. Rev. 71 (1947), 215-231, J.M.Blatt and V.F.Weisskopf

Theoretical Nuclear Physics, John Wiley and Sons,Inc., New York (1952) p. 74

L.H.Thomas

Phys.Rev. 41 (1935) 903-909 K.Huang and C.N.Yang

Phys.Rev. 105 (1957), 767- 775 K. Huang, C.N.Yang and J.M.Lattinger Phys.Rev.

102

(1957), 776- 784 T.D.Lee, K. Huang and C.N.Yang,

Phys. Rev. _1_06~ (1957), 1134- 1145 T.T.Wu

Phys.Rev. ~11?~~ (1959), 1390- 1404 K. Huang

Statistical Mechanics

John Wiley and Sons, Inc., New York (1963) G.S.Danilov, J.E.T.P. ~ (1961), 349 - 355

L.D. Faddeev, R.A.l!Iinlos, J.E.P.P. ~14 (1962) 1315- 1316 G. Flamand, Cargese Lectures in Theoretical Physics,

Gordon and Breach, N·ew York ( 1965) 24 7 - 287 S. Albeverio, J.E.Fenstad, R.H0egh-Krohn

Singular Perturbation and Non-Standard Analysis, to appear in Trans.Am.Math.Soc.

[sJ

A. Grossmann, R. H0egh-Krohn, H.J:viebkhout

A Class of explicitly soluble, many-center

Hamiltonians for one-particle quantum mechanics in two and three dimensions

CNRS, CPT, Marseille preprint to appear in J.J:viath.Phys.

191

l_ ...J L.E. Thomas

Scattering from point interactions to appear in J.Math.Phys.

l

¹

o-l

J .E. Avron - A. Grossmann - R. Rodriguez

L ...J

_N-1n.Phys. 103. 4 7 ( 1977)

Acknowledgement~ Vfe would like to thank the professors Louis Michel (IHES) and T.T.Wu, W. Thirring and L. Thomas for illuminating discussions.

We also thank CNRS and Faculte des Sciences de Luminy for financial support and Mrs.

s.

Cordtsen for

excellent typing.