Completeness of the Coulomb eigenfunctions

Completeness of the Coulomb eigenfunctions

Myles AkinCyclotron Institute, Texas A&M University, College Station, Texas

77843 University of Georgia, Athens, Georgia

Mentor: Prof. A.M. Mukhamedzhanov

NSF REU Summer Program 2005

Completeness of the eigenfunctions of a Hermitian Hamiltonian plays an important role in nuclear reactions and nuclear sructure calculations. Recently it has been used in the so-called Gamow shell model (N. Michel et. al.,Nucl. Phys. A752 335c (2005)).Completeness is useful in “path integrals” formulation of Quantum Mechanics-R. P. Feynman and A. R. Hibbs, Quantum Mechanics and Path Integrals, McGraw-Hill, N.Y. 1965.However, a formal proof of the completeness had been done long ago by R. Newton only for the short range potentials, R. Newton, J. Math Phys., 1, 319 (1960) .

The aim of this work is to present a proof of the completeness of the eigenfunctions of the two-body Hamiltonian with the repulsive Coulomb interaction for arbitrary angular orbital momentum.

Definition of completeness: set of eigenfunctions of the self-adjoined Hamiltonian forms a complete set in the Hilbert space if

0 0

*

00 0

*

0 0

( ) ( ) ( ) ( )

' ( ')[ ( ') ( ) ( ') ( )],

( ) ( ), ( ) ( ) ( )

N

n n kn

N

n n k kn

n n k

h r C r dk C k r

dr h r r r dk r r

C dr r h r C k dr r h r

{ ( ), ( )}n kr r

*

0

Norm ( ) ( )drh r h r

Newton proved the completeness of the eigenfunctions of the two-body Hamiltonian for particles interacting via potentials satisfying conditions:

0

| ( ) |dr r V r

2 2

0

| ( ) |ldr r V r

guarantees that a regular solution is entire function of ,

secures behavior at 0k

k

[ ] ( ' )r r

Newton’s method to prove the completeness. Consider the integral

( )

( ) 0

( ) ' ( ') ( , '; )l

c

I r dk dr h r G r r k

Here , is the Green function of the two-body system.

2( )h r L ( ) ( , '; )lG r r k

The integration contour is shown in Fig. 1c

Re k

Im k

R

cut

The idea of the Newton’s proof: according to Cauchy’s theorem the integral over a closed contour which does not contain any singularities of the integrand on or inside the contour is zero. The integral over gives , integral over disappears,and integral over the real axis can be reduced to

R ( ) ( 0)h r l

*

00 0

' ( ')[ ( ') ( ) ( ') ( )]N

n n k kn

dr h r r r dk r r

However, for the Coulomb interaction due to the complicated behavior of the Coulomb wave functions, especially singular solutions and Jost functions the proof has not beendone yet. One of the problems is the behavior of the singular solutions at , which is a singular point, and there is a cut in the complex plane going from to .

0k k

0k i

We use the Newton’s integral to prove the completeness for the Coulomb case for arbitrary angular momentum.

The integration contour in the complex plane consists of the different parts:

c k

1( ) ( ) ( ) ( )RI r I r I r I r

1 1 1

( )1

0

( )1

0

( ) ( ) ( )

( ) ' ( ') ( , '; )

( ) ' ( ') ( , '; )

l

l

I r I r I r

I r dk dr h r G r r k

I r dk dr h r G r r k

( )

0

( ) ' ( ') ( , '; )lI r dk dr h r G r r k

( )

0

( ) ' ( ') ( , '; )R l

R

I r dk dr h r G r r k

( ) ( , '; )lG r r k is the Coulomb Green’s function:

( ) ( )( )( )

( )( )

( , ) ( , ')( , '; ) .

( )

c cl l

l cl

k r f k rG r r k

L k

Here is the regular solution of the Schrödinger equation, is the singular solution and is the Jost function:

( ) ( , )cl k r ( )

( ) ( , )clf k r

( )( ) ( )clL k

( ) 12

( 1) ( 1)( ) ( )2 2( ) ( )

( , ) (2 )

(2 2) (2 2)( , ) ( , ) ,

( 1 ) ( 1 )

c ll

i l i lc cl l

k r e k

l le f k r e f k r

l i l i

( ) 2( ) , 1( , ) ( 2 ),cl i lf k r e W ikr

( ) 2 2( )

(2 2)( ) (2 ) .

( 1 )

i lc ll

lL k k e e

l i

Each integral then needs to be split further into two separate integrals:

( ) ( )( )

( )( )0

( , ') ( , )( ) ' ( ') .

( )

c crl l

R clR

k r f k rI r dk dr h r

L k

( ) ( )( )

( )( )

( , ) ( , ')( ) ' ( ') ,

( )

c cl l

R clR r

k r f k rI r dk dr h r

L k

The same splitting can be done for the other contours.

( ) ( ) ( ),R R RI r I r I r

( )

0

( ) ' ( ') ( , '; ) ,r

R l

R

I r dk dr h r G r r k First consider the integral over the large semicircle (we drop the subscript ) :

Asymptotic behavior of the integrand for

, .k R k

.k ( )( ) ( , )clf k r

( ) 2( ) , 1( , ) ( 2 ),cl i lf k r e W ikr

, 1

0

( 2 )( 2 ) 1 ,

( 1 ) ( 2 )

i likr it i l

i l

e ikr tW ikr dte t

l i ikr

1. Singular solution :

| arg( 2 ) | ,ikr / 2 / 2Im 0 .i i ik i e e e

R

Asymptotic behavior of regular solution:

( 1)( ) 1 ( ln(2 ))2

( 1) ( ln(2 ))2

(2 2)( , ) (2 ) [

( 1)

].

i lc l i kr krl k

i l i kr kr

lk r k e e

l

e e

Asymptotic behavior of the Jost function:

( ) 2( )

(2 2)( ) (2 ) .

( 1)

i lc ll

lL k k e

l

For and,k

( ) ( ln(2 ))( ) ( , ) .c i kr krl kf k r e

10

k

Then the Green’s function becomes

1 ( ') ( ' )( , '; ) (2 ) ( 1) .ik r r l ik r rlG r r k i k e e

Substituting it into the integral gives

( ) ( 1) ( ).2

lRI r h r

We evaluate each term separately using integration by parts. The first term containing tends to zero for .

The second term containing reduces to

1 ( ') ( ' )

0

( ) ' ( ') (2 ) ( 1) .r

ik r r l ik r rR

R

I r dk dr h r i k e e

( )RI r

( ')ik r re

( ' )ik r re

k

Integration over the real axis.

It consists of two separate integrals:

1

0

( ) ' ( ') ( , '; ),r

lI r dk dr h r G r r k

1

0

( ) ' ( ') ( , '; ).r

lI r dk dr h r G r r k

This integral can be rewritten as a sum:

1 0 00

( ) ' ( ') lim ( , '; ) lim ( , '; ) .r

l lI r dr h r dk k G r r k dk k G r r k

Using the property of the regular solution the integral can be written as:

( ) ( )( , ') ( , ')c cl lk r k r

It can also be shown that:

( ) ( )( ) ( )( )

1 ( ) ( )0( ) ( )0

( , ) ( , )( ) ' ( ') lim ( , ') .

( ) ( )

c crl lc

l c cl l

f k r f k rI r dr h r dk k k r

L k L k

( ) ( )( ) ( )

( ) ( )( ) ( )

( , ') ( , ').

( ) ( )

c cl l

c cl l

f k r f k r

L k L k

Then the integral reduces to

( ) ( )( ) ( )( )

1 ( ) ( )0( ) ( )0

( , ) ( , )( ) ' ( ') lim ( , ')

( ) ( )

c crl lc

l c cl l

f k r f k rI r dr h r dk k k r

L k L k

:( ) ( ) ( ) ( )( ) ( ) ( ) ( )( )

( ) ( )0( ) ( )0

( , ) ( ) ( , ) ( )' ( ') lim ( , ') ,

( ) ( )

c c c crl l l lc

l c cl l

f k r L k f k r L kdr h r dk k k r

L k L k

( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( , ) ( ) ( , ) ( ) 2 ( , ).c c c c cl l l l lf k r L k f k r L k ki k r

Thus we arrive at

( ) ( )2

1 ( ) ( )0( ) ( )0

( , ') ( , )( ) 2 ' ( ') lim .

( ) ( )

r c cl l

c cl l

k r k rI r i dr h r dk k

L k L k

Taking into account

2( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )c c cl l lL k L k L k

in the limit we get

( ) ( )2

1 2( )0 0 ( )

( , ') ( , )( ) 2 ' ( ') .

( )

r c cl l

cl

k r k rI r i dr h r dk k

L k

0

Now we must show that the contribution of the integral over the small semicirclearound the singularity at , is zero.0k

( )

0

( ) ' ( ') ( , '; ).r

lI r dk dr h r G r r k

First, we look at the singular solution, in the form:

( ) 2( )

0

2( , ) 1

( 1 )

i likrc t

l

e ikrf k r e dte

l i t

Letting the integral is:2

1ikr

zt

( ) 2( )

0

( , )( 1 )

ikrc t i l

l

ef k r e dte z

l i

From the following:

| | | | | |[ ]i ii i e i e

i ik k kz z z z e

cos sinsin cos| | | || | | |iik kk kz e z e

.

.

.

It follows that:

cossin | || | 1,i

kk kz z e

sin

sin| |

22 22

.4 4

k

kt

zt k rt k r

sin sin

2 2 2 22 20 0 2 22

1.4 44 4

k kt t

dt dtt k rt k rt k rt k r

The behavior of the integral near the lower limit at is 0k

1,lz

Then it follows that:

2( )( ) ( , ') .

( 1 )

ikr

cl

e ef k r

l i

Now consider the regular solution. According to Humblet, Ann. Phys.155, 461(1984), is an entire function of and is given by a series which absolutely and uniformly converges. Hence it is analytic at and around We can therefore write in a proximity of it as:

( ) ( , )cl k r k

0k

( ) ( , ') ( ).cl k r A r

Here is a some function. Now we consider the integral:( )A r

( ) ( )( )

( )( )0

( , ') ( , )( ) ' ( ') .

( )

c crl l

cl

k r f k rI r dkk dr h r

L k

0k

On the contour : 1

( )2 2

( )

( )( )

( , )1,

( 1 ) ( 1 )( )

cl

cl

f k r e e

l i l iL k

( ) ( , ') ( ).cl k r A r

So the integral can be majorized at as 0k J2 2

0

( ) ( ) ( ) 0.J dkkA r A r k d A r k

Hence the contribution over the contour is zero.

Since the contribution of is zero the integral become just:( )I r

1( ) ( ) ( ).RI r I r I r

Placing in the values:

Solving for : ( )h r

( ) ( )2

2( )0 0 ( )

( , ') ( , )( ) 2 ' ( ') ( 1) ( ) 0.

2( )

r c cll l

cl

k r k rI r i dr h r dkk h r

L k

( ) ( )2

2( )0 0 ( )

( , ') ( , )4( ) ( 1) ' ( ') .

( )

r c cl l l

cl

k r k rh r dr h r dkk

L k

The same procedure follows for the integral from to so:r

( ) ( )2

2( )0 ( )

( , ') ( , )( ) 2 ' ( ') ( 1) ( ) 0.

2( )

c cll l

cr l

k r k rI r i dr h r dk k h r

L k

( ) ( )2

2( )0 ( )

( , ') ( , )4( ) ( 1) ' ( ') .

( )

c cl l l

cr l

k r k rh r dr h r dk k

L k

So that adding the two integrals gives

( ) ( )2

2( )0 0 ( )

( , ') ( , )2( ) ( 1) ' ( ') .

( )

c cl l l

cl

k r k rh r dr h r dk k

L k

From this we can see that:

( ) ( )2

2( )0 ( )

( ') ( )2( 1) ( ').

( )

c cl lk lk

cl

r rdk k r r

L k

This is just the condition for completeness for continuum spectrum.Thus we have shown that the regular solutions to the Schrödinger equation with a pure Coulomb potential from a complete set.

2( )( )

2( ) ( 1)

( )

l

cl

kE

L k

reduces completeness condition to the form

( ) ( )( ) ( ') ( ) ( ').c clk lkd E r r r r

Using the spectral function

Applications.1. A formal proof of the completeness of the eigenfunctions of the Hamiltonian with therepulsive Coulomb interaction presented here along with the proof by Newton for pure nuclear case allows us to state that the eigenfunctions of the Hamiltonian with the Coulomb + nuclear potential also form a complete set.

2 Nuclear reaction theory. The Green function can be calculated using the spectral decomposition ( ) *( )( ) ( )

2ln ln2

0 0

( )( )

( )( )

( ') ( )( ') ( ) 2( ', ; ) ( 1) ,

2

( )( ) .

( )

c cb bNl lk lk

ln n

cc lklk c

l

r rr rG r r k dkk

kEE i

rr

L k

3. In work by T. Berggren, Nucl. Phys. A109, 265 (1968), the use of the resonant states has been proposed by including the resonant states into the complete set of the eigenfunctions. In such an approach a new complete set is formed by the bound states,resonances, which we pick up, scattering states at real energies and scattering statesat complex energies. But it has been done only for short range potentials. This work allows one to extend the Berggren’s method for the Coulomb + nuclear potential.

4. Inclusion of the resonant states into “complete” set requires extension of the Hilbert space, because the norm of the vectors (elements) of this space should be determined.

Resonance points

( ) ( ),{ , , }R complexn m k k new Berggren’s complete set

Here I will show how to determine the norm of the resonant wave functions using Zel’dovich’s regularization procedure (a special case of the general Abel regularization). For resonancesat ( ):

1 2( )( ) 1/ 2

1( , ) , .

i k ik rrR

i

ek r C C

r

A conventional scalar product then becomes:

2

*

2( ) ( ) *( ) ( ) 2

2 ( )0

| ( ) ( ) .R R

R k rR R R R

iR

edr r r C dr

r

To avoid the divergence of this product we use a dual basis for Hilbert space and and regularize it using Zel’dovich’s procedure:( ){ }R *( ){ }R

The integrand is oscillatory and 2 212 2

0 2 20

lim convergesR

i k r k

ir

re e

redr

0 1Rk k ik 0 / 2RE E i

21 22 2( ) ( ) ( ) ( ) 2

0 2 20

* | ( ) ( ) lim .R

R i k r k rR R R R

iR

ree e

dr r r C drr

Diverges!!!

Conclusions

1. A first proof of the completeness of the eigenfunctions of the two-body Hamiltonian with the repulsive Coulomb potential is presented.2. It justifies the spectral decomposition of the Green’s function in terms of the eigenfunctions for charged particles without screening of the Coulomb potential.3. The Berggren’s method of including the resonant states (nuclear + Coulomb interaction) into complete set of the eigenfunctions can be extended for charged particles.4. It is shown how to regularize the scalar product of the resonant Gamow functionsusing Zel’dovich regularization procedure.5. It justifies the application of the Gamow shell model for charged particles

Acknowledgements:This work is suported by the National Science Foundation REU summer programUnder Grant No. 463291-00001, National Science Foundation under Grant No. PHY-0140343, U.S. Department of Energy under Grant No. DE-FG03-93ER40773. I would also like to thank Dr. Zhanov for his help this summer.

1. Branch point singularity is not an isolated singularity.

( )f x x has two singularities: and . 0x

2. Hilbert space: complete linear vector space with respect to its norm. Completeness in this context means that any Cauchy sequence of elements of the space converges to an element in the space, in the sense that the norm of differences approaches zero.

3. The Green’s function definition is

2( )

2 2

1 ( 1)( ) ( , '; ) ( ' )

2 2 l

d l lV E G r r k r r

dr r

It gives for the Coulomb case ( ) ( )

( )( )( )( )

( , ) ( , ')( , '; )

( )

c cl l

l cl

k r f k rG r r k

L k

We need to show that the Green’s function can be written as

( ) *( )2

20

( )( )

( )

( ') ( )2( ', ; ) ( 1)

2

( )( )

( )

c cl lk lk

l

cc lklk c

r rG r r k dkk

kE i

rr

L k

equivalent to completeness

5. Gamow function

The bound state wave function

1( ) ,

rr e

r Cr

21 2Z Z e

Coulomb parameter for the bound state

2 bound-state wave number

Rik leads to the resonant pole

1i

Rk