Photodisintegration of in three dimensional Faddeev approach

Photodisintegration of in three dimensional Faddeev approach

3H

The 19th International IUPAP Conference on Few-Body Problems in Physics

S. BayeganM. A. Shalchi

M. R. Hadizadeh

University of Tehran

The 3N electromagnetic reactions

Nd capture 3N Photodisintegrations

1 1(0)N J

To calculate N-matrix we need to follow these diagrams.(Three body force is neglected)

N

1U 2U 3U

0 0 11N J P U

1 0 0 1U tG J tG P U

12 23 13 23P P P P P

Normally we can divide a free state to three sub states (Faddeev scheme). Using permutation operator, P, we can rewrite the free state in term of

the sub state in which nucleons 2 and 3 are in the sub system.

If we consider nucleons with their spins and isospins:

is anti symmetric under permutation of nucleons 2 and 3.

So is a fully anti symmetric state

0 2 3 2 3 1 1 1 2 3 1 2 3a apm m qm pqm m m

0

1

3p

q2

2 31 ( )2

p k k

1 2 32 1[ ( )]3 2

q k k k

0

0 01 P

By rewriting the free state in term of it’s sub states:

We can use symmetry properties as follow:

Then :

Which is equal to:

By introducing:

We have:

0 0 11 3 1N P J P U

01U P J PtG U

11 0 03 1 1U tG P tG P J

10 0 01 1 1 1N P tG P tG P J

10 0 01 1 1N tG tG P P J

0 01N tG U

By multiplying on the left side of the previous equation by

And using the properties of permutation operator:

And definition of:

we obtain the main equation which is:

After that we can calculate the N-matrix elements by:

1 1 121 1 12

P P

P P P

1U P U

01 1 1 1P U P P J P PtG U

1P

√

√

1 1;2 2 Tpq pq ls j I jI Jm t Tm

Why do we use the Three-Dimensional approach?

The 3D approach replaces the discrete angular momentum quantum numbers with continuous angle variables;– consequently it considers automatically all PWs.– The number of equations in this non-truncated 3D representation

is energy independent. – In higher energies the number of equations in 3D approach are

very smaller than the number of equations in PW approach

Therefore this non PW method is more efficient and applicable to the 3N and 4N scattering problems which consider higher energies, and consequently many PWs are needed to achieve convergence results.

We use these basic sates:

Which are assumed to be normalized as follow:

Completeness relation is:

By multiplying these basic states on the left side of U integral equation we have:

And the matrix element becomes:

Effect of permutation operator on our basic states is

So the first term in calculating of the N- matrix can be written as:

Where

And the second term:

1 2 3

1 2 3

3 3 3 31 2 3 1 2 3 0 1 2 3 1 2 3 0 1 2 3 1 2 3

1 2 3 1 2 3 1 2 3 1 2 3 ,

a a a a a

m m

am m m

pqm m m tG P U d p d q d p d q pqm m m tG p q m m m

p q m m m P p q mm m U p q

2 3 12 3 2 3 12 3

32 3 2 3 2 3 2 3 2 2

1 1 12 ,2 2

aa

m m mm m

d q pm m t q q m m U q q qq q q qE

m

Now we can write integral equation and N-matrix in our basic states as:

1 2 3

1 2 3

2 3 12 3 2 3 12 3

1 2 3 1 2 3

1 2 3 1 2 3 1 2 3 1 2 3 0

1 2 3 1 2 3

32 3 2 3 2 3 2 3 2 2

,

1

1

1 1 122 2

am m m

a a

a

aa

m m mm m

pqm m m U U p q

pqm m m P J pqm m m tG P U

pqm m m P J

d q pm m t q q m m Uq q q qE

m

,q q q

2 3 1 3 1 2

2 3 1 3 1 2

2 3 12 3 2 32 3

32 3 2 3 2 3 2 3 2 2

1 1 3 1 1 3 1, ,2 2 4 2 2 4 2

1 12

m m m m m m

aa

m m mm m

N U p q p q U p q p q

d q pm m t q q m m Uq q q qE

m

1

1 ,2q q q

The first term in the integral equation contains current and triton binding energy as follow:

So it is important to evaluate this term:

where

1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3

1 2 3 1 2 3

2 3 1 2 3 1

3 1 2 3 1 2

1

1 3 1, ,2 4 2

1 3 1, ,2 4 2

a a a

a

a

a

pqm m m P J pqm m m J pqm m m PJ

pqm m m J

p q p q m m m J

p q p q m m m J

1 2 3

1 2 3

3 31 2 3 1 2 3

,

1 2 3 1 2 3 1 2 3 1 2 3 ,

a

m

a

m m m

pqm m m J d p d q

pqm m m J p q m m m p q

1 2 3

1 2 3

1 2 3 1 2 3, am m m p q pqm m m

The current which we use contains of single nucleon and two body current as follow (three body current has been neglected):

Because of symmetry properties we can write:

For single nucleon current:

I IIJ J J

1 2 3I SN SN SNJ J J J

1,2 2,3 3,1IIJ J J J

1 2 3 1 2 3 1 2 3 1 2 31 3 1 1 2,3aa

SNpqm m m P J pqm m m P J J

2 2 3 3 2 22 3 32 2 3 3 2 2 32 3 22 1 1

1 1

1 2 3 1 2 3 1 2 3 1 2 31

1 2 ,2 3

a aSN

m m m m m m m m m m

p q m m m J pqm m m

q q Q p p p p J Q q

the single nucleon current which we have used has two terms; convection current and spin current :

So the matrix element of this single nucleon current in our basis can be written as:

1 11 11

2 2SN

E MN N

k k iJ G Q G Q k km m

11convec p p n n

E M MN

qJ G Q G Gm

1 12

2spin p p n n

M MN

QJ S G Gm

2 2 3 3 2 22 3 32 2 3 3 2 2 32 3 22 1 1

1 1

1 2 3 1 2 3 1 2 3 1 2 3

1 1 11

1

1 2 ,2 3

2 if:2

a a

m m m m m m m m m m

p pE M

N N

p q m m m J pqm m m

q q Q p p p p J Q q

p AND m AND mq QG Q Gm m

1

1 1 1 1

11 1 1 1

1 1 1 11

1 1 1 1

11

if

2 if:2

if

pE

N

n nE M

N N

nE

N

p AND m AND m

q G Q p AND m mm

n AND m AND mq QG Q Gn AND m AND mm m

q G Qm

1 1 1n AND m m

For the two body current, conservation of the momentum causes the following relation:

Where:

And:

1 1 1 11 2 3 2 3

2 3 2 3

1 2 3 1 2 3 1 2 3 1 2 3

,2 3

2,3

1 , ,3

a aII

II am m m m m m

p q m m m J pqm m m

q q Q J q q Q

2 3 2 3 2 3 2 3 2 3 3 2 3 2 2 3 3 2 3 2

2 3 2 3 2 3 2 3 2 3 3 2 3 2 2 3 3 2 3 2

,2 3 2 3 2 3 2 3 3 2, , , , , , , , , ,II a II II II II

m m m m m m m m m m m m m m m m m m m mJ q q Q J q q Q J q q Q J q q Q J q q Q

2 2 2

3 3 3

1212

q Q p p k k

q Q p p k k

2

3

1212

q Q p p

q Q p p

2 3

3 2

1212

q Q p p q

q Q p p q

The two body current which has been used contain and exchange currents:

This current is obtained using continuity equation to the NN force AV18.

2 3 2 32,3 , , , ,II exch exchJ J q q Q J q q Q

, , , ,

The exchange part can be written as:

Different spin and isospin operator parts of this current can be treated as follow:

2 3 2 3 3

2 33 2 2 2 2 3 3 3 3 2 2 2 3 32 2

2 3

, ,exch p nE EJ q q Q i G Q G Q

q qq v q q v q v q v q q qq q

1 , 12 2 m mm m m S m

†ˆ ˆ2 m m m mm

m q m q m D q D q

2 3 2 2 3 3 3 3 2 23

1 1T T T T T T T Ti i

The exchange operator is:

2 2 3

2 33 2 3 2 3 3 2 3 2 22 3

2 3

3 32 2 3 3 2 32 2

2 3

2 2 3

3 33 2 3 2 2 2 2 3 3 32 2

2 3

[

]

exch p nE E

s s

p nE E

J i G Q G Q

q q v q v q v q q v q qq q

v q v qq q q q

q q

i G Q G Q

v q v qq q q q q q

q q

The spin operator parts of this current can be simplified as a form which is suitable for our basis and can be treated easily:

2 3 3 3 2 3 3 2 3q q q

2 3 2 3 2 3 2 32z z S S S S

2 2 3 3 2 3 2 3 2 2 2 3. .q q q q q q

3 2 3 2 2 2 2 3 3 3

3 2 2 2 3 2 2 3 2 3 3 3 2 2 3 3 2 3 3 2

2 3 3 2 2 3.

q q q q q q

q q q q q q q q q q q q

q q q q

For evaluating the Triton wave function we need to make a relation between this wave function in our basic states to one which has been calculated before.

The wave function has been calculated in this basis:

Where:

If we introduce our spin parts of our basis as:Then we can relate these two states with Clebsch–Gordan coefficients .

This is very important to mention that the spin of the nucleons is quantized in direction of the z axis which in the calculation of wave function it has been chosen to be to the direction of q. but we have to consider the z axis along the direction of incident photon Q.

12 S pqpq s Sm pqX

1 12 2S Ts Sm t Tm

1 2 3 1 1 1mm m

g

x

Y

Z

p

Q

q

q

qp

So we should first rotate the spin of the nucleons in our basis to be settled in the direction of q axis. Then we should use Clebsch–Gordan coefficients to obtain the wave function in the calculated basis.

Where

And finally:

1 1 2 2 3 3

1 2 31

1 1 2 2 3 3

1 2 31

1 1 2 2 3 3

1 2 31

1 2 3 1 2 3

1 2 3 1 2 3

, , ,

, , ,

, , ,

m m q q m m q q m m q qm m m

m m q q m m q q m m q q pqm m m

m m q q m m q q m m q q pqm m m

pqm m m

pqm m m

D D D pq

D D D pqX

D D D pqX g

cos cos sin sin cospq p p p p pqX

1 2 3 1 2 3 1 2 3 1 2 3 1 3 2 1 3 212

a pqm m m pqmm m pqm m m

For numerical calculations, we need to write U as a function of real parameters which should be treated as grids in a suitable map.

For example:

In which:

1 2 3 1 2 3

1 2 3 1 2 3

, , , , , ,m m m m m m p q qU p q U p q Q

1 2 3 1 2 3

1 2 3 1 2 3

1 21 3 1, , , , , ,2 4 2m m m m m mU p q p q U A B Q

2 21 9 34 4 4

A p q p q 2 21

4B p q p q

1

1 32 4cos

z zp q

A

2

12cos

z zp q

B

1 22 2

1 2

cos cos coscos1 cos 1 cos

1 3 12 4 2cosp q p q

AB

For two body t –matrix singularity problem We have:

has a singularity in which is deuteron binding energy.

By definition of:

We can rewrite integral equation as:

1 2 3

1 2 3

2 3 2 3

2 3 2 3

2 3 12 3 2 3 12 3

1 2 3 1 2 3

2

32 2 2

, 1

1 3ˆ , ,2 4 12 ,

234

a

m m m

m m m mN

m m mm m

dN

U p q pqm m m P J

qt p q q Em

d q U q q qq q q q qE E i E im m

1 1 1 1

2 3 2 3

2 3 2 3

1 2 3 1 2 3 1 2 3 1 2 3 2 3 2 3 1 2 31 2 3

23 , ,4

a a a am m

m m m mN

pqm m m t p q m m m pm m t p m m q q

qt p p Em

1 1 1 1m mq q

2 3 2 3

2 3 2 3

, ,m m m mt p q Z

dZ E dE

2 3 2 3

2 3 2 3

2 3 2 3

2 3 2 3

ˆ , ,

, ,m m m m

m m m md

t p p Z

t p p ZZ E

Two body t matrices can be related to the one which calculated in helicity basis:

Where

And

0 0

1 2 1 2

0 0

2 3 2 3

,

2 3 2 3 2 3 0

2 3 0

1 1 14

1 1 1 1 1 1, , ,2 2 2 2 2 21 1 , , ,2 2

p p

a a

S ti

St

S S Stp p

p m m t pm m

e

C t C t C S m m

C S m m d d t p p z

, , , , ,

p pS

iN S SN p N p

St StN SS

e d dt p p z t p p z

d

cos cos cos sin sin cosp p p p p p

Two body t-matrix in helicity basis has been calculated before.

Where:

12 ,

1 0 11

12 ,

0 0 01

, , , ,

1 cos , , , , ,2

1 cos , , , , ,2

St St

St St

St St

T p p V p p

dp p d v p p G p T p p

dp p d v p p G p T p p

2

,

0

, , , ,

iSt Stv p p d e V p p

Real part of t-matrix in this basis for s=0,t=0 and positive parity

Real part of t-matrix in this basis for s=0,t=0 and positive parity

Real part of t-matrix for s=1 t=1 and positive parity

Imaginary part of t-matrix for s=1 t=1 and positive parity

squared t-matrix in free particle states

squared t-matrix in free particle states

Past & current research projects:– Two body t-matrix calculation by using Bonn-B potential– Two body t-matrix calculation using AV18 potential– Two body t-matrix calculation using chiral potential– NN and 3N bound state calculations with chiral potential

Future Plans:– cross section calculation of 3N photodisintegration and Nd capture by

using• two-body current• three-body current

– Electric and magnetic form factor calculations

33

I Thank you for your attention