Disappearance of Mott oscillations in the scattering of ... · Disappearance of Mott oscillations...

Disappearance of Mott oscillations in the scattering of identical charged particles

M.S. Hussein Universidade de São Paulo

Canto, Hussein, Mittig, Phys. Rev. C89 (2014) 024610 Canto, Donangelo, Hussein, Mod. Phys. Lett. A16 (2001) 1027

Quantum theory of scattering - discernible particles:

� ~22µ

r2 + V (r)

� (+)(r) = E (+)(r)

asymptotic boundary condition: in(r) + scatt(r)

Cross section:

A [ ]

scatt(r) = f(✓)eikr

r

in(r) = eikz

�(✓) ⌘ d�(✓)

d⌦= |f (✓)|2

(+)(r) !

identical bosons à symmetric wave function:

Consequence:

And thus,

P � T ⌘ r ! �r =) ✓ ! ⇡ � ✓

f (✓) ! f (✓) + f (⇡ � ✓)

(r) = (�r) =) (r) ! 1p2

[ (r) + (�r)]

symmetry around ⇡/2 !

�(✓) =���f (✓) + f (⇡ � ✓)

���2

c.m. q z

T

P

p - q

zP

T

q

c.m.

direct: detection of P

exchange: detection of T

f (q)

f (p - q)

�class(✓) =���f (✓)

��� +���f (⇡ � ✓)

���2

The  cross  sec)on  can  be  wri/en,  

Classical  term,  

Interference  (Q.M.)  term,  

�(✓) = �class(✓) + ⌅int(✓)

⌅int(✓) = 2Ren

f⇤(✓)⇥ f(⇡ � ✓)o

Symmetry around  π/2      à    σ (π/2)    is a maximum or a minimum

Which one ? Look at data!

Maximum !

Is this a general result ?

An problem with exact solution: Coulomb scattering

- discernible particles (Rutherford scattering)

fC (✓) = �a

2e2i�0 e�i⌘ ln(sin2 ✓/2)

Rutherford cross section:

�0 is the Coulomb phase shift for l = 0

⌘ = qP qT/~v is the Sommerfeld parameter

a is half the distance of c. a. in a head�on collision

�R(✓) =���fC(✓)

���2

=a2

4

1

sin4 (✓/2)

�

- identical particles (Mott scattering)

�R(✓) =���fC(✓) + fC(⇡ � ✓)

���2

⌅int(✓) ⌘⌅int(✓

⌅int(⇡/2)=

1

16

2

cos [2⌘ ln (tan(✓/2))]

sin

2(✓/2) cos2(✓/2)

�

�class(✓) ⌘�class(✓)

�class(⇡/2)=

1

16

1

sin

4(✓/2)

+

1

cos

4(✓/2)

�classical term (renormalized) – system & energy independent:

exchange term (renormalized) – system & energy dependent:

�class(⇡/2) = 1

!

⌅int(⇡/2) = 0

!

Behavior of σM(θ) around π / 2

�class(⇡/2) ! minimum (constant curvature )

�(⇡/2) ! competition : �class ⇥ ⌅int

• minimum for small ⌘

• maximum for large ⌘

⌅int(⇡/2; ⌘) ! maximum (curvature grows with ⌘)

40 60 80 100 120 140q (deg)

0

1

2

3

4

s (q

) / s

(90o

)

h = 0.2 (sclass dominates) h = 4.0 (Xint dominates)

• minimum at ⌘ = 0.2 !d2�(✓; ⌘)

d✓2

�

✓=⇡/2,⌘=0.2

> 0

• maximum at ⌘ = 0.2 !d2�(✓; ⌘)

d✓2

�

✓=⇡/2,⌘=0.2

< 0

A transition takes place at η = η0 , such that:

d2�(✓; ⌘)

d✓2

�

✓=⇡/2,⌘=⌘0

= 0

That is:

d2

d✓2

1

sin

4(✓/2)

+

1

cos

4(✓/2)

+ 2

cos [2⌘ ln (tan(✓/2))]

sin

2(✓/2) cos2(✓/2)

�= 0

Very complicated equation ….. but can be solved analytically by programs like Maple

The solution is ⌘0 =

p2

Thus, for this Sommerfeld parameter, The cross section should be very flat.

40 60 80 100 120 140(deg)

0

1

2

3

4/

= 0.2= 4.0

= 2

(Transverse isotropy - TI)

Experimental investigation of TI Nucleus-nucleus collisions are good candidates

VN (r) VC (r) Vtot (r)

VN (r) ~ 0

VB E

r

V(r)

Only if E < VB so that

Is this condition satisfied ?

VN ' 0 ) Vtot

' VC

0 2 4 6 8 10Z

0

5

10

15E 0

/ V

B

E0 / VB = 1

2H4He

6Li8Be

10B

12C

⌘ =Z2e2

~v �! E0 =e4Z4µ

2~2⌘20VB ' Z2e2

2r0A1/3

Ideal system: 4He + 4He, Ec.m. = 397 keV

4He – 4He elastic scattering

106

105

104

103

102

100

101

102

103

101

100

102103104

101

100

102103104

101

100

102103104

104

103

102

103

102

104

103

102

101

104

103

102

101

101

100

102103104

10-1

101

100

102

103104

10-1

10-3

101

103105

0 40 80 120 0 40 80 120 0 40 80 120

0 40 80 120 0 40 80 120 0 40 80 120

0 40 80 120 0 40 80 120 0 40 80 120

16012080400 16012080400

data at Ec.m. = E↵/2 � 1.0 MeV

• no data at Ec.m. ⇠ 0.4 MeV

• TI at Ec.m. ⇠ 1.5 MeV

Are there other ⌘0 solutions for

d2�(✓; ⌘)

d✓2

�

✓=⇡/2,⌘=⌘0

= 0

Abdullah et al. NPA 775 (2006) 1

• TI at Ec.m. ⇠ 1.0 MeV

Second derivative for pure Coulomb potential

0 0.5 1 1.5 2 2.5 3 3.5 4Ec.m. (MeV)

-2

-1

0

1

2

s'' (9

0o)

(b

arn/

rad2 ) only VC

0.397 MeV (h0 = ÷2 )

V = VC à TI only for Ec.m. = 0.397 MeV !!

Effect of nuclear forces?

Akÿuz-Winther interaction (popular in HI collisions)

VC �! VC + V A.W.N =) VB = 0.87 MeV

data at Ec.m. = 1 MeV > VB

Nuclear potential is not negligible !

Must be taken into account in the condition,

d2�(✓; ⌘)

d✓2

�

✓=⇡/2,⌘=⌘0

= 0

0 0.5 1 1.5 2 2.5 3 3.5 4Ec.m. (MeV)

-2

-1

0

1

2

s'' (9

0o)

(b

arn/

rad2 ) only VC

VC + VN

VB = 0.87 MeV

0.40 MeV1.10 MeV

VC + V A.W.N

• Ec.m. < 0.4 MeV : �00 < 0 ! maximum at 90

o

• Ec.m. = 0.4 MeV : �00 = 0 ! flat(TI)

• 0.4 < Ec.m. < 1.1 : �00 > 0 ! minimum at 90o

• Ec.m. = 1.1 MeV : �00 = 0 ! flat(TI)

• Ec.m. > 1.1 MeV : �00 < 0 ! maximum at 90

o

Ec.m. = 1.0 MeV Ec.m. = 1.5 MeV

min

max max max

max max

max max max

flat flat

Ec.m. = 1.92 MeV

Ec.m. = 2.63 MeV Ec.m. = 3.24 MeV Ec.m. = 3.74 MeV

Ec.m. = 4.44 MeV Ec.m. = 4.94 MeVmax

Ec.m. = 5.44 MeV

Ec.m. = 5.94 MeV Ec.m. = 6.15 MeV Ec.m. = 7.60 MeV

• 0.4 < Ec.m. < 1.1 : ! minimum

• Ec.m. = 1.1 MeV : ! flat

• Ec.m. > 1.1 MeV : ! maximum

Theory Experiment

• Ec.m. = 0.4 MeV : ! flat(TI)

• Ec.m. < 0.4 MeV : ! maximum

Cannot  be  checked  

About  right  

Wrong!  minimum                        at  1.92  MeV  

A more realistic potential for 4He + 4He scattering (AW is a Woods-Saxon. For 4He, gaussians are more realistic) The BFWC potential (sum of attractive and repulsive gaussians):

0 0.5 1 1.5 2 2.5 3Ec.m. (MeV)

-2

-1

0

1

2

''(barn/rad2)

VA = 123 MeV, RA = 2.13 fm

VR = 3.0 MeV, RR = 2.0 fmV FBWCN = � VA e�r2/R2

A + VR e�r2/R2R , with

Ec.m. = 1.0 MeV Ec.m. = 1.5 MeV

min

max max max

max max

max max max

flat flat

Ec.m. = 1.92 MeV

Ec.m. = 2.63 MeV Ec.m. = 3.24 MeV Ec.m. = 3.74 MeV

Ec.m. = 4.44 MeV Ec.m. = 4.94 MeVmax

Ec.m. = 5.44 MeV

Ec.m. = 5.94 MeV Ec.m. = 6.15 MeV Ec.m. = 7.60 MeV

0.5 1 1.5 2 2.5 3

-2

-1

0

1

2

s'' (9

0o)

(b

arn/

rad2 ) VBFWC + VC

s'' small --8 flatweak min.

stron

g m

ax.

Theory (BWFC potential) Experiment

Fully consistent !!

Concluding remarks •  Mott scattering is a good testing vehicle for effects arising from different

contributions, such as vacuum polarization, dipole polarization, color Van der Waals contribution etc. These are very small contributions to the predominantly Coulomb, nucleus-nucleus interaction, but their effects are quite visible in the interference term of the cross section

•  The Transverse Isotropy is a purely EM effect, but the correction arising

from the very weak nuclear interaction at sub-barrier energies is found to be very important to accurately account for the available data

•  Missing, is a measurement of the 4He + 4He cross section at Ec.m. about 0.4

MeV. We are hoping that the measurement is forthcoming.