Isospin symmetry, Saclay, April 2011 Measuring isospin mixing from E1 and E2 transitions P. Van Isacker, GANIL, France Isospin mixing from E1 transitions

Isospin symmetry, Saclay, April 2011

Measuring isospin mixingfrom E1 and E2 transitions

P. Van Isacker, GANIL, France

Isospin mixing from E1 transitionsCollaboration: York, Argonne, Edinburgh, Kolkata

Isospin mixing from E2 transitionsCollaboration: G. de France, E. Clément, A. Dijon,…

Isospin mixing from E1 or E2

Motivation: Determine (0hw) isospin mixing in N≈Z nuclei.

Measurement ofB(E1)s between states in a T=1/2 doublet and a

T=3/2 quadruplet.B(E2)s between states in a T=0 singlet and a T=1

triplet.

E1: See PRC 78 (2008) 024301.E2: See LoI-5 for SPIRAL-1 (11/03/2010, Dijon

& de France). For A=38.



Isospin selection rule for E1s

Internal E1 transition operator is isovector:

Selection rule for N=Z (Tz=0) nuclei: No E1 transitions are allowed between states with the same isospin.

€

ˆ T μ E1( ) = ekk =1

A

∑ rμ k( ) =e

2 k =1

A

∑ rμ k( ) ⎛

⎝ ⎜

CM motion1 2 4 3 4

+ 2 ˆ t z k( )k =1

A

∑ rμ k( ) ⎞

⎠ ⎟

isovector1 2 4 4 3 4 4

L.E.H. Trainor, Phys. Rev. 85 (1952) 962L.A. Radicati, Phys. Rev. 87 (1952) 521


E1 transitions and isospin mixing

B(E1;5-4+) in 64Ge from:

lifetime of 5- level;(E1/M2) mixing ratio of 5-4+ transition;relative intensities of transitions from 5-.

Estimate of minimum isospin mixing:

E.Farnea et al., Phys. Lett. B 551 (2003) 56

P T =1,5−( ) ≈P T =1,4+

( )

≈2.5%

Isospin analogue E1 transitions



Isospin symmetry is exact

If isospin is an exact symmetry, the B(E1) values satisfy the following equations:

Ten B(E1) values can be expressed in terms of four (J,T)-reduced matrix elements M11, M13, M31, M33.

€

B E1;JiTiTz →JfTfTz( ) =1

2Ji +1Jf ;TfTz

ˆ T 1( ) E1( ) Ji;TiTz

2

Jf ;TfTzˆ T 1( ) E1( ) Ji;TiTz = −( )

Tf −TzTf 1 Ti

Tz 0 Tz

⎛

⎝ ⎜

⎞

⎠ ⎟ Jf ;Tf

ˆ T 1( ) E1( ) Ji;Ti

≡ −( )Tf −Tz

Tf 1 Ti

Tz 0 Tz

⎛

⎝ ⎜

⎞

⎠ ⎟M2Tf ,2Ti


Isospin symmetry is broken

If isospin is broken, initial and final states in the Tz=±1/2 nuclei become

and

Two additional unknowns: mixing angles for initial and final states.

€

J1i = cosφi Ji;T =1/2 + sinφi Ji;T = 3/2

J2i = −sinφi Ji;T =1/2 + cosφi Ji;T = 3/2

J1f = cosφf Jf ;T =1/2 + sinφf Jf ;T = 3/2

J2f = −sinφf Jf ;T =1/2 + cosφf Jf ;T = 3/2



If isospin is broken, the equations for B(E1) values are modified.

For example, in Tz=+1/2 nucleus:

In Tz=±3/2 nuclei:

Etc.

€

B E1;J1i →J1f( ) =1

6 2Ji +1( )M11 cosφf cosφi( −

M33

10sinφf sinφi

+ M13 sinφf cosφi − M31 cosφf sinφi)2

€

B E1;Ji →Jf( ) =3

20 2Ji +1( )M33( )

2


Elimination of our ignorance

There are a possible ten measurable B(E1) values.

There are two unknown mixing angles and four reduced matrix elements.

Since the E1 matrix elements are difficult to calculate (our ignorance), we treat them as unknowns and try to eliminate them.


Application to A=31 & 35

Five transitions have ‘known’ B(E1) values.The reduced matrix elements can be

eliminated to give a relation between the two mixing angles:

with

€

3 M −1/2,1,1( ) + M +1/2,1,1( ) −1+ cos2φf + cos2φi( )[

+ M +1/2,1,2( )sin2φf + M +1/2,2,1( )sin2φi]

= M +3/2,1,1( )sinφf sinφi

€

M Tz,k, l( )2

= B E1;Ji,k →Jf,l( )

Schematic doublets & quadruplets


Real doublets & quadruplets


Isospin mixing correlation, A=31


Isospin mixing correlation, A=35



Isospin selection rule for E2s

E2 transition operator is isoscalar + isovector:

Rule for E2 transitions: E2 matrix elements vary linearly with Tz within an isospin multiplet.

€

ˆ T μ E2( ) = ekk =1

A

∑ rμ2 k( ) = eν

k∈ν

∑ rμ2 k( ) + eπ

k∈π

∑ rμ2 k( )

=eν + eπ

2ˆ T μ

0( ) E2( ) +eν − eπ

2ˆ T μ

1( ) E2( )

Isospin analogue E2 transitions



Isospin symmetry is exact

If isospin is an exact symmetry, the B(E2) values satisfy the following equations:

Four B(E2) values are expressed in terms of three (J,T)-reduced matrix elements M0

11, M1

11, M101.€

B E2;JiTiTz →JfTfTz( )

=1

2Ji +1Jf ;TfTz

ˆ T 0( ) E2( ) Ji;TiTz + Jf ;TfTzˆ T 1( ) E2( ) Ji;TiTz

2

Jf ;TfTzˆ T k( ) E2( ) Ji;TiTz = −( )

Tf −TzTf k Ti

Tz 0 Tz

⎛

⎝ ⎜

⎞

⎠ ⎟ Jf ;Tf

ˆ T k( ) E2( ) Ji;Ti

≡ −( )Tf −Tz

Tf k Ti

Tz 0 Tz

⎛

⎝ ⎜

⎞

⎠ ⎟MTf ,Ti

k



If isospin is broken, initial and final states in the Tz=0 nucleus become

and

One additional unknown: mixing angle between the 2+ states.

€

01+ = 0+;T =1,Tz = 0

21+ = cosφ 2+;T =1,Tz = 0 + sinφ 2+;T = 0,Tz = 0

22+ = −sinφ 2+;T =1,Tz = 0 + cosφ 2+;T = 0,Tz = 0



If isospin is broken, the equations for B(E2) values are modified:

€

Tz = −1: B E2;01+ →21

+( ) =

1

3M11

0 −1

6M11

1 ⎛

⎝ ⎜

⎞

⎠ ⎟2

Tz = 0 : B E2;01+ →21

+( ) =

1

3M11

0 cosφ +1

3M01

1 sinφ ⎛

⎝ ⎜

⎞

⎠ ⎟2

Tz = 0 : B E2;01+ →22

+( ) =

1

3M11

0 sinφ −1

3M01

1 cosφ ⎛

⎝ ⎜

⎞

⎠ ⎟2

Tz = +1: B E2;01+ →21

+( ) =

1

3M11

0 +1

6M11

1 ⎛

⎝ ⎜

⎞

⎠ ⎟2


Elimination of our ignorance

We treat the reduced matrix elements as unknowns and eliminate them. This leads to

with

€

tanφ =M1 0( ) − M 1 cosφ

M2 0( ) − M 1 sinφ⇒ φ ≈

M1 0( ) − M 1M2 0( )

M1 0( )2

= B E2;01+ →21

+( )

Tz =0

M2 0( )2

= B E2;01+ →22

+( )

Tz =0

M 1 =M1 −1( ) + M1 +1( )

2, M1 ±1( )

2= B E2;01

+ →21+

( )Tz =±1

Conclusion

E1: Applied to A=31 and A=35 but not enough E1 transitions were measured.

E2: See G. de France for status of A=38.


Documents

Isospin symmetry, Saclay, April 2011 Measuring isospin mixing from E1 and E2 transitions P. Van Isacker, GANIL, France Isospin mixing from E1 transitions