No measurement on a single decay reveals its angular momentum

(identifying the multi-pole nature of the ).

Need to take many measurements on a sample

counting the occurrence of E -decays

as a function of angle

source

We know, for fixed mj the radiation is given by the Poynting vector

)(4

1 *jmjmjm HE

cS

Re

where ijm

ijm

ijm

ijm

AH

ikAE

2

21

21

2

)]1()1(

)]1()1([

)1(2

1

mjj

mjjj

mjjj

jm

Ym

Ymmjj

Ymmjj

jjS

Same for E and M multipoles

60Co

I = 0+

60Ni GROUND

1333 keV

2506 keV

I = 2+

I = ?

60Co’s -decay to 60Ni is accompanied by two rapid gamma emissionsin succession (lifetimes of ~10-12 sec, make them seem simultaneous)

½ = 5.26 years

Obviously cascading to its ground state

1g7/2

1g9/2

2p1/2

1f 5/2

2p3/2

1f 7/2 81d3/2 42s1/2 21d5/2 6 281p1/2 21p3/2 41s1/2 2

4

60Co

I = 0+

60Ni GROUND

1333 keV

2506 keV

I = 2+

I = ?

60Co’s -decay to 60Ni is accompanied by two rapid gamma emissionsin succession (lifetimes of ~10-12 sec, make them seem simultaneous)

½ = 5.26 years

Assuming (trying) different values for I1

would lead you to expect

42 cos10cos2

15

2

5)( W

I1 = 0

4cos24

9

10

11)( W

I1 = 3

42 cos24

1cos

8

11)( W

I1 = 4

Jintial

Jfinal

mj1 = -j1, -j1+1, …, j1-1, j1

mj2 = -j2, -j2+1, …, j2-1, j2

unless special steps taken to orient nucleithe various mj states are equally populated

and mj Sjmj

= constant

Jintial = 1

Jfinal = 0

mj1 = -1, 0, +1

mj2 = 0

0 0

1 0

minitial mfinal

2sin

)cos1(2

1 2

With the three mj1 states equally populated:

)cos1(2

1

3

2 2 2sin3

13

2 Isotropic!

independentof .

Ji = 1

Jf = 0

mi = -1, 0, +1

mf = 0

mi = 0

mi = 1

mi = 1

Jf = 0 mf = 0

E+E

EEE

B-field

E = B but to compare atomic to nuclear splittinglook at one Bohr magneton and the nuclear magneton:

eN m

e

m

e

22

Howeverapplying a strong magnetic field

to a nuclear system at low temperature

can exploitthe Boltzmann distribution

kTEmem /)( PAt kT<<B

the nuclear spins are aligned with the external B-field

minitial = +1

/2 3/2 2

At T << B nuclear spins tend to be aligned with the external B

Yielding, for example for mi=+1

Description of Nuclear Polarization

                                                                                                                                                                                                                                                                                                                                  

Polarization

                                                                

m

kTBmm

kTBm

e

me

I

mTf /

/

1

)()(

At high temperatures ( 1 K !) occupation of the nuclear energy levels

are equal.

Low temperature Nuclear Orientation

For this a 3He:4He dilution refrigerator is widely used.

At lower temperatures (100 mK) the lower energy levels

are preferentially occupied.

Measurement of temperatureResistance thermometry

3He Melting curve thermometryNuclear orientation thermometry

Nuclear orientation is detected by the temperature-dependent change in

the pattern of emitted radiation from appropriate nuclei.

most easily measured for the radiation that exits the cryostat walls of the cooling chamber

(for low energy and radiation the detectors need to be inside the cryostat).

For 60Co, even at 1 K radiation, the pattern of radiation

is uniform in all directions.

At lower temperatures the radiation pattern becomes distorted

An external magnetic field may be needed to sweep out magnetic domains, ensuringall of the target nuclei are correctly aligned.

This is most easily detected by a -detector aligned with the sample axis

though frequently an azimuthal detector is also used.

                                                                                                                                                                                                                                                                                          

Nuclear orientation schematic for the example of 60Co.

For work below 0.3 K3He/4He dilution refrigerator

Capable of reaching ultra low temperatures down to 10 mK.

Ultra Low Temperature Thermometry

The equilibrium concentration of 3He/4He is temperature and pressure dependent.

The vapor pressure of 3He is higher than 4He

Manipulating the 3He/4He concentration(by pumping) can control the temperature.

Similar to evaporation techniques in your refrigerator.

The production of low temperaturesEvaporation refrigeration using liquid 3He (T~0.3 to 0.5 Kelvin)

                                                           

                                   

Dilution refrigerationcontinuous refrigeration to low temperatures

but low cooling power

Phase Separation Driven by osmotic pressure differences

when temperature is low enough~0.8 K

3He

4He

3He

4He

Dilution and CoolingDriving temperatures lower

T0.01 K

3He + 4He mixtureat T>0.8 K

“Sink” Mixing chamber

Still

Pumping chamber

draws 3He outof solution

3He/4Nemixture

3He

return line

High magnetic fields and superconducting magnets

Useful when there is a cascade of

successive radiations

source

Angular Correlation Technique

Ei-counter

Ef -counter

Ji

Jf

J1

2

After the 1st transition the orientation of atoms is no longer randomi.e., not all mj-values for the 2nd transition are equally probable!

Ji

Jf

J1

2

As a simple example, consider the special casewhere Ji = Jf = 0

with an intermediate state of J.

For the initial transition to the intermediate nuclear state mj = J, J-1, … (J-1), J

are equally likely.

Suppose 1 carries off angular momentum 1.

It must leave the intermediate nuclear state with mj = 1.??

Since mf can only = 0, it follows 2 = mj = 1

Nature’s randomly selected 1st step, fixes the nature of the 2nd step.

Ji

Jf

J1

2

For example suppose

J = 1mj = 1, 0, +1

The 1st transition emits 1

leaving the nucleus in one state of mj.

source

Ei-counter

Positioning a detector effectively fixes the z-axis, by selecting a direction!

Ef -counter

When the Ei-counter registers a hitit is selecting p1 as the z-axis.^

Any it detects can’t be froma ~sin2 distribution, only the

~ 1 + cos2.

The Ei-counter preferentially selects out the m=1 decays!

With 1 = 0, the Yjm=Yj

0 transition is not even detected. The remaining (equally likely) cases m = 1 have been selected out.

The distribution resulting from these two contributions of coincident radiation

1,1,)(

jjSSW

22

21

20

)1(

)1)(2(

)1(

2jjj

Yjj

jjY

jjY

So if Ji=0, J=1, Jf=0 (the dipole transition Krane uses as an example)

221

1

20 cos1)( YYWj

or if J=2

1cos3cos43

2

3

1)( 2422

2

21

2

20

2 YYYW