Absolute polarization measurement by elastic scattering

April 14, 2017

GoLuckyRyan experimental , , , , , , , , Leave a comment

The magnitude of proton polarization can be measured by NMR technique with a reference. Because the NMR gives the free-induction decay signal, which is a voltage or current. For Boltzmann polarization using strong magnetic field and low temperature, the polarization can be calculated. However, when a reference point is not available, the absolute magnitude of proton polarization can be measured using proton-proton elastic scattering. The principle is the nuclear spin-orbital coupling. That creates left-right asymmetry on the scattering cross section.

Because of spin-orbital interaction:

V_{ls}(r) = f(r) \vec{l} \cdot \vec{s} ,

where f(r) is the distance function, \vec{l} is the relative angular momentum, \vec{s} is the spin of the incident proton. In the following picture, the spin of the incident proton can be either out of the plane (\uparrow ) or into the plan (\downarrow). When the proton coming above, the angular momentum is into the plane (\downarrow ). The 4 possible sign of the spin-orbital interaction is shown. We can see, when the spin is up, the spin-orbital force repulses the proton above and attracts the proton below. That creates an asymmetry in the scattering cross section.

LS.PNG

 

The cross section is distorted and characterized using analysing power A_y. Analyzing power is proportional to the difference between left-right cross-section. By symmetry (parity, time-reversal) consideration, A_y = 1 + P sin(2\theta) (why?), in center of mass frame. In past post, the transformation between difference Lorentz frame. The angle in the A_y has to be expressed in lab angle. The cross section and A_y can be obtained from http://gwdac.phys.gwu.edu/ .

In scattering experiment, the number of proton (yield) is counted in left and right detectors. The yield should be difference when either proton is polarized. The yield is

Y(\theta, \phi) = L \epsilon \sigma_0 (1 + cos(\phi)A_y(\theta) P) ,

where L is the luminosity, \epsilon is the detector efficiency, \sigma_0 is the integrated cross-section of un-polarized beam and target of the detector, P is the polarization of either the target or beam. When both target and the beam are polarized, the cross section is

\sigma = \sigma_0 (1 + (P + P_T)A_y + P P_T C_yy),

where C_yy is spin-spin correlation due to spin-spin interaction of nuclear force.

Using the left-right yield difference, the absolute polarization of the target or the beam can be found using,

\displaystyle A_y P = \frac{Y_L - Y_R}{Y_L + Y_R} ,

where Y_L = Y(\phi =0) and Y_R = Y(\phi=\pi) .

 

 

 

Analyzing power for proton elastic scattering from the neutron-rich 6He nucleus

February 8, 2011

GoLuckyRyan experimental, papers reading, Working on , , , , , , , , , , , Leave a comment

DOI : 10.1103/PhysRevC.82.021602

This paper is based on the solid proton polarized target, and make improvement on the experiment. The proton polarization was monitored by NMR method and the absolute polarization is known. Therefore, the Analyzing power ( or the spin asymmetry) can be determined. The experimental result is different form the prediction of the t-matrix folding model.

____________________________________________

The 1st and 2nd paragraphs explain why the spin-asymmetry is important in nuclear research. a simple reason is, by manipulating the spin, we can reveal the spin-dependence of the nuclear-structure. For example, the orbital-spin coupling, which is the major factor in the nuclear potential.

the 3rd, 4th and 5th paragraphs talk about the solid proton polarization target and the advantage of operating under weak magnetic field. (already discussed in Here )

The 6th paragraph report the result from the 2003 experiment. and say that the unsatisfiable result on the spin-asymmetry.

The 7th to 11th paragraphs talk about the experiment procedures and conditions.

the 12th paragraph explains the angular distribution in center of mass frame of the differential cross section. the data of 6He is similar to the 6Li for angle less then 50 degrees, which is the forward angle in the lab frame. at the backward angle or angle more then 50 degree in C.M. frame, there are some different and it may be due to the halo structure of 6He.

the next paragraph turns the focus on the analyzing power, which can be determined at this time. the data is very different from 6Li data. the t-matrix folding method predicted that the analyzing power should be positive but the experiment result is different. this indicated that there is other mechanism to explain the nuclear structure of 6He.

The t-matrix stands for transition matrix.

the other mechanism may be the g-matrix folding model…..i am sorry, i don’t understand the following….

Optical Model II

February 7, 2011

GoLuckyRyan advance, Basic, Scattering, theory , , , Leave a comment

Last post on optical model, we did not include the spin. to introduced the spin, we just have to modify the wave function. For spin-½ case.

\begin {pmatrix} \psi_i \\ \psi_2 \end {pmatrix} \rightarrow Exp( i k r ) \begin {pmatrix} a_1 \\ a_2 \end {pmatrix} + \frac { Exp ( i k r) }{r} M \begin {pmatrix} a_1 \\ a_2 \end {pmatrix}

where the M is a matrix:

M = f + g \vec{ \sigma } \cdot \vec{n}

the f is for the spin-Independence part of the wave function. For the incident wave and the scattered plane wave.

\begin {pmatrix} a_1 \\ a_2 \end {pmatrix} = \begin {pmatrix} Exp( - i \phi_s /2 ) cos ( \theta_s /2 ) \\ Exp( i \phi_s /2 ) sin ( \theta_s /2 ) \end {pmatrix}

where \theta_s and \phi_s are the angle of spin . not the detector angle.

after calculation by routine algebra, we have the scattered spherical wave.

\chi = M \cdot \begin {pmatrix} a_1 \\ a_2 \end {pmatrix} = \begin {pmatrix} (f+g)Exp( - i \phi_s /2 ) cos ( \theta_s /2 ) \\ (f-g)Exp( i \phi_s /2 ) sin ( \theta_s /2 ) \end {pmatrix}

The expectation of the wavefunction, or the intensity of the spherical part will be:

I(\theta_s) = \chi^{\dagger} \chi = |f|^2 + |g|^2 + 2 Re( f^* g ) cos( \theta_s)

the beam polarization should be equal the intensity and normalized polarization.

I P_z = \chi^{\dagger} \sigma_z \chi = ( |f|^2+ |g|^2 ) cos ( \theta_s) + 2 Re(f^* g)

Thus, we have the induced polarization when incident beam is unpolarized:

P_z ( \theta_s = \pi /2 ) = \frac { 2 Re ( f^* g ) }{ |f|^2 + |g|^2 }

for a beam of many particle and formed an ensemble, the \theta_s is the average.

and Analyzing power, which is a short term for Polarization Analyzing Power , or the spin asymmetry, is given by

A_y=\frac { I(\theta_s = 0 ) - I( \theta_s = \pi ) } { I ( \theta_s = 0 ) + I ( \theta_s = \pi ) } = \frac { 2 Re( f^*g) }{ |f|^2 + |g|^2 } =P_z

Therefore, in order to get the spin asymmetry, we have to use 2 polarized beams, one is up-polarized, and another is down-polarized, to see the different between the scattering result.

However, to have 100% polarized beam is a luxury. in most cases, we only have certain polarization. thus, the spin-asymmetry is not equal to the analyzing power. the spin-asymmetry \epsilon is from the yield measurement.

\epsilon = \frac { I(\theta_s) - I(\theta_s) }{ I(\theta_s ) + I(\theta_s) }

since f and g only depend on the detector angle. and we can assume they are symmetry. Thus

\epsilon = \frac {2 Re( f^* g ) }{|f|^2 +|g|^2 } cos ( \theta_s) = A_y P

the P is the polarization of the target.

 

 

First experiment of 6He with a polarized proton target

February 4, 2011

GoLuckyRyan experimental, papers reading , , , , , , , , , , , , , , Leave a comment

DOI : 10.1140/epjad/i2005-06-110-5

this paper reported a first spin polarized proton solid target under low magnetic field ( 0.08 T ) and hight temperature ( 100K )

the introduction overview the motivation of a solid target.

  • a polarized gas target is ready on many nuclear experiment.
  • on the radioactive beam ( IR beam ), the flux of a typical IR beam is small, since it is produced by 2nd scattering.
  • a solid target has highest density of solid.
    • most solid target can only be polarized on low temperature ( to avoid environmental interaction to reduced the polarization )
      • increase the experimental difficult, since a low temperature should be applied by a cold buffer gas.
    • high field ( the low gyromagnetic  ratio ).
      • high magnetic field make low energy scattered proton cannot get out from the magnetic field and not able to detect.
  • a solid target can be polarized at high temperature and low magnetic field is very useful

the material on use is a crystal of naphthalene doped with pentacene.

the procedure of polarizing the proton is :

  1. use optical pumping the polarize the electron of pentacene
    • the population of the energy states are independent of temperature and magnetic field.
  2. by Dynamic Nuclear Polarization (DNP) method  to transfer  the polarization of the electron to the proton.
    • if the polarization transfer is 100% and the relaxation time is very long. the expected polarization of proton will be 72.8%

The DNP method is archived under a constant microwave frequency with a sweeping magnetic field. when the magnetic field and  microwave frequency is coupled. the polarization transfer will take place.

the next paragraph talks about the apparatus’s size and dimension, in order to fit the scattering experiment requirements.

the polarization measurement is on a scattering experiment with 6He at 71 MeV per nucleons. By measuring the polarization asymmetry \epsilon , which is related to the yield. and it also equal to the polarization of the target P_t  times the analyzing power A_y .

\epsilon = P_t \times A_y

with a reasonable guess of the target polarization. the analyzing power of  6He was found.

the reason why the polarization-asymmetry is not equal to the analyzing power is that, the target is not 100% polarized, where the analyzing power is defined. when the polarization of the target is 100%, both are the same.

in the analysis part. it used optical model and Wood-Saxon central potential to simulate the result. And compare the result from 6He to 6Li at same energy. the root mean square of 6Li is larger then 6He. it suggest the d-α core of 6Li may responsible for that.

they cannot go further discussion due to the uncertainly on the polarization of the target.