Deuteron

February 8, 2011

GoLuckyRyan no math, Porperties, Working on , , , , , , , , , , , Leave a comment

[last update 2020-08-09]

The deuteron is the nucleus that contains 1 proton and 1 neutron. The spin and isospin of proton and neutron are the same, both are equal to half.  It is the only stable state for 2 nucleons. Deuteron provides a unique place to study the inter-nuclear force. The strong force is believed to be charge independent. Thus, the strong force can be easy to study on deuteron due to the absence of other forces or eliminate from the Coulomb force, which is understood very much.

The mass of deuteron is 1876.1244 MeV. The binding energy is then 2.2245MeV. It was determined by the slow neutron capture of a proton. The emitted gamma ray is approximately equal to the binding energy and the deuteron mass was extracted.

Deuteron has no excited state. It is because any excitation will easily make the system break apart. When thinking of deuteron as one of the families of the NN system. Because of tensor force, which favors T=0 pn pair, thus only T=0, S=1 pn pair, which is deuteron is bounded. Any excitation will change the isospin from T=0 to T=1, which is unbound.

pn_isospin.PNG

Here is a list of the experimental fact of the deuteron:

  • The binding energy is 2.2245 MeV (reference?)
  • The total spin is 1. (reference? exp?)
  • The magnetic dipole moment is 0.857 \mu_N, where \mu_N \frac{eh}{2 c m_N} is nuclear magnetron. (reference? exp?)
  • The electric dipole moment is 0.00282 b. (reference? exp?) In other way to view that is from the mean square values of wave function along the z-axis and all axis, i.e. \left<z^2\right> and \left<r^2\right> = \left<x^2\right> + \left<y^2\right> + \left<z^2\right>, the ratio between these 2 are 1.14/3, instead of 1/3.
  • The radius is 2.1254(50) fm [Randolf Pohl et al., J. Phys. Conf. Ser. 264, 012008 (2011)]

The parity is positive from experiment (how? ref?). If we separate the deuteron wavefunction into 3 parts. The proton wavefunction, neutron wavefunction, and the orbital wavefunction. Under the only force, the strong force in this system, proton and neutron are the same nucleon with different states. Thus, the parity is the same for proton and neutron. So, the product of these 2 wavefunctions always has positive parity. The total parity then is solely given by the angular orbital.

Any orbital wave function can be represented by the spherical harmonic, Y(l,m) . The parity transform is changing it to

Y(l,m) \rightarrow (-1)^l Y(l,m)

So, the experimental face of positive parity fixed the angular momentum must be even.

Ok, we just predicted the possible angular momentum from parity.

The experimental fact on spin is 1. Since J = L + S, and the value of J can take every integer from |L-S| to L + S. and L must be even. The spin of proton and neutron is 1/2. Thus the possible S is 0 or 1 ( we are using L-S coupling scheme ). J = 1 = L + S , which tells us S must be odd to give out 1 for an even L. Thus S=1. So, the only possible L is 0 and 1. Thus, the possible state of the deuteron is (L,S) = (0,1) or (2,1). Therefore, a deuteron could ve a mixed state, if without any further argument.

The isospin can now be fixed by the law that 2 fermions state must be antisymmetry. The spatial state symmetry is even by L = 0 or 2. And for the state (L , S) = ( 0, 1 ), the spin state is symmetric. Thus, the isospin must be antisymmetric. Since the algebra for isospin and spin are the same. We use T = 0 for the isospin. Thus a complete wavefunction is ( L , S , T ) = ( 0 , 1, 0 ).

[20230228 updates]

The coupled equations for the radial function of the deuteron are in here.

The Argonne V18 potential for NN system is in here.

The deuteron pn-interaction from AV18 potential is here.

The deuteron radial wave function is here.

The deuteron rms mass radius, dipole moment, and quadrupole moment are here.

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….