Deuteron

[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.

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 , where 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. and , 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)]

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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, . The parity transform is changing it to

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 ).

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[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.

2p-2p decay of 8C and isospin-allowed 2p decay of the isobaric-analog state in 8B

DOI: 10.1103/PhysRevC.82.041304

this paper reports another 2 protons decay mode in 8C. They also discover an “enhancement” at small relative energy of 2 protons. They also reported that an isobaric analog state, 8C and 8B, have same 2-protons decay, which is not known before.

the 1st paragraph is a background and introduction. 2 protons decay is rare. lightest nucleus is 6Be and heaviest and well-studied is 45Fe. the decay time constant can be vary over 18 orders and the decay can be well treated by 3-body theory.

the 2nd paragraph describes the decay channel of 8C and 8B. it uses the Q-value to explain why the 2-protons decay is possible. it is because the 1-proton decay has negative binding energy, thus, it require external energy to make it decay. while 2-protons decay has positive binding energy, thus, the decay will automatic happen in order to bring the nucleus into lower energy state. it also consider the isospin, since the particle decay is govt by strong nuclear force, thus the isospin must be conserved. and this forbid of 1-proton decay.

it explains further on the concept of 2-protons decay and 2 1-protons decay. it argues that, in the 8C, the 2-protons decay is very short time, that is reflected on the large energy width, make the concept of 2 1-protons decay is a unmeasurable concept. however, for the 8Be, the life time is 7 zs (zepto-second ), the 8Be moved 100 fm ( femto-meter $latex 10^{-15} ), and this length can be detected and separate the 4-protons emission in to 2 2-protons decay.

the 3rd paragraph explain the experiment apparatus – detector.

the 4th paragraph explains the excitation energy spectrum for 8C, 6Be.

the 5th and 6th paragraphs explain the excitation energy spectrum for the 6Be form 8C decay. since the 2 steps 2 – protons decay has 4 protons. the identification for the correct pair of the decay is important. they compare the energy spectrum for 8C , 6Be and 6Be from decay to do so.

the 7th paragraph tells that they anaylsis the system of 2-protons and the remaining daughter particle, by moving to center of mass frame ( actually is center of momentum frame ) and using Jacobi T coordinate system, to simplify the analysis. the Jacobi T coordinate is nothing but treating the 2-protons the 2 protons are on the arm of the T, and the daughter particle is on the foot of the T.

Objects of Interest

Nuclear Physics is a study on nuclear matter which is fundamental building block of the world.

electron, proton , neutron, deuteron, tritium, etc… those are objects in nuclear, we call them “particle”. the most simple particle in here is electron, proton and neutron.

The different between nuclear and atom is:

Nuclear core (sit in the center) + Electrons (moving around) = Atom

the mass of atom is almost contributed by nuclear. This is because the mass of proton is about 1830 times bigger than electron, and neutron’s mass is only heavier a bit then proton.

There are many properties contained in each particle. there are electric charge, mass, spin, kinetic energy, etc… and the objective of nuclear physics is understand all these properties and how these properties affect the inter-reaction among them. for example, how a proton and neutron form a nuclear core in deuteron? how they attract each other?

these properties, some may say, are ASSIGNED to the particles. Basically, we can only measure the effect or the result from each interaction. we think, there is a FORCE to make particles able to INTER-ACT with each others. simple to say, when an electron meets another electron, they affect each other by ELECTROMAGNETIC force. but when consider an electron meet a neutron, they don’t interact by electromagnetic(EM) force. in order to distinguish these. we assign an electric CHARGE to electron, and no charge for neutron.

so, basically, Nuclear Physics is study the PROPERTIES of particles and the INTERACTION among them.

There are 3 major forces/interactions, Weak force, EM force and Strong force. Until this moment, we only know the weak and EM force and not fully understand the strong. We neglect the gravity in here, because it is very weak and do no observable effect.

Force	Strength	Range
Strong	10,000	10-15m
EM	1000	long
Weak	1	10-18m