6 Nuclear Force and its Properties-2

Sanjay Kumar Chamoli

Learning Outcomes

From this module students may get to know about the following:

The knowledge of ground state properties of deuteron.
The composition of the ground state of deuteron.
The various components of nuclear force.

1.1.3. Spin and parity of deuteron

The deuteron has non-zero ground state spin. The total spin of deuteron (or in general for a nucleus) is given by

I = Sn+Sp + l

As the neutron and proton spins can be either parallel or antiparallel, there are four ways to get the total spin as 1 ħ:

(a) Both neutron and proton spins are up, i.e. Sn and Sp parallel with l = 0

(b) Neutron and proton spins are antiparallel, i.e. Sn and Sp antiparallel with l = 1

(d) Both neutron and proton spins are parallel, i.e. Sn and Sp parallel with l = 2

Fig. 1: Parallel and antiparallel combination of neutron and proton.

Since the experimentally measured parity of deuteron is even so only l = 0 and l = 2 give the correct parity determined from experimental observations. It implies that two nucleons are not bound together if their spins are anti-parallel. Since the parallel spin state is forbidden by the Pauli Exclusion Principle in case of identical particles so there are no proton-proton or neutron-neutron bound states. Thus the nuclear force is spin dependent.

Magnetic dipole moment of deuteron:

If l = 0 is perfect description for deuteron, there should be no orbital contribution to the magnetic moment. Thus the total magnetic moment of deuteron will be the combination of the neutron and proton magnetic moments:

Where gsn = -3.826084 & gsp = 5.585691 are the spin g-factor experimentally, = z when spins have their maximum value (ћ), so

The observed value of μ is 0.8574376 ± 0.0000004 μN. The Observed value is very close but not in exact agreement with the calculated value. The disagreement between the measured and the calculated values can be due to variet of reasons like:

Modifications in the internal structure of proton and neutron in the bound state so that gsp & gsn get different than their free values. Which is extremely unlikely, as deuteron is only a loosely bound system.
There are contributions from charged (virtual) mesons exchanged between the proton and the neutron, and these have not been included. It is possible as in fact it has been shown that measonic currents are important in understanding magnetic dipole moments of odd-mass nuclei.
There is a small admixture of the 3D1-state in deuteron ground state.

As the experimental value of magnetic dipole moment of Deuteron, is very close to the calculated value, so admixture (if any) must be very small.

Electric quadrupole moment of deuteron:

For bare neutron & proton the electric quadrupole moment is zero (Q = 0), so any measured nonzero value of Q must be due to the orbital motion. The observed value of electric quadrupole moment is

The quadrupole moment of a nuclear state is defined as the expectation value of Qo in the substate of maximum M. i.e.,

As for any nuclear state with J < 1, Q0 = 0. So observation of non-zero quadrupole moment in deuteron implies that L>1 (direct evidence of 3D1-component) in deuteron ground state. Thus the mixing of l values is the deuteron is a direct proof of non-central (tensor) term in nuclear interaction

In calculating Q0, the biggest challenge is to get a proper d-state wave function. Calculations using d-state wave function obtained from realistic phenomenological potential suggest a few percent mix of d-state wave function in the ground state of deuteron.

Scattering experiments using deuterons as targets give d-state admixtures ~ 4%. This validates our conclusions drawn from the values of and Q.

2. Nuclear interaction potential :

From the study of the properties of deuteron, it can be concluded that the nuclear interaction unlike other interactions is a complicated function of various terms;

1st term: central component (dominant term, VC(r))

It doesn’t dependent on and and can be written as VC(r).

2nd term: Spin term (a function of s1, s2)

There are two orientations of spins for each nucleon. Arrows up (↑) and arrows down (↓) are used to denote spin orientations.

Symmetric spin wave functions for a two nucleon system are

The above three wave functions correspond to three magnetic substates, M = 1, -1, 0, respectively, of the spin S = 1 state resulting from the coupling of two spin 1/2 particles. Spin S = 1 coupling is referred to as the triplet.

There is only one antisymmetric spin function for a two nucleon system:

3rd term: Nuclear interaction has a non-central (Tensor) component

Deuteron has a non-zero electric quadrupole moment which means it is not a pure l = 0 state but also a small admixture of the l = 2 state. So, there is a non-central, ‘Tensor’ component. Thus the term must be of the form V(r) rather V (r). Tensor force between nucleons in the nucleus provides a more definite description of the average potential experienced by nucleons and the effective residual interaction between them.

As for a nucleon the only reference direction is its spin, so only terms like s . r and s x r, which relate r to the direction of s can contribute. For parity invariance, there must be an even numbers of factors of r potential must have terms like (s1 . r)(s2 . r) OR (s1 x r) . (s2 x r). Where (s1 x r) . (s2 x r) can be written as (s1 . r)(s2 . r) & (s1 . s2)

Fig. 2: Schematic illustration of the dependence of the sign of the tensor force between two nucleons on the orientation of the spins relative to the spatial coordinates.

The dependence of sign of tensor force between two nucleons in deuteron on the spins orientation relative to the spatial coordinates is shown in figure 2.

Summary

The measured properties like the ground state spin and parity, the magnetic dipole moment and the presence of electric quadrupole moment in deuteron suggest that the nuclear interaction is not a simple function, but is a complicated function of various terms. The first and the main contribution in nuclear interaction comes from the ‘central component’ which has the radial dependence (i.e. Vc (r)). The next important contribution comes from the spin dependent term. The strong interaction also gets significant contribution from other non-central terms like ‘tensor term’.

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References:

Introduction to Nuclear Physics – by Keneth S Krane.
Introductory Nuclear Physics – by Samuel S M Wong.
Nuclear Physics – by R R Roy & B P Nigam.
Handbook of Physics by Condon and Odishaw, TMH NewYork.
Introduction to Nuclear Physics, 2nd Edition, W.N.Cottingham & D.A. Greenwood.
Concept of Nuclear Physics by B L Cohen, McGraw Hill.
Nuclear Physics ; an Introduction by S.B. Patel.
The Origin of the Concept of Nuclear Force by L.M. Brown and Rechenberg.
Theoretical Nuclear Physics by John M. Blatt and Victor F. Weisskopf.
Experimental techniques in Nuclear Physics by Dorin N. Poenaru & Walter Greiner
Exotic Nuclear Excitation by S.C. Pancholi
Nuclear spectroscopy Part B, by Fay Ajzenberg- Selove
Theory and Problems of modern Physics (Schaum’s outline Series)
Basic Ideas & Concepts in Nuclear Physics – by K Heyde
The “Particles of Modern Physics” by J. D. Stranathan, Philadephia: Blakiston.
5. Nuclear Physics by Irving Kaplan, Narosa Publishing House.

Web Links

Did you know ?

A deuteron (2H) is a simple nucleus consisting of only a proton and a neutron.
It is simplest bound state of nucleons and therefore an ideal system to studying the nuclear interaction.
Deuteron is a weakly bound system and therefore doesn’t have an excited state.
Since deuteron is an isotope of hydrogen, the proton is in 1s state (L = 0) and therefore in ground state, it has zero orbital angular momentum.
Since the measured ground state spin of deuteron is 1, so it indicates that the proton and neutron inside the deuteron have parallel spins (Sn + Sp = ½ + ½ = 1).
The existence of deuteron with parallel neutron -proton spins justify the non –observation of proton-proton (anti-parallel spins) and neutron –neutron bound systems. The Pauli’s principle forbids the parallel spin state for identical particles and therefore such states will not be stable.
The observation of a stable deuteron (with parallel spins) and non-observation of a stable 2-proton and 2-neutrons is an indication of spin dependent nature of nuclear force.
The observation of significant value of ground state quadrupole moment in deuteron confirms that its ground state in non-spherical (1S0) i.e. not a pure S-state.
The little disagreement of the observed value from the predicted value of magnetic moment in Deuteron indicates that its ground state is a mixed state, i.e. a combination of a S-state and a D-state.
The S = 1 state configuration has the lowest energy, so the nuclear force must be more attractive for total spin S = 1 (parallel spin) than for S = 0 (anti-parallel spin).
The observed hole in the center of deuteron nucleus means that at very short distances, the nuclear force is repulsive i.e. the neutron and the proton wave functions do not overlap.

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