San José State University

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Thayer Watkins
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A Quantum Mechanical Analysis of
the Triteron, the H3 Nuclide

The triteron, consisting of two neutrons and one proton, will be presumed to be a neutron-neutron spin pair and a proton revolving about their center of mass. In order to keep matters simple and yet capture the essentials of the physics of the model it is assumed that the masses of the neutron and proton are equal. This mass is denoted as m. The axis of rotation is presumed to be parallel to the axis of the neutron-neutron spin pair. The rate of rotation is denoted as ω. The separation distance between the centers of the neutrons is a crucial parameter, but it shows up only in the form of one half of its value. This half distance will be denoted as u. The separation distance of the neutrons is established by the force of repulsion between them at the limit of the formation of spin pairs.

Angular Momentum

Let s denote the distance between the center of the proton and the midpoint of the line between the centers of the neutrons. This distance can be characterized as the width of the triteron. The radius of rotation of the proton is then (2/3)s and the radius of rotation of the neutron pair is (1/3)s. Therefore the angular momentum L of the triteron is

L = mω(2s/3)² + 2mω(s/3)² = mωs²(4/9 + 2/9) = (2/3)mωs²

Angular momentum is quantized so

L = (2/3)mωs² = nh

Thus

ω = (3/2)nh/(ms²)
and
ω² = (9/4)n²h²/(m²s4)

Dynamic Balance

The centrifugal force on the proton is

mω²(2s/3)

The distance between the proton and one of the neutrons, z, is given by

z² = s² + u²

The force F on the proton due to one neutron is

F = −(2/3)H·exp(−z/z0)/z²
and its radial component is
F(s/z) = −(2/3)H·exp(−z/z0)s/z³

The radial component of the force on the proton due to the two neutrons is therefore

Fr = −(4/3)H·exp(−z/z0)s/z³

Dynamic balance requires that

Fr + mω²(2s/3) = 0

Therefore

mω²(2s/3) = (4/3)H·exp(−z/z0)s/z³
which reduces to
ω² = 2H·exp(−z/z0)/(mz³)

Thus two expressions for ω² have been derived. When they are equated the result is

2H·exp(−z/z0)/(mz³) = (9/4)n²h²/(m²s4)
which reduces to
(s4/z³)exp(−z/z0) = (9/8)n²h²/(mH)
and further to
s(s/z)³exp(−z/z0) = (9/8)n²h²/(mH)
or, equivalently
z(s/z)4exp(−z/z0) = (9/8)n²h²/(mH)

Since z²=s²+u², s²=z²−u² and thus

(s/z) = (1−(u/z)²)½
and hence
(s/z)4 = (1−(u/z)²)²

Thus

z(1−(u/z)²)2exp(−z/z0) = (9/8)n²h²/(mH)

This is the quantization condition for z. In terms of ζ=z/z0 it is

ζ(1−(ζ0/ζ)²)2exp(−ζ) = (9/8)n²h²/(mz0H)

where ζ0=u/z0.

The form of the function on LHS of the quantization condition is:

In the above graph ζ0 is set at 1.5.

Let the above function be denoted as Γ(ζ) and thus the quantization condition for ζ is

Γ(ζ) = φn²

where φ is equal to (9/8)h²/(mz0H).

The quantization condition for the width s of the triteron can be found from the relation s²=z²−u². Better yet let σ=s/z0 and hence

σ²=ζ²−ζ0
and likewise
ζ = (σ² + ζ0)½

In the function Γ(ζ), ζ may be replaced by the above expression to give Π(σ) so the quantization condition for the width of the triteron is

Π(σ) = φn²

The shape of Π(σ) is shown below.

Once the quantized values of s are known the values for ω are found through the relation

ω = (3/2)nh/(ms²)

The kinetic energy K of the triteron is given by

K = ½mω²(2s/3)² + 2(½mω²(s/3)²
which reduces to
K = mω²s²(2/9 + 1/9) = (1/3)ω²s²

Since ω² is equal to (9/4)n²h²/(m²s4)

K = (3/4)n²h²/(ms²)

(To be continued.)


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