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A Quantum Mechanical Model of a Deuteron as a Rotor

Consider a system consisting of two particles of equal mass m separated by a variable distance s rotating
about their center of mass. The coordinates are the distance r of the masses from their center of mass and the
angle θ which the line between the masses makes with
an arbitrary line.

Let the rate of rotation be denoted as ω. The angular momentum of the system is then

L = 2m(ωr)r = 2mωr²

The radial momentum is

M = 2m(dr/dt)

The tangential kinetic energy K of the system is then

K = 2(½m(ωr)²) = mω²r² = m(ωr)²

The radial kinetic energy J is

J = 2(½m(dr/dt)²) = m(dr/dt)²

The force of attraction F between the particles is a function of the separation distance s=2r. Thus the potential energy
of the system is

V(s) = ∫_{s}^{∞}F(z)dz

The Hamiltonian Function for the System

It is necessary to express the kinetic energies in terms of the momenta. For the radial component this is easy.

J = M²/2m

For the tangential component note that (ωr)=L/(2mr). Thus

K = L²/(4mr²)

Thus the Hamiltonian function for the system is given by

H = L²/(4mr²) + M²/2m + V(2r)

The Hamiltonian operator for the system is then

H = (1/(4mr²)(∂²/∂θ²) + (1/(2m))(∂²/∂r²) + V(2r)

The time independent Schroedinger equation for the system is then

−h²[(1/(4m)((1/Θ)∂²Θ/∂θ²) + (r²/(2m))((1/R)∂²R/∂r²)] + r²[V(2r) − E] = 0
and a further division by
the coefficient of (1/Θ)∂²Θ/∂θ²)
results in
(1/Θ)∂²Θ/∂θ²) + (2r²/R)∂²R/∂r²)] − (4m/h²)r²[V(2r) − E] = 0

Gathering all functions of r on the left side of the equation and all functions of θ on the right side results in