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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

#### 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

## 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/(4mr²)(∂²Ψ/∂θ²) + (1/(2m))(∂²Ψ/∂r²)] + V(2r)Ψ = EΨ

Multiplying through by r² gives

#### −h²[(1/(4m)(∂²Ψ/∂θ²) + (r²/(2m))(∂²Ψ/∂r²)] + r²V(2r)Ψ = r²EΨ

Now consider separation of variables; i.e., Ψ=R(r)Θ(θ). This results in the equation

#### −h²[(R/(4m)(∂²Θ/∂θ²) + (r²Θ/(2m))(∂²R/∂r²)] + r²[V(2r) − E]RΘ = 0

Division by RΘ results in

#### −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

#### (2r²/R)∂²R/∂r²) − (4mr²/h²)[V(2r)−E] = ((1/Θ)∂²Θ/∂θ²)

This means that each side of the equation must be equal to a constant, say −μ².

This means that the equation

#### ((1/Θ)∂²Θ/∂θ²) = −μ² or, equivalently ∂²Θ/∂θ²) + μ²Θ = 0

must be satisfied. The solution to this latter equation is

#### Θ(θ) = exp(iμθ)

Since Θ must be single valued Θ(2π) = Θ(0) and hence μ must be an integer.

The equation for the radial function R(r) is

#### (2r²/R)∂²R/∂r²) − (4mr²/h²)[V(2r)−E] = −μ² or, equivalently ∂²R/∂r² = (2m/h²)[V(2r) − E + μ²/r²]R(r)

For large values of r the terms V(2r) and μ²/r² become insignificant. The equation reduces to

#### ∂²R*/∂r² = −(2m/h²)ER*(r)

which has the solution

#### R*(r) = [exp(±(-2mE)½r/h]

It is then useful to look for a solution of the form

#### R(r) = R*(r)U(r)

(To be continued.)