Consider two particles of equal mass m with their centers separated by a distance s. Let ω be the rate of rotation of the particles about their center of mass. The particles will travel in circular orbits of radius r equal to s/2. The angular momentum of the system is then 2m(ωr)r, which is equal to mωs²/2. Angular momentum is quantized so

mωs²/2 = nh

where n is an integer, called the principal quantum number, and h is Planck's constant divided by 2π. Thus the angular rate of rotation is given by

ω = 2nh/(ms²)
and, for later use,
ω² = 4n²h²/(m²s⁴)

The kinetic energy K of the system is 2(½m(ωr)²) which reduces to mω²s²/4. If the expression above for ω² is substituted into this formular the result is

K = m[4n²h²/(m²s⁴)]s²/4
which reduces to
K = n²h²/(ms²)

This formula will be used later.

The Balance of Centrifugal Force by the Central Force

For circular orbits of radius r the centrifugal force on each particle is mω²r which is equivalent to mω²s/2. Thus a balance requires

mω²s/2 = H*exp(−s/s₀)/s²
which means that
ω² = 2H*exp(−s/s₀)/(ms³)

The Quantization Conditions

Equating the two expressions found above for ω² gives

2H*exp(−s/s₀)/(ms³) = 4n²h²/(m²s⁴)
which reduces to
s*exp(−s/s₀) = n²h²/(mH)
or, upon dividing both sides by s₀
(s/s₀)exp(−s/s₀) = n²h²/(mHs₀)
which, letting σ denote s/s₀,
can be expressed as
σ*exp(−σ) = λn²

where λ = h²/(mHs₀). The function σ* exp(−σ) has the form

This is the quantization condition for σ which leads to the quantization of ω and the other variables.

The Physical Dimensions of the Deuteron

The separation distance s₁ in the physical deuteron is about 2.252 fermi (2.252×10^-15 m), Thus s₁/s₀ is equal to 1.4796, a pure number. When this value is substituted into the LHS of the above equation the result is

σ*exp(−σ) = 0.33695

This is essentially at the maximum of the function σ*exp(−σ) so n can only possibly have the value of 1.

Determination of the Force Constant H

Values for all quantities in the quantization condition are known except for the constant H.

If s₁ is the actual separation distance then

H =(2n²h²)/[ms₁*exp(−s₁/s₀)]
or, equivalently
H =(2n²h²*exp(s₁/s₀)/(ms₁)

The Potential Energy Function

The potential energy function V(s) is given for s₁ by

V(s₁) = −H∫_s1^∞[exp(−z/s₀)/z²]dz

If the above expression for H is substituted into this formula the result is

V(s₁) = −[2n²h²/(ms₁)]exp(s₁/s₀)∫_s1^∞[exp(−z/s₀)/z²]dz

The function exp(s₁/s₀) may be taken inside the integral to give

V(s₁) = −[2n²h²/(ms₁)]∫_s1^∞[exp(−(z-s₁)/s₀)/z²]dz

The variable of integration may be changed from z to y=z-s₁ to give

V(s₁) = −[2n²h²/(ms₁)]∫₀^∞[exp(−y/s₀)/(y+s₁)²]dy

Another change of the variable of integration to p=y/s₀ gives

V(s₁) = −[2n²h²/(ms₁)]∫₀^∞[exp(−p)/(s₀p+s₁)²](s₀dp)
which reduces to
V(s₁) = −[2n²h²/(ms₁s₀)]∫₀^∞[exp(−p)/(p+s₁/s₀)²]dp

Since K(s₁)=n²h²/(ms₁²)

V(s₁)/K(s₁) = −2(s₁/s₀)∫₀^∞[exp(−p)/(p+s₁/s₀)²]dp

The integral in the above equation can be approximated by numerical integration. For s₁/s₀ equal to 1.4796 its value is 0.24683. This makes the ratio V(s₁)/K(s₁) equal to −0.73042.

Conclusion

Since the magnitude of V(s₁) is less than the magnitude of K(s₁) it is physically impossible for the two particles to come together with a principlal quantum number of 1 without an input of energy. The base state has to be one of zero angular momentum. From this base state of principal quantum number of zero a state with a high principal quantum number would require an input of energy.