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 Quantization in a Central Force Field

This material derives the quantization scheme that exists for particles in a central force field when the orbital angular momentum is quantized. Relativistic conditions are assumed.

Consider a single particle of mass m traveling in a circular orbit of radius r. Let the central force be given by the formula

#### F = −Kf(r)/r²

This is still a completely general case although it is presumed that the force is carried by particles, such as photons or pi mesons, and hence there is a 1/r² dependence. K stands for a force constant k times a charge number q; i.e., K=kq. The charge might be electrostatic charge or mass. The charge number would be the number of electrostatic charges or it could be the number of nucleons or the number of particles. The quantity f(r) is a dimensionless quantity. For forces carried by a particle that does not decay such a the electrostatic force or gravity the quantity f(r)=1. For a force carried by a decaying particle such as a meson the quantity f(r) is of the form e-λr. The force may be a combination of nuclear force, electrostatic force and gravity. The force may be an attraction or a repulsion, depending upon the sign of k.

For a circular orbit the attractive force must balance the centrifugal force; i.e.,

#### mv²/r = Kf(r)/r² or, upon expressing the velocity v as βc mc²β²/r = Kf(r)/r² which reduces to mc²β² = Kf(r)/r

where c is the speed of light.

Angular momentum is equal to mvr so its quantization is

#### mvr = nh

where n is an integer and h is Planck's constant divided by 2π.

Expressing v as βc this quantization condition along with the condition for a circular orbit are:

#### mc²β² = kf(r)/r mcβ = nh/r

The division of the first equation by the second results in

#### β = (K/(hc))f(r)/n = (k/(hc))qf(r)/n

so that if the quantization of r is determined then the quantization of velocity is given by the above expression. The quantity (k/(hc)) is a dimensionless constant that may be denoted as γ, a characteristic of the force field. Thus the general relationship, which applies in both the relativistic and nonrelativistic case, is

###### β = γqf(r)/n

For the electrostatic force γ is the fine structure constant 1/137.036. An estimate of γ for the nuclear force is 1.5217246. For more on this quantity γ see Force Constants.

## The Quantization of Orbit Radius

Under relativistic conditions the mass of the particle is given by

#### m = m0/(1−β²)½

The angular momentum condition then takes the form

#### m0βcr/(1−β²)½ = nhor, equivalently β/(1−β²)½ = nh/(m0cr) or, upon squaring both sides β²/(1−β²) = [nh/(m0cr)]²

For convenience let nh/(m0cr) be denoted as D.

The previous equation may be solved for β² as

#### β² = D²/(1+D²) = 1/(1/D² +1)

On the other hand from the general expression previously found for β

#### β² = [γqf(r)/n]²

Equating the two expressions for β² gives

#### [γqf(r)/n]² = 1/(1/D²+1) so [γqf(r)/n]²(1/D²+1) = 1

Replacing D by its definition and carrying out some simple algebraic manipulation gives

#### γ²q²f²(r)[r² + n²(hc/(m0c²))²] = n4(hc/(m0c²))²

The quantity hc/(m0c²) has special significance but it has the dimensions of length. Let r0 be some significant unit of length. The variable r then can be expressed as r=zr0.

The quantization condition then takes the form

#### γ²q²f²(zr0)[z²r0² + n²(hc/(m0c²))²] = n4(hc/(m0c²))² which, upon division by r0², becomes γ²q²f²(zr0)[z² + n²(hc/r0(m0c²))²] = n4(hc/(r0m0c²))²

Now let (hc/(r0m0c²)=μ.

The quantization condition then takes the form

#### γ²q²f²(zr0)[z² + n²μ²] = n4μ²

The function f(r) could be a constant or it could be e−λr, in which case r0=1/λ. In that case f(zr0)=e−z. Let f(zr0) be denoted as φ(z). Then the quantization condition takes the form

###### γ²q²φ²(z)[z² + n²μ²] = n4μ²

This is the condition sought for the quantization of z and r.

The constant hc is equal 3.1616×10-26 kg m³/s². For a proton m0c² = 1.503353×10-10 kg m²/s² (joules). Then for r0=(0.5)1.522×10-15 m, the nondimensional constant μ = 27.635116 and μ² = 763.7. As stated previously an estimate of γ for the nuclear force is 1.5217246 and hence γ²=2.31564583.

For a given value of n the quantization condition may have one solution, no solution or more than one solution. Let N be the set of values of n for which the quantization condition has solutions and R the set {rn, n∈N}. Then the set of allowable β's B={βn, n∈N} is given by βn=f(rn)/(nhc).

The kinetic energies κn are given by

#### κn = m0c²[1/(1−βn²]½−1]

The potential energies are given by

#### Vn = ∫rn∞(Kf(r)/r²)dr

The change in potential energy from n to n-1 can be approximated by the trapezoidal rule; i.e., ΔV is approximately equal to ½K[f(rn-1)/rn-1² + f(rn)/rn²]Δr.

When a transition in n results in a lower potential energy some of that energy loss goes into an increase in kinetic energy but some of it goes into the energy of an emitted photon. The energy of the emitted photon is then −ΔV - ΔKE and its frequency ν is then given by

#### ν = (−ΔV - ΔKE)/h.

The spectrum of the system is the set of ν's for transitions from n1 to n2 for n1 and n2 belonging to N.

## Spectral Relationships

Some insights into the relationships the spectrum to changes in the angular momentum can be obtained by looking at the partial derivatives of the quantities with respect to n and using the approximation Δy=(∂y/∂n)Δn for Δn=±1.

What is known concerning the potential energy V is that

#### ∂V/∂n = (∂V/∂r)(∂r/∂n) = −(Kf(r)/r²)(∂r/∂n)

Likewise it is known that

#### ∂KE/∂n = (∂KE/∂β)(∂β/∂n) = −[m0c²β/(1−β²)3/2](∂β/∂n)

From the equation β=γqf(r)/n it follows that

#### ∂β/∂n = γq[(f'(r)/n)(∂r/∂n) − f(r)/n²]

All that remains is to evaluate (∂r/∂n). For this exercise it is convenient to put the quantization equation γ²q²φ²(z)[z² + n²μ²] = n4μ² into a simpler form by dividing through by n²μ² and defining (z/(nμ)) as ζ. This results in

#### γ²q²φ²(ζnμ)[ζ² + 1] = n²which in logarithmic form is 2ln(γ) + 2ln(q) + 2ln(φ(ζnμ)) + ln(ζ²+1) = 2ln(n)

Taking the derivative of this equation with respect to n gives

#### (2φ'(ζnμ)/φ(ζnμ))[nμ(∂ζ/∂n) + ζμ] + 4ζ(∂ζ/∂n)/(ζ²+1) = 2/n

A factor of 2 can be cancelled.

(To be continued.)

## The Nonrelativistic Case

For the case in which m=m0, the circular orbit condition gives

#### v² = Kf(r)/(m0r) and hence v = [Kf(r)/(m0r)]½

When this value of v is substituted into the angular momentum quantization conditon the result is

#### mvr = nhand hence rKf(r) = n²h²/m0

Division of both sides by hc gives

#### γqrf(r) = n²h/(m0c) and upon division by a unit length r0 γqzφ(z) = n²h/(r0m0c) or, more succinctly γqzφ(z) = n²μ or zφ(z) = n²(μ/(γq))

This is the quantization conditon for the nonrelativistic case.

Thus, if φ(z)=1 then

#### zn=(μ/(γq))n².

Then the potential energy is proportional to 1/n² and likewise the kinetic energy. So higher levels of the angular momentum quantum number correspond to greater orbit radii.

For the relativistic case the quantization conditions with φ(z)=1 reduces to

#### zn² = (μ/(γq))²n4 − n²μ² or, equivalently zn² = (μ/(γq))²n4[1 − 1/(n²γ²q²)] and hence zn = (μ/(γq))n²[1 − 1/(γ²q²n²)]½

So here also higher levels of the angular momentum quantum number correspond to greater orbit radii.

Also for higher values of n the inner particles may shield the charge seen by the particles thus effectively reducing the value of q and thus modifying the value of radius.

## Conclusion

A quantization scheme for particles in a central force field has been established based upon the quantization of angular momentum. This scheme applies for any formula for the central force. The quantization of angular momentum establishes quantization for potential and kinetic energy such that a change in state results in incompatible changes in potential and kinetic energy and the difference must be reconciled by the emission or absorption of a photon.