﻿ The Spectrum of Two Particle Systems
San José State University

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Thayer Watkins
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 The Spectrum of Two-Particle Systems

## General

A system has kinetic energy and potential energy. Let these quantities be denoted by K and V, respectively. When the properties of a two-particle system are quantized there is a principal quantum number n. When this number changes from a higher level to a lower level the kinetic and potential energy change. These changes do not necessarily preserve energy and the difference in energy is emitted as a photon. Let γ be the energy of the emitted photon. Then

#### γ = −(ΔK + ΔV) and hence γ + ΔK + ΔV = 0

If there exists two coefficients, α and β, such that

#### K = α/n² and V = −β/n²

where n is a positive integer, called the principal quantum number, then

#### γ = (β − α)Δ(1/n²)

This is the scheme that was found to explain the spectrum of the hydrogen atom.

## The Nonrelativistic Electrostatic Case

Two specific subcases will be analyzed first to acquaint the reader with the analysis before the general case is examined. Those two cases are: 1. Positronium 2. The Hydrogen Atom.

### Positronium

Positronium is an electron and a positron rotating about their center of mass. Let m be the common mass of an electron and positron. Let s be their separation distance and r their orbit radius, with r=s/2. Let ω be the rate of rotation of the system.

The electrostatic attraction experienced by the particles is -k/s², where k is the constant of the electrostatic force. A balance of centrigugal force and the electrostatic force for maintaining a circular orbit requires:

#### mω²r = mω²s/2 = k/s² which implies ω² = 2k/(ms³)

(By limiting the analysis to circular orbits the special case of zero angular momentum is missed. This is dealt with elsewhere.)

The angular momentum of each particle is m(ωr)r=mωr²=mωs²/4. The total angular momentum is then 2mωs²/4=mωs²/2.

Angular momentum is quantized; i.e.,

#### mωs²/2 = nhand hence ω = 2nh/(ms²)

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

For a background on the quantization of angular momentum see Bohr's Condition.

This quantization equation squared is

#### ω² = 4n²h²/(m²s4)

Setting equal the two expressions derived for ω² yields

#### 2k/(ms³)= 4n²h²/(m²s4) which reduces to s = 2n²h²/(mk) = (2h²/(mk))n²

The expression (h²/(mk)) arises often enough to justify giving it a special label. However, it is appropriate to first introduce the concept of the reduced mass for two particles. The reduced mass μ of masses m1 and m2 is given by

#### 1/μ = 1/m1 + 1/m2which reduces to μ = m1m2/(m1+m2)

Thus for a system of two equal masses of m the reduced mass is m/2. Therefore the previous equation

Let

#### σ = (h²/(μk))

It is a natural unit of length scale. It is twice the Bohr radius for electrons and thus the diameter of the innermost electron orbit in a hydrogen atom. Its value is 1.05835442×10-10 meters or 105,835.442 fermi.

Thus the quantization condition for the separation distance for positronium is

#### s = σn² or, equivalently (s/σ) = n²

This means the spatial dimensions of positronium are the same as those of the hydrogen atom.

Now consider the kinetic and potential energies.

The kinetic energy for the two particles is given by

#### K = 2(½mω²r²) = mω²s²/4 which is equivalent to K = m(2k/ms³)s²/4 = (k/s)/2 which can be expressed as K = (k/(σn²)/2 = ½(k/σ)/n²

The potential energy is given by

#### V = −k/s and thus V = −(k/σ)/n²

It is immediately clear that K=−V/2. Thus the total energy of the system is K+V=V/2. The spectrum is given by

This means that

#### ΔK = −½ΔV = γ and ΔV = −2γ

The frequency ν of the photon is given by

#### ν = γ/h = ½(k/(hσ))Δ(1/n²)

Before going on to the next case there are some observations to be made about the scale length σ. The constant for the electrostatic force k can be expressed as αhc, where α is the fine structure constant 1/137.06. Thus σ can also be expressed as

This means that

#### k/σ = αhc/((1/α)(hc))/(μc²) which reduces to k/σ = α²μc²

The energy of the γ ray emitted upon the formation of positronium corresponds to a change in 1/n² from 0 to 1. Thus

#### γ = (1/4)α²(rest mass energy of an electron) which evaluates to γ = 1.33081766×10-5*(rest mass energy of electron)

Since the rest mass energy of an electron is 0.511 MeV the energy of the γ ray emitted on the formation of positronium should be, according to the model, 6.8 eV. The increase in kinetic energy of the system then should also be 6.8 eV and the decrease in potential energy 13.6 eV. This is the same as the ionization energy of electrons for a hydrogen atom.

### The Hydrogen Atom

The analyis is basically the same as for positronium but now r=s. The mass of proton is about 1836 times that of the electron. The reduced mass of the system is virtually the same as the mass of the electron. In the standard treatment the center of mass is taken to be at the center of the proton and the angular momentum of the system is entirely in the electron and is equal to m(ωr)r, which is the same as mωs². Thus

#### mωs² = nhand hence ω² = n²h²/(m²s4)

The dynamic balance condition for a circular orbit, is now

Therefore

#### (k/m)/s³ = n²h²/(m²s4) and hence s = n²h²/(mk) or, equivalently s = σn²

The potential energy is then

#### V = −k/s = (k/σ)/n²

The kinetic energy of the electron is ½mω²s² and thus

#### K = ½m(n²h²/(m²s4)s² which reduces to K = ½n²h²/(ms²) which can be expressed as K = ½nk²h²/(mks²) = ½n²kσ/s² and hence K = ½n²kσ/(σn²)² which reduces to K = ½(k/σ)/n²

Again K=−½V. Thus, the spectrum is given by

#### γ = ½(k/σ)Δ(1/n²)

The value of σ for this case is one half of its value for positronium because the the reduced mass of the system is twice that for positronium. The value of γ is then 13.6 eV, the ionization energy. The increase in kinetic energy is then also 13.6 eV and the decrease in potential energy is 27.2 eV.

## The General Case

The hydrogen atom model presumes that the proton is is infinitely massive compared to the electron. Although the model gives excellent empirical results even though the mass ratio is only 1836 instead of infinity there are other exotic combinations such as a muon and proton where the finite mass ratio must be taken into account.

Let m1 and m2 the masses of two particles subject to elecrostatic attraction. The radii of their orbits about their center of mass are given by

#### m1r1 = m2r2hence r1 = (m2/(m1+m2))s r2 = (m1/(m1+m2))s

where s is their separation distance (1+2).

The electrostatic force on each particle is −k/s². Thus dynamic balance for circular orbits requires

#### m1ω²r1 = k/s² and m2ω²r1 = k/s² and hence m1m2/(m1+m2)ω²s = k/s²

The expression m1m2/(m1+m2) is just the reduced mass of the system, which is denoted as μ. Thus the previous equation is

#### μω²s = k/s² and hence ω² = (k/μ)/s³

The quantization of the angular momentum of the system requires

#### m1ω²r1²+m2ωr2² = nhor, equivalently (m1²r1²+m2r2²)ω = nh

The expression (m1²r1²+m2r2²) can be greatly simplified.

#### (m1²r1²+m2r2²) = (m1²m2²+m2m1²)s²/(m1+m2)² = m1m2((m1+m2)²)²/(m1+m2)² = m1m2/(m1+m2)s² = μs²

Thus the quantization of angular momentum condition can be expressed as

Hence

#### ω² = (k/μ)/s³ = n²h²/(μ²s4) which reduces to s = n²(h²/(kμ)) or, if (h²/(kμ)) is denoted as σ s = σn² or, equivalently (s/σ) = n²

The potential energy of the system is then

#### V = −k/s = −(k/σ)/n²

The kinetic energy is then

#### K = ½m1ω²r1² +½m2ω²r2² or, equivalently K = ½(m1²r1²+m2r2²)ω² which, from the previous reduction of (m1²r1²+m2r2²) to μs² means K = ½μω²s² and, upon substituting in the expression for ω² K = ½μ(n²h²/(μ²s4))s² which reduces to K = ½n²(h²/(μs²)which can be put into the form K = ½n²k(h²/(μk)/s²) or K = ½n²(kσ)/s²and hence K = ½n²(kσ)(σ²n4) = ½(k/σ)/n²

Once again K=−½V. Thus the spectrum of the system, in energy terms, is

## The Relativistic Case

### The Hydrogen Atom

The relevant factor for this case is that mass is given by

#### m = m0/(1−(v/c)²)½

where m0 is the rest mass of the electron.

The two equations which were previously derived for the hydrogen atom are:

#### mωs² = nh (quantization of angular momentum) mω²s = k/s² (dynamic balance for circular orbits)

The second equation can be put into a form which matches the LHS of the first equation; i.e.,

This means that

#### k/(ωs) = nhand hence ωs = k/(nh)

This is a quantization condition for the tangential velocity of the electron. This velocity relative to the speed of light c is then

The equation

The result is

#### m0(k/(nh))²/(1−β²)½ = k/s which may be solved for s as s = k(1−β²)½/(m0(k/(nh)²) which can be rearranged to s = (h²/(m0k)n²(1−β²)½or, replacing h²/(m0k) with σ s = σn²(1−β²)½(Note that β depends upon n.)

This is the quantization condition for the separation distance s.

The potential energy V is again given by

#### V = −k/s and thus V = −k/(σn²(1−β²)½) which is equivalent to V = −(k/σ)(1/(n²(1−β²)½))

The kinetic energy in relativistic terms is

#### K = m0c²[1/(1−β²)½ − 1]

It is completely determined by the value of β, whose quantized values have already been determined to be k/(nhc).

The first order approximation of K is K* = ½m(ωs)². This reduces to

#### K* = ½(m0/(1−β²)½)(k/(nh)² which simplifies to K* = ½k/(σn²(1−β²)½) or, equivalently K* = ½(k/σ)(1/n²(1−β²)½)

This reveals that K*=−½V. Thus the energy spectrum of the system is given approximately by

#### γ ≅ ½(k/σ)Δ(1/(n²(1−β²)½)

(To be continued.)

## The General Central Force Function (Nonrelativistic Case)

A general attractive central force between two particles can be represented in the form

#### F = −Hf(s)/s²

where s is the separation distance between the centers of the two particles, H is a positive constant and f(s) is a function such that f(0)=1. The major relevant cases for f(s) are: 1. The electrostatic case in which f(s)=1 for all s; 2. The nuclear force case in which f(s)=exp(−λs) where λ is a positive constant reflecting the rate of decay of the force carrying particles, the π mesons. The parameter λ can be expressed as 1/r0 so that f(s)=exp(−s/r0).

Consider the case in which the two particles have an equal mass of m. These is roughly the case of the deuteron.

The quantization of angular momentum equation is then

#### 2(m(ωr)r = 2mωr² = 2mω(s/2)² = mωs²/2 = nhand thus ω = 2nh/(ms²) and further ω² = 4n²h²/(m²s4)

Dynamic balance for circular orbits requires

#### mω²r = mω²s/2 = Hf(s)/s² and hence ω² =2Hf(s)/(ms³)

Equating the two expressions for ω² yields

#### 2Hf(s)/(ms³) = 4n²h²/(m²s4) which reduces to sf(s) = 2n²h²/(mH)

Now is the time to note that the reduced mass μ of the system is m/2. Thus the previous equation is

#### sf(s) = n²h²/((m/2)H) = n²h²/(μH) or, denoting h²/(μH) as σ sf(s) = σn²

This gives the allowable values of the separation distance s. From the quantized values of s the quantized values of ω are determined.

The kinetic energy is given by

#### K = 2(½mω²s²/4) = mω²s²/4

The quantity ω²s² can be represented as

#### ω²s² = 2Hf(s)/(ms) = 2(H/m)(f(s)/s)

This means the kinetic energy can be respresented as

#### K = m(2(H/m)(f(s)/s)/4 = ½H(f(s)/s)

The quantity ω²s² can also be represented as

#### 2Hsf(s)/(ms²) = sf(s)[H/(μs²)] but sf(s) is equal to σn² so ω²s² = σn²[H/(μs²)]

Thus kinetic energy can be expressed as

#### K = m(σn²H/(μs²))/4 or, equivalently K = μ(σn²H/(μs²))/2 which reduces to K = ½σn²H/s²

If s is approximately proportional to n² then K is approximately inversely proportional to n². For a further investigation of the dependences of s and K on n see n dependence.

Another way of representing the kinetic energy is to note that if sf(s)=σn² then s²=σ²n4/f²(s). Thus

#### K = ½σn²/[σ²n4/f²(s)] which reduces to K = ½f²(s)H/(σn²) = ½f²(s)(H/σ)/n²

Since σ=h²/(μH) and H can be expressed as ζhc, where ζ is the structure constant for the nuclear force,

Thus

#### K = ½f²(s)ζ²μc²/n² or, equivalently K/(μc²) = ½f²(s)ζ²/n²

An estimate of ζ as 1.58 is given in Force Constants.

The increase in kinetic energy from the case of 1/n²=0 to 1/n²=1 is then

#### ΔK = ½f²(s0)ζ²

The potential energy V is given by

#### V(s) = ∫s∞(Hf(z)/z²)dz an application of integration-by-parts puts V(s) into the form V(s) = −Hf(s)/s + ∫s∞(Hf(z)/z)dz

Note that it was found above that one representation of kinetic energy was as ½H(f(s)/s). Thus, when s goes from ∞ where V=0 to s0

#### γ = −(ΔV + ΔK) = −(−½Hf(s0)/s0 + ∫s0∞(Hf(z)/z)dz) which reduces to γ = H(½f(s0)/s0 + ∫s0∞(f(z)/z)dz))

This means that if γ and s0 are known, along with f(s), then an estimate of H can be obtained as

## Empirical Investigations for the Nuclear Force FormulaF = -H*exp(-s/r0)/s²

Some invesigation of empirical values is appropriate at this point. Estimates of the parameters of = -H*exp(-s/r0)/s² are available. The values are H=1.92570×10−25 kg·m³/s² and r0=1.522×10-15=1.522 fermi. The reduced mass of a proton and neutron is 8.368746×10−28 kg.

Since h²=1.112121×10−68 m4 kg2 / s2 this means

#### σ = 1.112121×10−68/(8.368746×10−28*1.92570×10−25) which evaluates to σ = 6.900857×10-17 meters or σ = 0.06900857 fermi

The charge radius of the deuteron as determined by scattering has been found to be 2 fermi ±0.1 fermi. Thus the separation distance s is about 4 fermi.

The change in kinetic energy when the separation distance s goes from s=∞ to s=4 fermi=4×10-15 m is then

#### ΔK = 4.15280632×10-13 joules = 2.59198×106 eV = 2.59198 Mev

Thus, according to the model and the parameter estimates, the mass deficit of a deuteron would be 2.59198+2.22457=4.81655 MeV rather than 2.22457 MeV. For a separation distance of 4.2 fermi the value of ΔK would be 2.35100 fermi. This makes the energy of the gamma ray and the increase in kinetic energy roughly equal, which is what occurs for the electrostatic force.

If the separation distance were 5 fermi instead of 4 the value of ΔK would be 1.6888 MeV and thus the mass deficit of a deuteron would be 3.88344 MeV rather than 2.22457 MeV, as is conventionally assumed.

## The Mass of the Neutron

The mass of the neutron is not directly measurable; instead it is estimated from the measured masses of the deuteron and the proton and the energy of the gamma ray involved in the formation or disassociation of a deuteron. The traditional analysis derives the mass of a neutron by presuming that the energy of the gamma ray is exactly equal to the mass deficit of the deuteron. If the conversion of mass goes partially into the energy of the gamma ray and the rest goes into an increase in the kinetic energies of the proton and neutron then the accepted estimate of the mass of the neutron in error by being less than the actual mass. A mass deficit for a deuteron of 3.88344 MeV, corresponding to a separation distance in the deuteron of 5 fermi, would mean the mass of the neutron is underestimated 1.6888 MeV; i.e., it is 941.2548 MeV rather than 939.566 MeV.

Using the conventional estimate of the mass of the neutron, the beryllium isotope having four protons and one neutron has a negative mass deficit (mass surplus) of 0.75 MeV. The mass deficit is also called the binding energy. This means that the Be5 supposedly has a negative binding energy. This value may just be the result of an underestimate of the mass of the neutron. This lends plausibility to the above estimates.

## The Separation Distance of the Nucleons in a Deuteron

In the preceding investigation an empirical estimate of the separation distance of the proton and neutron was used. The model itself implies a value for the separation distance. It is the solution to the equation

#### s*exp(−s/r0) = σ which can be put into a better form for solution as (s/r0)exp(−s/r0) = σ/r0or z*exp(−z) = σ/r0where z=s/r0

The estimate of r0 as 1.522 fermi is available in Nuclear Force Formula.

Since the estimate of σ is 0.069 the transcendental equation to be solved is

#### z*exp(-z) = 0.0453407

The solution is z=0.0475487 and thus s0=1.522*0.0475487=0.07237 fermi. This is too small, indicating that the value of H used to compute σ was too large.

If s0=4.2 fermi is taken as the correct solution then

#### z = 4.2/1.522 = 2.759527 and hence σ/r0 = 2.759527*exp(-2.759527) = 0.174737975 and thus σ = 0.174737975*1.522 = 0.265951198 fermi

Thus the value of H is given by

#### H = h²/(μ*σ) = 4.99677×10-26 kg·m³/s²

which is about one fourth the magnitude of the estimate of H used previously.

The value of the increase in kinetic energy is given by

#### ΔK = 3.76672×10-13 joules = 2.351 MeV

This means the mass deficit of the deuteron is 2.22457+2.351=4.57557 MeV rather than 2.22457 Mev. The decrease in potential energy is divided 51.4% into an increase in kinetic energy and 48.6% into the energy of the emitted gamma ray. It also means the mass of the neutron is 941.917 MeV instead of 939.566 MeV. This also means that the mass deficit (binding energy) of the Be5 nuclide is 1.6 Mev instead of −0.75 MeV.

The kinetic energy of the system in terms of the tangential velocity v is

#### 2(½mv²) = mv² = 3.76672×10-13 joules and therefore v² = 3.76672×10-13/1.672621×10-27 = 2.25198655×1014 m²/s² and thus v = 1.50066204×107 m/s and β = v/c = 0.05 (1-β²)½ = 0.998749218

This means that relativistic adjustments will amount to only about 1/8 of 1 percent.

The relativistic form of the model is not needed for numerical significance but would be worthwhile for further theoretical analysis.

## The General Central Force Function (Relativistic Case)

Here relativistic means only that mass is given by

#### m = m0/(1−β²)½where β=v/c, the particle velocity relative the speed of light

The conditions for the quantization of angular momentum and the dynamic balance can be reduced to

#### μωs² = nh μω²s = Hf(s)/s²

where μ is the reduced mass of the particles.

From these it can deduced that

#### ωs = Hf(s)/(nhor, upon expressing H as ζhc ωs = ζf(s)c/n

The tangential velocity of the particles is then ½ωs. Thus

The equation

#### μω²s = Hf(s)/s² may be rewritten as μ(ωs)² = Hf(s)/s and, since ωs = Hf(s)/(nh) μ(Hf(s)/(nh)which reduces to μsf(s) = h²n²/H

Noting that μ=μ0/(1-β²)½ and H=ζhc the previous formula may be expressed as

#### sf(s)/(1-β²)½ = hc/(ζμ0c²) which upon squaring both sides, clearing the fractions and making the substitution β=½ζf(s)/n becomes [s²f²(s)(μ0c²)² (1/4)f²(s)(hc)²/n²]ζhc)²

This may be further reduced to

#### ζ = 1/(f(s)[s²(μ0c²/(hc))² + ¼(1/n²]½)

If s is known for a specific value of n, say n=1, then the above formula can be used to estimate ζ.

For the nuclear force formula f(s)=exp(−s/r0) where r0=1.522 fermi. Using the conventional masses of the proton and neutron μ=8.36310818×10-18 kg. The constant hc = 3.1637145×10-26 joules-meters (m³ kg/s²).

Using s=4.2 fermi for n=1 then gives

#### ζ = 1.5784662 H = 4.9938164×10-26 m³ kg/s² β = 0.0499519585

From these values the increase in kinetic energy created from the formation of of the deuteron is

#### ΔK = 2.34463 MeV and the decrease in potential energy (and mass deficit) is −ΔV = mass deficit = γ + ΔK = 2.22457 MeV + 2.34463 MeV = 4.56910 MeV

The mass of the neutron is then 941.910 MeV rather than the conventional 939.566 MeV.

Since the conventional mass of the neutron was used in computing the reduced mass of the system the computation must be repeated until a solution is found for H, ζ, ΔK and mn.

After a small number of iterations the solutions for s=4.2 fermi converge to:

#### ζ = 1.578496333 β = 0.04997655 H = 4.990457×10-26 m²kg/s ΔK = 2.352441517 MeV

The mass deficit of the deuteron is then 4.5749 MeV, which means that the mass of the neutron is

#### mn = 941.9184415 MeV

This would mean that the mass of the neutron is greater than than of the proton by 7.136 electron masses instead of the conventional 2.532. It means that the mass deficit or binding energy of the He4 (alpha particle) would be 33 MeV instead of 28.3 MeV. This value corresponds very closely with the figure of 32.94963 MeV that was found in one statistical analysis of the binding energies of 2932 nuclides. For more on this investigation see The Enigma of Nuclear Mass Deficits.

The value of s=4.2 may correspond to the limits of the charge distance rather than the separation of the centers of the nucleons. The diameters of the proton and neutron are thought to be about 1 fermi. Thus the separation of the centers would be 4.2-0.5-0.5=3.2 fermi. For a value of 3.2 fermi the solutions values are:

#### ζ = 1.073017248 β = 0.0655351 H = 3.392372×10-26 m²kg/s ΔK = 4.054286374 MeV

This would mean the mass deficit of the deuteron would be 6.2789 MeV and hence the mass of the neutron would be

#### mn = 943.6202864 MeV

This exceeds the mass of the proton by 9.89 electron masses. For this value of neutron mass the mass deficit (binding energy) of the alpha particle would be 36.4 MeV.

For comparison consider what the solutions would be if s=5.2 fermi.

## Further Analysis to be Completed

• The Zero Angular Momentum Case was not included in the analysis but may represent the ground state of the two-particle systems of interest.
• Only circular orbits were considered. Elliptical orbits may be of interest.
• The analysis should be recast in terms of Schrödinger's Equation and standard Quantum Mechanics.

## Conclusions

• The spectrum of two particle systems under a central force is given exactly or approximately, in energy terms, in the form

#### γ = αΔ(1/n²)

where n is the principal quantum number of the system.

• The nuclear force is of the form

#### F = −H*exp(−s/r0)/s²

where s is the separation distance of the nucleons. The values of the parameters are r0=1.522 fermi and H is in the range of 3×10-26 to 5×10-26 m²kg/s.

• The mass deficit of a nuclide is equal to the decrease in potential energy resulting from its formation. This means that

#### mass deficit = γ + ΔK

where γ is the energy of the photon emitted upon its formation and ΔK is the increase in kinetic energy of the nucleons resulting from its formation.

• The mass deficit of the deuteron is underestimated by ΔK and thus the mass of the neutron is underestimated by the same amount.

(To be continued.)

APPENDIX

## Dependence of the Separation Distance s Upon the Principal Quantum Number n

The dependence of s on n can be elucidated by considering d(ln(s))/d(ln(n)); i.e.,

#### sf(s) = 2σn² and hence ln(s) + ln(f(s)) = ln(2σ) + 2ln(n) and thus [1 + (f'(s)/f(s))(ds/d(ln(s))]d(ln(s))/d(ln(n)) = 2

Since ds/d(ln(s)) is equal to s

#### d(ln(s))/d(ln(n)) = 2/[1 + f'(s)s/f(s)]

Thus, of course, if f'(s)=0 then d(ln(s))/d(ln(n))=2. If f(s)=exp(−λs) then d(ln(s))/d(ln(n))=2/(1−λs). Thus For values of λs which are small relative to 1, d(ln(s))/d(ln(n)) is only slightly larger than 2. This means that the kinetic energy is approximately of the form

#### K = σn²/(ρ(n²)²) = (σ/ρ)/n²

where ρ is a constant.

However, λ can be considered to be the reciprocal of a scale length s0; i.e. λ=1/s0. The previous analysis indicated that for s<<s0 s is approximately proportional to n², but for s>s0 the dependence is drastically altered since