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A Generalization of the Quantum
Mechanical Analysis of a Hydrogen Atom

The successful application of Schrödinger's wave mechanics to the hydrogen atom was a key to the acceptance of wave mechanics as the proper model of atomic reality. It was recognized that that analysis equally applied to any hydrogenic atom; i.e., any atom or ion with but a single electron. This is a further generalization to any system consisting of two objects of opposite electrostatic charges. This would include hydrogenic atoms but also positronium (an electron/positron pair), muonium and tauonium hydrogenic atoms, and any atom involving an antiproton and a positively charged lepton.

Let the two subatomic objects and their properties be labeled 0 and 1. The force between them is proportional to the product of their charges q0 and q1 and inversely proportional to the square of their separation distance s; i.e.,

F = Jq0q1/s²

where J is a constant. Since q0 and q1 are of opposite sign the force is an attraction. The force can be expressed as

F = −Q/s²

In the subsequent analysis it is only Q=−Jq0q1 that is relevant. The potential energy V is given by

V(s) = −Q/s

The kinetic energy K of the system is given by

K = ½m0v0² + ½m1v1²
or, in terms of
the momenta pj=mjvj
K = p0²/2m0 + p1²/2m1

The Hamiltonian (total energy) function for the system is then

H = p0²/2m0 + p1²/2m1 −Q/s

The time-independent Schrödinger equation for the system is then

h²[(∂²ψ/∂r0² (∂²ψ/∂r1²] −Q/s

But this is not a convenient form for analysis. What is needed is a coordinate system suited to the geometry of the system.

Formulation of a Convenient Coordinate System

Let the origin of the coordinate system be located at the center of mass for the two objects. Then

m0r0 = m1r1
and
r0 + r1 = s

It is easily shown that

r0 = (m1/(m0+m1))s
r1 = (m0/(m0+m1))s

Thus there are not two "radial" variable but only one, s. Likewise there are not four angle variables but only two.

The Momenta and the Components of Kinetic Energy

The Radial Components

The two radial momenta are

ps0 = m0(dr0/dt)
= m0(m1/(m0+m1))(ds/dt)
which can be put into the form
ps0 = [1/(1/m0+1/m1)](ds/dt)

The reduced mass μ of the system is defined by

1/μ = 1/m0+1/m1

Thus

ps0 = 1/(1/μ)(ds/dt) = μ(ds/dt)

Likewise ps1 = μ(ds/dt).

Therefore the radial component of kinetic energy is

Ks = ps0²/(2m0) + ps1²/(2m1)
= μ²(ds/dt)²/(2m0) + μ²(ds/dt)²/(2m1)
= (μ²/2)(ds/dt)²[1/m0 + 1/m1]
= (μ²/2)(ds/dt)²(1/μ) = ½μ(ds/dt)²

Define ps as μ(ds/dt) and thus

ps² = μs²(ds/dt)²
and hence
Ks = ps²/(2μ)

The Angular Momenta and Tangential Components of Kinetic Energy

First of all note that

m0r0 = m1r1 = (m0m1/(m0+m1))s = μs

The kinetic energy due to the tangential velocities is

Kφ = ½m0(r0ω)² + ½m1(r1ω)²
where ω=(dφ/dt).

This can be represented as

Kφ = ½((m0r0)²/m0)ω² + ½((m1r1)²/m1)ω)²
which reduces to
Kφ = ½(μ²s²ω²)(1/m0 + 1/m1) = μs²ω²

Now define pφ as

pφ = μs(sω) = μs²ω
and thus
pφ² = μ2s4ω2

This means that

Kφ = pφ²/(2μs²)

The Hamiltonian and Schrödinger Equation

From the previous material, the Hamiltonian is given by

H = ps²/(2μ) + pφ²/(2μs²) − Q/s

The process of solving the Schrödinger equation for the system is intricate and involved. It is easy in that process to lose sight of the objective. What is needed is a fast-track solution for a simplified case without every step being derived.

The Solution for a Special Case

Consider the case in which the mass one of the objects is relatively large compared to the mass of the other. In effect, the mass of one object is infinite compared to the other. The center of mass is then at the center of the larger object and the separation distance is the same as the orbit radius for the smaller object. In this case the variable s is replaced by r and the coordinate system is spherical.

The Hamiltonian function for the system is then

H = p²/2μ − Q/r

The time-independent Schrödinger equation for the system is then

−(h²/2μ)∇²ψ −Qψ/r = Eψ

The Laplacian operator ∇² for spherical coordinates (r, φ, θ) is

(1/r)(∂²(rψ)/∂r² + (1/r²)[(1/sin θ)(∂(sin θ (∂ψ/∂θ))/∂θ + (1/sin² θ)(∂²ψ/∂φ²)]

From this Laplacian it can be shown that −(h²/2μ)∇²ψ is equivalent to the operator for pr²ψ+pφ²ψ. Furthermore it is shown at that pφ²ψ is quantized to h²l(l+1).

If ψ(r, θ, φ) is assumed to be of the form R(r)Ylm(θ, φ) the analysis is separated into two parts.

The function Ylm(θ, φ) is called a spherical harmonic and it will be dealt with later. The function R(r) must satisfiy the equation

−(h²/2μ)((1/r)∂²(rR)/∂r²) + [h²/2μl(l+1)/(2μr²) − Q/r − E]ψ = 0

The first step toward a solution is to let rR(r) be denoted as u(r). The resulting equation for u is

−(d²u/dr²) + [l(l+1)/r² − (2μ/h²)(Q/r) − (2μE/h²)]u = 0

The analysis can be further simplified by introducing some nondimensional variables; i.e.,

κ² = 2μ|E|/h²
ρ = 2κr
λ = Q/(h²κ)

The equation for u(r) then simplifies to

(d²u/dr²) − (l(l+1)/ρ²)u + (λ/ρ − 1/4)u = 0

As ρ increases without bound the equation for u asymptotically approaches the equation

(d²U/dr²) − (1/4)U = 0

The solution to this equation is of the form

U(ρ) = A*exp(−ρ/2) + B*exp(ρ/2)

where A and B are constants. The only solutions of this form that are bounded as ρ→∞ are those for which B=0.

As ρ goes to zero the equation for u asymptotically approaches the equation

(d²V/dr²) − [l(l+1)/ρ²]V = 0

If V(ρ) is of the form ρβ then

β(β-1)ρβ-2 − [l(l+1)/ρ²]ρβ
which reduces to
β(β-1) = l(l+1)
which has the solution
β = (l+1)

This suggests a solution for u(ρ) of the form

u(ρ) = exp(−ρ/2)ρl+1F(ρ)

where F(ρ) is a polynomial in ρ that is finite everywhere.

When the proposed solution is substituted into the equation for u the result is that F(ρ) must satisfy the following equation.

ρ(d²F/dr²) + (2l + 2 − ρ)(dF/dρ) − (l + 1 − λ)F(ρ) = 0

If

F(ρ) = Σ0 Cjρj

then the coefficients Cj must satisfy the condition

Cj+1 = Cj[(j+ l +1 − λ)/(j+1)(j+2l+2)]

The polynomial F(ρ) will be of finite order if and only if there is an integer q such that the numerator of the fraction in the above condition is equal to zero; i.e.,

q + l + 1 = λ

This means that λ must be an integer and that q=λ−(l+1). Usually λ is called the principal quantum number and it is denoted as n. The definitions of of κ and λ imply a quantifization of energy; i.e.,

κ² = 2μ|E|/h²
n = λ = Qμ/(κh²)
hence
|E| = h²κ²
and
κ² = (Qμ)²/(n²h4)
and therefore
|E| = Q²μ²/(n²h²)

The quantization condition for energy is then

E = −Q²μ²/(n²h²)

The polynomials represented by F(ρ) are known as Laguerre polynomials. They depend upon an integral parameter q, which is equal to (n−(l+1)). The first few are:

qLq(ρ)
0(1/0!)1
1(1/1!)(1-ρ)
2(1/2!)(2-3ρ+ρ²)
3(1/3!)(6-18ρ+9ρ²-ρ³)

The radial function is then

Rnl(ρ) = exp(-ρ/2)ρl+1Ln-l-1(ρ)

The probability density function is proportional to the square of this radial function. The shapes of R and R² for several values of n and l are shown in the display below. For historical reasons the values of l are coded as letters: s=0, p=1, d=2.


from Introductory Quantum Mechanics by Richard L. Liboff, p. 191.

The Angular Equation

The angular equation is dealt with in Angular.

The Full Analysis

The time-independent Schrödinger equation for the system is

h²[(∂²ψ/∂s²)/(2μ) + (∂²ψ/∂φ²)/(2μs²)] −(Q/s)ψ = Eψ

where ψ is the wave function for the system and E is its total energy.

It is now assumed that the wave function is of the form

ψ(s, φ, θ) = R(s)Ψ(φ, θ)

where φ is the latitudinal angle and θ the longitudinal angle of the line connecting the centers of the two objects.

When derivatives are denoted with subscripts the Schrödinger equation becomes

h²[RssΨ/(2μ) + RΨφφ/(2μs²)] −(Q/s)RΨ = ERΨ
which, upon division
by RΨ, becomes
h²[(Rss/R)/(2μ) + (Ψφφ/Ψ)/(2μs²)] −(Q/s) = E
and after multiplication by s²
h²[(s²Rss/R)/(2μ) + (Ψφφ/Ψ)/(2μ)] −(Qs) = Es²

The term (Ψφφ/Ψ)/(2μ) is not a function of s whereas all of the other terms are functions of s. The above equation may be rearranged to the form

−(Ψφφ/Ψ) = s²Rss/R + (2μ/h²)[Qs + Es²]

Since one side of this equation is a function of s and the other is not, both sides must be equal to a constant, say λ². Thus

Ψφφ/Ψ = −λ²
which has as solution
Ψ = A(θ)cos(λφ) + B(θ)sin(λφ)

In order for Ψ to be a single-valued function of φ, λ must be an integer. Now let λ be designated as n.

The equation for the radial function is

s²Rss/R + (2μ/h²)[Qs + Es²] = n²
or, equivalently
Rss + [(2μ/h²)(Q/s + E) −n²/s²]R = 0

(To be continued.)


Reference:

Richard L. Liboff, Introductory Quantum Mechanics, Holden-Day Inc., San Francisco, 1980.