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By a ‘one-dimensional hydrogen atom’ has to be understood an electron and a proton locked together into a one-dimensional box. Depending on the distance r of the electron to the proton and due to Coulombic attraction the electron possesses potential energy E

_{p}(r) = - e

^{2}/(4π ε

_{0}r) with e the charge of the electron (proton) and ε

_{0}the vacuum permittivity.

What follows is a ‘semi-heuristic’ derivation of the wave functions, electron densities and energy quantisation of such a quantum system.

The wave functions of a quantum system are solutions to the time-independent form of the Schrödinger Equation:

**ψ = Eψ**

*H*With ψ the amplitude of the wave function,

**the Hamiltonian Operator and E the total energy of the quantum system (E = E**

*H*_{k}+ E

_{p}).

For the one dimensional hydrogen atom the Schrödinger Equation becomes:

**- δψ”(r) - ρψ/r = Eψ**

With:

• r the distance between the electron and the proton (assume the latter to be completely stationary)

• ψ (i.e. ψ(r)) the amplitude of the wave function as a function of r

• - ρ ψ/r the potential energy term resulting from the coulombic attraction between the electron and the proton

• δ and ρ two material constants that can be found in any decent physics book. Both are positive.

• E the total energy which in this case is always negative (E < 0), reaching 0 only for the unbound electron (r = ∞)

Dividing everything by - δ and then multiplying by r gives: rψ” + (ρ/δ) ψ = - (E/δ)ψr or rψ” = [-(E/δ)r - (ρ/δ)]ψ

Substituting: ψ = y, r = x the differential equation (DE) becomes: xy”= (α

^{2}x - β)y

with α

^{2}= - (E/δ)

and β = ρ/δ

We know the system is quantised and a colleague (‘Watson.fawkes’) at the ScienceMadness.org forum proposed β = 2 α for the ground state.

Let’s look at this proposed ground state energy:

α = β/2

Square both sides:

α

^{2}= β

^{2}/4

which works out as E

_{1}= - ρ

^{3}/(4 δ

^{2}), the ground state energy of the electron, which is indeed ≠ 0 and negative as it should be.

The DE can then be re-written as xy”= (α

^{2}x - 2α)y

The proposed solution to this DE is y = xe

^{-αx}and it does indeed fit the equation xy”= (α

^{2}- 2α)y. Plugging y = xe

^{-αx}into the DE shows it is a solution for it.

Let’s now look at the first excitation:

It was proposed that for the first excitation α = β/4, so E2 = - ρ

^{3}/(16 δ

^{2})

The factors 4 and 16 would suggest quantisation as (2n)2 with n the quantum number, so that:

E

_{n}= - ρ

^{3}/((2n)

^{2}δ

^{2}) = E

_{1}/n

^{2}

The DE for n = 2 would then be:

xy”= (α

^{2}x - 4α)y

Well, assuming the energy quantisation is indeed E

_{n}= - ¼ (ρ

^{3}/δ

^{2})(1/n)

^{2}, then it can be shown easily with α

^{2}= - (E/δ) that α = ½ (ρ/δ)

^{3/2}(1/n)

Plugging y = x(1- αx)e

^{-αx}into the DE shows that it is a solution for β = 4α and presumably other solutions exist for β = 6α, β = 8α etc.

So that a function:

y

_{n}= f(x,n)e^(- ½(ρ/δ)

^{3/2}(1/n)x)

would appear a likely general solution, with f(x,n) a suitable polynomial.

**Regarding the quantisation:**

E

_{n}= - ρ

^{3}/((2n)

^{2}δ

^{2})

And from above α

_{2}= - (E/δ), E = - α

^{2}δ

Thus: α

_{n}= ½ (ρ/δ)

^{3/2}(1/n)

Set α

_{1}= ½ (ρ/δ)

^{3/2}

Then α

_{2}= α

_{1}/2

α

_{3}= α

_{1}/3

...

α

_{n}= α

_{1}/n

So y

_{2}= x(1 - ½ α

_{1}x )e^(- ½ α

_{1}x)

And the second excitation was found to be:

y

_{3}= x(1 - 2αx + 2/3.α

^{2}x

^{2})e

^{-αx}

With β = 6α and α = α

_{1}/3

**A General solution?**

Assume that: y

_{n}= x F(x, n, α

_{n})e^(-α

_{n}x)

With F = 1 + c

_{1}x + c

_{2}x

^{2}+ … + c

_{n-1}x

^{n-1}

Or F = [1+ (i from 1 to n-1) ∑ c

_{i}x

^{i}]

Setting up the polynomial using F = 1 + ∑c

_{i}x

^{i}(from i = 1 to n - 1) was surprisingly easy. First y was transformed to y = Ge

^{-αx}with G = x + ∑c

_{i}x

^{i+1}.

y” can now be written as e

^{-αx}[G” - 2αG’ + α

^{2}G]

The x and the factor e

^{-αx}in the equation:

xy” = (α

^{2}x - 2nα)xFe

^{-αx}

then drop out and the equation becomes:

G” - 2αG’ + α

^{2}G = (α

^{2}x - 2nα)F

After deriving G twice and a bit of rearranging, the full polynomial is then obtained. All its coefficients need to be zero to satisfy the DE. The recursing is really easy because it turns out that c

_{i}is always a simple function of c

_{i-1}:

c

_{i}= 2/(i(i+1)) . α (i-n) c

_{i-1}and c

_{0}= 1

For i=1, c

_{1}= α(1-n) which fits for n=1 (c

_{1}= 0) and n=2 (c

_{1}= -α) and for n=3 (c

_{1}= -2α), all as found above.

For i=2, c

_{2}= 1/3 α

^{2}(2-n)(1-n)

Which for n=3 gives c

_{2}= 2/3 α

^{2}, as found above.

c

_{3}= 1/18 . α

^{3}(3-n)(2-n)(1-n)

Etc, etc for each c

_{i}.

Note also that in all cases α = α

_{1}/n

The general formula for c

_{i}appears to be:

c

_{i}= 2i / i!(i+1)!) α Π ((i-n)(i-1-n)(i-2-n)…(2-n)(1-n))

with Π from i to 1 and α = α

_{1}/n

The coefficients c

_{i}alternate in sign (+ or -) in such a way that the polynomial x.F always has n real roots (zero points),

Plugging c

_{i}into F = 1 + ∑c

_{i}x

^{i}then yields the general solution for the wave functions:

y

_{n}= x Fe^(-α

_{n}x)

This leaves in essence only one pesky little problem, that of the normalisation of the wave functions. The Schrödinger Equation is a second order DE and so far only one integration constant has been deployed, i.e. the quantum number n.

But there is a second ‘boundary condition’ that has to be imposed on the system. In quantum physics it is well understood that the probability of finding the electron at position r, P(r), is given by:

P(r) = ψ

^{2}(r)

We also know that the total probability of finding the particle between r=0 and r=∞ must be 1, so that:

(from 0 to ∞)∫ ψ

^{2}(r) dr = 1, known as the ‘normalisation condition’.

To satisfy it we multiply the non-normalised wave function ψ with a normalisation constant A, so that ψ

_{norm}= Aψ and thus with some rearranging:

A

^{2}= 1 / [(from 0 to ∞)∫ψ

^{2}(r)dr]

Needless to say, given the complexity of ∫ψ

^{2}(r)dr, this is no walk in the park and I didn’t attempt to find a general expression for A

_{n}. Instead I numerically integrated ψ

^{2}dr for n = 1, 2 and 3 (for ψ

_{1}the normalisation constant is in fact easily calculated as A

_{1}= 2α

_{1}

^{3/2}).

Below are the normalised wave functions and electron probability distributions for the ground state and the first two excitations of a one-dimensional hydrogen atom: