Saturday, 31 December 2011

The One-Dimensional Hydrogen Atom: Quantum Solutions

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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 Ep(r) = - e2/(4π ε0 r) with e the charge of the electron (proton) and ε0 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:

Hψ = Eψ

With ψ the amplitude of the wave function, H the Hamiltonian Operator and E the total energy of the quantum system (E = Ek + Ep).

For the one dimensional hydrogen atom the Schrödinger Equation becomes:

- δψ”(r) - ρψ/r = Eψ

• 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”= (α2x - β)y

with α2 = - (E/δ)
and β = ρ/δ

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

Let’s look at this proposed ground state energy:

α = β/2

Square both sides:

α2 = β2/4

which works out as E1 = - ρ3/(4 δ2), 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”= (α2x - 2α)y

The proposed solution to this DE is y = xe-αx and it does indeed fit the equation xy”= (α2 - 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 = - ρ3/(16 δ2)

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

En = - ρ3/((2n)2 δ2) = E1/n2

The DE for n = 2 would then be:

xy”= (α2x - 4α)y

Well, assuming the energy quantisation is indeed En = - ¼ (ρ32)(1/n)2, then it can be shown easily with α2 = - (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:

yn = 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:

En = - ρ3/((2n)2 δ2)

And from above α2 = - (E/δ), E = - α2δ

Thus: αn = ½ (ρ/δ)3/2 (1/n)

Set α1 = ½ (ρ/δ)3/2

Then α2 = α1/2

α3 = α1/3


αn = α1/n

So y2 = x(1 - ½ α1x )e^(- ½ α1x)

And the second excitation was found to be:

y3 = x(1 - 2αx + 2/3.α2x2)e-αx

With β = 6α and α = α1/3

A General solution?

Assume that: yn = x F(x, n, αn)e^(-αnx)

With F = 1 + c1x + c2x2 + … + cn-1xn-1

Or F = [1+ (i from 1 to n-1) ∑ cixi]

Setting up the polynomial using F = 1 + ∑cixi (from i = 1 to n - 1) was surprisingly easy. First y was transformed to y = Ge-αx with G = x + ∑cixi+1.

y” can now be written as e-αx [G” - 2αG’ + α2G]

The x and the factor e-αx in the equation:

xy” = (α2x - 2nα)xFe-αx

then drop out and the equation becomes:

G” - 2αG’ + α2G = (α2x - 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 ci is always a simple function of ci-1:

ci = 2/(i(i+1)) . α (i-n) ci-1 and c0 = 1

For i=1, c1 = α(1-n) which fits for n=1 (c1 = 0) and n=2 (c1 = -α) and for n=3 (c1 = -2α), all as found above.

For i=2, c2 = 1/3 α2(2-n)(1-n)

Which for n=3 gives c2 = 2/3 α2, as found above.

c3 = 1/18 . α3(3-n)(2-n)(1-n)

Etc, etc for each ci.

Note also that in all cases α = α1/n

The general formula for ci appears to be:

ci = 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 = α1/n

The coefficients ci alternate in sign (+ or -) in such a way that the polynomial x.F always has n real roots (zero points),

Plugging ci into F = 1 + ∑cixi then yields the general solution for the wave functions:

yn = x Fe^(-αnx)

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) = ψ2(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 ∞)∫ ψ2(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 ψnorm = Aψ and thus with some rearranging:

A2 = 1 / [(from 0 to ∞)∫ψ2(r)dr]

Needless to say, given the complexity of ∫ψ2(r)dr, this is no walk in the park and I didn’t attempt to find a general expression for An . Instead I numerically integrated ψ2dr for n = 1, 2 and 3 (for ψ1 the normalisation constant is in fact easily calculated as A1 = 2α13/2).

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: