The Hydrogen Atom

In general the Couloumb potential of an electron in field of nucleus Z · e is given by V(r) = −Ze²/(4peor) where r is the distance between electron and nucleus (for H-atom Z = 1, for He+ Z = 2, for Li++ Z=3, etc.). The Schroedinger equation for an electron in central symmetrical field V(r) is as follows:

(−h²/2µΔ + V(r)) y = E y | (1)

where µ is the reduced mass of nucleus-electron system, i.e. µ = mEmK/(mE+mK) with mE is the electron mass and mK is the nuclear mass. Now we would like to find possible solutions for y and corresponding energy values E in order to find out states where the bounded electron can be. For instance, in the H-atom. Since we have spherical symmetry system here then we will use polar coordinates to solve the above-mentioned equation. Only Δ is changed when tranforming to polar coordinates since potential V(r) has been already given in polar coordinates:

Δ = ∂²/∂x² + ∂²/∂y² + ∂²/∂z²

= 1/r² ∂/∂r(r² ∂/∂r) −1/r²h²L²

where operator L²(J,j) is the squared kinetic moment operator (L² = Lx²+ Ly²+ Lz²) in polar coordinates and depends only on angles but not on radius r. The Schroedinger equation for electron in central syymetrical field is given in polar coordinates:

−h²/2µ.1/r².¶/¶r·(r² ¶/¶ry) + V(r) y + 1/2µr²·L²(J,j) y = E y

L² = - h² [1/sin J¶/¶J(sin J¶/¶J) + 1/sin²J¶²/¶j²]

As for a particle in three-dimensional well we choose the variable separation here: y(r,J,j) = R(r)·Y(J,j) because the squared kinetic moment operator L²(J,j) depends only on angles rather than r. Substituting y = R·Y in the Schroedinger equation for an electron in the central symmetrical field and then making corresponding equations separation we will obtain the following expression:

1/R(r) [−h²/2µ.1/r².¶/¶r(r² ¶/¶r) + V(r)] R(r) + 1/2µr².1/YL² Y = E | (4)

We have derived the eigenvalues and eigenfunctions for L² in the corresponding chapters devoted to kinetic moment (derivation of kinetic moment eigenvalues, rotational energy and kinetic moment eigenfunctions) and therefore we can substitute them into our expression and then obtain ODE for the radial part of wavefunction:

[−h²/2µ.1/r².¶/¶r(r² ¶/¶r) + V(r) + h²l(l+1)/2µr²] R(r) = E · R(r) | (5)

We will talk about this ODE solution a bit later.

We will talk about wavefunctions Y(J,j) and kinetic moment eigenvalues in more details in the next chapter.