The Hydrogen Atom

In general the Couloumbic potential of an electron in the field of the nucleus  Z · e is given by V(r) = −Ze²/(4peor) where r is the distance between the electron and the 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 the corresponding energy values so we can locate the electron. This can be used, for example, in the H-atom. Since we have a spherically symmetrical system, 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 not only on angles but also the radius r. The Schroedinger equation for an electron in the middle of a syymetrical field is given below in polar coordinates:

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

L² = - h² [1/sinJ/¶J(sinJ/¶J) + 1/sin²J²/¶j²]

As we did for a particle in three-dimensional well, we choose to seperate variables 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 center of the symmetrical field and then separating the equation we obtain the following expression:
 

 1/R(r) [−h²/2µ.1/r²./r(r² /r) + V(r)] R(r) + 1/2µr².1/YL²Y  =  E 
   (4)

We have derived the eigenvalues and eigenfunctions for L² in the chapters devoted to kinetic moment (derivation of kinetic moment eigenvalues, rotational energy and kinetic moment eigenfunctions) we will now substitute them into our expression and then obtain an ODE for the radial part of the 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 detail in the next chapter.

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