Hydrogen Schrodinger Equation

1. Schrödinger time-independent
equation for the electron:

	
The electron in the hydrogen atom moves around the proton that 
maintain a potential energy V(r) = - e2/4πεo r .

  
The Schrödinger time-independent equation: 
- (ℏ 2/2m)∂2ψ(r)/∂r2 + V(r) ψ(r) = E ψ(r)    (1.1)


For the electron in the Hydrogen atom, It takes the form:
  
- (ℏ 2/2m)∂2ψ(r)/∂r2  - (e2/4πεo r) ψ(r) = E ψ(r)    (1.2)


It is necessary to work on the problem with spherical polar coordinates. 
Refer to spherical coordinatesto find the Laplacian: 
Δ = (1/r2)∂/∂r(r2 ∂/∂r) + 
(1/r2sinθ)∂/∂θ(sinθ∂/∂θ) + 
(1/r2sin2θ)∂2/∂φ2) 
    (1.3)

The Schrödinger time-independent equation becomes: 

- (ℏ 2/2m) Δψ(r) - (e2/4πεo r) ψ(r) = E ψ(r)    (1.4)

Now, let's separate the wave-function as follows:
ψ(r) = ψ(r, θ φ) = P(r) O(θ) M(φ)    (1.5)

Substituting this expression in the relationship (1.3) yields:

Δ ψ(r) = O(θ) M(φ) (1/r2)∂/∂r(r2 ∂P(r)/∂r) + 
P(r) M(φ) (1/r2sinθ)∂/∂θ(sinθ∂O(θ)/∂θ) + 
P(r) O(θ) (1/r2sin2θ)∂2M(φ)/∂φ2)     (1.5)

So, the equation  (1.4) becomes: 
- (ℏ 2/2m) [O(θ) M(φ (1/r2)∂/∂r(r2 ∂P(r))/∂r) + 
P(r) M(φ) (1/r2sinθ)∂/∂θ(sinθ∂O(θ)/∂θ) + 
P(r) O(θ) (1/r2sin2θ)∂2M(φ)/∂φ2)]
 - (e2/4πεo r) P(r) O(θ) M(φ) - E P(r) O(θ) M(φ)= 0     (1.6)
 
Dividing by  ψ(r) = P(r) O(θ) M(φ), we get:
- (ℏ 2/2m) [(1/P(r)) (1/r2)∂/∂r(r2 ∂P(r))/∂r) + 
(1/O(θ)) (1/r2sinθ)∂/∂θ(sinθ∂O(θ)/∂θ) + 
(1/M(φ)) (1/r2sin2θ)∂2M(φ)/∂φ2)] - (e2/4πεo r)  - E = 0 

Multiplying by - r2 2m/ℏ 2, we get:

(1/P(r)) ∂/∂r(r2 ∂P(r))/∂r) + 
[2m/ℏ 2][(re2/4πεo) + r2 E] +
(1/O(θ)) (1/sinθ)∂/∂θ(sinθ∂O(θ)/∂θ) + 
(1/M(φ)) (1/sin2θ)∂2M(φ)/∂φ2)  = 0      (1.7)

The first two terms contain only the variable r. These two radial terms 
must be equal to a constant "A" . This separates out the radial equation. 
The two remaining terms contain angular variables θ and φ. 
Therefore, the sum of these angular terms must  be equal to the negative of 
that constant "-A". The two angular terms must separately equal to a constant. 
This separates the two angular parts into the colatitude and 
azimuthal equations.

That is:

(1/P(r)) ∂/∂r(r2 ∂P(r))/∂r) + 
[2m/ℏ 2][(re2/4πεo) + r2 E] = A
(1/O(θ)) (1/sinθ)∂/∂θ(sinθ∂O(θ)/∂θ) + 
(1/M(φ)) (1/sin2θ)∂2M(φ)/∂φ2)  = - A  

Therefore, the colatitude part and azimuthal part must have negative values.

 Remark that the dimension of A is the one of r2 m2 v2/ℏ 2, that is the dimention of 
the square of orbital momentum L2/ℏ 2 
We will write: A = L2/ℏ 2.

1. The azimuthal equation:

Separated alone, the azimthal equation (1/M(φ)) ∂2M(φ)/∂φ2) 
must be equal to a negative constant. That is: (1/M(φ)) ∂2M(φ)/∂φ2) = - B, or 
∂2M(φ)/∂φ2) + B M(φ) = 0

We write the constant B = = ml2.  "l" stands for orbital and 
the saquare m2 is necessary to have the equation of the form: 
∂2 y/∂x2 + ω2 x = 0) . So, the separated azimuthal equation becomes:

∂2M(φ)/∂φ2) + ml2 M(φ) = 0     (1.8)

Which has the solution of the form:

M(φ) = A exp{i ml φ}
 

Normalized to 1, gives ∫M(φ)M*(φ) d(φ) [from 0 to 2π] = 1, that is 
1 = A2 2π, then A = 1/(2π)1/2

Finally, the solution of the azimuthal equation is :

M(φ) = (1/(2π)1/2)exp{i ml φ}
 

2. The colatitude equation:

With the azimuthal part, it is written as:

(1/O(θ)) (1/sinθ)∂/∂θ(sinθ∂O(θ)/∂θ) - ml2/sin2θ  = - L2/ℏ 2

Or:

(1/sinθ)∂/∂θ(sinθ∂O(θ)/∂θ) + [L2/ℏ 2 - ml2/sin2θ] O(θ) = 0
    (1.9)

Let's write: 
x = cos θ. Then θ = arccos x, therefore:
dx = - sin θ dθ and  dθ =  - dx / sqrt(1 - x2), so 
d2θ/dx2 = - x/(1 - x2)3/2

Then:
(1/sinθ)∂/∂θ(sinθ∂O(θ)/∂θ) = 
(1/sqrt(1 - x2)) ( - sqrt(1 - x2)) ∂/dx[(- sqrt(1 - x2)2 ∂O(θ)/dx)] = 
 ∂/dx[(1 - x2) ∂O(θ)/dx] =
 - 2x ∂O(θ)/dx + (1 - x2) ∂2O(θ)/dx2

The equation (1.9) becomes:

(1 - x2) d2O(x)/dx2 - 2x dO(x)/dx +
[L2/ℏ 2 - ml2/(1 - x2)] O(x) = 0    (1.10)

This equation has a solution, only when L2/ℏ 2 = l(l + 1), 
and L2/ℏ 2 - ml2 >= 0, plus l > |ml| where l is a positive integer 
called quatum orbital number..

Once the three conditions are satisfied, the solution are Legendre 
polynomial expressions.

If m = 0, we have le following Legendre equation:.

(1 - x2) d2O(x)/dx2 - 2x dO(x)/dx + l(l + 1) O(x) = 0    (1.11)

which has the set of the following solutions, depending 
on  the quantun number "l" , called the Legendre polynomial 
of order n, and denoted by Pl(x) or 
Pl(cosθ). Here are some firsts of them:

P0(x)	=	1	
P1(x)	=	x	
P2(x)	=	1/2(3x2-1)	
P3(x)	=	1/2(5x3-3x)	
P4(x)	=	1/8(35x4-30x2+3)	
P5(x)	=	1/8(63x5-70x3+15x)	
P6(x)	=	1/(16)(231x6-315x4+105x2-5)


If m ≠ 0, we have le following Legendre equation: (equation 1.10).

(1 - x2) d2O(x)/dx2 - 2x dO(x)/dx + 
[l(l + 1) - ml2/(1 - x2)]O(x) = 0    (1.11')

which has the set of the following solutions, depending on l and ml, called 
the asspciated Legendre functions:

Plml (x) = (1 - x2)m/2 dmlPl(x)/dxml


Its normalisation factor is :
N = [(2l + 1)(l - m)!/2(l + m)!]1/2

The associated legendre functions, therefore are the solution of the 
radial equation of the Schrodinger equation for the hydrogen atom:

Plml (x) = [(2l + 1)(l - m)!/2(l + m)!]1/2 (1 - x2)m/2 dmlPl(x)/dxml


Here some first solutions:

P00(x)	=	1	
P10(x)	=	x	
P11(x)	=	-(1-x2)(1/2)	
P20(x)	=	1/2(3x2-1)	
P21(x)	=	-3x(1-x2)(1/2)	
P22(x)	=	3(1-x2)	
P30(x)	=	1/2x(5x2-3)	
P31(x)	=	3/2(1-5x2)(1-x2)(1/2)
P32(x)	=	15x(1-x2)	
P33(x)	=	-15(1-x2)(3/2)

3. The spherical harmonics:

They are the product of the two angular functions:

Ylml = Olml(θ) x Mml(φ) 

Mml(φ) = (1/(2π)1/2)exp{i ml φ} 
Olml(θ) = [(2l + 1)(l - m)!/2(l + m)!]1/2 Plml(cos θ)

Plml (cos θ) are the associated Legendre polynomials.
 


Here are some first spherical harmonics:

Y00(θ,φ)	=	1/2(sqrt(2π))	
Y1(-1)(θ,φ)	=	1/2sqrt(3/(2π))sin θ(-iφ)	
Y10(θ,φ)	=	1/2sqrt(3/π)cos θ
Y11(θ,φ)	=	-1/2sqrt(3/(2π))sin θe(iφ)	
Y2(-2)(θ,φ) =	1/4sqrt((15)/(2π))sin2 θe(-2iφ)	
Y2(-1)(θ,φ)	=	1/2sqrt((15)/(2π))sinθcos θe(-iφ)	
Y20(θ,φ)	=	1/4sqrt(5/π)(3cos2 θ-1)	
Y21(θ,φ)	=	-1/2sqrt((15)/(2pi))sin θcos θe(i φ)	
Y22(θ,φ)	=	1/4sqrt((15)/(2pi))sin2 θe(2iφ)	
Y3(-3)(θ,φ)	=	1/8sqrt((35)/π)sin3 θe(-3iφ)	
Y3(-2)(θ,φ)	=	1/4sqrt((105)/(2pi))sin2 θcos θe(-2iφ)	
Y3(-1)(θ,φ)	=	1/8sqrt((21)/π)sin θ(5cos^2 θ-1)e(-iφ)	
Y30(θ,φ)	=	1/4sqrt(7/π)(5cos3 θ-3cos θ)	
Y31(θ,φ)	=	-1/8sqrt((21)/π)sin θ(5cos2 θ-1)e(iφ)	
Y32(θ,φ)	=	1/4sqrt((105)/(2pi))sin2 θcos θe(2iφ)	
Y33(θ,φ)	=	-1/8sqrt((35)/π)sin3 θe(3iφ).

4. The radial equation:

The first two terms of the equation (1.7) is the radial function. It is: 

(1/P(r)) ∂/∂r(r2 ∂P(r)/∂r) + 
[2m/ℏ 2][(re2/4πεo) + r2 E] 
 = L2/ℏ 2 = l(l + 1)      (1.7')

 or: 
 
∂/∂r(r2 ∂P(r)/∂r) + 
[2m/ℏ 2][(re2/4πεo) + r2 E - l(l + 1)] P(r) = 0      (1.7')

Let's write: α2 = 2mE/ℏ 2, 
ρ = αr, and 
β = 2me2/4πεoℏ 2
(1.7') becomes:

∂/∂ρ(ρ2 ∂P(ρ)/∂ρ) + [ρβ/α + (ρ/α)2 - l(l + 1)] P(ρ) = 0 

Let's change P(ρ) to R(ρ), we have:

ρR(ρ) = exp {- ρ/2} ρl+1 L2l+1β-l-1 (ρ)
 
R(ρ) = exp {- ρ/2} ρl L(2l+1,β-l-1)(ρ)