Quantum Mechanics

Schrodinger equation

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# Solving the Radial part of the Schrodinger Equation

### 1. New form of the Schrodinger Hydrogen atom radial equation

```

∂/∂r(r2 ∂R(r))/∂r) +
[2m/ℏ 2][(re2) + r2 E]R(r) = l(l + 1)R(r)

(∂/∂r(r2 ∂R(r))/∂r) + (2m/ℏ 2) [(re2) + r2 E - l(l+1)ℏ 2/2m]R(r) = 0

We have also:
∂/∂r(r2 ∂R(r))/∂r = 2r ∂R(r))/∂r + r2 ∂2R(r))/∂r2 = r ∂(rP(r))∂r

Therefore:

r ∂2(rP(r))∂r2 + (2m/ℏ 2) [(re2) + r2 E - l(l+1)ℏ 2/2m]R(r) = 0

Dividing by r2 , yields:

(1/r) ∂2(rR(r))∂r2 + (2m/ℏ 2) [(e2/r) +  E - l(l+1)ℏ 2/2m r2]R(r) = 0

Let' write: u(r) = rR(r), then:

∂2(u(r))∂r2 + (2m/ℏ 2) [(e2/r) +  E - l(l+1)ℏ 2/2m 2 r2]u(r) = 0

By substituting ρ = αr, r2 = ρ2/α2, dρ = αdr , dr2 = dρ/α2, We obtain:

α2∂2(u(ρ))∂ρ2 +  (2m/ℏ 2)(αe2/ρ) +  E (2m/ℏ 2) - l(l+1)α2/ρ2]u(r) = 0

Let's write:
ℏ2/me2 = ao
λ /2 = 1/αao = me2/αℏ2

That gives:
α2∂2(u(ρ))∂ρ2 +  (λα2/ρ) +  E (2m/ℏ 2) - l(l+1)α2/ρ2]u(r) = 0
Dividing by α2 yields:
∂2(u(ρ))∂ρ2 + λ/ρ +  E (2m/ℏ 2)α2 - l(l+1)/ρ2]u(r) = 0
With:
ℏ 2 α2/2m = - 4E

We have: ∂2(u(ρ))∂ρ2 + λ/ρ -  1/4  - l(l+1)/ρ2]u(r) = 0

∂2 u(ρ)/∂ρ2 - l(l+1)/ρ2 u(ρ) +
(λ/ρ) u(ρ) - (1/4) u(ρ) = 0

```

### 2. Solving the Radial part of the Schrodinger Equation

```
u(x)= e - x/2 x (k + 1)/2 Lkm = A x B x C

du/dx = A'BC + AB'C + ABC'
d2/dx2 = A"BC + A'B'C + A'BC + A'B'C + AB"C + AB'C'
+ A'BC' + AB'C' + ABC"
= A"BC + AB"C + ABC" + 2(A'B'C + A'BC' + AB'C')

The derivatives give:
A' = (-1/2)A, A" = A/4
B' = (k + 1)B/2x, B" (k2 -1)B/4x2

Substituting yiels:
d2/dx2  = ABC/4 + ABC(k2 -1)/4x2/4x2 + ABC" +
2[-ABC(k + 1)/2x - ABC'/2 + ABC'(k + 1)/2x] =
ABC[1/4 + (k2 -1)B/4x2 - (k + 1)/2x] + ABC" + AB[-1 + (k + 1)/x]C'

Rearranging:

d2/dx2 = ABC" +  AB[-1 + (k + 1)/x]C' + ABC[1/4 + (k2 -1)/4x2 - (k + 1)/2x]

The Schrodinger equation becomes:

2∂2 u(x)/∂x2 +
- l(l+1)/x2 u(x) + λ/x u(x) - (1/4) u(x) = 0
or:
ABC" +  AB[-1 + (k + 1)/x]C' + ABC[1/4 + (k2 -1)/4x2 - (k + 1)/2x
- l(l+1)/x2  + λ/x  - (1/4) ] = 0

That is:
C" +  [- 1 + (k + 1)/x]C' + C[(k2 -1)/4x2 - (k + 1)/2x
- l(l+1)/x2  + λ/x] = 0
Or
x C" +  [- x + (k + 1)]C' + C[(k2 -1)/4x - (k + 1)/2
- l(l+1)/x + λ  ] = 0

C =  Lkm
Then, the equation:
x Lkn" +  [(k + 1) - x]Lkm' + [(k2 -1)/4x - (k + 1)/2
- l(l+1)/x + λ  ]Lkm = 0
is an associated Laguerre's equation if (k2 -1)/4x - (k + 1)/2
- l(l+1)/x + λ   = n
Or:
n =   (k2 - 1  - 4l(l+1))/4x  - (k + 1)/2  + λ . To zero he first term, we
have: (k2 - 1  - 4l(l+1)) = 0. That is (k2  =  4l(l+1) + 1 = 4l2 + 4l + 1 =
(2l + 1)2

k = 2l + 1

Therefore:
(k + 1)/2 = l + 1

To satisfy the Laguerre's equation, it remains:
λ - (k + 1)/2 = λ -(l+1 )
λ must be an integer, Let λ  = n, then

λ  = n

The expression of u(x) is then:

u(x) = e-x/2 x (k + 1)/2 Lkm = e-x/2 x (l + 1) L2l + 1n - l - 1 )

R(r) = u(r)/r

We have finally the expression of the radial function:

R(r) = exp{- (α r)/2} (αr)l L2l + 1n - l - 1(r)

n is the principal quatum number
l is the orbital quantum number

```

### 3. Normalisation of the radial function

```
The normalisation is done by:
∫ N2|R(r)|r2dr = 1  [r : from 0 to ∞]
That is
N2 ∫ e-r r 2l+2 L2l + 1n - l-1 (r) L2l + 1n - l-1 (r) dr = 1

We have: ∫ e-x xk+1 Lkn Lkn dx = (n + k)!(2n + k + 1)/(n)!  [x : from 0 to ∞]

With k = 2l + 1, and replacing n by n - l - 1, we get:
∫ e-x x2l + 2 L2l + 1n - l-1  L2l + 1n - l - 1  dx = (n - l - 1 +  2l + 1)!(2n - 2l - 2 +  2l + 1 + 1)/(n - l - 1)! =

∫ e-x x2l + 2 L2l + 1n - l - 1  L2l + 1n - l- 1  dx = (n + l )!(2n )/(n - l - 1)!
Therefore:
N2 ∫ e-r r 2l + 2 L2l + 1n - l - 1 (r) L2l + 1n - l - 1 (r) dr = N2 (n + l )!(2n )/(n - l - 1)! = 1

Hence:
N2  = (n - l - 1)! /(n + l )!(2n )

N = [(n - l - 1)!/2n (n + l)!]1/2

The final expression of the radial function for the hydrogen atom
satisfying the radial art of the Schrodinger equation is:

R(r) = [(n-l-1)!/2n (n+l)!]1/2
exp{-(α r)/2} (αr)l L2l+1n-l-1(αr)

α = 2/nao

The corresponding energy is:

E = - ℏ 2 α2/8m

With: λ /2 = n/2 = 1/αao = me2/αℏ2, we have: α = 2m e2/nℏ2

E = - ℏ 2 [4m2 e4/n2ℏ4]/8 =  - [m e4/n2ℏ2]/2

E = - m e4/2n2ℏ2

```

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