Quantum Mechanics
Schrodinger equation
Quantum Mechanics
Propagators : Pg
Quantum Simple Harmonic
Oscillator QSHO
Quantum Mechanics
Simulation With GNU Octave
© The scientific sentence. 2010
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| Many-electron atoms: Repulsion
I = ∫ exp{- 2(r1 + r2)} dr1 dr2 /|r1 - r2|
from 0 to ∞
|r1 - r2| = [r12 + r22 - 2 r1 r1 cos θ]1/2
I = ∫ exp{- 2(r1 + r2)} dr1 dr2 /[r12 + r22 - 2 r1 r2 cos θ]1/2
from 0 to ∞
θ is the angle between the vectors r1 and r2.
I = ∫ exp{- 2 r2} J(r2) dr2
from 0 to ∞
where
J(r2) = ∫ exp{- 2 r1} dr1 /[r12 + r22 - 2 r1 r2 cos θ]1/2
from 0 to ∞
dr1 = r12 dr1 sin θ dθ dφ
J(r2) = 2 π ∫ exp{- 2 r1} r12 sin θ dr1 dθ /[r12 + r22 - 2 r1 r2 cos θ]1/2
r1 from 0 to ∞
θ from 0 to π
Let
u = cos θ, then du = - sin θ dθ
Therefore
J(r2) = 2 π ∫ exp{- 2 r1} r12 dr1 ∫ du /[r12 + r22 - 2 r1 r2 u]1/2
r1 from 0 to ∞ and u from - 1 to +1
First let's find
L = ∫ du /[r12 + r22 - 2 r1 r2 u]1/2
u from - 1 to +1
= - [r12 + r22 - 2 r1 r2 u]1/2/ r1r2
u from - 1 to +1
= - [r12 + r22 - 2 r1 r2 ]1/2/ r1r2
+ [r12 + r22 + 2 r1 r2 ]1/2/ r1r2 =
([r1 + r2] - |r1 - r2|) / r1r2
= 2/r1 if r1 > r2
= 2/r2 if r1 < r2
L = 2/r1 if r1 > r2
L = 2/r2 if r1 < r2
Hence
J(r2) = J1 + J2 , where:
J1 = 2 π ∫ exp{- 2 r1} r12 dr1 (2/r2) = 4 (π/r2) ∫ exp{- 2 r1} r12 dr1
from 0 to r2;
and
J2 = 2 π ∫ exp{- 2 r1} r12 dr1 (2/r1) = 4 π ∫ exp{- 2 r1} r1 dr1
from r2 to ∞
J(r2) = J1 + J2 , where:
J1 = 4 (π/r2) ∫ exp{- 2 r1} r12 dr1
from 0 to r2;
J2 = 4 π ∫ exp{- 2 r1} r1 dr1
from r2 to ∞
Let
K1 = ∫ exp{- α x} x2 dx and
K2 = ∫ exp{- α x} x dx
1. K2 = ∫ exp{- α x} x dx
Let
u = x , then du = dx and
dv = exp{- α x} dx , then v = - exp{- α x}/α
An integral by part ∫ udv = uv - ∫ vdu, gives:
K2 = - x exp{- α x}/ α + ∫ dx exp{- α x}/ α =
- x exp{- α x}/ α - (1/α2) exp{- α x} =
- (exp{- α x}/ α2)[ α x + 1]
Hence
∫ exp{-αx}x dx =
- (exp{- α x}/α2)[αx + 1]
2. K1 = ∫ exp{- α x} x2 dx
With u = x2 and dv = exp{- α x} dx. An integral by part
∫ udv = uv - ∫ vdu, gives:
K1 = -(x2/α) exp{- α x} + (1/α) ∫ exp{- α x} 2 x dx =
-(x2/α) exp{- α x} + 2 (1/α) ∫ exp{- α x} x dx =
-(x2/α) exp{- α x} + 2 (1/α) K2 =
= -(x2/α) exp{- α x} + 2 (1/α) (- (exp{- α x}/ α2)[ α x + 1]) =
= - (exp{- α x} /α3){ x2 α2 + 2α x + 2}
Hence
∫ exp{- α x} x2 dx =
- (exp{- α x}/α3)[x2 α2 + 2α x + 2]
Rearranging, we get:
J1 = 4 (π/r2) ∫ exp{- 2 r1} r12 dr1 =
- 4 (π/r2) (exp{- 2 r1}/8){ 4 r12 + 4 r1 + 2}
from 0 to r2
=
- 4 (π/r2){(exp{- 2 r2}/8){ 4 r22 + 4 r2 + 2} - (2/8) }
= - 4 (π/r2){(exp{- 2 r2}/8){ 4 r22 + 4 r2 + 2} - (1/4 }
and
J2 = 4 π ∫ exp{- 2 r1} r1 dr1
= - 4 π (exp{- 2 r1}/ 4)[ 2 r1 + 1]
from r2 to ∞
= + 4 π (exp{- 2 r2}/ 4)[ 2 r2 + 1]
J(r2) = J1 + J2 =
4 (π/r2){ - (exp{- 2 r2}/8){ 4 r22 + 4 r2 + 2} + 1/4
+ (exp{- 2 r2}/ 4) [ 2 r22 + r2] } =
(π/r2){ - (exp{- 2 r2}){ 2 r22 + 2 r2 + 1} + 1
+ (exp{- 2 r2}) [ 2 r22 + r2] } =
(π/r2)(exp{- 2 r2}){ - 2 r22 - 2 r2 - 1 + (exp{ 2 r2})
+ [ 2 r22 + r2] } =
- (π/r2)(exp{- 2 r2}){ r2 + 1 - (exp{ 2 r2}) }
J(r2) = - (π/r2)(exp{- 2 r2}){ r2 + 1 - (exp{ 2 r2})}
J(r2) = - (π/r2)(exp{-2r2})[r2 + 1 - exp{2r2}]
Now let calculate I:
I = ∫ exp{- 2 r2} J(r2) dr2 = 2π ∫ exp{- 2 r2} J(r2) r22 dr2 sin θ dθ
from 0 to ∞
∫sin θ dθ = 2
from 0 to π
I = 4π ∫ exp{- 2 r2} J(r2) r22 dr2
from 0 to ∞
I = - 4 π2 ∫ exp{- 4 r2} r2 dr2 { r2 + 1 - (exp{ 2 r2})}
from 0 to ∞
Let's write:
I = - 4 π2 (I1 + I2 + I3)
I1 = ∫ exp{- 4 r2} r22 dr2
I2 = ∫ exp{- 4 r2} r2 dr2
I3 = ∫ exp{- 2 r2} r2 dr2
•
I1 = ∫ exp{- 4 r2} r22 dr2 =
- (exp{- 4r2} /43){ 16 r22 + 8 r2 + 2}
•
I2 = ∫ exp{- 4 r2} r2 dr2 = - (exp{- 4r2}/ 42)[ 4r2 + 1]
•
I3 = ∫ exp{- 2 r2} r2 dr2 = - (exp{- 2r2}/ 4)[ 2r2 + 1]
Therefore
I1 + I2 + I3 = - (exp{- 4r2} /43){ 16 r22 + 24 r2 + 6}
+ (exp{- 2r2}/ 4)[ 2r2 + 1]
and
I = 4 π2{
(exp{- 4r2} /43){ 16 r22 + 24 r2 + 6}
- (exp{- 2r2}/ 4)[ 2r2 + 1]}
from 0 to ∞
= 10 π2 /16
∫exp{- 2(r1 + r2)} dr1 dr2 /|r1 - r2| = 5 π2/8
0 → ∞
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