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© The scientific sentence. 2010

Free Particle in an infinite box (infinite well) of length L


The Schrödinger time-dependant equation is :
i ∂ψ(r,t)/∂t = - ( 2/2m)∂2ψ(r,t)/∂r2 + V(r) ψ(r,t)    (2.1)

Its solution is: 
ψ(r,t) = ψ(r) x ψ(t) 
Where ψ(r) = A cos(kr) + B sin (kr), and  ψ(t) = D exp {i ωt} 

Let's take a space of one dimension, and write: 
ψ(x) = A cos(kx) + B sin (kx)

We have two regions:
1. Where the particle is free:  0 < x < L , V(x) = 0
2. Where the particle is not allowed to be present: - ∞ < x< 0 and 0 < x < ∞

The wavefunction ψ(x) must be null at the walls: ψ(0) = ψ(L) = 0. 
These boundary conditions constrain the form of the wavefunction ψ(x) as 
fllows:
- At x = 0, ψ(0) = 0 then A = 0
- At x = L, ψ(L) = 0 then sin (kL) = 0 therefore: 
 
ψ(x) = B sin (kx) ;  k = nπ/L 
where n is positive integer. We find here the expression of harmonic 
waves that form standing waves.

The probability for finding the particle within the box is 
equal to 1. The condition for normalization is then:
∫ ψ*(x) ψ(x) dx = 1 from 0 to L. Tha is:

B2 ∫ sin2(kx) dx = 1 from 0 to L = 
(1/2) B2 ∫ [1 - cos (2kx)] dx = 1 from 0 to L = 

(1/2) B22[L - (1/2k)(sin(2kL))]= 1 = 
(1/2) B22[L - (1/k)(sin(kL) cos (kL)]= (L/2) B2 = 1 
Therfore:
 
B = [2/L]1/2
Finally,
ψ(x) = [2/L]1/2 sin (kx) ;  k = nπ/L 

The spatial wave function for a free particle in an infinite box of 
length L has the following expression:

 
ψn(x) = [2/L]1/2 sin(nπx/L)      n = 0, 1, 2, 3, ... 





  


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