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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Free Particle in an infinite box (infinite well) of length L
The Schrödinger time-dependant equation is :
iℏ ∂ψ(r,t)/∂t = - (ℏ 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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