JavaPsi - Simulating and Visualizing Quantum Mechanics (english)

3 SOLVING SCHRÖDINGER’S EQUATION 12
Interestingly to note the transmission coefficient D is still greater than
0 if the wave packet hits the wall with a total energy E < V0! I.e. the
probability that a particle with E < V0 is measured after the wall is greater
than 0 — the particle can tunnel the wall.
3.2.5 Harmonic oscillator
Figures 6a until 6c show that a wave packet (3.21) in a harmonic oscillator
potential (c.f. section 3.1.5) oscillates from the left to the right again and
again. That is the behavior of the particle here is very similar to the classical
description.
3.3 Two-dimensional, time-independent case
Schrödinger equation :
− 2
2m ∇2
+ V (x, y) Ψ(x, y) = EΨ(x, y)
Solution: centered space (CS) discretization and relaxation yields
ψ(x, y) =
with F (x) from section 3.1.1.
ψ(x + h, y) + ψ(x − h, y) + ψ(x, y + h) + ψ(x, y − h)
4 − h 2 F (x, y)
3.4 Two-dimensional, time-dependent case
Schrödinger equation :
− 2
2m ∇2
∂Ψ(x, y, t)
+ V (x, y, t) Ψ(x, y, t) = i
∂t
or
∂ 2 Ψ(x, y, t)
∂x 2
+ ∂2 Ψ(x, y, t)
∂y 2
+i 2m
∂Ψ(x, y, t)
∂t
− 2m
V (x, y, t)Ψ(x, y, t) = 0
Solution: forward time, centered space (FTCS) algorithm. Discretization
and solving the obtained system yields recursion for Ψ(x, y, t + ∆t) =
Ψr(x, y, t + ∆t) + iΨi(x, y, t + ∆t)
Ψr(x, y, t + ∆t) = c2V (x,y,t)Ψ∂2 r
Ψi(x, y, t + ∆t) = c2V (x,y,t)Ψ∂2 i
Ψ ∂2
(x,y,t)+c1Ψ∂2 i (x,y,t)+c2 1Ψr(x,y,t)−c1c2V (x,y,t)Ψi(x,y,t)
c2 1 +c2 ,
2V (x,y,t)2
(x,y,t)−c1Ψ∂2 r (x,y,t)+c2 1Ψi(x,y,t)−c1c2V (x,y,t)Ψr(x,y,t)
c2 1 +c2 2V (x,y,t)
r (x, y, t) = Ψr(x+∆x,y,t)+Ψr(x−∆x,y,t)+Ψr(x,y+∆y,t)+Ψr(x,y−∆y,t)−4Ψr(x,y,t)
(∆x) 2
Ψ∂2 Ψi(x+∆x,y,t)+Ψi(x−∆x,y,t)+Ψi(x,y+∆y,t)+Ψi(x,y−∆y,t)−4Ψi(x,y,t)
i (x, y, t) = (∆x) 2
with c1 = − 2m
∆t and c2 = 2m
2 .
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