Notes: Particle in a Box By Random Walks
— Phys690 HW5 (Spring ’07)
Shiwei Zhang
The goal is to solve for the ground state of a particle in a box. We set ¯h=m= 1.
The Hamiltonian is H=−1
2
d2
dx2+V(x), where V(x) is 0 when xis on (−1,1) and ∞
otherwise.
Note that this is the drunkard problem, which we studied earlier, in disguise. As we
have shown, the solution to
−∂
∂t ψ(x, t) = Hψ(x, t)
leads to the ground state of Hat large t:ψ(x, t → ∞)→ψ0(x). In the drunkard
problem, the probability distribution of the drunkard at large time is a smooth function
which vanishes at the two bars. According to the argument above, this distribution is the
ground-state wave function of the quantum-mechanical problem of a particle in a box! Our
strategy to solve the quantum mechanical problem is therefore to simulate the motion of
many drunkards in order to obtain their distribution at large time. Of course, because
of trapping at the bars, there is a finite probability of losing the drunkard. We would
have to multiply ψ(x, t) by a constant to make up for that loss, so as to ensure that the
normalization of ψ(x, t) stays a constant.
Now we need to figure out how to simulate the drunkard motion. The zeroth-order
description is that it is just diffusion (Gaussian). But boundary effect must be properly
accounted for. In the lattice version the drunkard either “lands” exactly at the bar, or
not. In the continuum version, however, the situation is less straightforward: there is
always a finite probability that the drunkard will land at the bar (and hence be absorbed).
Therefore we need to modify the Gaussian to take this effect into consideration. Below is
a more formal description.
We need to repeatedly operate exp(−τH) on an initial wave function. (τis small.)
Recall that exp(−τH)≡g(x, x′) is called the short-time Green’s function. It can be
approximated by:
g(x, x′) = g0(x, x′)−g0(x, I x′),if x&x′inside box
0,otherwise ,(1)
where g0is the free-particle Green’s function: g0(x, x′) = 1/√2πτ exp[−(x−x′)2/2τ] , and
Ix′denotes the mirror image of x′with respect to the closer side of the box. It is easily
verified that g(x, x′) indeed: (i) satisfies −∂g/∂τ =Hg, (ii) is δ(x−x′) at τ= 0 and, (iii)
(approximately) satisfies the boundary condition at the sides. Note that, except for near
the two sides, g(x, x′) is essentially the free Green’s function g0(x, x′).
We want to propagate
ψ(x) = eτ E0Zg(x, x′)ψ(x′)dx′.(2)
The constant eτE0ensures normalization when ψis the ground-state wave function, i.e.,
when the propagation has reached equilibrium. We can rewrite Eq. (2) as:
ψ(x) = eτ E0ZK(x, x′)g0(x, x′)ψ(x′)dx′,(3)
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