One of the questions of some physical interest is “how can we create a qunatum-mechanical state that behaves as much as a classical particle as possible?” From the uncertainty principle,
|
ΔxΔp ≥ {1\over
2}ℏ,
| (10.13) |
this must be a state where Δx and Δp are both as small as possible. Such a state is known as a “wavepacket”. We shall see below (and by using a computer demo) that its behavior depends on the Hamiltonian governing the system that we are studying!
Let us start with the uncertainty in x. A state with width Δx = σ should probably be a Gaussian, of the form
|
ψ(x,t) =\mathop{ exp}\nolimits \left (−{{(x − {x}_{0})}^{2}\over
2{σ}^{2}} \right )A(x).
| (10.14) |
In order for ψ to be normalised, we need to require
|
|A(x){|}^{2} = \sqrt{ {1\over {
σ}^{2}π}}.
| (10.15) |
Actually, I shall show below that with
|
A(x) = \root{4}\of{ {1\over {
σ}^{2}π}}{e}^{i{p}_{0}x∕ℏ},
| (10.16) |
we have
|
\langle \hat{x}\rangle = {x}_{0},\quad \langle \hat{p}\rangle = {p}_{0},\quad Δx = σ,\quad Δp = ℏ∕σ
| (10.17) |
The algebra behind this is relatively straightforward, but I shall just assume the first two, and only do the last two in all gory details.
Thus
Let \hat{p} act twice,
Doing all the integrals we conclude that
|
\langle {p}^{2}\rangle = {p}_{
0}^{2} + {{ℏ}^{2}\over
2{σ}^{2}}.
| (10.21) |
Thus, finally,
|
Δp = \sqrt{\langle {p}^{2 } \rangle −\langle {p\rangle }^{2}} = {ℏ\over
σ}
| (10.22) |
This is just the initial state, which clearly has minimal uncertainty. We shall now investigate how the state evolves in time by usin a numerical simulation. What we need to do is to decompose our state of minimal uncertainty in a sum over eigenstates of the Hamiltonian which describes our system!