Question

ψ(x) is said to be an optimum state of position and momentum if for this state:
∆x∆p = ~/2

That is, for an optimum state, the product of the uncertainties of position and momentum is the smallest value
allowed by Heisenberg’s Uncertainty Relation:
∆x∆p ≥ ~/2

Show that the ground state of the harmonic oscillator is an optimum state.

Answer 1

Answer:

Heisenberg's Uncertainty Principle, at least an approximate form of it, can be stated as follows: delta(x)delta(p) > h, where delta(x) and delta(p) are the respective uncertainties of the particle's position and momentum and h is Planck's constant. The symbol > means greater than.