**Problem 1:**

Use the uncertainty relation to find an estimate of the ground state
energy of the harmonic oscillator.
The energy of the harmonic oscillator is E = p^{2}/(2m) + ½mω^{2}x^{2}.

Solution:

- Concepts:

The uncertainty principle - Reasoning:

We are asked to use the uncertainty relation, Δx Δp ≥ ħ, to estimate of the ground state energy of the harmonic oscillator. - Details of the calculation:

Assume the uncertainty in the position of the electron is Δx about x = 0.

Then the uncertainty in its momentum is Δp = ħ/Δx about p = 0.

The average kinetic energy is on the order of T = (Δp)^{2}/2m,

and the average potential energy is on the order of U = ½mω^{2}(Δx)^{2}.

The average total energy is E = T + U ~ ħ^{2}/(2m(Δx)^{2}) + ½mω^{2}(Δx)^{2}.

The ground state has the lowest energy, so let us minimize E with respect to Δx.

dE/d(Δx) = -ħ^{2}/(m(Δx)^{3}) + mω^{2}Δx = 0, (Δx)^{4}= ħ^{2}/(m^{2}ω^{2})

Inserting (Δx)^{2}= ħ/(mω) into the equation for E yields E = ħω.

**Problem 2:**

Electrons of kinetic energy 10 eV travel a distance of 2 km. If the size
of the initial wave packet is 10^{-9} m, estimate the size at the end
of their travel.

- Concepts:

The uncertainty principle - Reasoning:

We use the uncertainty principle to estimate the uncertainty in the momentum of the electrons. This translates into an uncertainty in the velocity. This Δv causes the wave packet to broaden with time. - Details of the calculation:

The electron energy is E = 10 eV = 1.6*10^{-18 }J.

Its speed is v = (2E/m)^{1/2 }= 1.88*10^{6 }m/s.

It travels 2 km in ~10^{-3 }s.

The initial uncertainty in its position is Δx = 10^{-9 }m.

Δp = ħ/Δx, Δv = ħ/(mΔx) = 1.1*10^{5 }m/s.

After 10^{-3}s the uncertainty in its position is on the order of Δv10^{-3 }s = 110 m.