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 = p2/(2m) + ½mω2x2.
The ground state energy of the harmonic oscillator is on the order of ħω.
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.Solution:
Its speed is v = (2E/m)1/2 = 1.88*106 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*105 m/s.
After 10-3s the uncertainty in its position is on the order of Δv10-3 s = 110 m.