The uncertainty relations

The Heisenberg uncertainty relations:
Δx Δk ≥ 1,  Δp = ħΔk,   Δx Δp ≥ ħ.
Quantum Mechanics implies that it is impossible to know, at any given time, the position and the momentum of a particle to an arbitrary degree of accuracy.
For a plane wave Ψ(x,0) = A exp(ik0x) we have Δk = Δp = 0,  g(k) = δ(k - k0),  Δx = ∞ .

Similarly:
Δt Δω ≥ 1,  ΔE = ħΔω,   ΔE Δt ≥ ħ.
Quantum Mechanics implies that it is impossible to observe a particle for a finite time interval and know its energy to an arbitrary degree of accuracy.

Example:
An excited state of an atom decays emitting an electron.  If you are reasonably sure that the electron will be emitted in a time interval Δt < 10-8 s after you fired a picosecond laser, the energy of the electron must have an uncertainty
ΔE ≥ ħ/Δt = (6.6*10-34 Js)/(2π*10-8 s) ≈ 10-26 J ≈ 6*10-8 eV.
If the energy of the ground state is known exactly, then the energy of the excited state must be uncertain by ΔE.

We may write the Fourier transform of Ψ(x,0) in the following way:
Ψ(x,0) = (2πħ)-1/2∫Ψ(p) exp(ipx/ħ) dp  and  Ψ(p)  = (2πħ)-1/2∫Ψ(x,0) exp(-ipx/ħ)dx.
Then   dP(x) = (1/C)|Ψ(x,0)|2dx is the probability of finding the particle at a position between x and x + dx at t = 0,   
and dP(0) = (1/C)|Ψ(p)|2dp is the probability of finding the particle with momentum between p and p + dp. 
The constant C is the same in both expressions.  This follows from the Bessel-Parseval relation
∫-∞∞|Ψ(x,0)|2dx = ∫-∞∞|Ψ(p)|2dp.

Example:
Consider a dust particle (m ~ 10-15 kg, diameter ~ 10-6 m, v ~ 10-3 m/s).  If you can practically measure its position to an accuracy of 10-8 m, how accurately can you determine its momentum?
Δx ≈10-8 m,  Δp ≥ ħ/Δx = (10-34 Js)/(10-8 m) = 10-26 Js/m.
The momentum of the particle is on the order of p = mv = 10-18 Js/m.  There are no practical restrictions on how accurately the particle’s momentum can be measured.

Example:
Consider the Bohr model of the atom.  It is assumed that the electron’s angular momentum is quantized (pr = nħ), the electron must move in an orbit which is consistent with one of the allowed values for the angular momentum.  Does it make sense to speak of a classical trajectory or orbit?
To have a well defined trajectory we need Δp << p,   Δx << r,  Δp Δx << pr, or Δp Δx << nħ.
But the uncertainty principle requires Δp Δx ≥ ħ.  We therefore need ħ << nħ, which only holds if n >> 1.  The uncertainty principle rejects the semi-classical Bohr model with Bohr orbits except for high Rydberg states.

Problem:

Consider the Hydrogen atom, i.e. an electron in the Coulomb field of a proton.  Use the uncertainty relation to find an estimate of the ground state energy of this system.

Solution:

Problem:

Assume that virtual π mesons are emitted and absorbed by a nucleus.  From this assumption, and the π meson mass, and the uncertainty principle, estimate the range of the nuclear potential r0. 

Solution: