**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(ik_{0}x)
we have Δk = Δp = 0, g(k) = δ(k - k_{0}), 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.

ΔE ≥ ħ/Δt = (6.6*10

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)|^{2}dx
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)|^{2}dp
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)|^{2}dx
= ∫_{-}_{∞}^{∞}|Ψ(p)|^{2}dp.

Δx ≈10

The momentum of the particle is on the order of p = mv = 10

**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:

- 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 hydrogen atom.. - Details of the calculation:

The potential energy of an electron in the field of a stationary proton is

U ≈ -e^{2}/r, e^{2}= q_{e}^{2}/(4πε_{0}) in SI units.

Let us assume a spherically symmetrical wave function with mean radius r_{0}.

Then x_{ }≈ r_{0}and U ≈ -e^{2}/r_{0}. For the ground state we have E = T + U = E_{min}. The uncertainty principle requires

Δx Δp ≥ ħ, Δp ≥ ħ/r_{0}, T_{min }= (Δp)^{2}/(2m) ≥ ħ^{2}/(2mr_{0}^{2}), E_{min}≥ ħ^{2}/(2mr_{0}^{2}) - e^{2}/r_{0}.

To estimate r_{0}we let

dE_{min}/dr_{0}= -ħ^{2}/(mr_{0}^{3}) - e^{2}/r_{0}^{2}= 0, r_{0}= ħ^{2}/(me^{2})_{, }E_{min}= -me^{4}/(2ħ^{2}).

The quantitative agreement is accidental, only qualitative agreement should be expected.

Classically: U ≈ -e^{2}/r_{0}, T = e^{2}/(2r_{0}), E = T + U = -e^{2}/(2r_{0}), E_{min}= -∞ at r_{0 }= 0.

**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 r_{0}.

Solution:

- Concepts::

The uncertainty relation, ΔE Δt ≥ ħ - Reasoning:

Assume that the π meson is the mediator of the nuclear force, the nucleus emits and absorbs π mesons. To have a range r_{0}, the π meson must exists for a time interval Δt, such that r_{0 }~ vΔt.

Its speed is related to its energy E = mc^{2}/(1 - v^{2}/c^{2})^{1/2}. The uncertainty in its energy is its energy, because it either exists or does not exist. We minimize the product ΔE Δt with respect to v and set this minimum equal to ħ to find the range of the nuclear potential. - Details of the calculation:

For the π meson mc^{2}~ 140 MeV. Assume the π meson exists for a time interval Δt.

It has energy E = mc^{2}/(1 - v^{2}/c^{2})^{1/2}.

For an order of magnitude estimate we use:

(ΔE)^{2}~ m^{2}c^{4}/(1 - v^{2}/c^{2}),^{ }Δt ~ r_{0}/v,

(ΔE)^{2}(Δt)^{2}~ [m^{2}c^{4}/(1 - v^{2}/c^{2})](r_{0}/v)^{2}= m^{2}c^{6}r_{0}^{2}/[(c^{2 }- v^{2})v^{2}].

To find the maximum range r_{0}, we minimize the expression 1/[(c^{2 }- v^{2})v^{2}]) with respect to v.

(d/dv)(1/[(c^{2 }- v^{2})v^{2}]) = 0, 2/(c^{2 }- v^{2}) - 2/v^{2}= 0, (c^{2 }- v^{2}) = v^{2}, v^{2}= c^{2}/2.

Then

(ΔE)^{2}(Δt)^{2}~ ħ^{2}--> 4m^{2}c^{2}r_{0}^{2}~ ħ^{2}, r_{0}~ ħ/(2mc), r_{0}~ 0.7*10^{-15 }m.

(Rule of thumb: To estimate the range of a force, divide ħ by the mass of the particle that carries it times c, R = ħ/mc).

Even simpler argument:

ΔE ~ mc^{2}since the mediator particle either exists or does not exist.

The mediator particle can propagate a distance no larger than R = cΔt in a time interval Δt.

If we insert Δt ~ ћ/ΔE, we have R ~ ћ/mc, or m ~ ћ/Rc.