__The uncertainty principle, ΔxΔp > ħ__

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 V = ½mω^{2}(Δx)^{2}.

The average total energy is E = T + V ~ 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. The ground state energy of the harmonic oscillator is on the order of ħω.

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.

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)^{½ }= 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.

Before the neutron was discovered one model assumed the
atomic nucleus to be made of protons and electrons. Show that the observation
that the characteristic size of a nucleus is several times 10^{-15} m
and that the average binding energy of a particle in the nucleus is less than 10
MeV makes this model inconsistent with basic principles of quantum mechanics.

Solution:

- Concepts:

The uncertainty principle: Δp_{x}Δx ~ ħ. - Reasoning:

Confinement of the electron leads to a large uncertainty in energy. - Details of the calculation:

Δp_{x}Δx ~ ħ, Δx ~10^{-15}m for a neutron, Δp ~ 10^{-19}kgm/s.

Δpc ~ 3*10^{-11}J ~ 200 MeV.

For the electron E_{kin}(min) ~ Δpc ~200 MeV.

The the potential well that confines the electron must be deeper than 200 MeV.

But it takes less than 200 MeV to take a small nucleus completely apart.

For a quantum mechanical point particle in a 1-dimensional harmonic potential
U(x) = ½mω^{2}x^{2}

(a) find the minimum (or “zero-point”) energy using the lower limit of
Heisenberg's uncertainty principle, ΔxΔp ≥ ħ/2.

(b) For this zero-point energy of the particle, find the probability
distribution by solving the time-independent Schroedinger equation for ψ(x).

Solution:

- Concepts:

The uncertainty principle, the Schroedinger equation - Reasoning:

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

Then the uncertainty in its momentum is Δp = ħ/(2Δx) 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 + V ~ ħ^{2}/(8m(Δx)^{2}) + ½mω^{2}(Δx)^{2}.

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

We then solve the second-order differential equation

∂^{2}ψ(x)/∂x^{2}- (m^{2}ω^{2}x^{2}/ħ^{2})ψ(x) + (2mE_{min}/ħ^{2})ψ(x) = 0 for ψ(x). - Details of the calculation:

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

Inserting (Δx)^{2}= ħ/(2mω) into the equation for E yields E = ħω/2. The ground state energy of the harmonic oscillator is on the order of ħω.

(b) Time-independent Schroedinger equation:

∂^{2}ψ(x)/∂x^{2}- (m^{2}ω^{2}/ ħ^{2})x^{2}ψ(x) + (mω/ħ)ψ(x) =0

Try a solution ψ(x) = Nexp(-a^{2}x^{2}). Then

4a^{4}x^{2}– 2a^{2 }= (m^{2}ω^{2}/ ħ^{2}) - (mω/ħ), or a^{2}= mω/(2ħ)

The corresponding probability distribution is P(x) = N^{2}exp(-mωx^{2}/ħ), with N^{2}determined using

∫_{-∞}^{∞}P(x,t)dx = 1. N^{2}= (mω/πħ)^{½}.

P(x) = (mω/πħ)^{½}exp(-mωx^{2}/ħ).

Consider thermal neutrons in equilibrium
at temperature T = 300 K.

(a) Calculate its deBroglie wavelength. State
whether a beam of these
neutrons could be diffracted by a crystal, and why?

(b) Use Heisenberg's Uncertainty principle to
estimate the kinetic energy (in MeV) of a
nucleon bound within a nucleus of radius 10^{−15}
m.

__The uncertainty relation, ΔE Δt > ħ__

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})^{½}. 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})^{½}.

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.

The transition rate of electrons from the first excited state of the hydrogen
atom to the ground state is ~10^{8}/s. What is the minimum range of
energies of the resulting photons that are emitted?

Solution:

- Concepts:

The uncertainty relation, ΔE Δt ≥ ħ - Reasoning:

The excited states only exists for a finite amount of time, Δt ~ 10^{-8 }s. Therefore its energy cannot be exactly known. - Details of the calculation:

Uncertainty principle: ΔEΔt ≥ ħ/2, ΔE ≥ ħ/(2Δt), Δt ~ 10^{-8 }s, ΔE ~ 3.3*10^{-8}eV.

E = 13.6 eV * ¾ = 10.2 eV.

The minimum range of energies is ~ 3.3*10^{-8}eV around 10.2 eV.

Assume a particle with a mass of 10^{5} MeV is the carrier of some
interaction. Estimate the range of this interaction.

Solution:

- Concepts:

The uncertainty principle ∆E∆t ~ ħ - Reasoning:

For the force carrier particle we have ∆E ~ mc^{2}since the particle either exists or does not exist. The virtual particle can propagate a distance no larger than R = c∆t in a time interval ∆t. If we insert ∆t ~ ħ/∆E from above, we have R ~ ħ/mc, - Details of the calculation:

R ~ ħ/(mc) = hc/(2πmc^{2}) = (1240 eV nm)/(2π*10^{11}eV) = ~ 2*10^{-9 }nm = 2*10^{-18}m.

__Generalized uncertainty principle__

Let A and B be two observables (Hermitian operators). In any state of the
system

ΔAΔB ≥ ½|<i[A,B]>|.

(a) Prove this generalized uncertainty principle.

[Hint: Let |ψ> be any state vector and let A_{1 }= A - <A>I and B_{1
}= B - <B>I.

Let |Φ> = A_{1}|ψ> + ixB_{1}|ψ> with x an arbitrary real
number. Use <Φ|Φ> ≥ 0.]

Now consider a single particle in an eigenstate of L^{2} with wave
function Ψ(**r**,t).

(b) Calculate the commutators [sinφ, L_{z}] and [cosφ, L_{z}],
where φ is the azimuthal angle.

(c) Use these commutation relations and the result from part (a) to obtain
uncertainty relations between sinφ, L_{z} and cosφ, L_{z}.

Note: You can complete parts (b) and (c) without completing part (a).