### The uncertainty principle

The uncertainty principle, ΔxΔp > ħ

#### Problem:

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.

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 = ½mω2(Δx)2.
The average total energy is E = T + V ~ ħ2/(2m(Δx)2) + ½mω2(Δx)2.
The ground state has the lowest energy, so let us minimize E with respect to Δx.
dE/d(Δx) = -ħ2/(m(Δx)3) + mω2Δx = 0,  (Δx)4 = ħ2/(m2ω2)
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.

#### 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 ≈ -e2/r,  e2 = qe2/(4πε0) in SI units.
Let us assume a spherically symmetrical wave function with mean radius r0.
Then x ≈ r0 and U ≈ -e2/r0.  For the ground state we have E = T + U = Emin.  The uncertainty principle requires
Δx Δp ≥ ħ,  Δp ≥ ħ/r0,  Tmin = (Δp)2/(2m) ≥ ħ2/(2mr02),  Emin ≥ ħ2/(2mr02) - e2/r0.
To estimate r0 we let
dEmin/dr0 = -ħ2/(mr03) - e2/r02 = 0,  r0 = ħ2/(me2)Emin = -me4/(2ħ2).
The quantitative agreement is accidental, only qualitative agreement should be expected.
Classically: U ≈ -e2/r0,  T = e2/(2r0),  E = T + U = -e2/(2r0),  Emin = -∞  at r0 = 0.

#### Problem:

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:
• 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*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.

#### Problem:

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: ΔpxΔx ~ ħ.
• Reasoning:
Confinement of the electron leads to a large uncertainty in energy.
• Details of the calculation:
ΔpxΔx ~ ħ,  Δx ~10-15 m for a neutron,  Δp ~ 10-19 kgm/s.
Δpc ~ 3*10-11J ~ 200 MeV.
For the electron Ekin(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.

#### Problem:

For a quantum mechanical point particle in a 1-dimensional harmonic potential U(x) = ½mω2x2
(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 = ½mω2(Δx)2.
The average total energy is E = T + V ~ ħ2/(8m(Δx)2) + ½mω2(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
2ψ(x)/∂x2 - (m2ω2x22)ψ(x)  + (2mEmin2)ψ(x) = 0 for ψ(x).
• Details of the calculation:
(a)  dE/d(Δx) = -ħ2/(4m(Δx)3) + mω2Δx = 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:
2ψ(x)/∂x2 - (m2ω2/ ħ2)x2ψ(x)  + (mω/ħ)ψ(x) =0
Try a solution ψ(x) = Nexp(-a2x2).  Then
4a4x2 - 2a2 = (m2ω2/ ħ2) - (mω/ħ), or a2 = mω/(2ħ)
The corresponding probability distribution is P(x) = N2exp(-mωx2/ħ), with N2 determined using
-∞P(x,t)dx = 1.  N2 = (mω/πħ)½.
P(x) = (mω/πħ)½exp(-mωx2/ħ).

#### Problem:

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

#### 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:

• 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 r0, the π meson must exists for a time interval Δt, such that r0 ~ vΔt.
Its speed is related to its energy E = mc2/(1 - v2/c2)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 mc2 ~ 140 MeV.  Assume the π meson exists for a time interval Δt.
It has energy E = mc2/(1 - v2/c2)½.
For an order of magnitude estimate we use:
(ΔE)2 ~ m2c4/(1 - v2/c2),  Δt ~ r0/v,
(ΔE)2(Δt)2 ~ [m2c4/(1 - v2/c2)](r0/v)2 = m2c6r02/[(c2 - v2)v2].
To find the maximum range r0, we minimize the expression 1/[(c2 - v2)v2]) with respect to v.
(d/dv)(1/[(c2 - v2)v2]) = 0,  2/(c2 - v2) - 2/v2 = 0,  (c2 - v2) = v2,  v2 = c2/2.
Then
(ΔE)2(Δt)2 ~ ħ2  -->  4m2c2r02 ~ ħ2,  r0 ~ ħ/(2mc),  r0 ~ 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 ~ mc2 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.

#### Problem:

The transition rate of electrons from the first excited state of the hydrogen atom to the ground state is ~108/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.

#### Problem:

Assume a particle with a mass of 105 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 ~ mc2 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/(2πmc2) = (1240 eV nm)/(2π*1011 eV) =  ~ 2*10-9 nm = 2*10-18 m.

Generalized uncertainty principle

#### Problem:

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 A1 = A - <A>I and B1 = B - <B>I.
Let |Φ> = A1|ψ> + ixB1|ψ> with x an arbitrary real number.  Use <Φ|Φ>  ≥  0.]

Now consider a single particle in an eigenstate of L2 with wave function Ψ(r,t).
(b)  Calculate the commutators [sinφ, Lz] and [cosφ, Lz], where φ is the azimuthal angle.
(c)  Use these commutation relations and the result from part (a) to obtain uncertainty relations between sinφ, Lz and cosφ, Lz.
Note:  You can complete parts (b) and (c) without completing part (a).