The Heisenberg uncertainty
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 =
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.
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
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
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
accurately can you determine its momentum?
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.
no practical restrictions on how accurately the particles momentum can be measured.
Consider the Bohr model of the atom. It is assumed that the electrons 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 <<
But the uncertainty principle requires Δp Δx ≥ ħ.
We therefore need ħ << nħ, which only holds if n >> 1.
uncertainty principle rejects the semi-classical Bohr model with Bohr orbits except for
high Rydberg states.
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.
The uncertainty principle
We are asked to use the uncertainty relation, Δx Δp
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
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
Classically: U ≈ -e2/r0, T = e2/(2r0),
E = T + U = -e2/(2r0), Emin = -∞ at r0
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.
The uncertainty relation, ΔE Δt ≥ ħ
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
Its speed is related to its energy E = mc2/(1 - v2/c2)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 mc2 ~ 140 MeV. Assume the π meson
exists for a time interval Δt.
It has energy E = mc2/(1 - v2/c2)1/2.
For an order of magnitude estimate we use:
(ΔE)2 ~ m2c4/(1
- v2/c2), Δt
(ΔE)2(Δt)2 ~ [m2c4/(1
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.
(ΔE)2(Δt)2 ~ ħ2
~ 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.