San José State University
Thayer Watkins
Silicon Valley
& Tornado Alley

The Satisfaction of the Uncertainty Principle
by a Physical System More General
than a Harmonic Oscillator

Time-Spent Probability Distributions

For a particle undergoing periodic motion the probability of finding it in an interval dx at a randomly chosen time is proportional to the time it spends in that interval in its periodic motion. That time dt is equal to dx/|v(x)|, where v(x) is the particles velocity at location x.

The probability density function for the particle's location is therefore

PX(x) = 1/(T|v(x)|

where T=2∫dx/|v(x)| is the time period of the motion. The factor of 2 arises from the particle traveling from a minimum to a maximum and then back down to the minimum.

There is also a probability density function for the particle's velocity ; i.e.,

PS(v) = 1/(T|a(v)|

where a(v) is the acceleration of the particle when its velocity is v. The time period for the period of the velocity is the same as the time period of the motion.

The General Problem

The particle is presumed to be moving in a potential field given by V(x). The potential is presumed to be such that V(0)=0 and V(−x)=V(x).

The energy E of the particle is given by

E = ½mv² + V(x)

where m is the mass of the particle.

The force F on the particle at location x is given by

F(x) = −(∂V/∂x)

Hence the acceleration of the particle at x is

a(x) = F(x)/m = −(∂V/∂x)/m

Limits of the Variables

The limits of x are where all of the energy is potential and none of it is kinetic;, i.e.,

V(xm) = E

The positive solution for this condition is the maximum value of and the minimum value for x is −xm.

Likewise there is a maximum value for velocity. It occurs where all of the energy is kinetic and none of it is potential; i.e.,

½mvm² = E
and thus
vm = (2E/m)½

The minimum velocity is −vm.

The Mean Values of the Variables

The symmetry of V(x) is such that the mean or expected value of x is zero. Likewise the expected value of velocity is zero.

The Variances of the Variables

The variance of a variable z is defined as

VarZ = ∫ (z−E{z})²PZ(z)dz

Where E{z}=0 this reduces to


Therefore the quantities sought are

VarX = ∫−xmxm(x²PX(x))dx
VarS = ∫−vmvm(v²PS(v))dv

Because of the symmetry these reduce to

VarX = 2∫0xm(x²PX(x))dx
VarS = 2∫0vm(v²PS(v))dv

From the formulas for PX and PS these further reduce to

VarX = 2∫0xm[(x²/(T|v(x))|]dx
VarS = 2∫0vm[(v²/(T|a(x))|]dv

Consider first VarS. Since

½mvm = E
vm = (2E/m)½
and hence
pm =mvm = (2Em)½

In the formula for VarS consider a change in the integration variable from v to x; i.e.,

VarS = 2∫0vm[v²/(T|(dv/dt)|)]dv
= 2∫0xm[v²/(T|(dv/dx)(dx/dt)|)](dv/dx)dx
= 2∫0vm[v²/(T|v|]dx = 2∫0xm(v(x)/T)dx )

The crucial quantity for the Uncertainty Relation is

VarX VarP = [2∫0xm[(x²/T|v(x))|]dx] m²[2∫0xm[v(x)/T]dx]
= (4m²/T²)[∫0xm[(x²/|v(x))|]dx]∫0xm[v(x)dx]

The Schwartz Inequality

Let f and g be two complex functions over the variable x. The Schwartz Inequality is then

[∫|f|²dx]·[∫|g|²dx] ≥ |∫fgdx|²

In the Schwartz Inequality let f(x)=(x/v½) and g(x)=v½. Then from the Schwartz Inequality

VarX VarP ≥ (4m²/T²)[∫0xmxdx]²
and thus
VarX VarP ≥ (4m²/T²)[½xm²]² = (m/T)²[xm4]


σXσP≥ (m/T)xm²

This is a general relationship.

The Harmonic Oscillator

Consider the special case of a harmonic oscillator, where V(x)=½kx². For this case v(x)=(2/m)½(E−½kx²)½. The term ½k may be factored out to give

v(x) = (k/m)½(2E/k−x²)½

But (2E/k) is equal to xm² so

v(x) = (k/m)½( xm²−x²)½


T = 2∫−xmxm(m/k)½dx/( xm²−x²)½

A change in the variable of integration to sin(θ)=x/xm results in

T = 2π(m/k)½

The oscillation frequency ω for the oscillator is (k/m)½. Therefore the above relation is

T = 2π/ω

Thus for this case

σXσP≥ (mω/2π)(2E/k)
which reduces to
σXσP≥ ω(E/(k/m)/π = ω(E/(ω²π)
which reduces to
σXσP≥ E/(ωπ)

The minimum energy for the system is hω, where h is Planck's constant. Thus E/ω=h and hence

σXσP≥ h/π = 4(h/4π)

This is four times the value of h/(4π) required to satisfy the Uncertainty Principle. Thus a harmonic oscillator satisfies Uncertainty Principle.


More generally

V(xm) = E
and hence
xm = V−1(E)
xm² = [V−1(E)]²

Here the inverse is for only the right half of V(x); i.e., for x≥0.


V(x) = ½kx²+ λx4

This is equivalent to a quadratic equation in x²; i.e.,

λx4 + ½kx² − V = 0

For λ=0 this is the case of a harmonic oscillator.

The inverses of V(x) are then

x² = (−½k ± (¼k² + λV)½)/(2λ)
and therefore,
taking the positive root
xm² = (−½k + (¼k² + 4λE)½)/(2λ)
or, equivalently
xm² = (−½k + ½k(1 + 16λE/k²)½)/(2λ)

For small values of z,

(1+z)½ ≅ 1 + z/2

Therefore for small values of E

xm² ≅ (½k(16λE/(2k²))/(2λ)
which reduces to
xm² ≅ (16λE/(4k))/(2λ) = 2E/k


(m/T)xm² ≅ (1/T)2E/(k/m))
but (k/m)=ω²
and 1/T=ω/(2π) so
(m/T)xm² ≅ (E/ω)/π

However the minimum energy is hω where h is Planck's constant so (E/ω)=h. Therefore

σXσP≥ (m/T)xm² ≅ h/π = 4(h/4π)

Thus, as in the case of a harmonic oscillator and regardless of the value of λ, the time-spent probability density distributions for a particle moving in a potential field of V(x)=½kx²+ λx4 with sufficiently small energies satisfy the Uncertainty Principle. There is no problem of systems with large energies satisfying the Uncertainty Principle. Therefore the Uncertainty Principle has no implication of the immateriality of a particle at the quantum level.

(To be continued.)

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