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

P_X(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.,

P_S(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(x_m) = E

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

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.,

½mv_m² = E
and thus
v_m = (2E/m)^½

The minimum velocity is −v_m.

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

Var_Z = ∫ (z−E{z})²P_Z(z)dz

Where E{z}=0 this reduces to

∫z²P_Z(z)dz

Therefore the quantities sought are

Var_X = ∫_−xm^x_m(x²P_X(x))dx
Var_S = ∫_−vm^v_m(v²P_S(v))dv

Because of the symmetry these reduce to

Var_X = 2∫₀^x_m(x²P_X(x))dx
and
Var_S = 2∫₀^v_m(v²P_S(v))dv

From the formulas for P_X and P_S these further reduce to

Var_X = 2∫₀^x_m[(x²/(T|v(x))|]dx
and
Var_S = 2∫₀^v_m[(v²/(T|a(x))|]dv

Consider first Var_S. Since

½mv_m = E
v_m = (2E/m)^½
and hence
p_m =mv_m = (2Em)^½

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

Var_S = 2∫₀^v_m[v²/(T|(dv/dt)|)]dv
= 2∫₀^x_m[v²/(T|(dv/dx)(dx/dt)|)](dv/dx)dx
= 2∫₀^v_m[v²/(T|v|]dx = 2∫₀^x_m(v(x)/T)dx )

The crucial quantity for the Uncertainty Relation is

Var_X Var_P = [2∫₀^x_m[(x²/T|v(x))|]dx] m²[2∫₀^x_m[v(x)/T]dx]
= (4m²/T²)[∫₀^x_m[(x²/|v(x))|]dx]∫₀^x_m[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/v^½) and g(x)=v^½. Then from the Schwartz Inequality

Var_X Var_P ≥ (4m²/T²)[∫₀^x_mxdx]²
and thus
Var_X Var_P ≥ (4m²/T²)[½x_m²]² = (m/T)²[x_m⁴]

Therefore

σ_Xσ_P≥ (m/T)x_m²

This is a general relationship.

Consider the condition determining the maximum deviation from equilibrium, x_m

V(x_m) = E

Because of the symmetry conditions imposed V(x) must have the form

V(x) = ½kx²+ λx⁴ + μx⁶ + …

The ½ is included in the first term to make comparisons with the case of a harmonic oscillator more convenient.

Thus the equation determining x_m is

½kx_m²+ λx_m⁴ + μx_m⁶ + … = E

For small values of x_m the terms involving the higher powers of x_m are insignificant compared with the first term½ kx_m². Therefore as E→0 the value of x_m asymptotically approaches the solution to

½kx_m² = E
namely
x_m = (2E/k)^½

This means that

(m/T)x_m² = (2/T)(E/(k/m))
and, since (1/T)=ω/(2π)
and k/m=ω²
(m/T)x_m² = (ω/π)(E/ω²)
(m/T)x_m² = = (1/π)(E/ω)

The smallest level of energy is hω, where h is Planck's constant. Thus (E/ω)=h. Therefore

σ_Xσ_P≥ (m/T)x_m² = h/π = 4(h/4π)

Thus the time-spent probability density distributions for a particle moving in a potential field of V(x) 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.)