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Probability Distribution of a Physical System Operating in a Potential Field in 2D Space |
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Consider a physical with a Hamiltonian function H(p,z), where z is the polar coordinates (r, θ) for a location for the system and p is its momentum vector. The Hamiltonian is composed of the Kinetic energy function K(p) and the Potential energy function V(z); i.e.;
The Hamiltonian function H(p, z) is converted into the Hamiltonian operator H^(∇², z)
by a set of substitutions;
i.e., p → (h/i)∇ and z→ z.
The time independent Schrödinger equation is then
where E is the total energy of the system and ψ(z) is its wave function. The squared wave function |ψ(z)|² is the probability density function.
The Kinetic energy function K(p) has to have certain properties. If there is no momentum there is no kinetic energy; i.e., K(0)=0. Also K(−p)=K(p). This means that the Maclaurin series for the Kinetic energy operator is of the form
The Potential energy function may also be assumed to satisfy the conditions V(0)=0 and V(−z)=V(z).
The crucial relationship is the one between Kinetic energy and velocities. Under nonrelavistic conditions for a particle of mass m_{0}
Therefore
The time-spent probability density is proportional to 1/|v| and hence inversely proportional to (K(z))^{½}.
On the other hand, under relativistic conditions, mass is a function of relative speed; i.e.,
where β=v/c.
Relativistic kinetic energy is given by
Let K/(m_{0}c²) be denoted as α.The above relationship is more conveniently expressed as
Thus
As α→0 the fraction [(α + 1)/(α + 2)^{½}] goes to 1/√2. This means the probability density asymptotically becomes inversely proportional to α^{½}, Since α is K/(m_{0}c²) probability density, as K→0, is asymptoticallyn inversely proportional to K^{½}, just as for the nonrelativistic case as it should be.
For the case as K and α→∞ note that
So, as α→∞, (1/|v|)→(1/c); i.e. |v|→c.
(To be continued.)
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