﻿ The Quantum Analysis of a Free Electron
San José State University

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 The Quantum Analysis of a Free Electron

## Background

I am working on a project to numerically solve the Schroedinger equation for an electron in a hydrogen atom as an alternative to the conventional solutions based upon the separation-of-variables assumption. For that I needed some initial condition. I considered using what is called the plane-wave solution for a free electron, an electron in a space without any external field. The plane-wave solution has the wave function of Schroedinger's equation consisting of a sinusoidal function that extends across an infinite space with the perpendicular to the planes being the direction of motion of the electron. This solution in conventional quantum analysis always puzzled me because it seemed to mean that an electron, or at least its probability density function, was spread over all of space and thus the electron could quantum jump from one anywhere to any other anywhere in space. This seems improbable to say the least.

While the plane wave solution doesn't make any sense in the conventional Copenhagen Interpretation it makes perfect sense when the probability density function from Schroedinger's equation is interpreted as the time-spent probability density function. A free electron travels across space at a constant velocity which in quantum terms is a pattern of fast-slow-fast-slow movement, what Schroedinger called "zitterbewegung" (trembling motion). Its path extends from the infinite past to the infinite future so the extension of the PDF over an infinite range of space is no puzzle. However it also means that the plane wave cannot be taken as an initial condition for the trajectory of a particle. Nevertheless this reinterpretation appears to be another step forward in properly reinterpreting quantum anaysis.

## The Quantum Analysis of a Free Electron

The Hamiltonian function for a free electron is

#### H = p²/(2m)

where p is the mometum of the electron and m is its mass.

If a Cartesian coordinate system is used with the x-axis coinciding with the momentum vector then the Hamiltonian operator H^ for a free electron is

#### H^ = −(h²/(2m))(∂²/∂x²)

The time-independent Schroedinger equation for the free electron is then

#### −(h²/(2m))(∂²Φ/∂x²) = EΦ or, equivalently (∂²Φ/∂x²) = −(2mE/h²)Φ

There are no momenta in the directions of the y-axis or the z-axis so no second derivatives with respect to y or z appear in in the Hamiltonian operator. Likewise since the potential function is zero everywhere there are no y or z variables in the Hamiltonian operator.

The above equation has the solution

#### Φ(x) = A·cos(k(x−x0) for −∞<x<+∞ and y=0 and z=0

where k=(2mE/h²) and A and x0 are constants.

The probability density function P(x) is then formally

#### P(x, y, z) = |Φ(x)|² = A²·cos²(k(x−x0) if y=0 and z=0 P(x, y, z) = 0 y≠0 or z≠0

The normalization of P(x) would result in it being zero everywhere. The function cos²(k(x−x0) could be called the relative probability density function.

The conventional treatment of the plane wave solution makes the probability density at any point orthogonal to the momentum vector equal to that on the momentum vector axis. Obviously this is not sensible, but if it is not true then the solution is not a plane wave. The width of the nonzero probability density function (PDF) is unresolved but a reasonable speculation is that is equal to the width of the particle. Let ρ be the radius of the particle. Then the relative density function for a free particle would be

#### P(x, y, z) = |Φ(x)|² = A²·cos²(k(x−x0) if y≤ρ and z≤ρ P(x, y, z) = 0 if y>ρ or z>ρ .

This is more in the nature of a stacked disks solution.

The treatment of P(x) as an intrinsic probability density function under the Copenhagen Interpretation of quantum theory is obviously problematical. But as a time-spent probability density function there is no problem. The velocity function for the electron is just

#### v(x) = B/P(x) = B/cos²(k(x−x0)

where B is a constant. This is just the fast-slow pattern of quantum motion.

As E increases without bound the fluctuations in v(x) become more rapid and hence the spatial average of velocity reduces to a constant as in classical analysis.

(To be continued.)