Journal of Kunming University of Science and Technology

Correctness of Coupling Theory on the Angular Momentum by the Motion of a Spinning Particle

Dai Meng-kun

Published Jan 1, 2000• 0 citations

Abstract

We discuss the motion of a particle which spin is 1/2 in a central force field in two different methods. We can get the wave function of a particle by matrix operating and vector coupling theory of the angular momentum. The results is the same, so that the coupling theory of the angular momentum is correct.

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Journal of Kunming University of Science and Technology

Correctness of Coupling Theory on the Angular Momentum by the Motion of a Spinning Particle

Dai Meng-kun

Published Jan 1, 2000• 0 citations

OpenAlex

Abstract

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{\hat{L}}_{z} = {\hbar\over{\imath}}\,(x{\partial\over{\partial y}} - y{\partial\over{\partial x}} )

is not that it is the operatorfor the

z

-component

L_{z}

of the angular momentum

{\mathbf{L}}

, but rather the expression of the operator for the angular momentum

{\mathbf{L}}

when it is aligned with the

z

-axis. Hence what we are used to note (erroneously) as

{\hat{L}}_{z}

is not a scalar but a vector operator.The same applies{\em mutatis mutandis} for the spin operators. In the correct interpretation, the wholealgebraic formalism is just the group representation theory for the rotationsof three-dimensional Euclidean geometry. It is thus mere,elementary high-school mathematics (in a less usual, more technical guise) and as such totally exempt of any physics, let alone quantum mysteries.The change of interpretation has no impact on the algebraic results, such that they remain in agreement with experimental data and the algebra quantum mechanics remains correct. It is all only a matter of the correct geometrical meaning of the algebra.All these statements are proved within the framework of the group representation theory for SO(3) and SU(2) which is the basic tool used to describe rotational motion in quantum mechanics.

\frac{S}{{r}^{2}}

, although the solutions for negative energy for $0\ensuremath{\geqq}S\ensuremath{\geqq}\frac{\ensuremath{-}{h}^{2}}{32{\ensuremath{\pi}}^{2}\ensuremath{\mu}}$ satisfy the usual boundary conditions, they can not be admitted because the Hamiltonian is not Hermitian in these solutions. This corresponds to taking

{(l+\frac{1}{2})}^{2}

in place of

l(l+1)

as the analogue of the square of the classical angular momentum. If we do this, we get a complete analogy between the classical and quantum mechanically allowed solutions, with no quantization. The solutions involve Bessel functions of both real and imaginary orders with both real and imaginary arguments.