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The Classical and Quantum Mechanics of a Thin Ring Spinning About Two Axes |
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Consider a thin ring of mass M and radius R. (Here thin means that it is a line.) The linear mass density of the ring is ρ=M/(2πR). The moment of inertia I_{1} for spinning about an axis perpendicular to the plane of the ring which passes through the ring center is MR². There is another spin of a ring which is like the flipping of a coin. This is a rotation about an axis which is a diameter of the ring. The moment of inertia for this type of spinning is given by
where the integration is from 0 to 2π. The integral of cos²(θ) from 0 to π is ½ so the integral from 0 to 2π is 1 and hence
This flipping rotation is of particular interest for the structure of nuclei. Experimental measurement indicate that that nuclei have at least approximately spherical shapes. If a circular band of nucleons rotates in this flipping fashion the dynamic appearance of the nucleus would be that of a sphere.
A spin can also take place perpendicular to the diameter considered above. The moment of inertia about that axis is same as that of I_{2}. It unnecessarily complicates the analysis and will not be included now.
The angular momentum of the ring spinning at an angular rate of ω about the axis perpendicular to its plane is L=MR(Rω)=MR²ω. This means that
For the spin about a diameter the angular momentum Λ is found as follows.
Thus
The kinetic energy of the spinning ring is then
and there is no potential energy. This means that the ring will continue spinning at the rates ω and Ω indefinitely. These rates can have any real values. This is the classical solution.
To get the quantum mechanical solution the energy is expressed as
The momentum associated with θ is (∂E/∂θ) which is I_{1}(dθ/dt). Let this be denoted as p_{θ}. Thus (dθ/dt)=p_{θ}/I_{1}. Likewise the momentum associated with φ is p_{φ}=(dφ/dφ)/I_{2}. Therefore the Hamiltonian function for the the spinning ring is
and the Hamiltonian operator is
If it is assumed that the wave function ψ is of the form Θ(θ)Φ(φ) then above equation becomes
This equation can be expressed as
The LHS is independent of φ and the RHS independent of θ. Therefore their common value must be a constant, say k². This means that
The solution is
where A and θ_{0} are constants. By proper choice of the coordinate system θ_{0} can be made equal to zero. Then k(2π) must be an integral multiple of 2π. Therefore k must be an integer.
The other equation is
From the previous case it is found that the coefficient of Φ must be a squared integer; i.e.,
Thus the energy of the spinning ring is quantized such that it is equal to a multiple of of the sum of squared integers. The wave function then has the form
This means that the probability density function P(θ, φ) is given by
When a second rotation axis is included the quantization condition for energy involves the sum of the squares of three integers.
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