On a ring of circumference

PHYSICS · Multi-part · Fermi University Physics · Mathematical Methods for Physics · Quantum Mechanics · The formalism of quantum mechanics

The Structure of Quantum Mechanics, Ch Observables, spectra and the spectral theorem

Problem

On a ring of circumference L=2.0 nmL=2.0\ \mathrm{nm}, the momentum operator p^=−iℏ d/dx\hat p=-i\hbar\,d/dx is made self-adjoint by imposing ψ(L)=eiθψ(0)\psi(L)=e^{i\theta}\psi(0), giving eigenvalues pn(θ)=(2πn+θ)ℏL,n∈Z.p_n(\theta) = \frac{(2\pi n+\theta)\hbar}{L}, \qquad n\in\mathbb{Z}. Consider the two self-adjoint momentum operators built this way with θ=0\theta=0 and θ=π\theta=\pi. Someone argues: "Both are self-adjoint versions of 'the' momentum operator, built from the same differential expression on the same ring, so they must have exactly the same set of allowed values; the boundary phase is just a relabelling of the same eigenvectors." ((a)) Compute the two lowest non-negative eigenvalues (n=0,1n=0,1) for each choice of θ\theta, in kg m s−1\mathrm{kg\,m\,s^{-1}}. ((b)) Decide whether the two spectra share any common value, and hence whether the claim is correct.

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