Spectrum of a Free Particle

What is the spectrum of the free-particle Hamiltonian (redefined up to a constant)

H ^ = d 2 d x 2 ? \hat{H} = - \frac{d^2}{dx^2}\:?

The set of real numbers The set of perfect squares The set of non-negative integers The set of non-negative real numbers

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1 solution

Matt DeCross
May 10, 2016

The eigenfunctions of this Hamiltonian are the plane waves e i k x e^{ikx} ; each wavenumber k k corresponds to the eigenvalue k 2 k^2 , a positive real number. There also exists exactly a zero energy solution ψ 0 1 \psi_0 \propto 1 , which has no degeneracy and eigenvalue 0 0 . Note that none of the free-particle states are normalizable, but this is not important for this question -- free particle states are only approximations in QM. The full spectrum of eigenvalues is thus the non-negative reals.

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