What is the spectrum of the free-particle Hamiltonian (redefined up to a constant)
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
The eigenfunctions of this Hamiltonian are the plane waves e i k x ; each wavenumber k corresponds to the eigenvalue k 2 , a positive real number. There also exists exactly a zero energy solution ψ 0 ∝ 1 , which has no degeneracy and eigenvalue 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.