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So if you take the free particle without spin, which has 2 eigenvectors for each energy value greater than 0, then say the dimension is 1+2*infinity(E>0). Energies less than 0 are tossed out because they don't belong to the Hilbert space.

Now put it in a square well. The eigenvalues become discrete at energies below the wall potential. So you'd think that the space spanned by the eigenvectors of this Hamiltonian would be smaller than the free-particle case. However, now energies below 0 are admissible, but those energies are still discrete.

My question is are the spaces spanned by the eigenvectors of both cases equal in dimension?

Apologies if what I'm saying doesn't make sense (I'm self-studying with Shankar's book).