The maximum # of mutually orthonormal eigenvectors of a Hermitian operator must be equal to 2*(number of coordinate basis vectors), where the 2 is for say spin 1/2. Now when solving the eigenvalue problem for the Hamiltonian operator, one of the boundary conditions is that the vector obtained belongs to the Hilbert space (normalizable to unity or Dirac delta). So we discard all the vectors which don't belong to the Hilbert space. However, because of this, the space spanned by our energy eigenvectors is now smaller than 2*(number of coordinate basis vectors), so our world evidently has shrunk - some combinations of position and spin cannot be obtained with linear combinations of the energy eigenvectors that we kept.(adsbygoogle = window.adsbygoogle || []).push({});

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).

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# Maximum # of mutually orthonormal eigenvectors

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