Take a Hilbert space comprising of all square-integrable functions.
If we demand that each permissible function must have a decomposition in the Hamiltonian eigenbasis, does this ban some functions that were previously allowed in this Hilbert space?
I'm asking this because the Hamiltonian eigenbasis is usually discrete. I'm not convinced that every square-integrable function is producible by combining the eigenvectors linearly. It's like we have "too few" eigenvectors.
Take another situation where we discretise the eigenvectors. If we demand that each wave-function must have a discrete Fourier series expansion, then we ban non-periodic wave-functions.
I was also thinking the same about other operators like Spin, Angular momentum, etc. For quantum mechanics to work, we demand that each allowed wave-function must be expressible as a linear combination of the eigenvectors of each of these operators.
For instance, for particles with spin, is every square-integrable wave-function of the form $\psi (x,y,z,s)$ allowed?
Best Answer
It does not imply any restriction on the admissible vectors in the Hilbert space. Every vector in a Hilbert space is a (generally infinite) superposition of eigenvectors of a selfadjoint operator with pure point spectrum. If a Hamiltonian, or every other operator, like a component of the angular momentum, has pure point spectrum this is the case.
Yes, it is allowed. Actually you need $2S+1$ such wavefunctions if the spin is $S$ and the integer $s$ varies from $-S$ to $S$.