# Infinite vs Finite dimensional Hilbert space

Let us consider an electron in an infinite square well. As we know that the electron has a spin=$$1/2$$ . The spin operator ($$z$$-direction) has two eigenvectors which span the vector space. But if we solve for the eigenvectors of Hamiltonian, We get infinite basis vectors which span the space. But in linear algebra, we cannot have two basis sets with different number of elements.

• Have you considered that you are looking at two different vector spaces? – Aaron Stevens Feb 10 at 4:47
• I might be missing something elementary, but can't we write any vector as a linear combination of spin eigenstates, even if they are energy eigenstates, meaning that we can produce any vector which is produced by energy eigenstates. – Jay Feb 10 at 5:11
• The full state space can be decomposed in two infinite dimensional orthogonal factors, one containing all spin 1/2 states, and one with all spin -1/2 states. – doetoe Feb 10 at 6:54
• In other words, there are two eigenvalues, but each with an infinite dimensional eigenspace – doetoe Feb 10 at 6:57
• You can't specify the state of the particle just with the energy OR the spin, you need both. If you're considering the 1D square well, you will have degeneracy of two states for each energy eigenvalue (the same energy for both values of spin) and if you think about the degeneracy for the spin eigenvalues that should be (countable) infinity. What you should be looking for is a Complete Set of Commuting Observables, just like the $n, l, m$ and $s$ quantum numbers in atomic states. en.wikipedia.org/wiki/Complete_set_of_commuting_observables – S V Feb 10 at 7:42

The spin operator $$S_z$$ has two eigenvalues, and its eigenvectors span the whole state space, but that doesn't mean it has two eigenvectors.
• @Jay This total space can be realized as the tensor product of the spin state space and the position state space. The spin eigenspaces can be written as $\mathbb{C}|\uparrow\rangle\otimes\mathcal H$ and $\mathbb{C}|\downarrow\rangle\otimes\mathcal H$. Each position eigenfunction (whatever that exactly is) when tensored with a spin eigenvector, is simultaneously an eigenvector of spin and position. – doetoe Feb 10 at 17:59