# Self-orthogonal states in Quantum Theory, 2

This is my second try on the same question -- see Self-orthogonal states in Quantum Theory. Suppose one considers a hypothetical "non-classical Quantum Theory": we work in a Hilbert-like space $k^n$ ($k$ some field) with Hermitian form $\langle \cdot \vert\cdot \rangle$, which is not necessarily an inner product, etc. The nonzero vectors in $k^n$ correspond to states. What would be the physical meaning of states $\vert \psi \rangle$ for which $\langle \psi \vert \psi \rangle = 0$, that is, self-orthogonal states ?

• This would imply that, given a system in state $|\psi\rangle$, there is no possibility that it is in the state $|\psi\rangle$. By allowing this type of product you, quite clearly, lose whatever physical meaning quantum mechanics has. – By Symmetry Jun 1 '17 at 19:02
• Related answers – By Symmetry Jun 1 '17 at 19:04
• In a physical theory, there are no such states - why do you expect them to have a physical meaning? (You might be coming from Gupta-Bleuler quantisation of QED, here we have to use virtual self-orthogonal photon states in our calculation. However, the actual Hilbert space is a quotient space where these states are removed.) – Noiralef Jun 1 '17 at 19:04
• @ Noiralef: How is this quotient space formally defined ? – THC Jun 7 '17 at 15:56
• @BySymmetry: so as I understand it, there is no possibility that we will find the system in the same state $\vert \psi \rangle$ after some measurement. Couldn't such states have some "singular meaning" ? – THC Jun 7 '17 at 16:10