Suppose one considers "classical Quantum Theory": we work in a Hilbert space $\mathbb{C}^n$ with standard inner product $\langle \cdot \vert\cdot \rangle$, etc. The nonzero vectors in $\mathbb{C}^n$ correspond to states. What is the physical meaning of states $\vert \psi \rangle$ for which $\langle \psi \vert \psi \rangle = 0$, that is, self-orthogonal states ?

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    $\begingroup$ In a "Hilbert space with standard inner product", there are no self-orthogonal states by definition. $\endgroup$ – Noiralef May 23 '17 at 13:00
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    $\begingroup$ The only "self-orthogonal" vector in a vector space endowed with a sesquilinear form that is non-degenerate is the zero vector. Any Hilbert space is by definition a inner product space with a non-degenerate sesquilinear form. $\endgroup$ – yuggib May 23 '17 at 13:38
  • $\begingroup$ To make an additional clarification, the zero vector in a Hilbert space is not the vacuum vector, but the unphysical vector representing the trivial state "with zero probability". $\endgroup$ – yuggib May 23 '17 at 13:42
  • $\begingroup$ @yuggib: a sesquilinear form can have nontrivial self-orthogonal vectors, in general. I guess you mean a sesquilinear form which is at the same time an inner product ? $\endgroup$ – THC May 23 '17 at 14:34
  • $\begingroup$ @THC I assumed the sesquilinear form to be non-degenerate. And non-degenerate in this context usually means that $\langle x,x\rangle=0\Rightarrow x=0$. If you interpret the non-degeneracy as "for any $x\in X$ there exists a vector $v\in X$ such that $\langle x,v\rangle \neq 0$", then you should add the aforementioned condition explicitly, yes. Anyways, it is always assumed when defining an inner product. $\endgroup$ – yuggib May 23 '17 at 15:05

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