4

I know that states in quantum mechanics are positive trace class operators acting on a separable complex Hilbert space $\mathcal H$ and having trace = 1. Specifically, pure states are one-dimensional orthogonal projections that of course can be identified with the one-dimensional subspace itself. So the set of pure spaces is the projective Hilbert space $\mathcal P(\mathcal H)$. My question is about the term referring to the Hilbert space $\mathcal H$ itself. Is it "state space", "structure space" or something else?

Wikipedia writes:

Specifically, in quantum mechanics a state space is a complex Hilbert space in which the possible instantaneous states of the system may be described by unit vectors.

I also found the term "state space" e.g. in this answer. However I haven't found this term in any quantum mechanics book except in the book of Claude Cohen-Tannoudji referred to in the Wikipedia article, not even the other book referenced there (Griffiths). Is "state space" a generally accepted term? References are welcome.

mma
  • 644
  • 3
    I won't make this an answer because I don't have much of a reference, but I think most physicists just call it "Hilbert space". – Níckolas Alves Oct 29 '22 at 06:50
  • The Hilbert space is the foundation of the solution theory of the Schroedinger equation. It is, however, not the physical state in which "the system" lives. You can never find a quantum mechanical system in a superposition. It is either in a well defined state (because we just performed a preparation or measurement on it) or we simply don't know which state it is in. This is compounded by the fact that we can perform a unitary transformation which "turns" pure states in one choice of base into superpositions in another. That is a purely mathematical operation that nature doesn't care about. – FlatterMann Oct 30 '22 at 06:39
  • @FlatterMann I am not sure to understand. Every state vector eigenvector of some observable $A$ is always a superposition of state vectors of another observable $B$ which is incompatible with $A$. In this sense we find always the system both in a superposition of states and in a definite state simultaneously. (Disregarding the case of a mixture of pure ststes.) – Valter Moretti Oct 30 '22 at 08:26
  • What we cannot see is the Schroedinger cat which is not alive nor dead if we measure the observable $A$ dead/alive. But if we measure an observable $B$ that is incompatible with $A$, we find the cat in a state where it is not alive nor dead in general. (A difficult point is that the cat is a macroscopic system so that it seems that all observables are compatible.) – Valter Moretti Oct 30 '22 at 08:35
  • @ValterMoretti The only physical "observables" of an individual quantum system that we have access to are energy, momentum, angular momentum and charges. These are selected by nature for us because they are the microscopically conserved physical properties of quantum fields. These are the properties that ATLAS and CMS are measuring. All of the other "observables" in von Neumann's Hilbert space formalism are derived properties of the quantum mechanical ensemble. Nature doesn't care about the latter, even though, they are comfortable tools for the theorist. – FlatterMann Oct 30 '22 at 08:36
  • So? Please conclude your reasoning. – Valter Moretti Oct 30 '22 at 08:37
  • @ValterMoretti Schroedinger's cat is the most horrible intellectual nonsense. The irreversible process that kills the cat is over as soon as that alpha particle is a few fm away from the nucleus. Everything else in that Gedankenexperiment is simply inedible garnish. I don't know if Schroedinger knew this and just didn't care because he wanted to be polemical anyway or he just didn't realize that he was basically warming up a stale dish made from the leftovers of Poincare's recurrence theorem. – FlatterMann Oct 30 '22 at 08:42
  • Why are you so rude and aggressive? – Valter Moretti Oct 30 '22 at 08:43
  • Let us continue this discussion in chat. – FlatterMann Oct 30 '22 at 08:43
  • Absolutely not! – Valter Moretti Oct 30 '22 at 08:43
  • @ValterMoretti My loss, I suppose. Peace. – FlatterMann Oct 30 '22 at 08:48

1 Answers1

8

I sometime use “state space”, but literally this name is wrong. The pure states are the rays of the Hilbert space, namely the unit vectors up to phases. And it is true only when all bounded selfadjoint operators represent observables. A better terminology would be “the Hilbert space of the system”.

Valter Moretti
  • 71,578