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We observe an electron within an atom and its orbit is computable. The location of any particle is within an envelope in the Euclidean space. I would like to know if there is a mapping between this envelope and the Hilbert space. As the Hamiltonian is on the entire Hilbert space, it seems that we cannot map the envelope in the Euclidean space where the particle is observed to a subspace of the Hilbert space where the particle resides.

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  • $\begingroup$ "The location of any particle is within an envelope in the Euclidean space." What does this mean? Any obvious interpretation is just wrong, since e.g the energy eigenfunctions of the hydrogen atom are nowhere vanishing. $\endgroup$
    – ACuriousMind
    Jun 19 at 11:56
  • $\begingroup$ Could you please clarify? A hydrogen atom is observed locally. $\endgroup$ Jun 21 at 13:27

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The physical position of the particle in real space (let's say Euclidean as you said, assuming non-relativistic quantum mechanics) is simply the eigenvalue of the position operator $\hat{\vec{r}}$ when it acts on a quantum state that is an element of the Hilbert space. That operator alongside the momentum operator $\hat{\vec{p}}= -i \hbar \vec{\nabla}$ are going to be the generators of the group of all ordered combinations of 3 real numbers i.e. the Euclidean plane $\mathbb{E}^{3}$, and they must satisfy the algebra:

\begin{equation} [\hat{\vec{r}} , \hat{\vec{p}}] = i\hbar \end{equation}

More specifically, the target space is going to be a vector space in and of itself, since you may also define spatial translations there. You could then say the particles implicitly are elements of the Euclidean vector space $\mathbb{R}^3$. In essence, the entire formulation of quantum mechanics is a mapping from the Hilbert space of quantum states to the real physical space.

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  • $\begingroup$ The state vectors in Hilbert space don't have a geometry. The Hamiltonian operates on the entire Hilbert space. We observe the particle locally within a boundary in the real space but there is no corresponding boundary for it in the Hilbert space. Why? $\endgroup$ Jun 21 at 13:33
  • $\begingroup$ What do you mean by "state vectors don't have a geometry"? I'm not sure I understand what you are saying. Also, there isn't necessarily a boundary in real space either. If the mapping is from the Hilbert space to the surface of a sphere, obviously the physical space we end up on won't have a boundary. In general, the existence of a boundary isn't something that needs to be in parity between the starting and target vector spaces. $\endgroup$
    – rhomaios
    Jun 21 at 16:26
  • $\begingroup$ The state vectors are not bounded in the sense that we cannot answer the question 'where in the Hilbert space are these vectors?'. Even if we map to a surface of a sphere, we can locate the vectors in the target vector space. We observe the particles within a location (e.g. an atom) in the real space. The information about the particle namely the state vectors are in the entire Hilbert space. Where is the particle? $\endgroup$ Jun 22 at 4:14
  • $\begingroup$ I don't see how this statement has much to do with boundaries. But you can ask (and can answer) where on the Hilbert space the vectors are. Any $n$-dimensional Hilbert space is isomorphic to $\mathbb{C}^{n}$, so you can represent it as an $n$-dimensional complex plane. $\endgroup$
    – rhomaios
    Jun 22 at 5:32
  • $\begingroup$ It is a good point that the state vectors can be represented in $\mathbb C^{n}$. But it is the same information about the particle that is already in the state vectors in the Hilbert space. The particle is strictly observed within a boundary in the real space and probabilities of finding the particle at a point within the boundary is predictable. But it is unclear if the particle is actually in the real space where it is observed or in the Hilbert space. If it is in the Hilbert space, where in the Hilbert space and how to compute its boundary there is the question. $\endgroup$ Jun 22 at 15:25

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