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Given an observable $Â$, any state can always be written as a linear combination of its eigenvectors, in other words its eigenvectors form a basis of the Hilbert space of all possible states.

I know that in finite dimension, this comes from the spectral theorem (for hermitian operators). But for infinite dimension, this theorem only applies on compact operators.

But, for example, the Hamiltonian is not always compact (for the particle in a box or the quantum harmonic oscillator, the energy can be arbitrarily large). So how do we know that its eigenvectors always form a basis ?

Qmechanic
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quantum_unicorn
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    The spectral theorem also applies for non-compact operators. Physicist, usually (in a first course on QM), ignore most mathematical issues and simply state that all self-adjoints operators admit a complete "basis", which however then can include "continuous" vectors, cf. the position operator.... Regarding your question, see e.g. this. – Tobias Fünke Feb 16 '23 at 20:29
  • Because we are forcing observables to be self-adjoint operators. Why? Because only then we have the guarantee that the whole spectrum is real, thus can be mapped to concrete measurement results. – DanielC Feb 16 '23 at 20:32
  • thank you @TobiasFünke – quantum_unicorn Feb 16 '23 at 22:58
  • You're welcome. Actually, the example above concerning the position operator is an example for an operator which as no eigenvectors in a strict sense (physicists will call these things continuous eigenvectors). An example for an operator with a "mixed" spectrum is the Hamiltonian of the hydrogen atom you will encounter in a first QM course; this operator has "discrete" and "continuous" eigenvectors that span the space, cf. this and the links therein. – Tobias Fünke Feb 16 '23 at 23:16

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