A space is said to be complete is all the cauchy sequence of elements belonging to the space coverge to some element which is also within the space. A complete inner product space is called Hilbert Space.

I understand why a (complex) inner product space is needed in quantum mechanics. But I don't get how exactly is this completeness property is being needed for quantum mechanics.

Also, in J.J.Sakurai's "Modern Quantum Mechanics" (sec1.2), it is mentioned that when the observables have a continuous spectra (nondenumerably infinite), the vector space required would be a Hilbert space. So, if the spectra is finite or denumerably infinite, just a (complex/real) inner product space is enough?

  • $\begingroup$ Completeness is used in QM anywhere you do a basis transform. $\endgroup$ Commented Feb 3, 2020 at 10:28
  • $\begingroup$ Defining the derivative of a state (with respect to time say) would be a lot more fiddly (where it was even possible) if you could not assume completeness $\endgroup$ Commented Feb 3, 2020 at 10:58
  • $\begingroup$ What you mention from Sakurai's book doesn't have to do with the 'completeness' property of a Hilbert space but with the fact that the space must be infinite dimensional in order for an observable to have continuous spectrum (numerable infinity is enough, as for separable Hilbert spaces). In finite dimension the spectrum is always discrete (and made of eigenvalues). $\endgroup$
    – lcv
    Commented Feb 3, 2020 at 11:57

1 Answer 1


Even to a more basic level, the Hilbert spaces assure you that, given a selfadjoint operator, you will always find a base of eigenvectors, hence you can write for a generic wave function $\psi$ $$ \psi = \sum_{n=0}^\infty c_n \psi_n,$$ with $\psi_n$ an eigenfunction of the operator you are considering.


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