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Do the eigenstates of the number operator in an arbitrary Hilbert-space form a complete basis? For simplicity I will restrict myself to the case of just one mode. Given we have 2 operators $\hat{a}$ and $\hat{a}^\dagger$, which satisfy the commutation relations $[ \hat{a}, \hat{a}^\dagger] = 1$, $[ \hat{a}, \hat{a}] = 0$ and $[ \hat{a}^\dagger, \hat{a}^\dagger] = 0$, We can derive that there must be a vacuum state $|0\rangle$ with $\hat{a} | 0 \rangle = 0$ and that the states $|n \rangle = (\hat{a}^\dagger)^n |0 \rangle$ are eigenstates of $\hat{a}^\dagger \hat{a}$.

Is there any hint that the states $|n\rangle$ do form a complete basis? What additional assumptions have to be made to derive this?

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    $\begingroup$ Comment to the title question (v2): Do the eigenstates of the number-operator in a Fock space form a complete basis? Yes, essentially by definition of a Fock space. $\endgroup$ – Qmechanic Aug 25 '17 at 8:38
  • $\begingroup$ I'll have to reformulate my Question in that case. What I want to know about is the case where I don't know anything about the Hilbert space. My bad. $\endgroup$ – Quantumwhisp Aug 25 '17 at 8:44
  • $\begingroup$ Your Hilbert space is isomorphic to $L^2$. You are asking whether the eigenfunctions of the quantum harmonic oscillator form a complete set, i.e., whether $H_n(x)\mathrm e^{-x^2}$ span $L^2$. $\endgroup$ – AccidentalFourierTransform Aug 25 '17 at 9:15
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    $\begingroup$ @AccidentalFourierTransform All Hilbert spaces are isomorphic, but two different representations of an operator algebra (such as the one of canonical commutation relations) may not be isomorphic or even homomorphic. $\endgroup$ – yuggib Aug 25 '17 at 9:20
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The "number operator" is a densely defined closed positive form only in the Fock irreducible representation and its unitarily equivalent ones (such as the Schrödinger and Bargmann-Fock representations in quantum mechanical systems, or the Q-space representation for free scalar boson fields). On any other irreducible representation of the canonical commutation relations it is still a closed and positive form, but it is not densely defined (see Bratteli and Robinson's book, second volume, for a proof).

Therefore the eigenstates of the number operator, counting multiplicity, can be a basis only in the Fock space.

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  • $\begingroup$ I'm sorry, but I'm not familiar with the terms "densely defined" and "close". Also, what is a irreducible representation? $\endgroup$ – Quantumwhisp Aug 25 '17 at 11:13
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    $\begingroup$ @Quantumwhisp The terms "densely defined" and "closed" are common in the theory of operators in Hilbert spaces, and of the corresponding quadratic forms. The first means defined in a set of vectors dense in the Hilbert space, while the second means roughly speaking that the graph of the operator is a closed manifold on the product of the Hilbert space with itself (with the so-called graph norm). $\endgroup$ – yuggib Aug 31 '17 at 7:59
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    $\begingroup$ Finally, an irreducible representation is a representation of an abstract C*-algebra as an algebra of (bounded) operators on a Hilbert space, such that the only subspaces of the Hilbert space invariant under the action of the whole algebra are the vector zero and the whole space itself. $\endgroup$ – yuggib Aug 31 '17 at 7:59

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