# Do the eigenstates of the number operator in an arbitrary Hilbert space form a complete basis?

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?

• 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. – Qmechanic Aug 25 '17 at 8:38
• 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. – Quantumwhisp Aug 25 '17 at 8:44
• 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 $L^2$. – AccidentalFourierTransform Aug 25 '17 at 9:15
• @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. – yuggib Aug 25 '17 at 9:20