# 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. 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. 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 $L^2$. 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. Aug 25 '17 at 9:20