# 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, 2017 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, 2017 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, 2017 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, 2017 at 9:20

## 1 Answer

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.

• I'm sorry, but I'm not familiar with the terms "densely defined" and "close". Also, what is a irreducible representation? Aug 25, 2017 at 11:13
• @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). Aug 31, 2017 at 7:59
• 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. Aug 31, 2017 at 7:59