Short maths version at end

Consider the self-adjoint operator (the occupation number operator that forms the Hamiltonian of the quantum harmonic oscillator) $$N = a^\dagger a$$ Where $a$ and $a^\dagger$ are the "creation" and" annihilation" operators satisfying $[a, a^\dagger ]=1$. It can be shown that the eigenvalues of this operator are $n \in \mathbb{N} $ and the associated eigenstates are denoted $|n\rangle$.

Now in physics we appeal to the spectral theorem, so that the eigenstates of a self-adjoint bounded operator are complete in the Hilbert space so that, for example, the identity admits a decomposition $$I = \sum_{n=0}^\infty |n\rangle \langle n|, $$ or an arbitrary state can be written as a linear combination of the eigenstates $$|a\rangle = \sum c_n |n\rangle $$ for coefficients $c_n$.

However as far as I can tell, the number operator is not bounded - there is no real number $K$ such that $$| \, N|n\rangle \, | < K|\, |n\rangle\, | ~~ \forall n$$ where the vertical lines indicate the norm on the Hilbert space. This is because the eigenvalues of the operator have no upper bound and the states are normalised to unity.

So why does the spectral theorem apply? Is there a generalisation to unbounded operators with nice (which?) properties? Or am I mistaken and the operator is bounded after all?

Mathematicians' version:

Suppose there is a linear self-adjoint operator $N: H \rightarrow H$ mapping a vector space over $\mathbb C$ to itself with eigenvectors $v_n$ satisfying $N v_n = n v_n$ for n a natural number. It seems N is unbounded yet a version of the spectral theorem holds so that the eigenvectors are complete.

  • $\begingroup$ This question has, I believe, heavy physics notation (for operators, inner product..or whatever). I'm afraid not many mathematicians will understand this, so if you will change to mathematics notation...or perhaps better: try in the physics network. $\endgroup$
    – DonAntonio
    Jul 14, 2019 at 8:36

1 Answer 1


The number operator is indeed unbounded; but there is a spectral theorem for general unbounded self-adjoint operators from the 1930's going back to Stone and von Neumann. The core statement reads as follows (compare for example Thm.10.4 in the book "Quantum Theory for Mathematicians" (2013) by B. Hall):

Let $A$ be a self-adjoint operator on a Hilbert space $\mathcal H$. Then there is a unique projection-valued measure $\mu^A$ on the spectrum $\sigma(A)\subseteq\mathbb R$ with values in the bounded operators $\mathcal B(\mathcal H)$ such that $$ \int_{\sigma(A)}\lambda\,d\mu^A(\lambda)=A\,.\tag{1} $$ [For more precision on how this integral is to be understood check the following paragraph & (2).]

Edit: The only difference between the spectral theorem for bounded and the one for unbounded self-adjoint operators is the sense in which the above integral has to be understood. While (1) for bounded $A$ holds in the usual norm sense, for unbounded $A$ one has to consider a sequence of bounded Borel functions $h_n(x)$ with $h_n(x)\to x$ as $n\to\infty$ for each $x$ and $|h_n(x)|\leq|x|$ for all $x$ and $n$. Then for all $\psi$ in the domain of $A$, eq.(1) changes into $$ \lim_{n\to\infty}\int_{\sigma(A)} h_n(\lambda)\,d\mu^A(\lambda)\psi=A\psi\,.\tag{2} $$ For more details on this topic, refer to "Methods of Modern Mathematical Physics. I: Functional Analysis" (1980) by Reed & Simon (Theorem VII.2 & VIII.5)

Now the number operator is a special case because it is a discrete operator, i.e. the spectrum of $N$ consists only of eigenvalues which do not accumulate anywhere. Thus its spectral measure---loosely speaking---corresponds to an orthonormal basis of $\mathcal H$, and not "just" a resolution of the identity in terms of an integral $I=\int 1\,d\mu^A$, cf. also here.

  • $\begingroup$ Your integral here is essentially the spectral decomposition of the operator, right? $\endgroup$
    – lux
    Jul 14, 2019 at 9:40
  • $\begingroup$ Yes, that is a generalization of the spectral decomposition for hermitian matrices or self-adjoint compact operators $T=\sum_{j\in J\subseteq \mathbb N} \tau_j |g_j\rangle\langle g_j|$---because the spectrum is not necessarily discrete anymore, the sum in general turns into an integral. $\endgroup$ Jul 14, 2019 at 9:43
  • $\begingroup$ So essentially the spectral theorem for bounded operators generalises to unbounded operators? What additional conditions on the operator are required aside from that it be self-adjoint? $\endgroup$
    – lux
    Jul 14, 2019 at 11:49
  • $\begingroup$ The theorem generalizes without additional requirements. The only difference lies within the type of integral convergence. I added a paragraph ("Edit:") on that to my original answer. $\endgroup$ Jul 14, 2019 at 12:32
  • 3
    $\begingroup$ Keep in mind that self adjointness is much more finicky for unbounded operators. $\endgroup$ Jul 14, 2019 at 12:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .