"To each real-valued function $f$ on the classical phase space, there is associated a self-adjoint operator $\hat{f}$ on the quantum Hilbert space. In almost all cases $\hat{f}$ is unbounded." As an example, position and momentum operators are unbounded in general. They are bounded provided a number of suitable conditions are imposed on the underlined Hilbert space. Identity operator is bounded in all Hilbert space. We have examples of self-adjoint operators, bounded on all Hilbert spaces, which may not have any physical significance.

My question is if there is a self-adjoint operator other than the identity operator with a physical importance that is bounded on every Hilbert space. If they exist, please provide some examples. If they do not exist, please justify their non-existence.

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    $\begingroup$ How can an operator on every Hilbert space be defined? Isn't the Hilbert space the operator is acting on part of the definition of the operator? If we restrict ourselves to single particle Hilbert spaces (say, a single electron), spin should be a bounded operator. $\endgroup$
    – Photon
    May 19, 2017 at 4:47
  • $\begingroup$ Identity operator is defined and bounded on every Hilbert space. Whenever we are considering single particle Hilbert space, we are actually imposing a restriction on the Hilbert space. Omitting this restriction, the spin operator may be undefined or unbounded. $\endgroup$
    – Supriyo
    May 19, 2017 at 5:07
  • $\begingroup$ But I think, even the identity operator, to be well-defined, needs to be specified with the space it is acting on. At least the identity operator can be written down without explicitly mentioning the space it is acting on, but how would you even write down any other operator without specifying the space it is acting on first? $\endgroup$
    – Photon
    May 19, 2017 at 6:12
  • $\begingroup$ This post (v2) seems like a list question. Start with projection operators. $\endgroup$
    – Qmechanic
    May 19, 2017 at 8:58

2 Answers 2


An operator is, by definition, a linear map from some given linear space $V$ to itself. And the definition of a linear map cannot, in general, be independent of the linear structure if the space it acts upon.

There are, however, algebraic objects that can be defined abstractly, and that they are always represented as bounded operators on some (actually many) Hilbert space(s). These algebraic objects are the ones forming involutive algebras, in particular C*-algebras. It is standard to consider the set of observables of a given physical system as forming an involutive algebra.

In non-relativstic quantum mechanics, the C* algebra of interest is fixed (up to *-isomorphisms), and it is uniquely (up to unitary isomorphisms) irreducibly represented in a Hilbert space. This irreducible representation is the so-called Schrödinger representation on $L^2(\mathbb{R}^d)$, where position and momentum are respectively the self-adjoint unbounded operators of multiplication and $1/i$-times differentiation with respect to the variable $x\in\mathbb{R}^d$. It is in this representation that the pseudodifferential calculus, that associates to (almost) any function (actually distribution) on the phase space $\mathrm{T}^*\mathbb{R}^d$ an operator from $\mathscr{S}(\mathbb{R}^d)$ to $\mathscr{S}'(\mathbb{R}^d)$ (that can, in some case, be restricted to a densely defined operator on $L^2(\mathbb{R}^d)$), is set.

Clarified all that, there are plenty of examples of physically relevant bounded self-adjoint operators in quantum mechanical systems. The foremost example is the spectral family $(P_\lambda)_{\lambda\in \mathbb{R}}$ of a given self-adjoint (possibly unbounded) observable $A$. Many people consider the spectral resolution much more physically relevant than the observable itself (there is a $1-1$ correspondence anyways), for $\lVert (P_{\lambda_1}-P_{\lambda_2})\psi\rVert^2$ is the probability that a measurement of the observable $A$ in the system in the state $\psi$ would yield a value in the interval $[\lambda_2,\lambda_1]\subseteq \mathbb{R}$ (assuming $\lambda_2 < \lambda_1$). The spectral family consists entirely of bounded (self-adjoint) operators (orthogonal projections, that each satisfy $P^2_{\lambda}=P_\lambda$).


What about the operator for the energy of a particle in a box? It's easy to see that it is bounded below. It corresponds to discrete sinusoid wave functions with half a wavelength. Different half-wavelength functions differ a little in energy (the ones with the highest energy are the ones with are perpendicular to two opposite walls.

The wavefunctions have un upper limit in the sense that they can't have wavelengths which correspond to energies higher than the energy necessary to make virtual particle pairs real. Well, these wave functions do of course exist, but then we're talking about a different situation than that of a single particle in a box.

It doesn't really matter what Hibert space you choose because the base functions for every space make up the same Hilbert space [like you can find an infinite number of three orthogonal (or non-orthogonal) three-vectors that span three-dimensional space], and the wavefunctions stay the same but expressed in different base functions.

  • $\begingroup$ This is basically taking an unbounded operator and truncating the Hilbert space, so that on the restricted (finite dimensional) space the operator is bounded. $\endgroup$
    – fqq
    May 19, 2017 at 9:54
  • $\begingroup$ The operator is just a bounded Hermitian operator, connected with a finite dimensionalHilbert space (because it's the case that for this system there is not an infinite set of eigenfunctions). In every Hilbert space (composed out of different base states) connected with the system (what does it matter if it's a truncated space) the wave functions will be the same, though as expressed in different base functions. $\endgroup$ May 20, 2017 at 12:37

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