# Does every hermitian operator represent a measurable quantity?

In Quantum mechanics, observables are represented by hermitian operator. But does every hermitian operator represent a observable? If not , how do we know that whether a hermitian operator represent observable or not? What is the precise definition of the term "observable"?

Given a quantum system with associated Hilbert space $\mathcal H$, the set of all self-adjoint bounded operators is $\newcommand{\bh}{\mathcal B(\mathcal H)_\text{sa}}\bh$. In general, only a small subset of $\bh$ will represent physically observable operators. For infinite-dimensional systems, $\bh$ is huge and there's no hope ever finding experiments for all its members; even in finite-dimensional systems it is very challenging to find experimental schemes sensitive to even a vector-space basis for $\bh$.

The physical approach to this is to begin with a finite set of operators which you know you can measure. For a single free particle, for example, you'd take position and momentum; for a finite set of spins you'd take all their Pauli matrices. You then form the set $\mathcal A$ of all operators that can be formed from them via products and linear combinations, which has the structure of a $\mathcal C^\ast$ algebra, and that is your set of physical observables. The $\mathcal C^\ast$algebra itself is the really fundamental description of the system; the Hilbert space is simply one possible representation.

In this formalism, states are functionals on $\mathcal A$: they are functions $$\rho:\mathcal A\rightarrow \mathbb C$$ that take an observable and give its measured value (or probable measured value, etc.) in that state. (In a Hilbert space representation, each such functional is associated with a density matrix $\hat\rho$, a trace-class positive operator such that $\rho(A)=\text{Tr}(\hat\rho\hat A)$ for $\hat A$ the Hilbert space operator associated with an arbitrary $A\in\mathcal A$.

Edit: As joshphysics and WetSavannaAnimal rightly point out, this works as stated only for bounded operators and not for unbounded ones like position or energy. I'm afraid I don't know well enough how this extends to that class of operators - that needs someone with much stronger functional analysis chops than mine.

• @dj_mummy I should think the possibility of extra observables can never be conclusively ruled out. However, as they would be fully quantum observables, they would not help in resolving EPR paradoxes: they are either local, and thus within Bell's treatment, or entangling, in which case the locality assumption is broken. – Emilio Pisanty Aug 27 '13 at 15:06
• @EmilioPisanty I'm confused. In standard courses on quantum mechanics, we usually consider certain unbounded, self-adjoint operators to be observables, but presumably $\mathcal B(\mathcal H)_\mathrm{sa}$ does not contain these beasts. What am I missing? – joshphysics Aug 27 '13 at 18:13
• @Emilio : to "each such functional is associated with a density matrix" : this is true only for normal states, see ncatlab.org/nlab/show/state+on+an+operator+algebra – jjcale Aug 27 '13 at 20:25
• @joshphysics : each normal (maybe unbounded) operator can be written as an integral over projection operators (which are bounded). So if you know the expectation values of the projection operators you know also the expectation value of the normal operator. – jjcale Aug 27 '13 at 20:35
• Emilio and @joshphysics . It would be great to get a reference: am I missing something (probably!)? Isn't resolving an operator into its unbounded-weight sum of always bounded projectors spectral decomposition itself - so your comment seems a bit like question begging. I don't doubt you're likely to be right (I've seen a good number of your posts!) - it's just that my (and probably joshphysics's) conception of spectral theory is of something that gets lots of fiddly bits when we wander into unbounded operator lands. I would love to see better versions and it sounds like you've glimpsed them! – Selene Routley Oct 4 '13 at 0:29