# Why are the charge operator $Q$ and the baryon number operator $B$ unbounded?

A friend recommended me to read PCT, Spin and Statistics, and All That written by R. F. Streater and A. S. Wightman. In page 5 to 6, here's what the authors of this book have to say:

[...] In fact the operators $Q$ and $B$, being unbounded, do not lie in $\theta^{\prime}$, but the associated projection operators, which project onto the states of various possible values of $Q$ and $B$, do. [...]

Can someone please explain to me as to why the operators $Q$ and $B$ are unbounded? Also, what could be understood by the term associated operator to an unbounded operator? Here $Q$ is an operator for charge and $B$ is an operator for baryon number.

• What is $\theta'$? – Ryan Unger Mar 4 '15 at 22:31
• Observe that, if you can have one baryon, you can also have two, and three, and four, and so on. That looks unbounded to me. – ACuriousMind Mar 4 '15 at 22:32
• Furthermore, there is nothing really preventing you from having more charge. – Ryan Unger Mar 4 '15 at 22:35
• @0celo7: The authors say that if we consider the set of all bounded operators which commute with all the observables; this is a set $\theta^{\prime}$ which is called the commutant of $\theta$. – Janus Boffin Mar 4 '15 at 22:39
• @ACuriousMind: Care to expand on your comment "[...] if you can have one baryon, you can also have [...]", please? – Janus Boffin Mar 4 '15 at 22:41

$\newcommand{\ket}[1]{\lvert #1 \rangle}$The charge and baryon number operators are not bounded because you can create states of arbitrary charge and baryon number:

Let $a^\dagger$ be any creation operator that creates a bosonic charged particle state (let's say with unit charge), and let $\ket{\Omega}$ be the vacuum. Then,

$$\ket{n_e} = (a^\dagger)^{n_e}\ket{\Omega}$$

is, for every $n_e \in \mathbb{N}$, a state with charge $n_e$. If you only have fermionic charged states, you need to choose $n_e$ different momenta $p_i$ and use $\prod_{i = 1}^{n_e}a^\dagger(p_i)$, but the idea is the same.

The same reasoning works for baryons - if you can create a state with one baryon, then you can create a state with $n_b$ baryons.

This shows that the number and charge operators are unbounded, since their eigenvalues $n_e,n_b \in \mathbb{N}$ can be chosen arbitrarily high.

Projection operators, on the other hand, are always bounded, because their highest eigenvalue is 1.

For the unboundedness see ACuriousMind's answer.

About the associated projections, for unbounded operators there is the notion of affiliation. An unbounded, closed and densely defined operator $A$ is affiliated with a von Neumann algebra $M$ if all its spectral projections are in $M$, that is $$A = \int\lambda\text dE(\lambda),$$ with the spectral measure $E$ taking values in the projections of $M$. In the book you have referenced $\theta'$ is the commutant of the C*-algebra of all the observables, and therefore a von Neumann algebra by von Neumann's bicommutant theorem.