# Number operator in interacting quantum field theory

When treating a quantum field, say the real scalar field, it's totally clear to me how to define a (global) number operator:

$$\hat N = \intop \text d ^3 \mathbf p \hat a ^\dagger (\mathbf p )a(\mathbf p ).$$ This turns out to commute with the hamiltonian and the 3-impulse of the system, therefore the physical interpretation of states with a definite number of particles with a definite total 4-impulse is straightforward. In particular, one could define the vacuum as $\hat N \rvert 0 \rangle =0$.

Now, consider a field interacting with itself, for example:$$\mathscr L =\frac{1}{2}\partial _\mu \phi \partial ^\mu \phi -\frac{1}{2}m^2 \phi ^2 -\frac{\lambda}{4!} \phi ^4.$$

In this case one still talks (in a sense which is not clear to me) of states with a definite number of particles. In a proof, my professor wrote, for a generic state $\alpha$:$$\lvert \alpha \rangle = \lvert \alpha \rangle _0 + \lvert \alpha \rangle _1 + \lvert \alpha \rangle _2 + ...$$where the pedices denote the number of particles in each state of the expansion. Now, this equation implicitly says that there is a certain observable $\hat N$ such that $\hat N \lvert \alpha \rangle _n = n\lvert \alpha \rangle _n$.

Question. Is there a theorem which guarantees the existence of such an operator for every physical field theory? Is it possible to construct explicitly $\hat N$?

• And interacting field theory does not have particle states, and doesn't have a number operator. Are you sure these "interacting particle states" don't like in the asymptotic Fock space instead? Commented Mar 17, 2016 at 17:56
• No, it is not always possible to construct the number operator; it depends on which representation of the CCR you're in. The canonical commutation relations for QFT admit uncountably many inequivalent irreducible representations. Technically speaking, given the algebra of the CCR and a state on it, it is possible to define a self-adjoint number operator on its GNS representation only if the state is normal with respect to the Fock representation. Commented Mar 17, 2016 at 22:21
• @ACuriousMind , I'm indeed confused about it. In the particular example I'm mentioning, those states are taken to be eigenstates of the full (interacting) 4-impulse, but since they must be also eigenstates of some number operator, from what you and yuggib say, I deduce that they must be linear combinations of Fock space $n$-particle kets. Commented Mar 18, 2016 at 11:00
• Anyway, if you or @yuggib want to turn your comment into an answer, I'll be glad to accept it. Commented Mar 18, 2016 at 11:00
• @pppqqq The point is that nobody at the moment knows what is the right Hilbert space (i.e. representation of the CCR) for any relativistic interacting QFT in $(3+1)$-dimensions. It may as well be the representation induced by a Fock-normal state, and therefore with a number operator. There are some suggestions that this should not be the case, but it's an open question. Anyways, the answer to your question is "there are representations of the CCR where it is not possible to define a self-adjoint number operator; whether these representations correspond or not to physical theories is not known" Commented Mar 18, 2016 at 14:34

The notion of a number operator needs the concept of particle number. For this concept one first needs to distinguish among the different degrees of freedom. There are internal degrees of freedom (spin, color, charge, ...) and then there are spacetime degrees of freedom. Once one has identified these, one can add to them the particle number degree of freedom. It is the latter that the number operator (also creation and annihilation operators, quadrature operators, etc.) is associated with.

In QFT, the input and output states generally have fixed numbers of particles. The interactions defined by the Feynman rules are also associated with interactions among specific individual particles. In a Feynman diagram the lines are associated with single particles. So, one can see that the notion of particle number is well-defined in QFT. By implication, the number operator composed of the creation and annihilation operators is also well-defined.

This does not constitute a proof for all possible QFTs, because one can imagine that it would in principle be possible to rederive QFT in terms of a different basis in the particle number degrees of freedom (perhaps the quadrature basis - I don't know) that does not have specific particle numbers associated with them. In such a case the definition of the number operator may not be well-defined. However, for QFT as it is currently used in the standard model, number operator should be well-defined.