Operators "carrying" momentum and particle number

Question

TL;DR: What does it mean for a operator in QFT to "carry momentum"?

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During a QFT lecture (discussing the real scalar field) my Prof. stated that the operator $$P^\mu := \int \frac{d^d\vec{p}}{(2\pi)^d 2\,\omega(\vec{p})} p^\mu(\vec{p})a^\dagger(\vec{p}) a(\vec{p})$$ "carries no momentum, since" $$[P^\mu,P^\nu]= [P^\mu,N] =0,\tag{$\diamondsuit$}$$ where $$N := \int \frac{d^d\vec{p}}{(2\pi)^d 2\,\omega(\vec{p})} a^\dagger(\vec{p}) a(\vec{p}).$$ I don't really understand what this means from a physical point of view and I don't really understand the condition either... Does the RHS of eq. ($\diamondsuit$) have to be $0$, or just independant of $\vec{p}$? What about the LHS, why are we looking at the commutation relation with the number operator?

He actually even stated that operators of the form $n(\vec{p}):=a^\dagger(\vec{p})a(\vec{p})$ conserve momentum and particle numbers since $$[H, n(\vec{p})]=[P^\mu, n(\vec{p})] = [N,\vec{p}]=0.$$ Why are we now looing at the commutation relation with the Hamiltonian in addition to $P^\mu$ and $N$? What exactly does the particle number have to do with the Hamiltonian? And what does it even mean to carry "particle number"?

Answer 1

Your lecturer is trying to express some useful intuition, but in a vague way.

To make it as simple as possible, let's go back to the simple harmonic oscillator. Recall that $$[a, a^\dagger] = 1, \quad [H, a^\dagger] = a^\dagger$$ and let's work throughout with units $k = m = 1$. Now suppose that some state has energy $n$, $$H |\psi\rangle = n |\psi \rangle.$$ Then the state $a^\dagger |\psi \rangle$ has energy $n+1$, because $$H(a^\dagger |\psi \rangle) = (a^\dagger H + [H, a^\dagger]) |\psi \rangle = a^\dagger (H+1) |\psi \rangle = (n+1) (a^\dagger |\psi \rangle).$$ The fact that $a^\dagger$ increases the energy eigenvalue by $1$ is a direct consequence of the fact that $[H, a^\dagger]$ is equal to $1$ times $a^\dagger$. We could summarize this by saying that the operator $a^\dagger$ carries unit energy.

You can also get intuition for this in Heisenberg picture. Because of the commutation relation $[H, a^\dagger] = a^\dagger$, the operator $a^\dagger(t)$ rotates in phase with unit angular velocity. But we also know that over time, the state $|\psi \rangle$ rotates in phase with angular velocity $n$. So the state $a^\dagger |\psi \rangle$ should rotate with angular velocity $n+1$ in Schrodinger picture, which it does. (If you don't like this, it's just another way of putting the concrete math above into words, so you can ignore it.)

Your professor is just extending this logic. An operator $A$ "carries" a quantity $a$ of a physical observable $\mathcal{O}$ if, when it acts on an eigenstate of $\mathcal{O}$, it yields another eigenstate whose eigenvalue is higher by $a$, which is equivalent to $[\mathcal{O}, A] = a A$.

Answer 2

I never heard the phrase "carry momentum" or "carry particle number". I think this is just how the lecturer tries to express something which is not easy to put into words. I think he just means that the commutator is zero. That is to say, the operator does not change the value of an eigenstate.

A conserved quantity does not change under time evolution, that is to say it does not change under the action of the Hamiltonian. This is again expressed by the vanishing commutator.