# Creation/annihilation operators question

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Antarres
I've recently stumbled upon something that looked kind of silly, but I still find myself a bit confused by it. Namely in quantum field theory, when we quantize a scalar field, we impose commutation relations on creation and annihilation operators that correspond to momenta in their mode expansion.
$$[a_{\textbf{k}},a^\dagger_{\textbf{k}'}] = \delta^{(3)}(\textbf{k} - \textbf{k}')$$

This continuum case of the usual creation/annihilation commutator makes a ##\delta^{(3)}(0)## factor when we act on a certain state vector with its annihilation operator. When we work with expected values of operators and this factor appears, we often treat it as the volume of space enclosing the system, and then draw conclusions in terms of densities corresponding to this volume. However, what if we're interested in looking exactly at a state vector, not bilinear forms that would be operators, nor their expected values.

For example, let's say we define a vector ##\lvert 1_{\textbf{k}}>##, as a vector which has occupation number 1 for 3-momentum ##\textbf{k}##, that is, it has one particle of this momentum. We act on it with annihilation operator:
$$a_{\textbf{k}}\lvert 1_{\textbf{k}}> = a_{\textbf{k}}a^\dagger_{\textbf{k}}\lvert 0> = (a^\dagger_{\textbf{k}}a_{\textbf{k}} + \delta^{(3)}(0))\lvert 0> = \delta^{(3)}(0)\lvert 0>$$

What sense do we make of this resulting state, if it has this diverging factor in front of it? I assume it may have to do with our assumption that the momentum of the particle is innitially completely defined, but this looked like the treatment that was usual so far in QFT textbooks, just they never consider this case as far as I've noticed.

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Gold Member
2022 Award
You stumbled over an issue that is usually not so carefully treated and then you run into trouble. That's the way QFT is usually taught and to some extent that's because there's no simple way to formulate it mathematically rigorously (at least not interacting field theories).

The problem is that you build the Fock space for a particle in an infinite 3D space and thus your momentum components are observables with an entirely continuous spectrum. Indeed the momentum "eigenvalues" are just ##\vec{p} \in \mathbb{R}^3##.

From ordinary quantum mechanics you already know that the momentum eigenfunctions are ##u_{\vec{p}}(\vec{x})=N \exp(\mathrm{i} \vec{p} \cdot \vec{x})## are not really "allowed functions", because they are not square integrable and you can only normalize them to ##\delta## distributions ##\label \vec{p}|\vec{p}' \rangle=\delta^{(3)}(\vec{p}-\vec{p}')##.

Now all this is cured in some sense by putting the particles into a finite volume. You could try it with a "cavity", i.e., a potential cubic box with infinite walls. That's however not a clever choice when you want to study finally indeed scattering theory for particles in the entire infinite-volume space, because there you like to have well-defined momentum operators also for the finite-volume box (which you don't have with rigid boundary conditions at the boundaries of the box).

That's why it's more suitable to consider a finite cube of length ##L## and impose periodic boundary conditions on the wave functions,
$$\psi(\vec{x}+\vec{n} L)=\psi(\vec{x}) \quad \text{for all} \quad \vec{n} \in \mathbb{Z}^{3}.$$
Then it's easy to see that you still have a well-defined self-adjoint momentum operator (in the 1st-quantization formalism), ##\hat{\vec{p}}=-\mathrm{i} \vec{\nabla}## and thus momentum eigensolutions ##u_{\vec{p}}=N \exp(\mathrm{i} \vec{p} \cdot \vec{x})##, but now due to the boundary conditions you have only a discrete set of momentum
$$\vec{p} = \vec{n} \frac{2 \pi}{L}, \quad \vec{n} \in \mathbb{Z}.$$
Also the eigenmodes are normalizable and you have
$$\langle \vec{p}|\vec{p}' \rangle=\int_{[0,L]^3} \mathrm{d} ^3 x u_{\vec{p}}^*(\vec{x}) u_{\vec{p}'}(\vec{x}) = \delta_{\vec{p},\vec{p}'},$$
where the ##\delta## here is a usual Kronecker ##\delta## which is simply 1 for ##\vec{p}=\vec{p}'## and 0 otherwise.

Now you can build your Fock space without the troubles you recognized above. To get the "infinite-volume limit" you have to define useful physical quantities in such a way that the limit ##L \rightarrow \infty## makes sense, and then the sum over momentum eigenstates translates into an integral, i.e., in an operator sense
$$\frac{1}{V} \sum_{\vec{p}} \rightarrow \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 p}{(2 \pi)^3}.$$

The first encounter with infinities is when looking at total energy and momentum, where in the relativistic case you get an infinity (even in the finite volume), because a naive expression for the total energy operator (the Hamiltonian of the free QFT) is
$$\hat{H}=\sum_{\vec{p}} E(\vec{p}) [\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}) + \hat{b}(\vec{p}) \hat{b}^{\dagger}(\vec{p}).$$
Apply this to the vacuum state, you get from the 2nd term
$$\hat{H} |\Omega \rangle=\sum_{\vec{p}} E(\vec{p}) |\Omega \rangle,$$
but since ##E(\vec{p})=\sqrt{m^2+\vec{p}^2}## the sum diverges.

On the other hand the energy is only defined up to an arbitrary additive constant, and you can simply use the commutation relations of the creation and annihilation operators to write
$$\hat{b}(\vec{p}) \hat{b}^{\dagger}(\vec{p}) = [\hat{b}(\vec{p}),\hat{b}^{\dagger}(\vec{p})] + \hat{b}^{\dagger}(\vec{p}) \hat{b}(\vec{p})=\hat{1} +\hat{b}^{\dagger}(\vec{p}) \hat{b}(\vec{p}) ,$$
which means you Hamiltonian gets
$$\hat{H}=\hat{H}'+ \hat{1} \sum_{\vec{p}} E(\vec{p})$$
with
$$\hat{H}'=:\hat{H}:=\sum_{\vec{p}} E(\vec{p}) [\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}) + \hat{b}^{\dagger} (\vec{p})].$$
Now ##\hat{H}'## has all the commutation properties with any operators as ##\hat{H}## but the infinite "vacuum energy" is subtracted and any proper Fock state (which are Fock states with a finite total number of particles!) has a definite finite energy.

Now let's discuss the infinite-volume limit. First note that from the translation rule of the momentum sums to an integral it makes sense to rescale the annihilation operators
$$\hat{a}^{(\infty)}(\vec{p}) \rightarrow \sqrt{V} \hat{a}(\vec{p}),$$
because then you can write
$$\hat{H}'=V \sum_{\vec{p}} E(\vec{p}) [\hat{a}^{(\infty)\dagger}(\vec{p}) \hat{a}^{\infty}(\vec{p}) + \hat{b}^{(\infty)\dagger} (\vec{p}) \hat{b}^{\infty}(\vec{p})]$$
and now take the limit to the integral for ##L \rightarrow \infty##, which gives
$$\hat{H}'=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} [\hat{a}^{(\infty)\dagger}(\vec{p}) \hat{a}^{\infty}(\vec{p}) + \hat{b}^{(\infty)\dagger} (\vec{p}) \hat{b}^{\infty}(\vec{p})].$$
For ##L \rightarrow \infty## now the commutator of the new annihilation and creation operator indeed goes to ##\infty## for ##\vec{p}=\vec{p}'## but stays 0 otherwise. The rescaling is such that you get
$$[\hat{a}(\vec{p}),\hat{a}^{\dagger}(\vec{p}')]=(2 \pi)^3 \delta^{(3)}(\vec{p}-\vec{p}').$$
where I skipped the ##(\infty)## supersecripts again, and the additional factor ##(2 \pi)^3## is put into make everything consistent with all momentum integrals always coming with the meaure ##\mathrm{d}^3 \vec{p}/(2 \pi)^3## which is the common convention among HEP physicists.

nrqed
Antarres
Hmm, that is clear. I am actually aware that, if we take a finite volume of direct space, that makes the momentum space discrete. So the solution to this would be to switch to discrete momentum space and then take the thermodynamic limit where volume tends to infinity, which would yield continuum of momentum states?

Or, taking this interesting rescaling of creation/annihilation operators, apply the rescaling to the commutation relation, after which I treat the factor I get from commutator as identity operator instead of this undefined constant? I think I understand what you're saying, but I'm checking if I got it right. Thank you for the response.

vanhees71
Keith_McClary
Antarres
Gold Member
2022 Award
Well, you don't need that level of rigor to use QFT as a physicist. Of course, if you like to learn how much is known towards a rigorous formulation of QFT, you can study all the attempts towards one. AFAIK there's none yet for physically relevant theories, let alone the Standard Model.

From a physicist's point of view a very good combination is to start with a good introductory book (my favorite for that is Schwartz, QFT and the Standard Model) then go on with Weinberg, Quantum Theory of Fields (vol. 1+2 for the standard model; vol. 3 supersymmetric QFT) and fill the gaps with Duncan, The conceptual framework of QFT. The latter has a very nice section on Haag's theorem, which answers the question of this thread very carefully and with enough rigor for any practical purposes.

Antarres
Antarres
Thank you both for responses. I'll try to read that review article. I actually have formal background in QFT(Schwartz, Peskin), and I'm going through Weinberg as a self-study, since it's not really used as a university textbook. The reason I labeled this as intermediate, is because I figured something like this I was first taught in my bachelor's. I stumbled upon this issue while trying to solve some problem with definition of Hilbert space in curved space-time, that is part of my current course.
It is something that I never wondered about while I was being introduced to QFT, but here it appeared in such a form that i wouldn't be solved by just erasing it(informally speaking), like we do when we have it inside Hamiltonian, for example, so I wondered how to do it. Either way, it is somewhat clearer now, I think. I just like to be as clear as possible when I do my notes, so I couldn't handwave that factor out, even though I think it shouldn't be there.

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vanhees71