# Creation and annihilation operators in QFT

As I said before, I'm not a QFT expert but I'm trying to understand the basics of its rigorous formulation.

Let's take Dimock's book, where the foundation of QM and QFT is discussed. If we consider, say, two particles, one living in a Hilbert space $$\mathcal{H}_{1}$$ and the other in another Hilbert space $$\mathcal{H}_{2}$$, the description of the state of the two-particle system is given in terms of the tensor product $$\mathcal{H}^{(2)}=\mathcal{H}_{1}\otimes \mathcal{H}_{2}$$. Of course, we could go furhter and study a system $$\mathcal{H}^{(N)}=\mathcal{H}_{1}\otimes \cdots \otimes \mathcal{H}_{n}$$. If all the particles are identical, then $$\mathcal{H}_{1}=\cdots = \mathcal{H}_{n} \equiv \mathcal{H}$$ and we must take into account symmetric and anti-symmetric subspaces of $$\mathcal{H}^{(N)}$$, which correspond to the fact that particles may be either bosons or fermions, respectivelly. At this point, one defines symmetrization and anti-symetrization operators. The next step is to consider a system of an arbitrary number of particles. At this point, one defines Fock spaces $$\mathcal{F}^{\pm}(\mathcal{H}) = \bigoplus_{n=0}^{\infty}\mathcal{H}_{n}^{\pm}$$ for bosons and fermions. Also, one defines creation and annihilation operators $$a(h)$$ and $$a^{\dagger}(h)$$ on $$\mathcal{F}^{\pm}(\mathcal{H})$$.

Now, as far as I understand, this is all quantum mechanics, not QFT. However, these ideas seem to find analogues in QFT, and this is the point where I get confused.

On section I.5 of Feldman, Trubowitz and Knörrer's book there is a quick discussion on (fermionic) QFT and it is stated that, in this context, creation and annihilation operators are special families $$\{\varphi^{\dagger}(x,\sigma):\hspace{0.1cm} x \in \mathbb{R}^{d}, \hspace{0.1cm} \sigma \in \mathcal{S}\}$$ and $$\{\varphi(x,\sigma):\hspace{0.1cm} x \in \mathbb{R}^{d}, \hspace{0.1cm} \sigma \in \mathcal{S}\}$$ on a Hilbert space $$\mathcal{H}$$. This is very different than the creation and annihilation operators mentioned above. For instance, these are now families of operators indexed by $$x$$ and $$\sigma$$. I believe this is a reflection of the fact that we passed from QM to QFT. But I'm really lost here and I don't know what's the difference between these two constructions and definitions. Can anyone help me, please? I'm mainly interested in understanding the second approach, since the first one I believe I understand (at least sufficiently well). If, in addition, you could suggest some reference where these ideas of Feldman, Trubowitz and Knörrer are discussed in more details and with rigor, I'd appreciate!

ADD: Based on Feldman, Trubowitz and Knörrer's book, it seems to me that the understanding of these objects (to be more precise, the objects they briefly describe in the first 2 pages of section I.5) is fundamental to understand the formulation of a bunch of QFT models (at least for fermions). Thus, I'd appreciate if someone could elaborate a little more on the structure behind these creation and annihilation operators and its connections to the quantum case that is needed to understand the rest of the discussion on FTK's book. In other words, I think I just need to better understand these first definitions (and how are they connected with the usual quantum case I (seem) to know) to be able to understand the rest of the text.

• Morally speaking, a linear functional on a function space is an integral. So $a(h) = \int a(x) h(x) dx$. Replace $a$ by $\varphi$ and make $h(x)$ multicomponent-valued to account for spin to get the FTK notation. If you shift the focus from rigor to motivation and historical context, you can't go wrong with Dirac's classic Principles of Quantum Mechanics (1930), §59–65. Aug 15, 2020 at 18:28
• The general idea is that in QFT -unlike QM- the wave functions themselves are "upgraded" to operators. If you wish to understand the basics, i suggest you first focus on the radiation field, that is QED (quantum electrodynamics). Among all field theories this is the first to be formulated (historically) and maybe the only one which has a rigorous formulation. Aug 16, 2020 at 11:35
• Mandl's book: archive.org/details/… (especially the first few chapters), could provide a descent starting point. Aug 16, 2020 at 11:36
• @IgorKhavkine this is something that came to my mind, too. But this is just a different way to represent these operators, right? I mean, is it all? And, besides, why doing that? (Sorry if my questions are really basic, I'm still getting acquainted with all this). Aug 16, 2020 at 15:40
• @KonstantinosKanakoglou thank you so much for the comments! I'm going to take a look at this book right now. Aug 16, 2020 at 15:41

The connection can be seen by taking $$H = L^2(\mathbb{R}^3)$$ in the first explanation. This is the Hilbert space of a nonrelativistic, spinless, three-dimensional particle. By direct summing the symmetric (antisymmetric) tensor powers of $$H$$ we get the Hilbert space of an ensemble of noninteracting Bosonic (Fermionic) nonrelativistic, spinless, three-dimensional particles, known as Fock space. The $$n$$th tensor power represents the states in which $$n$$ particles are present.

Now we have "creation" and "annihilation" operators which take states in the $$n$$th tensor power into the $$(n \pm 1)$$st tensor power. For each state $$h$$ in the original Hilbert space $$H$$ there is a creation operator which tensors with $$h$$ and symmetrizes (antisymmetrizes), taking the $$n$$th tensor power into the $$(n+1)$$st, and its adjoint which goes in the opposite direction and removes a tensor factor of $$h$$.

In the physics literature one usually works with idealized creation/annihilation operators for which the state $$h$$ is a fictional Dirac delta function concentrated at some point of $$\mathbb{R}^3$$. This is what is described in your second explanation. As is usual in physics, the Hilbert space is unspecified, but in the case of free fields it corresponds to the Fock space in the first explanation.

Fock space is inadequate to model interacting fields (indeed, here the mathematical issues become deep and fundamentally unresolved). However, it is not trivial; for instance, one can study free quantum fields against a curved spacetime background and derive Hawking radiation, the Unruh effect, etc. Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics by Wald is an excellent, mathematically rigorous explanation of this setting.

In QFT the intuition is that one has a separate Hilbert space at each point of space, and one takes their tensor product to get the Hilbert space of the entire field. I indicated how, intuitively, the Fock space models a "measurable tensor product" of a family of harmonic oscillators (Bosonic case) or two-state systems (Fermionic case) indexed by all the points of space in my answer here. See Section 2.5 of my book Mathematical Quantization for a full explanation.

• very nice! Bravo Nik. And thanks for filling us in with a more accurate pciture Aug 16, 2020 at 16:50
• Thank you Mirco! Aug 16, 2020 at 17:00
• @NikWeaver thank you so much for the amazing answer. This clarifies the things to me. Let me ask you something. You said that my second representation arises when the state $h$ is taken to be a delta. Dimock's book discusses many particle systems where $\mathcal{H}=L^{2}(\mathbb{R}^{3})$. In this case, $\mathcal{F}^{\pm}=\bigoplus_{n=0}^{\infty}\mathcal{H}^{\pm}$ , where $\mathcal{H}^{\pm} = L^{2}_{\pm}(\mathbb{R}^{3n})$ are, respectivelly, symmetric and anti-symmetric subspaces of $L^{2}(\mathbb{R}^{3n})$. continues... Aug 17, 2020 at 14:54
• At some point, Dimock states: "If $\psi \in L^{2}_{\pm}(\mathbb{R}^{3n})$ is a continuous function, then in $a(h)\psi$ we can take $h$ to be a $\delta$-function and define an operator $a(x)$ by $(a(x)\psi)(x_{1},...,x_{n-1}) = \sqrt{n}\psi(x,x_{1},...,x_{n-1})$." Question: according to you answer, this is precisely what my $\varphi^{\dagger}(x,\sigma)$ are, right? The only difference is that, in this case, the spins are not being taking into account, so that $\sigma$ is omitted. Aug 17, 2020 at 14:58
• Yes, I believe that is what the $\phi^{\dagger}(x,\sigma)$ are. Just be careful about $\psi$ needing to be continuous --- this makes the operator $a(x)$ unbounded on each summand, whereas $a(h)$ for $h \in L^2(\mathbb{R}^3)$ would be bounded on each summand ... Aug 17, 2020 at 15:51

Disclaimer: I am not a mathematical physicist.

Even with one Hilbert space, namely the quantum harmonic oscillator, you can define "creation-annihilation" operators, except that in this case they simply raise or downgrade the energy level of the single particle system.

Now, you consider the Fock space $$\mathcal{F}^{\pm}(\mathcal{H}) = \bigoplus_{n=0}^{\infty}\mathcal{H}_{n}^{\pm}$$ the way you describe above: it is actually a functor, hence the infamous dictum that second quantization is a functor.

Therein, you define again the two operators, but you re-interpret them as ladder operators which, from the ground state, create and destroy particles. Formally they behave very much as with the toy harmonic oscillator, and that analogy is far-reaching:

basically it tells you that the quantum field described by the Fock functor can get "excited": particles are excitations of the void (in fact there are some beautiful pictures of quantum fields as (infinite) ensembles of (coupled ) harmonic oscillators, see here).

What has this to do with the second definition? If the quantum field creates and annihilates particles, it can do it at each point of your ambient space. Hence the indexes...

• Regarding whether each quantum field can be represented using creation and annihilation operators on a Fock space -- the answer is no. Aug 16, 2020 at 5:15
• Hello again Mirco! And thanks for the answer! You said that the Fock space formulation is already QFT, right? This is confusing to me. See, I know that, rigorously, a QFT theory is defined by Wightman axioms, right? But intuitively (since I'm not an expert), this seems really like a QM problem. What is you understanding on this? Aug 16, 2020 at 15:37
• @Iamwill you see the beaty of MO? You try to answer a question, and sometimes you mak a fool out of yourself :) . Now, assuming that one can take it (and I can), it is GOOD. Why? because you learn what you do not know. My picture was an oversimplification, and was based on the erroneous assumption that every QFT can somehow be "represented" as a Fock space, in other words, that the Hilbert space which is assumed by the Wightman axioms can be realized as a Fock tensor product. Aug 16, 2020 at 16:21
• Robert disabused me of this. So, here i my new understanding (perhaps also faulty): some QFT can be so realized, but actually they are in a way the silly ones. Now, the exact picture still escapes me Aug 16, 2020 at 16:21
• @MircoA.Mannucci I think to recognize one's mistake is as important as learning from others. In my opinion, this is a very nice humble behavior of yours! This is well done science for sure! Aug 16, 2020 at 16:30