# Overarching Quantum Field Theory Question

I've been self-studying several qft books and I am having a little trouble reconciling the original purpose for introducing quantum fields with the function of these fields as operators. I understand that it is desirable to introduce quantum operators as functions of spacetime in order to reconcile Quantum Mechanical measurements with relativity, as the vanishing of the commutator for quantum operators related via spacelike interval implies causality. Based on this fact, it clearly follows that we want to introduce quantum fields. However, I am a little unclear on what motivation there is for ultimately associating these operators with particle creation/annihilation at point x, in the case of theories that can be solved via cannonical quantization. Is the harmonic oscillator treatment a simple result of field equations of motion, for instance for the Klein-Gordon Equation, looking like an oscillator in momentum space, or is the interpretation of the field operators more fundamental for the theory.

• Related, as an interesting background to fields in general, not an answer though: archive.is/7p5mu
– user167453
Sep 29 '17 at 6:41
• Have you read up on "Second Quantization?" It is the main step in going from QM to QFT. Its main merit is solving the problem of how to deal with antiparticles. Dirac's original anti-electron theory (as part of relativistic QM) necessitates a non-observed sea of negative energy states. The creation/annihilation operators, in conjunction with the normal ordering prescription, eliminate this nonsensical notion. See e.g. Itzykson/Zuber, section 3-3-2, eq. (3-164), for the resulting properties of charge and energy: "charge is no longer a positive square norm". Sep 29 '17 at 12:05

Have you read Weinberg's QFT Volume 1?

Creation/annihilation operator decomposition is apparently not the only way to do it, but it makes it easy to satisfy Cluster Decomposition principle which is one of the axioms that is used for QFT.

Given several particle groups ($\alpha_i$, $\beta_i$) and S-matrix, when one moves one group very far from each other - S-matrix is expected to decompose:

$$S_{\beta_1 \alpha_1 \beta_2 \alpha_2 \cdots} \to S_{\beta_1\alpha_1} S_{\beta_2\alpha_2 \cdots}$$

This in turns implies that the connected part of S-matrix must vanish when some particles are moved away to infinity in position space. Vanshing connecting part implies that the connected part of S-matrix must have just one 4-dimensional momentum-space delta function:

$$S^C_{\beta\alpha} = \delta( E_\beta - E_\alpha) \delta^3( p_\beta - p_\alpha) C_{\beta\alpha}$$

(Where $C$ has no further delta functions).

There is a theorem that says that S-matrix will have this form if the Hamiltonian could be expressed in terms of products of creation/annihilation operators with a single momentum-space delta function:

$$H = \sum_{NM} \int d^3p_1 \cdots d^3p_N \, d^3q_1 \cdots d^3q_M \, a^\dagger(p_1) \cdots a^\dagger(p_N) a(q_1) \cdots a(q_M) h_{NM}(p, q)$$

Where $h_{NM}$ has at most one 4-dimensional delta-function.

This is why we used creation/annihilation operators in fields.

I still recommend referring to Weinberg, as he offers elaborated insight into this

• The modern approach is indeed to start from annihilation and creation operators (and from quantum states before that by looking at representations of the Poincaré algebra), and then to build fields out of them, and Weinberg is indeed the reference here. Shame on the -1 but I guess you were expected to explain in more details!
– user154997
Sep 29 '17 at 10:40
• I agree, this needed more explanation. Please let me know if there is anything wrong in this rather brief answer. Sep 29 '17 at 11:55
• This is correct but now I'd suggest you make the extra effort to connect $H$ to the $S$-matrix and to the fields! Sure, this is all in Weinberg, but in there it is split between a handful of chapters, and your answer could make a nice synthetic map for potential readers who don't have a lifetime to spend on the Weinberg (I mean the cumulated time of my reading it must run into the order of half a year, and I am only talking about the bulk of the first book!)
– user154997
Sep 29 '17 at 13:23
• Thanks for checking it! I'm really not sure if further elaboration on this would be productive at this point. There are lots of careful arguments and discussions in Weinberg, omitting those would be a mistake in my opinion. Sep 29 '17 at 13:47
• Thank you for the response! I will use weinberg to complement your discussion. Sep 30 '17 at 1:24

I'll try to explain this in the way I found better for me. Your question is:

However, I am a little unclear on what motivation there is for ultimately associating these operators with particle creation/annihilation at point x, in the case of theories that can be solved via cannonical quantization.

Which I believe boils down to: why on earth do we decompose a field into exponentials and postulate that the expansion coefficients become annihilation and creation operators in a certain Fock space?

The thing is: if you follow the canonical quantization line of reasoning, it is quite clear that given a system described by a field $\phi(x)$ with conjugate momentum $\pi(x)$ you want to impose

$$[\phi(x),\phi(x')]=[\pi(x),\pi(x')]=0,\\ [\phi(x),\pi(x')]=i\delta(x-x').$$

Now there is a subtle thing in QFT which is not always mentioned. In QM, given the position and momentum operators, if you want them to satisfy

$$[Q_i,P_j]=i\delta_{ij}$$

There is a theorem called "Stone-Von Neumman theorem" which ensures there is a unique construction which gives a Hilbert space of states on which $Q_i$ acts as multiplication by the eigenvalue and $P_i$ acts as the corresponding derivative as usual.

This allows us to talk about one abstract space of states $\mathcal{E}$ and be sure that the $Q_i$ representation do exist, and this is what is usually done in QM.

This isn't the case in QFT. The commutation relations above do not give rise to a unique construction of a Hilbert space. In Mathematical terms: there are infinitely many unitarily inequivalent representations of the Canonical Commutation Relations (CCR).

Now, the fact is that you must build a representation: a Hilbert space on which the relations hold. It turns out that one way to do it is exactly the Fock Space route.

Essentialy you identify a single particle space of states $\mathfrak{h}$. This one is required to satisfy a simple requirement: it must carry a unitary representation of the Poincare group. In this way we implement the symmetries of Minkowski spacetime and furthermore, we get the existence of, for example, the momentum operator in $\mathfrak{h}$.

Then you build the symmetric Fock Space associated to $\mathfrak{h}$. In that setting you get $\mathfrak{H}=\mathcal{F}_S(\mathfrak{h})$. The latter one comes with creation and annihilation operators. With this in place you then construct the fields, loosely speaking, as

$$\phi(x)=\int \dfrac{d^3p}{(2\pi)^3} \dfrac{1}{\sqrt{2\omega_p}}(a_p e^{-ipx}+a_p^\dagger e^{ipx})$$

So lets keep track: you don't derive this for the quantum field. You define it. You start with the CCR, then since you can't use Stone-Von Neumann you must build the representation directly. Then one shows that starting with a single particle space, building the Fock space and defining the quantum fields like that do define a representation of the CCR, plus since this is the form of the solution for the classical field using Fourier transform, this automatically satisfies the equations of motion.

Your question might be: if there are infinitely many inequivalent representations, why to pick this one? And well, the most convincing answer I've found myself is that this one is physically reasonable because it directly gives you the interepretation of a theory of relativistic particles which agrees with observations, and furthermore, it is directly hinted by the solution of the classical field in Fourier space.

The motivation is the same one you already have in QM: you never experimentally observe "half of an electron" so your field must also have a particle-like nature. Moreover in QFT, due to $E=mc^2$, the number of these particles is not conserved, allowing for decays and related phenomena.