When I learned Newtonian mechanics I found a vast variety of computations that I could do and that was so interesting. And it was so when I learned Maxwell theory. When I started learning QFT I hoped to find much more variety of computations and it's so disappointing when I see only two! Of course I'm not a fan of computations but every new computation is a new window to understanding the theory and appreciating it's beauty and power.

  • $\begingroup$ This question might be better as a discussion. It's true that classical mechanics and electromagnetism are at least initially mathematically richer than what you first encounter in QFT. However, all of your intuition from classical field theory carries over into quantum field theory, but is also supplemented by special quantum effects that you will study in more detail in more advanced courses. For now, focus on physical reasoning, and appreciate the amazing fact that you can deduce so much considering only simple local scattering. $\endgroup$
    – TLDR
    May 29, 2016 at 2:25
  • $\begingroup$ Nothing stops you from learning to calculate the path integral for the hydrogen atom, it's just a demanding calculation that requires a lot of stamina and technical skill but that, at the end of the day, doesn't teach you much about anything. $\endgroup$
    – CuriousOne
    May 29, 2016 at 3:37
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    $\begingroup$ Your assertion that we "only" learn the computation of cross sections and decay rates is simply wrong. You learn that first because it's a) the most straightforward application of perturbative QFT and b) what it was designed to do. The most obvious other thing you can compute are expectation values of arbitrary operators. There are subfields (non-equilibrium QFT, condensed matter,...) that do not "just" compute cross sections or decay rates. This question cannot really be answered because its premise is just false. $\endgroup$
    – ACuriousMind
    May 29, 2016 at 10:04
  • $\begingroup$ so I will be grateful if you give a short or long list of "other computations" in page? of text? I already know chapter 4 of SCHWARTZ's text where the Lamb shift of hydrogen atom is computed. texts will be good but as i'm graduated in math i can't understand advanced physics articles! thanks. $\endgroup$
    – moshtaba
    May 29, 2016 at 17:27

2 Answers 2


The small number of "conceptually independent types of processes and calculations" is exactly a symptom of the theory's being fundamental! Even in classical physics, all calculations could have been mathematically reduced to the calculation of the final state that evolves from an initial state (or a state that is stationary etc.).

In quantum mechanics, this must be replaced by the calculations of the probabilities that an initial state evolves to a final state. According to QFT, all objects in the world may be described by a Hilbert space with some particle excitations (creation operators).

All the dynamical transformations are included in the probabilities to transform an initial state of particles to a final state of particles. For the calculation to be nontrivial, the initial state contains at least 1 particle.

If it contains 1 particle, the only nontrivial process that may occur is the decay of the particle. If it contains 2 particles, they may do something and the probability is unavoidably described in terms of a cross section because the probability depends on the flux of the beams etc.

If the initial state contains at least 3 particles, it becomes extremely unlikely - at least in the empty space - that all the particles interact simultaneously. Instead, some interaction of 2 particles occurs first, and that interaction may be reduced to the cross section calculation from the previous paragraph.

So basically all processes may be reduced to probability amplitudes of the two kinds. This is a big victory of reductionism.

It doesn't mean that QFT doesn't allow one to calculate everything you could have calculated in classical physics – or other theories less complete than QFT. These calculations are just hard and, in some sense, they are not fundamental or elementary.

In classical mechanics, one may compute some behavior of a machine with lots of wheels and gears etc. This is clearly an example of applied, not fundamental, physics, and researchers of QFT generally do not do applied physics.

People usually study relativistic QFT because they want to learn the fundamental laws of physics and that's why they're not deliberately focusing on more complex and composite "exercises". But lots of them are possible. In principle, the behavior of all composite objects may be probabilistically predicted using QFT. And advanced papers using QFT surely do use lots of concepts that differ from decay rates and cross sections, i.e. viscosities (in the AdS/CFT correspondence applied to ion physics etc.) and new "emergent" methods to calculate them.

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    $\begingroup$ "According to QFT, all objects in the world may be described by a Hilbert space with some particle excitations (creation operators)." That seems a rather general statement that is not true. The particle excitations only exist in the asymptotic free states - you cannot describe the actual states of the interacting theory as such particle excitations, because you don't have the notion of creation operators. Non-perturbative computations in e.g. lattice field theory or non-eq QFT do certainly not consider "particle states". $\endgroup$
    – ACuriousMind
    May 29, 2016 at 9:59
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    $\begingroup$ Well. the states may still be obtained by acting with functions - effectively polynomials - of the fields, and those may be decomposed and interpreted as some particle basis. One may have on-shell an off-shell Green's functions; the former just generalize the latter. They're not the same but I guess that they wouldn't change anything about the OP's feeling that the "number of types of things" calculated in a QFT is very limited. $\endgroup$ May 29, 2016 at 16:49
  • $\begingroup$ @ACuriousMind: Two interacting particles in Classical Mechanics (a bound state or a scattering process at finite time) can be equally represented as two "free" and "independent" quasi-particles - the center of mass motion and the relative motion with a certain energy each. The same is valid for quasi-particles in QFT. An external filed may "couple" the quasi-particle equations and change the corresponding occupation numbers. $\endgroup$ May 30, 2016 at 18:03

One can calculate a lot more in quantum field theory if one goes beyond asymptotic computations into thermal field theory. I recommend that you look at the book ''Nonequilibrium Quantum Field Theory'' by Calzetta and Hu.


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