# Why does the QFT Lagrangian not already use operators?

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• Gerenuk
In summary, canonical quantization involves transforming a Lagrangian to a Hamiltonian and then "putting hats" on the fields to make them operators. This allows for the derivation of equations of motion for the Hamiltonian. However, it is not always possible to put hats on the Lagrangian directly, and other methods such as the Faddeev-Popov formalism may be necessary. Additionally, a systematic approach using group theory can also be used to determine the operators for a quantum theory.

#### Gerenuk

TL;DR Summary
Is it mathematically possible to have a working QFT Lagrangian based on operators without canonical quantization? (and without path integrals)
I've learned that in canonical quantization you take a Lagrangian, transform to a Hamiltonian and then "put the hat on" the fields (make them an operator). Then you can derive the equations of motion of the Hamiltonian.

What is the reason that you cannot already put hats in the QFT Lagrangian? Therefore write the Lagrangian with operators and go straight to Euler-Lagrange equations without any additional steps like canonical quantization. You would have to add an operator which transforms the operator expression into a scalar for the Lagrangian. So it it possible to put all that into the Lagrangian alone and do straightforward Euler-Lagrange equations from that? (without introducing additional machinery like path-integrals!)

Of course you can do that, because the Lagrangian defines all the quantities you need to do the canonical-quantization prescription (in the Heisenberg picture). You start from the Lagrangian and then calculate the canonical conjugate field momenta as well as the Hamiltonian and then "put hats" on the field symbols, defining their algebra by the canonical equal-time commutation (bosons) or anti-commutation (fermions) relations.

Note that this prescription does not always work as expected. E.g., for the electromagnetic field and QED there are characteristic problems with the formalism due to gauge invariance. In this case there are two ways out: (a) you work in a completely fixed gauge (e.g., Coulomb gauge), which is not manifestly covariant, write down the Lagrangian for the corresponding gauged theory and do the canonical prescription or (b) use the socalled Gupta-Bleuler formalism, where you can work, e.g., in the Lorenz gauge and use weak conditions to constrain the states to the physical Hilbert space such that you get a unitary S-matrix for these physical states and gauge-invariant observables (and observables always must be gauge invariant of course).

Alternatively the path-integral formalism is much more convenient, particularly for gauge theories. The Faddeev-Popov formalism is much more straight forward than the operator formalism, particularly for non-Abelian gauge theories. For the latter the manifestly covariant operator formalism has been developed only after the Faddeev-Popov formalism was known and BRST invariance discovered. Within the path-integral formalism the socalled "background field gauge" is however even more convenient, because instead of Slavnonv-Taylor identities of the general Fadeev-Popov formalism you get back a gauge-invariant effective quantum action and thus more simple Ward-Takahashi identities very similar to the Abelian QED case.

• topsquark
vanhees71 said:
Of course you can do that, because the Lagrangian defines all the quantities you need to do the canonical-quantization prescription (in the Heisenberg picture). You start from the Lagrangian and then calculate the canonical conjugate field momenta as well as the Hamiltonian and then "put hats" on the field symbols, defining their algebra by the canonical equal-time commutation (bosons) or anti-commutation (fermions) relations.
Hmm, in my question I rather ask if you can "put hats" before doing any conjugate momenta. In fact, never doing conjugate momenta at all and no Hamiltonian either. Just - if mathematically possible - "put hats" on the Lagrangian and do Euler-Lagrange equations only, to get all results for the dynamics.

"Putting hats" after transforming the Lagrangian in a particular way seems rather artificial (even though it works).

Gerenuk said:
Hmm, in my question I rather ask if you can "put hats" before doing any conjugate momenta.
If you don't know which operators are conjugate momenta, then you don't know the commutation relations between operators. And if you don't know the commutation relations, then how do you know that they are operators?

So it seems to me that mathematically you cannot "put hats" from the beginning.

• topsquark and vanhees71
Exactly. Just putting hats doesn't help. You need to know the operator algebra, which generates the observable algebra of your theory. One heuristic way is "canonical quantization", and for that you need the Hamiltonian formalism. The heuristics is that Poisson brackets between observables in the classical theory become commutators among the corresponding operators (modulo a factor ##1/\mathrm{i}## of course).

• topsquark and Demystifier
Demystifier said:
If you don't know which operators are conjugate momenta, then you don't know the commutation relations between operators. And if you don't know the commutation relations, then how do you know that they are operators?

So it seems to me that mathematically you cannot "put hats" from the beginning.
You construct the commutations such that you still get the same results for QFT. It may be more complex than just guessing [q,p], but the question is if there is any way at all to use operators in the Lagrangian and define commutations (the algebra) such that you get identical results, but skip all the Hamiltonian formalism. No path integral either. Just the usual Lagrangian mechanics, but with more complex algebraic objects.

If cannot be done, then there must be a proof that it's algebraically impossible?

Gerenuk said:
You construct the commutations such that you still get the same results for QFT. It may be more complex than just guessing [q,p], but the question is if there is any way at all to use operators in the Lagrangian and define commutations (the algebra) such that you get identical results, but skip all the Hamiltonian formalism. No path integral either. Just the usual Lagrangian mechanics, but with more complex algebraic objects.

If cannot be done, then there must be a proof that it's algebraically impossible?
As far as I know nobody has done something like that, but I'm not aware of any theorem saying that it's impossible.

• topsquark
Demystifier said:
As far as I know nobody has done something like that, but I'm not aware of any theorem saying that it's impossible.
Ok, Thanks! I'm only starting with QFT and I assume that the math has been studied a lot. So probably it is not possible to use Lagrangian formalism alone to construct the correct dynamics of whatever objects you need. I just have no idea how to find a source that mathematically discusses where exactly the issues appear.

Of course, "canonical quantization" is only a heuristic way to quickly make an educated guess for a quantum theory from the corresponding classical theory. A more systematic approach is to use group theory. In the case of relativistic QFT you look for the irreducible unitary representations of the Poincare group to define the quantum theory of elementary particles. This, together with the assumption of causality and locality, is the most convincing argument for the form the QFT used to formulate the Standard Model of elementary particle physics.

• topsquark and Demystifier
vanhees71 said:
Of course, "canonical quantization" is only a heuristic way to quickly make an educated guess for a quantum theory from the corresponding classical theory. A more systematic approach is to use group theory. In the case of relativistic QFT you look for the irreducible unitary representations of the Poincare group to define the quantum theory of elementary particles. This, together with the assumption of causality and locality, is the most convincing argument for the form the QFT used to formulate the Standard Model of elementary particle physics.
From this particles-first point of view, which is used in the textbooks by Weinberg and Duncan, a particle is a "more fundamental" object than a field. Namely, particles are physical objects that one really wants to describe, while the field is only a technical tool for that. From this point of view, the notion of "second quantization" actually makes sense, see Sec. 6.3 in the Duncan's book.

The canonical quantization, on the other hand, is natural from the fields-first point of view, which is used in most other high-energy QFT textbooks. From that point of view, the notion of "second quantization" does not make much sense.

Of course, each of the two points of view has some advantages and disadvantages. They are complementary to each other. They differ philosophically, but at the end they lead to the same measurable results.

During my life, I had phases in which I preferred particles-first approaches, as well as phases in which I preferred fields-first approaches. At the moment I am in the fields-first phase, but it may change in the future. Last edited:
• gentzen and topsquark
Well, although you are right that Weinberg and also Duncan take a "particles-first approach", at the end they discuss, of course, the only hitherto successful type of relativistic QT, i.e., local (microcausal) relativistic QFT. At the end, from a physical point of view, even starting from a "particles-first approach" you end up with QFT, and that's, in my opinion, not by accident but quite "natural" in relativistic physics.

I don't think that point particles are a well-defined concept for relativistic physics at all. It's just that in Newtonian physics the model of interacting point particles is the most simple dynamical model describing successfully an amazing range of phenomena (including the motion of bodies in astronomy, which was among the main motiviations for forcing Newton to write his "Principia"). In relativistic physics point particles make only a lot of trouble, which has become clear very early already before Einstein's seminal paper of 1905 in Lorentz's classical electron theory which already then lead to the study of models of electrons as (rigid or elastic) bodies by Lorentz himself and Abraham (and for sure some more physicists at the time). Within the fully established SR, I'd say Poincare had the best description. Today we know that indeed the point-particle concept is highly problematic. Working with an action principle, one can show that a Poincare invariant Hamiltonian formalism for interacting point particles leads only to the trivial case of non-interacting particles. The model that goes as far as one can get is that of a single particle moving in an external field, but then (using electromagnetism as the most simple model for interactions) you run into the issue of the infamous "radiation-reaction problem", which seems not to have a fully consistent formulation, and the best one can do is to use the "Landau-Lifshitz approximation" to the Lorentz-Abraham-Dirac equation. So from a classical point of view a point-particle approach is highly problematic, while field-theoretical descriptions (relativsitic hydrodynamics, transport equations, etc.) don't show all these pathologies. In this sense it seems to me that a field-theoretical approach also to relativistic QT is more "natural" also from a heuristic point of view.

Of course, no matter how you start, you end up with local relativistic QFT as the only so far successful description of "interacting matter" (to use a neutral word for what we are studying in this theory). The price to pay is that we have only rather abstract definitions of "elementary particles" as particular quantum states of the various quantized fields (asymptotic free Fock states).

Of course, if philosophers try to clarify this already complicated subject, all hope is lost to understand anything ;-)).

vanhees71 said:
Of course, if philosophers try to clarify this already complicated subject, all hope is lost to understand anything ;-)).
Clarifications are always welcome, but emphasis on confusion has its merits too. • 