In both classical and quantum physics Lagrangians play a very important role. In classical physics, paths that extremize the action $S$ are the solutions of the Euler-Lagrange equations, and the action is given by the integral of a Lagrangian. $$ S[q] = \int\mathrm{d}t L(q, \dot{q})$$ $$ \delta S[q] = 0 \iff q(t) \in \{Solutions~to~EoM\}$$ For fields the picture is similar, with the action coming from the integral of a lagrangian density: $$ S[\phi] = \int\mathrm{d}^4x \mathcal{L}(\phi, \partial_{\mu} \phi)$$ $$ \delta S[\phi] = 0 \iff \phi(t, \mathbf{r}) \in \{Solutions~to~EoM\}$$

I learnt then that in QFT we see that this as a consequence of amplitudes being path integrals of the exponential of the action, for particles and fields respectively:

$$K(t_f, t_i, q_f, q_i) = \int Dq~\exp i S[q]$$ $$K(t_f, t_i, \phi_f, \phi_i) = \int D\phi~\exp i S[\phi]$$

Where in both cases the action is an integral of the corresponding lagrangian or lagrangian density, and the classical paths come from saddle point integration (correct me if I'm wrong here, I am by no means an expert in QFT, also I'm using units where $\hbar = 1$)

The Lagrangian is, if I understand correctly, for the most part, "free" to be whatever you want it to be if it fits empirical results. Of course, it must be Lorentz invariant, local, etc, etc, because the universe is Lorentz invariant, local, etc.

But I can imagine, maybe, that there could be some strange theory that had solutions to its equations of motion that could not come from a lagrangian. In this strange theory $S[q]$ can not be written as $\int L$, instead it is some other complicated functional.

My question is as twofold:

1) Can actions always be written in that form if we allow the lagrangian to depend on all derivatives, and I'm wrong?

2) Is there anything specifically odd with the theory having an action that can't be written as $\int L$ ? Is my theory maybe necessarily non-local? Maybe it can't be Lorentz invariant?

I understand that it would be "odd" since current theories all come from lagrangians, it would also probably be difficult to deal with, since the usual machinery of Noether's theorem would be unavailable (probably?) but a priori I don't see anything particularly natural about a Lagrangian, and I would like to understand if there is anything that makes it philosophically or intuitively appealing to have theories whose actions can be written in that form.

  • $\begingroup$ By the way, even if we are talking about Lagrangians, the answers to this question are inevitably quantum. So perhaps the tag "classical-mechanics" is not necessarily relevant. Just a remark, I'm not suggesting to remove it. $\endgroup$
    – MannyC
    Commented Mar 20, 2020 at 2:19

2 Answers 2


Lagrangian theories are indeed obiquitous because those are the ones we can understand better and ultimately those with which we can do computations. However, sadly, they only make sense when the theory admits a weakly coupled description. The reason is the very existence of the couplings. The fact that your theory has couplings means that there is a fine tuning that leads me to a free theory, where all the couplings are turned off. This is not a general property of QFTs. So there are two points that should be made

  1. Not all theories have a Lagrangian description in terms of weak interactions and weakly coupled fields.
  2. Those that have such a description must have an action functional of the form $S = \int \mathcal{L}$.

For point 1. there are several examples that can be constructed in supersymmetry for which one can prove that no Lagrangian description can exist. I am talking about $(2,0)$ six dimensional superconformal field theories. A non supersymmetric scenario which is unlikely but still viable is that of interacting seven dimensional conformal field theory (CFT). In that case it is obvious that no Lagrangian can exist since all interactions are irrelevant in 7d.

Non-Lagrangian constructions appear mostly in the realm of CFTs because, due to their rigidity, we can grasp their properties even if we essentially cannot compute anything. Suppose there exist two CFTs which are both strongly coupled and one flows into the other under renormalization group. Such a scenario would imply a QFT with no Lagrangian description anywhere in the energy range. And it is plausible to imagine that this might happen somewhat frequently.

A reference on the six dimensional CFT I mentioned is $[1]$. A paper where evidence for the existence of 7d CFTs is $[2]$.

As for 2. the reason is essentially locality. It is not known how to make a theory that respects the principle of microcausality if we start from an action that is not the integral of a local Lagrangian. In fact, it is believed that it is impossible to have a causal theory if the starting point is a nonlocal Lagrangian. By local I mean that $\mathcal{L}$ can be written as single spacetime integral of a polynomial of the fields at the same point with a finite number of derivatives. Sadly, I don't have references for this, perhaps some comments can improve this part of the discussion.

$[1]\;$ Christopher Beem, Madalena Lemos, Leonardo Rastelli, Balt C. van Rees The $(2,0)$ superconformal bootstrap 1507.05637;
See also Clay Cordova, Thomas T. Dumitrescu, Xi Yin Higher Derivative Terms, Toroidal Compactification, and Weyl Anomalies in Six-Dimensional $(2,0)$ Theories and refs. [10-16] therein. 1505.03850

$[2]\;$ Clay Cordova, G. Bruno De Luca, Alessandro Tomasiello $\mathrm{AdS}_8$ Solutions in Type II Supergravity 1811.06987

  • 1
    $\begingroup$ So if I understand correctly there is a correspondance "QFT + Weakly Coupled + Local => Action can be written as the integral of a Lagrangian" and there is a generally believed correspondance "Causal => Local" I think you clarified a lot with your paragraph on locality, my question is a bit confused in that regard. $\endgroup$
    – Ignacio
    Commented Mar 20, 2020 at 2:44

Dijkgraaf-Witten theories (Chern-Simons theory for a finite gauge group) are an example of a quantum theory without a Lagrangian, which nevertheless has an action (an exponentiated action to be precise). It is neither odd nor difficult to deal with, in fact, toy models like these are often used precisely because they are easier to handle than the quantum field theories we see in more familiar situations. The reason there is no Lagrangian here is because the gauge group is discrete, so there is no gauge field. Nevertheless, the quantum theory is perfectly well defined.

Lagrangians are particularly useful when we want to take a classical theory and get a quantum theory out of it, but if you look at any of the sets of axioms that go into defining a quantum field theory on its own, the Lagrangian does not really play much of a role.

  • $\begingroup$ How would you define "a quantum field theory" in the context you describe? If you say, a theory of path integrals and summed histories that result in amplitudes and probabilities, aren't you more or less forced to the concept of the lagrangian density (or action) as you need something that takes the field configurations and outputs amplitudes for you to sum? $\endgroup$
    – BjornW
    Commented Jan 18, 2023 at 12:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.