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It is an usual practice that any quantum field theory starts with a suitable Lagrangian density. It has been proved enormously successful. I understand, it automatically ensures valuable symmetries of physics to be preserved. But nevertheless the question about the generality of this approach keeps coming to my mind. My question is how one can be sure that this approach has to be always right and fruitful. Isn't it possible, at least from the mathematical point of view that a future theory of physics does not subscribe to this approach?

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Actually, physicists advance first physical equations, and then, if possible, write the corresponding Lagrangian. Doing in the opposite direction is funny but does not guarantees equations describing some physics. Moreover, the equations obtained for the principle of least action may have unphysical solutions. –  Vladimir Kalitvianski Jan 21 '11 at 21:48
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Vladimir, I think what your wrote is what physicists SHOULD do, but in actual fact they write down Lagrangians based on observed symmetries from experiments. Mathematically, the typically used method is glorified curve fitting. –  Carl Brannen Jan 21 '11 at 23:40
    
Vladimir, Carl, I think what Vladimir wrote is what physicist don't do anymore (at least in fairly fundamental physics) and shouldn't do. An equation falls from the sky as least as arbitrarily as a Lagrangian density. And equations are not that easily quantised, but alas, the world seems to be quantum, so we want to jump to quantum theories as fast as possible. (Of course, if we are doing completely classical stuff such as environmental physics, we might me happy with just equations.) –  Turion Jun 5 '12 at 9:18

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up vote 23 down vote accepted

That's an excellent question, which has a few aspects:

  1. Can you quantize any given Lagrangian? The answer is no. There are classical Lagrangians which do not correspond to a valid field theory, for example those with anomalies.

  2. Do you have field theories with no Lagrangians? Yes, there are some field theories which have no Lagrangian description. You can calculate using other methods, like solving consistency conditions relating different observables.

  3. Does the quantum theory fix the Lagrangian? No, there are examples of quantum theories which could result from quantization of two (or more) different Lagrangians, for example involving different degrees of freedom.

The way to think about it is that a Lagrangian is not a property of a given quantum theory, it also involves a specific classical limit of that theory. When the theory does not have a classical limit (it is inherently strongly coupled) it doesn't need to have a Lagrangian. When the theory has more than one classical limit, it can have more than one Lagrangian description.

The prevalence of Lagrangians in studying quantum field theory comes because they are easier to manipulate than other methods, and because usually you approach a quantum theory by "quantizing" - meaning you start with a classical limit and include quantum corrections systematically. It is good to keep in mind though that this approach has its limitations.

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Thanks for your brilliant answer. Thanks a lot. –  user1355 Jan 21 '11 at 17:56
    
Perfect answer, Moshe, +1. –  Luboš Motl Jan 21 '11 at 19:25

As far as you're wondering about 'quantum' field theories, all bets are off - just take a look at the arXiv or on Google. However, most of those theories seem (to me) less well studied than regular QFT. They do have a lot of structure in common with normal field theory (you can still have a Hamiltonian, for example).

In terms of classical fields, I might be giving some kind of trivial answer, but you could consider a (classical) field that obeys a differential equation that can not be derived from a Lagrangian through the Euler-Lagrange equations. Maybe someone with a background in PDE's can elaborate on this.

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I was confused by this too.
Maybe I'm wrong, but at some point I made my mind in a following way:

  1. The most general way to describe 'quantum anything' is a Feynman path integral
  2. In this picture for each 'path' we have to have an action
  3. If we want causality and are agree that our world is Poincaré invariant, then the action must be local.
  4. which means that action must be an integral of "something" over the space-time.
  5. This "something" is a Lagrangian

Therefore questioning generality of the Lagrangian you are actually:

  • either questioning the generality of Feynman path integral
  • or questioning the validity of Poincaré invariance
  • or the necessity of casuality

Repeat -- I can be wrong. And I would be happy if someone points out my mistake.

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It is (1), which is the most general way to "quantize" - start with a classical limit and include quantum corrections systematically. There are inherently quantum theories which have no classical limits, and they fall outside (1) almost by definition. –  user566 Jan 21 '11 at 17:05
    
I thought causality and locality are two different things. How does causality and Poincare invariance imply locality? –  user7757 Jun 28 at 12:11

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