# Why are Lagrangian densities and actions in Quantum Field Theory always Lorentz invariant?

Newtons laws of motion are Galilean invariant. But the Newtonian Lagrangian and Newtonian action for a particle are not Galilean invariant. Similarly we want the Euler-Lagrange (EL) equations in quantum field theory to be Lorentz invariant. But how can we say that the Lagrangian (I mean Lagrangian density) should always be Lorentz invariant? Is there any deep reason for that?

One reason I thought of is since we quantize the classical relativistic Lagrangians which are already invariant we get invariant Lagrangians. But then I am not sure why even in classical field theory the Lagrangians are invariant.

I found in some places that Lagrangian must be invariant for the EL equations to be invariant. But that is not obvious. In some places it is said that action must be invariant and since the volume element $$d^4x$$ is invariant so must Lagrangian densities. But then why should action be invariant?

Edit: This answer explained it using group theory which I haven't studied yet. If there is an answer without using Group theory you can write it otherwise close this question.

• You can do non-relativistic quantum field theory where Lorentz symmetry doesn't play a role. There are at least three different questions here with different answers: "Why is QFT often relativistic?" or "Why do we want a Lorentz invariant action?" or "Why does invariance of the action/Lagrangian imply invariance of the equations of motion?". The last question is a duplicate of physics.stackexchange.com/q/51327/50583 and perhaps also solves the second. Which of these questions do you really want to ask? Jun 28 at 15:47
• I believe that to appreciate this story completely you should study Weinberg's "The Quantum Theory of Fields" Volume 1 up to the chapter on canonical quantization. In the first chapters Weinberg will discuss exactly how to construct a relativistic quantum mechanical theory. The lesson one learns in the end is that without starting from a Lorentz invariant lagrangian density it is remarkably difficult to do it. In a sense, Weinberg shows that the usage of quantum fields and Lorentz invariant lagrangians appears as the simplest way to construct relativistic quantum mechanical theories.
– Gold
Jun 28 at 15:55
• @ACuriousMind I know that non-relativistic quantum field theory doesn't need Lorentz invariant equations of motion. I was asking about relativistic QFT. I do not have any doubt regarding "Why is QFT often relativistic?" and "Why does invariance of the action/Lagrangian imply invariance of the equations of motion?". I am only asking "Why do we want a Lorentz invariant action?". Jun 28 at 16:09
• @Gold thanks I will check that. Jun 28 at 16:09
• @SolubleFish I am not asking whether or not equations of motion are Lorentz invariant. Jun 28 at 16:28

Actually, the Lagrangian for Newtonian mechanics is Gallilean invariant. While it does "change" under a tiny boost $$x \mapsto x + \epsilon t$$, it changes by a total time derivative. When a Lagrangian changes by a total derivative under a transformation, we still say the Lagrangian is "invariant" because adding a total derivative to a Lagrangian doesn't change its dynamics.
Your point is essentially that, whereas we recognise $$\tfrac12m\dot{x}^2-V(x)$$ changes under $$x\mapsto x+vt$$, interest in $$\tfrac12\partial^\mu\phi\partial_\mu\phi-V(\phi)$$ hinges on its invariance under Lorentz transformations of $$x^\mu$$. (For now, let's not go to more elaborate alternatives for either.) The comparison here is flawed, though, and the issue isn't that Galilean invariance is inaccurate; we could replace the first Lagrangian with a relativistic alternative, and it wouldn't address the point I'm about to make.
Note that $$x^\mu$$ in the second example is actually analogous to $$t$$ in the first; going in the other direction, the equivalent of $$x$$ is $$\phi$$. The role of Lorentz transformations is to covariantly transform $$\partial_\mu$$, which is analogous to $$\frac{d}{dt}$$, which is invariant under Galilean transformations. By contrast, the $$\phi$$ Lagrangian isn't invariant under $$\phi\mapsto\phi+v_\mu x^\mu$$.