# I Why are free-field Lagrangians quadratic in fields?

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1. May 1, 2017

### Frank Castle

What is the intuitive reasoning for requiring that a Lagrangian describing a free-field contains terms that are at most quadratic in the field?

Is it simply because this ensures that the EOM for the field are linear and hence the solutions satisfy the superposition principle implying (at least in the classical) sense, that wavepackets do not interfere with one another as they propagate past one another, i.e. they are free-fields?!

Furthermore, what is the motivation for including the term $\frac{1}{2}m^{2}\phi^{2}$ in the free-field case? I get that the parameter $m$ is attributed to the mass of the field a posteriori, but is the reason for the inclusion of such a term in the first place? Is it simply because a priori there is no reason not to - one should include all possible terms up to quadratic order?! Or is there also some physical motivation as well, in that from quantum mechanics, the wave function of a relativistic particle (of mass $m$) should satisfy the Klein-Gordon equation?!

2. May 1, 2017

### stevendaryl

Staff Emeritus
Yes, if you include higher-order terms in the Lagrangian, such as $\phi^3$, that will lead to terms in the equations of motion corresponding to self-interactions. In the Feynman diagrams, there would be processes where a single particle splits into two or more particles.

There are two different motivations for Lagrangians: one is to work backwards from the equations of motion. So the Klein Gordon equation implies a certain Lagrangian. The other motivation is to just look for all possible Lagrangians that have certain desirable properties, such as being renormalizable, being Lorentz-invariant, being gauge-invariant, etc. Since the mass term works for both motivations, you might as well include it.

3. May 1, 2017

### Frank Castle

But what is it about linearity that makes the equations interaction free? Is it simply because it ensures that the solutions satisfy the superposition principle and as such implying that there are no interactions. Is this motivated from the classical case in which waves satisfying linear equations of motion are free in the sense that they propagate "through one another" without interacting, since each wave satisfies the equation of motion independently of the other (implied by the superposition principle), which means that we can consider the waves independently of one another. This would certainly not be true if they interact since the interaction of one wave with the other would couple their respective equations of motion?!

Last edited: May 1, 2017
4. May 1, 2017

### ChrisVer

A term let's say in the Lagrangian of the form $\phi^n~,~n\ge 3$ would end up in vertices with $n\ge 3$ legs.... obviously this is an interaction!

5. May 1, 2017

### Frank Castle

True. But would it be correct to say (before introducing Feynman diagrams, etc.) that the reason why Lagrangians of free-fields are quadratic is because this leads to linear equations of motion which guarantees that the field can be treated independently of other fields, since the superposition principle applies. If the equation of motion isn't linear then this isn't the case and the dynamics of the field are effected by the presence of other fields.

6. May 1, 2017

### Orodruin

Staff Emeritus
Yes and no. The superposition principle applies, which means that the field configurations based on any sources may be treated independently. There need not be other fields in order for a field theory to be interacting, the most basic example probably being $\phi^4$ theory.

It is more than this, the field configurations based on any sources for the same field are independent. For example, in electromagnetism, you can superpose the EM fields of any charge configurations and obtain a resulting field that is the field configuration for the total charge configuration. Yet there is only the electromagnetic field.

7. May 1, 2017

### Frank Castle

So is the point that linearity guarantees that the superposition principle applies such that different field configurations can be treated independently, implying that they are not interacting, regardless of whether or not it is self-interaction or with another type of field?!

Also, does a particular field configuration simply correspond to the solution to the equation of motion for that field with particular boundary conditions? Or is it simply a particular assignment of field values, $\phi(t,\mathbf{x})$, at each spacetime point?!

8. May 1, 2017

### Orodruin

Staff Emeritus
Yes.

This depends on whether it is an on-shell or off-shell configuration.

9. May 1, 2017

### Frank Castle

Ok cool. I feel like I have a better understanding of the reasoning why quadratic Lagrangians are used for free-fields.

So what does the term "field configuration" mean in a more general sense?

10. May 1, 2017

### Orodruin

Staff Emeritus
An assignment of a field value to every point in space-time. Generally, such a field configuration does not necessarily satisfy the equations of motion.

11. May 1, 2017

### Frank Castle

Ah ok, that makes sense. In the case of self interacting fields, does this mean that different configurations of the same field are no longer independent, they become coupled, such that their equations of motion are no longer independent?

12. May 1, 2017

### Frank Castle

Is this why the classical Maxwell equations are linear, because the EM fields produced by different charge configurations are non-interacting, i.e. they are independent of one another, such that they superpose to obtain a net EM field at any given point due to the total charge configuration. The point being that this net EM field configuration is a sum of independent (i.e. non-interacting) field configurations sourced from the respective charge configurations.