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During lectures we've often written down 'the most general' Lagrangians in QFT with the reasoning 'Because why else?'.

Suppose we are looking at a theory with real scalar fields $\phi(x)$ in arbitrary number of dimensions. I'm looking for a explanation why the most general Lagrangian (which is Lorentz invariant, has a canonical kinetic term and polynomial interactions) is given by $$\mathscr{L} = \frac{1}{2}\partial^\mu \phi \partial_\mu \phi - \frac{\lambda_2}{2!}\phi^2 - \frac{\lambda_3}{3!}\phi^3 - \frac{\lambda_4}{4!}\phi^4 - \frac{\lambda_5}{5!}\phi^5 - ...$$ where the dots represent higher order interactions.

Or is the one given here not the most general one?

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    $\begingroup$ That's not the most general Lagrangian. For example, you could put in $\phi^2 \partial_\mu \phi \partial^\mu \phi$. $\endgroup$
    – knzhou
    May 13, 2023 at 23:01
  • $\begingroup$ I'm assuming a canonical kinetic term, which is quadratic in the fields and derivatives... $\endgroup$ May 13, 2023 at 23:03
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    $\begingroup$ There is also a matter of selecting physically meaningful interactions. I can't give an exhaustive answer, but there are stability problems with the solutions to equations of motion with certain terms, for example cubic terms in derivatives. Many resources can be found in physics.stackexchange.com/q/670965 $\endgroup$ May 13, 2023 at 23:08
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    $\begingroup$ The term proposed by @knzhou is not a kinetic term since it is quartic. It is one derivative interaction vertex coupling four fields. In particular you could have all sorts of interactions in which there are arbitrarily many derivatives appropriately distributed among the fields and contracted with one another. You could also have quadratic terms with more derivatives. These remarks are just to point out that what you have written is simply not the most general possible Lagrangian built out of a scalar field. $\endgroup$
    – Gold
    May 14, 2023 at 0:07
  • $\begingroup$ @Gold Could you then maybe write down what the most general Lagrangian is? And why would we not have considered these derivative terms? $\endgroup$ May 14, 2023 at 9:07

2 Answers 2

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Some things were already mentioned in the comments but I figured a recap might not hurt.

So generally the answer depends on a lot of factors. Some of which I will discuss in the following.

The first information one needs to settle on in the manifold $M$ on which you want your physics to happen, together with the metric on that manifold. Now for your case, this will likely be $\mathbb{R}^{d,1}$, i.e. Minkowski space. However, note that generally, this may differ e.g. if you are looking to couple your theory to gravity the metric will be determined dynamically.

I will assume $\mathbb{R}^{d,1}$ from now on.

The next parameter to fix is probably going to be the fields you want to couple (here the real scalar $\phi$ ) which will determine the particle content of your final theory.

Now there are (as far as I can see) two more things to determine: The first is the global symmetry group (or the introduction of gauge symmetries) your theory should contain. Lorentz invariance should almost always be contained in your choice of the symmetry group, however otherwise this really depends on your physics. E.g. for a lot of systems, a $\mathbb{Z}_2$ symmetry or $O(n)$ symmetry is introduced.

Now the next step is to write down all terms you can build from your fields and (appropriate) derivatives thereof that are compatible with your symmetry group.

E.g. for your theory, this would be all terms of the form $\prod_{i} \partial_\mu^{n_i} \phi^{m_i}$, where all partial derivatives have to be contracted in order for the theory to be Lorentz invariant. The Lagrangian would then be any arbitrary sum of these terms. (I think this is all of them at least but sometimes in more complicated theories there are non-obvious topological terms or the like that are easy to miss).

Now the last thing you have to think about is kind of subtle. This has to do with the renormalization of your theory, something I am not sure you have learned about yet. The essence of this idea is that your low-energy physics should not depend on the high-energy details of your lagrangian, though the way to see what effect this actually has is a little complicated so I'm not going to explain it here.

The essence however is that your terms will split into two subsets:

Renormalizable and non-renormalizable terms. The effect of this depends on the situation. Certain theories one only thinks of being valid at certain energies, these are called effective field theories and here you can just ignore the renormalizability of your terms and essentially freely pick which terms your lagrangian includes. (e.g. just compare with experiment which terms give the best scattering amplitudes).

If your theory however should be viewed as a true high-energy description of your physics you have to exclude all non-renormalizable terms from your Lagrangian. Which terms these correspond to depends on the dimension of your Manifold and the exact choices of fields as this has to do with the energy/mass dimension of terms e.g. $[\phi] = (d-1)/2$ and $[\partial] = 1$ allows you to calculate the mass dimension of $\partial^n \phi^m$ terms. Only terms with dimensions smaller than $-[d^{d+1}x]$ should be included. Which reduces the number of allowed terms considerably. (This is e.g. the reason why higher derivative terms do not show up in $3+1 d$ for your theory)

As such the question if your Lagrangian is the most general depends on the dimension and symmetry group, as well as its application.

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  • $\begingroup$ Thanks for the answer! You are spot on: my question is in fact in terms of an exercise where we are looking for which space-time dimensions $d$ a theory is renormalizable. We have to start from the most general Lagrangian which is Lorentz invariant, has a canonical (i.e. quadratic in the fields and the derivatives) kinetic term and only includes polynomial interactions of degree less than or equal to 6. The wording of 'most general' just seems a bit vague to me as we've only discussed lagrangians as the one in my post. $\endgroup$ May 14, 2023 at 14:02
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The most general Lagrangian density $𝔏(Ο†,βˆ‚Ο†)$ that is a Lorentz-invariant function of fields with components $Ο†^a$, and their gradients $βˆ‚_ΞΌΟ†^a$, where $0 ≀ ΞΌ < n$, and $0 ≀ a < N$ (where $n > 1$ and $N > 0$) is a function $𝔏\left(Ο†^a, I^{ab}\right)$ of the Lorentz-invariants $Ο†^a$ and $I^{ab} = Β½ g^{ΞΌΞ½} βˆ‚_ΞΌΟ†^a βˆ‚_Ξ½Ο†^b$; where $g_{ΞΌΞ½}$ a Lorentzian-metric with inverse $g^{ΞΌΞ½}$ and $βˆ‚_ΞΌ = βˆ‚/βˆ‚x^ΞΌ$. The Einstein summation convention is being used on the indices.

The derivatives of the Lagrangian density comprise the constitutive coefficients of the field theory: $$𝔉_a = \frac{βˆ‚π”}{βˆ‚Ο†^a},\quad ΞΊ_{ab} = \frac{βˆ‚π”}{βˆ‚I^{ab}},$$ and are in general, themselves, functions of the Lorentz invariants that satisfy the differential equations: $$ \frac{βˆ‚π”‰_a}{βˆ‚Ο†^b} = \frac{βˆ‚π”‰_b}{βˆ‚Ο†^a},\quad \frac{βˆ‚π”‰_a}{βˆ‚I^{bc}} = \frac{βˆ‚ΞΊ_{bc}}{βˆ‚Ο†^a},\quad \frac{βˆ‚ΞΊ_{ab}}{βˆ‚I^{cd}} = \frac{βˆ‚ΞΊ_{cd}}{βˆ‚I^{ab}}.$$ The Euler-Lagrange equations have a form $$βˆ‚_ΞΌ\left(ΞΊ_{ab} g^{ΞΌΞ½} βˆ‚_Ξ½Ο†^b\right) = 𝔉_a$$ whose Lagrangian-dependence is completely encapsulated in the coefficients, but is in all other respects independent of the Lagrangian density.

If the "kinetic term" is "canonical", as you call it, the coefficients $ΞΊ_{ab}$ are independent of the field and its gradients; i.e. $$\frac{βˆ‚ΞΊ_{bc}}{βˆ‚Ο†^a} = 0,\quad \frac{βˆ‚ΞΊ_{ab}}{βˆ‚I^{cd}} = 0.$$ Therefore, $$\frac{βˆ‚π”‰_a}{βˆ‚I^{bc}} = \frac{βˆ‚ΞΊ_{bc}}{βˆ‚Ο†^a} = 0,$$ so that the coefficients $𝔉_a$ are functions of the $Ο†^a$ alone. The total differential for the Lagrangian density reduces to the form: $$d𝔏 = 𝔉_a(Ο†) dΟ†^a + d\left(Β½ ΞΊ_{ab} g^{ΞΌΞ½} βˆ‚_ΞΌΟ†^a βˆ‚_Ξ½Ο†^a\right),$$ and since $$\frac{βˆ‚π”‰_a}{βˆ‚Ο†^b} = \frac{βˆ‚π”‰_b}{βˆ‚Ο†^a},$$ we have $$𝔉_a = -\frac{βˆ‚π”˜}{βˆ‚Ο†^a},$$ for some function $π”˜(Ο†)$. Thus, $$𝔏 = Β½ ΞΊ_{ab} g^{ΞΌΞ½} βˆ‚_ΞΌΟ†^a βˆ‚_Ξ½Ο†^a - π”˜(Ο†) + 𝔏_e,$$ where $𝔏_e$ is a function of whatever else is involved in the dynamics.

You said you wanted $π”˜(Ο†)$ to be a polynomial, so it's a polynomial, and that's your answer. The coefficients $ΞΊ_{ab}$ and coefficients of the polynomial function $π”˜(Ο†)$ may depend on whatever other fields are in the dynamics.

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