Theories whose behavior would qualitatively change if their couplings were not fine-tuned to particular values are often dismissed as "unnatural" (in high-energy physics) or "unrealistic" (in condensed-matter physics), while theories whose couplings are constrained by symmetry requirements are happily accepted. Why is this? A symmetry constraint can be thought of as just a collection of fine-tunings that has some unifying pattern.

For example, a complex scalar field Lagrangian $$\mathcal{L}_1 = \partial_\mu \varphi^\dagger \partial^\mu \varphi - m^2 \varphi^\dagger \varphi - \lambda \left( \varphi^\dagger \right)^2 \varphi^2 \tag{1}$$ with a global $U(1)$-symmetry $$\varphi \to e^{i \theta} \varphi$$ can be thought of as a general Lagrangian $$\mathcal{L}_2 = \partial_\mu \varphi^\dagger \partial^\mu \varphi - m^2 \varphi^\dagger \varphi - m'^2 \left( \left( \varphi^\dagger \right)^2 + \varphi^2 \right) - \lambda \left( \varphi^\dagger \right)^2 \varphi^2 - \lambda' \left( \left( \varphi^\dagger \right)^4 + \varphi^4 \right) - \dots\tag{2}$$ in which the primed couplings have all been fine-tuned to zero. (Things are admittedly more complicated in the case of gauge quantum field theories, because in that case the gauge symmetry requires you to modify your quantization procedure as well). This kind of "fine-tuning" is a little less arbitrary than the usual kind, but arguably it's not much less arbitrary.

  • $\begingroup$ Fun fact, the second lagrangian is also "fine-tuned", due to the presence of the symmetry $\varphi\leftrightarrow\varphi^\dagger$. $\endgroup$ – Adam Sep 4 '16 at 8:53
  • $\begingroup$ Fine tuning requires order by order matching of loop corrections coming to m in terms of $\lambda$ (which gets correction itself). A symmetry group which can handle a proper cancellation for both fermionic and bosonic loop to every order must show this symmetry. This strong condition requires odd graded elements in your symmetry algebra which requires an extension of symmetry to supersymmetry to avoid fine tuning. $\endgroup$ – ved Sep 4 '16 at 14:16
  • $\begingroup$ @Adam No, I think that symmetry comes from the requirement that the Lagrangian be real/the Hamiltonian be Hermitian. $\endgroup$ – tparker Sep 4 '16 at 16:53
  • $\begingroup$ @tparker: isn't that some fine tuning too ? ;-) $\endgroup$ – Adam Sep 4 '16 at 17:22
  • $\begingroup$ @Adam I would consider the unitarity of the theory to be a fundamental postulate of quantum mechanics, necessary for Born's probabilistic interpretation to make any sense. $\endgroup$ – tparker Sep 4 '16 at 17:57
  1. If one has a theory $S[\alpha]$ that depends on some parameters $\alpha$, one can always introduce new artificial parameters $\beta$, and claim for free that the theory $S[\alpha,\beta]:=S[\alpha]$ has a symmetry $\beta\to \beta+ b$. This is of course not a very interesting symmetry.

  2. A symmetry, say $\alpha \to\alpha + a$, is only interesting if different values of $\alpha$ are physically meaningful/realizable/accessible. In contrast, if there is a physical principle/superselection rule that completely fix $\alpha$ to a certain value, then we are essentially back to pt. 1.

  3. How do we quantify naturalness? It seems relevant to mention 't Hooft's definition of technical naturalness, cf. Refs. 1 & 2. At energy scale $\mu$, two small parameters of the form $$m^{\prime 2}~\sim~ \varepsilon^{\prime} \mu^2 , \qquad \lambda^{\prime}~\sim~ \varepsilon^{\prime}, \qquad|\varepsilon^{\prime}|~\ll~ 1, $$
    in the Lagrangian density $${\cal L}_2~=~{\cal L}_1 - m^{\prime 2} \left( \left( \varphi^{\dagger} \right)^2 + \varphi^2 \right) - \lambda^{\prime} \left( \left( \varphi^{\dagger} \right)^4 + \varphi^4 \right)$$ is technical natural, since the replacement $\varepsilon^{\prime}=0$ would increase the symmetry of the system, namely it would restore the $U(1)$-symmetry $\varphi \to e^{i \theta} \varphi$, cf. the Lagrangian density ${\cal L}_1$.

  4. Similarly, at energy scale $\mu$, two small parameters of the form $$m^2~\sim~ \varepsilon \mu^2 , \qquad \lambda ~\sim~ \varepsilon , \qquad|\varepsilon |~\ll~ 1, $$
    in the Lagrangian density $${\cal L}_1~=~ \partial_{\nu}\varphi^{\dagger} \partial^{\nu} \varphi - m^2 \varphi^{\dagger} \varphi - \lambda \left( \varphi^{\dagger} \varphi \right)^2$$ is technical natural, since the replacement $\varepsilon=0$ would increase the symmetry of the system, namely it would restore the translation symmetry $\varphi \to \varphi+a$.


  1. G. 't Hooft, Naturalness, Chiral Symmetry, and Spontaneous Chiral Symmetry Breaking, NATO ASI Series B59 (1980) 135. (PDF)

  2. P. Horava, Surprises with Nonrelativistic Naturalness, arXiv:1608.06287; p.2.

  • $\begingroup$ What do you think of fine-tuning/naturalness arguments in physics? $\endgroup$ – innisfree Sep 20 '16 at 14:33

I would argue exactly the opposite : picking a symmetry requires much less fine-tuning.

Indeed, when picking a symmetry, your are fine-tuning one "parameter", the fact that the model has a given symmetry, but effectively killing an infinite number of coupling in one sweep.

On the other end, standard fine-tuning forces you to put an infinite number of of coupling constants to zero. Since one is much smaller than infinity, I think that choosing a symmetry amounts to much less fine-tuning.

  • $\begingroup$ Well in field theories (other than ones that are being explicitly considered as effective theories with a known energy cutoff), we generally only consider renormalizable couplings, so fine-tuning actually only eliminates a finite number of couplings. I would argue that whether that finite number is "much" greater than one is largely a matter of opinion. $\endgroup$ – tparker Sep 4 '16 at 16:58
  • $\begingroup$ @tparker: you can require (old school) renormalizability, but that's not necessary for this discussion. Especially in a condensed matter context, where there are generically an infinite number of coupling constants (even in presence of symmetries). $\endgroup$ – Adam Sep 4 '16 at 17:25
  • $\begingroup$ In condensed matter, the microscopic lattice Hamiltonian may have an infinite number of couplings, but once you coarse-grain to a continuum theory all but a finite number are irrelevant under RG, so there are only a finite number of important couplings for explaining the low-energy physics. $\endgroup$ – tparker Sep 4 '16 at 18:01
  • $\begingroup$ @tparker: Not necessarily, if the system is not close to a fixed point (which is generically not the case, unless you fine tune the system ;-) ). And anyway, in a Wilsonian point of view, all (allowed) coupling constants are finite at a non-trivial fixed point. $\endgroup$ – Adam Sep 4 '16 at 18:10

Why is imposing a symmetry on a theory considered more “natural” than fine-tuning its couplings?

Take a one carat diamond (200 milligrams ofcarbon). It can be described by a simple symmetry


Unit cell of the diamond cubic crystal structure

Which is "simpler" : imagining the build up of the crystal by using the symmetry or listing all the (x,y,z,t) coordinates of the atoms, even the few ones in this unit cell?

Symmetries "simplify" concepts. Even your argument about "fine tuning of the primed couplings" falls into the category of simplification. One does not have to impose it to each of them, the symmetry does it.

I suppose if we were computers it would make no difference, but man is a pattern recognition animal, and finding patterns and repetitions of patterns is inherent in our tools of accumulation of knowledge, but this is outside physics.

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    $\begingroup$ How is fine-tuning related to a symmetric unit cell? $\endgroup$ – Statics Sep 4 '16 at 13:01
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    $\begingroup$ @Statics The "discovery" that a unit cell repeated described crystal symmetries is an analogue of symmetriew reducing fine tuning; if one sees the infinity coupling constants as uncorrelated as individual atom position might be seen as uncorrelated and needing individual determination of coordinate. It is an analogy. $\endgroup$ – anna v Sep 4 '16 at 13:53
  • $\begingroup$ I certainly agree that imposing a symmetry is simpler than fine-tuning many couplings in that you need to specify less information, but without explaining a physical mechanism that imposes the symmetry, it doesn't really seem much more natural to me. It just shifts the question from "why are the couplings fine-tuned to certain values?" to "why do the couplings have this funny symmetry pattern?" $\endgroup$ – tparker Sep 4 '16 at 17:03
  • $\begingroup$ @tparker Yes. You must have realized by now that ultimately physics does not answer "why" questions. Rather it answers "why" questions by pushing them to postulates and laws in as series of "whys". Naturalness is one of those traits, imo. It is, like Ocams razor, a preference we have as analyzers of nature. $\endgroup$ – anna v Sep 4 '16 at 17:24

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