A symmetry can be explicitly broken by adding terms in the Lagrangian that aren't compatible with the symmetry, and we say the symmetry is softly broken if all these terms have positive mass dimension. This is important for, e.g. SUSY phenomenology because all the symmetry breaking terms are relevant, so the symmetry "reappears" at high energies and solves the hierarchy problem.

However, I don't understand what soft symmetry breaking physically means. In the usual effective field theory picture, we simply write down every term in the Lagrangian that is compatible with the given symmetries. If a symmetry is broken, then we have to write down all terms that break it, not just the relevant ones.

I suspect the physical picture is this: we start with the symmetry intact, spontaneously break it at a high scale $\Lambda$, and then integrate out whatever fields are responsible for the spontaneous symmetry breaking. As a result, the effective theory does not respect the symmetry, not even nonlinearly. However, if we work at a low scale $\Lambda' \ll \Lambda$, the irrelevant terms are too small to be seen, so we only get relevant (soft) terms. Is this the right way of thinking about what's going on?

  • $\begingroup$ In the beginning, you talk about explicit breaking, whereas in your last paragraph about spontaneous breaking. Is it correct? $\endgroup$
    – apt45
    Commented Aug 9, 2018 at 16:57
  • $\begingroup$ @apt45 Yes, that's intentional. My question is whether the particular kind of explicit symmetry breaking we are allowing in soft symmetry breaking is actually the result of spontaneous symmetry breaking we've integrated out. $\endgroup$
    – knzhou
    Commented Aug 9, 2018 at 16:58
  • $\begingroup$ There is some widely held intuition that at high enough energy, all (anomaly-free, 0-form) global symmetries are restored. This makes sense in a thermodynamic picture, where eventually everything melts into a featureless soup of quark plasma or whatever. However, I don't know how to explain it with irrelevant operators, since there are so many of them (and many break the symmetries), but only a measure zero subset of irrelevant perturbations to an effective field theory will flow to a QFT in the UV (where our intuition works). I guess it must be these special operators preserve all symmetries. $\endgroup$ Commented Aug 9, 2018 at 18:26
  • $\begingroup$ Aren't there always infinitely many irrelevant terms that respect a given symmetry? What does it mean to write down all of them? $\endgroup$
    – tparker
    Commented May 24, 2019 at 3:30
  • $\begingroup$ @tparker Well, we can stop at some order where the additional irrelevant terms are irrelevant at the energies and accuracies we work at. $\endgroup$
    – knzhou
    Commented May 24, 2019 at 15:40

1 Answer 1


This is a matter of terminology and the physics behind it is very simple.

How do we break a symmetry explicitly? As you know, that's enough to add symmetry-breaking operators to the Lagrangian: relevant, marginal and irrelevant operators do the job.

Why should we break a symmetry explicitly? I would say because of the evidence. Phenomenology tells us what are symmetries (accidental or not) in a given energy regime. If the symmetry is believed to be approximate (because its exactness may lead to a non-observed mass spectrum), then we need to modify the IR.

The term "soft breaking" means you are introducing modifications to the theory which do not spoil the initial UV behaviour. It is an infrared statement realized by introducing relevant deformations (e.g. mass terms). In other words, by breaking the symmetry through relevant deformations, you are deforming the IR behaviour of your theory but not the UV properties. Equivalently, if the soft breaking terms comes with a coupling $\lambda$, then the high energy property of the theory should not change as $\lambda \rightarrow 0$.

Notice that the IR is very often accessible to experiments and it's the energy regime where we can establish whether a theory is correct or not. From an EFT point of view, where you don't have access to the UV detail of the theory, it does not make any sense to ask what mechanism would have produced these soft breaking terms. Unless you are interested in finding the UV complete theory :)

I'll make two examples, hoping not to confuse you more.

Two examples

  • Baryon and Lepton numbers in the Effective Standard Model. Besides anomalies and non-perturbative effects, these accidental symmetries may be broken by irrelevant operators. This means that you are performing an ultraviolet (UV) modification of the theory. In other words, you are selecting different kinds of UV completions which may lead, in the infrared (IR), to symmetry-breaking irrelevant operators. For example, consider a massive scalar leptoquark $\phi$, namely a scalar field which couples to quarks $q$ and leptons $\ell$ through the coupling $g\bar{q} \phi \ell$. In this context, $g$ is a coupling and I don't care about chiralities. Now, unless the UV theory has some very particular structure, this interaction will highly produce irrelevant operators $\sim g^2 (\bar{q} \ell)^2$ which break the (accidental) symmetries of your IR theory.

  • The theory of a Goldstino (a supersymmetric companion of a Goldstone boson) which arises when you break $\mathcal{N}=1$ SUSY spontaneously. The EFT of a consistent interacting Goldstino $\psi$ starts with operators of dimension-8 (dim-6 operators are forbidden by the non-linear realization of supersymmetry) $$ \mathcal{L} = i\bar{\psi}\gamma^\mu \partial_\mu \psi + (\bar\psi\gamma^\mu\partial_\mu\psi)^2 + ... $$ The theory is still invariant under supersymmetry transformations: SUSY is just spontaneously broken and non-linearly realized. However, a mass term for $\psi$ does break the symmetry explicitly and this corresponds to an IR modification of the theory. It is not relevant here which mechanism has generated the mass term. Whatever it is (Higgs-like mechanism), it surely breaks supersymmetry. This is the case of soft breaking: IR symmetry-breaking modification of the theory. You can see it also as a modification of the pole structure of the $2\rightarrow 2$ scattering amplitudes involving the Goldstino. Ask me more, if you want.

I also suggest you this wonderful answer by Luboš Motl.


  • Consider the simple verion of the Chiral Lagrangian arising from spontanously breaking of $U(1)$. None here is telling you what the UV-complete theory is. In order to perform IR computations, what you need to know is the symmetry breaking pattern. In the IR you have a goldstone boson which interacts through irrelevant operators $$ \mathcal{L} = \frac{1}{2}(\partial \pi)^2 + \frac{a}{f_\pi^4}(\partial\pi)^4 + ... $$ and the theory is invariant under $\pi \rightarrow \pi + c$ where $c$ is a constant. Now, the theory looks nice, it has a different number of properties and you can make predictions. The only problem is that in the IR we don't see scalar particles. This means that the symmetry we were considering before must be explicitly broken. To break the symmetry below the cut-off (you are sensible to physical effects below the cut-off), you need a relevant deformation, i.e. a mass term. Adding a mass term physically means considering a new and completely different theory!! An example of mechanism which may produce (in the UV-completed theory) such relevant deformation in the IR is an Higgs-like mechanism and in Nature you see such mechanism. See next item.

  • Take the chiral lagrangian from $SU(2)_L \times SU(2)_R\rightarrow SU(2)_V$. The situation is completely analogous to the previous case: in the IR you have the massless pions. Now, the pions do have a mass. This means that, from and EFT point of view the $SU(2)_L\times SU(2)_R$ symmetry we started from had to be explicitly broken. You can realize this fact in the IR by adding a mass term for the pions. Now, we know the UV theory. The mass of the pions depends on the mass of the quarks which are generated through the Higgs mechanism. The yukawa couplings explicitly break the chiral symmetry (you have different yukawas for up and down-type quarks).

  • $\begingroup$ My question is about what the physical origin of soft symmetry breaking is. You state: "It is not relevant here which mechanism has generated the mass term." This appears to me to be just avoiding answering the question! What kind of mechanism would generate only relevant symmetry breaking operators? $\endgroup$
    – knzhou
    Commented Aug 9, 2018 at 18:12
  • $\begingroup$ Also, I think you have misunderstood what I'm proposing. Consider the SM. Symmetries forbid a mass term for the electron. Now let the Higgs break electroweak symmetry, integrate everything out except for the electron and photon, and go to low energies. You're left with QED, which does allow a mass term for the electron. That seems to me to be a prototype of soft symmetry breaking. The mass term for the electron explicitly breaks electroweak symmetry, if we look at QED in isolation. $\endgroup$
    – knzhou
    Commented Aug 9, 2018 at 18:15
  • $\begingroup$ Along with the relevant electron mass, there are also irrelevant operators generated, but they are too small to see if we are working at low energies. That's the motivation for my proposal: I think soft symmetry breaking is just an abbreviation for what I said above: spontaneous symmetry breaking at a high scale integrated out. $\endgroup$
    – knzhou
    Commented Aug 9, 2018 at 18:18
  • $\begingroup$ The question "which kind of mechanism?" is too general and depends on the UV completion you deal with. As far as you ask for the mechanism, you are also demanding to know the UV theory. From an EFT point of view, soft breaking is synonymous with relevant deformation. Physically, it means you are modifying the IR behaviour of the theory. What mechanism would do that, depends on the UV-completed theory. $\endgroup$
    – apt45
    Commented Aug 9, 2018 at 18:18
  • $\begingroup$ Could you give an example of such a mechanism? $\endgroup$
    – knzhou
    Commented Aug 9, 2018 at 18:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.