So in the journey of trying to understand more about the strong interaction I have encountered some passages linking mass with strong interaction. Like from Greiner and Müller Quantum Mechanics - Symmetries

Clearly, all pion masses (energies) are nearly equal. In analogy to the mass difference between proton and neutron, this near equality may be interpreted as meaning that the strong interaction (which determines the dominant part of the mass) is invariant in isospin space, and that the small mass differences of a few MeV are caused by the electromagnetic or other interactions.

Also from Griffiths Introduction to Elementary Particles I read

Apart from the differences in quark masses, the strong interactions treat all flavors equally. Thus isospin is a good symmetry because the effective u and d masses are so nearly equal (which is to say, on a more fundamental level, because their bare masses are so small); the Eightfold Way is a fair symmetry because the effective mass of the strange quark is not too far from that of the u and d. But the heavy quarks are so far apart that their flavor symmetry is severely broken.

Now I do not really understand why or how the mass depends on the strong interaction or vice versa. From Griffiths it seems, that the reason why isospin is not a good symmetry for higher mass differences is, that only "apart from the mass differences" strong interaction treats the quarks equally. I was not aware that strong interaction cares about mass. How come? What exactly is the relation?

Actually I have thought that strong interaction doesn't care at all about mass and so I find these passages really strange. In the case that someone could clear up that no, indeed strong interaction does not depend on mass, than I would like to have another explanation of why isospin is a worse symmetry for huge mass differences.

Personally I had the feeling that it is just because exchanging the particles with very different mass makes the Lagrangian not really symmetric anymore, but there is no word on that in these passages I have quoted.


1 Answer 1


If the $SU(2)_f$ or $SU(3)_f$ symmetry were exact, then all members of the multiplets would be exactly equally massive. The neutron would be exactly as heavy as the proton, and perhaps the whole baryon octet, and similarly for the mesons. It would have to be true because the rest mass is nothing else than the energy in the rest frame and energy's operator (the Hamiltonian) commutes with the symmetry generators, by the definition of a symmetry.

This is not quite correct which means that the isospin $SU(2)_f$ symmetry and the analogous $SU(3)_f$ symmetry aren't exact symmetries – we mean that they're not exact symmetries of the laws of physics. $SU(2)_f$ is slightly broken while $SU(3)_f$ adding the strange quark is broken much more strongly. The original source of the breaking – the terms in the Hamiltonian that refuse to commute with the symmetry generators – are the mass terms which aren't "characteristic terms" of the strong interaction. However, they're still parts of the Hamiltonian so the whole Hamiltonian is only approximately symmetric.

Now, an important triviality you have to appreciate is that the separation of the Hamiltonian to various terms is just something in our minds and any realistic calculation depends on more or less all the terms.

So the fundamental terms in the Hamiltonian have many consequences which depend both on the mass terms as well as the flavor-blind, characteristic terms of the strong interaction (imagine loop diagrams that are affected both by the strong interaction vertices and by the masses). Sometimes we interpret these manifestations as manifestations of the strong force (only) because the strong force is the "more nontrivial" cause of these phenomena (mass terms are trivial, if you wish). So they're consequences of the strong force that nevertheless violate the flavor symmetries because the mass terms that always do matter violate these symmetries.

At sufficiently short distances or, equivalently, high energies, the strong interaction terms are more important than the mass terms. That's where the $SU(2)_f$ or $SU(3)_f$ symmetries are more accurate. The breaking by the mass terms may also be said to arise at long distances. This pattern is rather general in particle physics: symmetries tend to be good approximations at short distances and they break at distances longer than some critical "symmetry breaking scale" distance.

The breaking of the $SU(2)_f$ is caused just by the electromagnetism between the differently charged quarks which is substantially weaker than the strong force (the electromagnetic fine-structure constant is just 1/137.036, smaller than its strong counterpart); and by the 7-4 = 3 MeV mass difference between the up- and down-quarks. This 3 MeV is much smaller than the typical QCD mass scale, 150 MeV or so, so the isospin symmetry is very close to being an exact one.

The $SU(3)_f$ symmetry involves the strange quark that is about 150 MeV heavier than the light u,d-quarks. This is comparable to the QCD scale so the $SU(3)_f$ symmetry is just "marginally useful". It's still good enough to organize the multiplets in the eightfold way etc.

  • $\begingroup$ "(the Hamiltonian) commutes with the symmetry generators, by the definition of a symmetry." I'm sad to say, I was not aware of that! Can you recommend something (article, pdf file, book) that would give me more insight? So can I say, that really the strong force does not see mass? Also you say that the SU(2) is broken by electromagnetism, but the charges would be different even is the mass terms were the same, so would the symmetry still be approximate even if u and d had the same mass? I want to understand all of this more deeply. Literature? Thank you a lot for your answer, much appreciated! $\endgroup$
    – Jack
    Jun 30, 2013 at 12:34
  • $\begingroup$ Oh and was what I said about an (approximately) invariant Lagrangian (with respect to the exchange of e.g. u and d quarks) crazy talk? $\endgroup$
    – Jack
    Jun 30, 2013 at 12:47
  • $\begingroup$ Dear @Jack, I am confident that every introduction to symmetry/groups in physics should immediately explain to you that a symmetry is defined as something that doesn't change the action or the Hamiltonian. Buy e.g. amazon.com/Lie-Algebras-Particle-Physics-Frontiers/dp/… For a continuous symmetry generated by a generator such as $J_z$, the infinitesimal variation of another operator such as $H$ is $\delta H = i \cdot\delta\phi [J_z,H]$ so if the commutator is zero, the hamiltonian is invariant which means that it is symmetric under $J_z$. $\endgroup$ Jun 30, 2013 at 17:16
  • $\begingroup$ Yes, electromagnetism also violates the flavor symmetry, whether $SU(2)_f$ or $SU(3)_f$, on top of the mass terms, but again, the electric charge is effectively $e^2\sim 4\pi/137$ which is much smaller than $g_{QCD}^2 \sim 4\pi/3$ so inside hadrons, the strong interaction is just stronger which is also why it's called this way. ... I am not sure you said anything "crazy" aside from things I tried to clarify - the answer may be No or Yes. $\endgroup$ Jun 30, 2013 at 17:18
  • $\begingroup$ Note that $[J_z,H]=0$ may be interpreted as the invariance of $H$ under symmetries (rotations) generated by $J_z$; but it may also be interpreted as the symmetry of $J_z$ under the transformations (translations in time) generated by $H$, i.e. $\delta J_z = i\delta t\cdot [H,J_z]/ \hbar$. The latter invariance is the angular momentum conservation law - so we derive "Noether's theorem", the relationship between symmetries and conservation laws, simply from the two possible interpretations of the commutator. $\endgroup$ Jun 30, 2013 at 17:21

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