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Initially we were introduced to isospin as a way of abusing the approximate symmetry of the strong force, e.g. we would define an $I=\frac{1}{2}$ group of $p$ with $I_3 = +\frac{1}{2}$ and $n$ with $I_3 = -\frac{1}{2}$ and use it to define combined-particle-wavefunction states (with multiple protons and neutrons) with well-defined isospin as leading order approximations to the wavefunction. This worked quite well since $p$ and $n$ have very similar masses.

At the time we would also define a separate $I=1$ group for the pions, which again (for writing wavefunction containing multiple pions) worked well since they have similar masses (even though they have quite different masses to protons and neutrons)

However, later in the course, when determining whether certain reactions were allowed we would invoke isospin conservation but mix-and-match the isospin of different groups. Does using isospin like this arise from $SU(3)$ flavour symmetry rather than $SU(2)$ isospin symmetry? And therefore is it a worse approximation than just using isospin to write wavefunctions purely including particles within a single group? (I would expect so since the approximation that the particles have identical masses is much worse in this case)

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  • $\begingroup$ My question ultimately I guess is, is isospin conservation in reactions (as applied due to isospin being a subgroup of flavour) a worse approximation than when using isospin to find multi-particle wavefunction states just between, say, a proton and neutron. (Because the approximation of identical particles is worse in the former since the masses are more different) $\endgroup$
    – Alex Gower
    Apr 9, 2021 at 17:30
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    $\begingroup$ Are you thinking about a reaction involving pions and nucleons? $\endgroup$
    – ragnar
    Apr 9, 2021 at 18:37
  • $\begingroup$ Yes I am @ragnar. $\endgroup$
    – Alex Gower
    Apr 9, 2021 at 18:50

2 Answers 2

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Here is how I understand your question. An action like $$L=\bar Ni\gamma^\mu \partial_\mu N+ \bar N m_N N+\frac{1}{2}\partial_\mu\pi_a\partial^\mu\pi_a +\frac{m^2_\pi}{2} \pi_a\pi_a $$ is invariant under the two isospin charges $$Q_a=\int\!\! d^3x \,\,N^\dagger\frac{\tau_a}{2}N ,$$ and $$Q'_a= -\int\!\! d^3x \,\, \dot\pi_b(\epsilon_{abc})\pi_c,$$ each of which obeys the same su(2) lie algebra (commutation relations), but are otherwise disconnected. In fact, the nucleons N are in the doublet representation (I=1/2), and the pion in the triplet one (I=1, adjoint).

However, pions and nucleons interact, $$L=\bar Ni\gamma^\mu \partial_\mu N + \bar N m_N N+\frac{1}{2}\partial_\mu\pi_a\partial^\mu\pi_a +\frac{m^2_\pi}{2} \pi_a\pi_a +g\bar N( i\gamma_5\pi_a \tau_a)N,$$ and this interacting lagrangian is not invariant under Q and Q', but only under their sum, $$I_a=\int d^3x \,\,[N^\dagger\frac{\tau_a}{2}N+\dot\pi_b(-\epsilon_{abc})\pi_c], $$ which also satisfies the very same commutation relations.

You should check each of the three terms in the Lagrangian has $I_1$, $I_2$, $I_3$, eigenvalues zero. The underlying reason is that the flavor SU(2) subgroup of the flavor SU(3) you mention describes the u and d quarks, which are not equal in mass at all (!), but they are both much much smaller than the scale of the strong interactions, Λ, ~ 200 MeV, which dominates all strong amplitudes. So, for all practical purposes, the quark masses, 2 and 5 MeV, are negligible, and the symmetry is "good".

Symmetry breaking terms are computable systematically, but possibly are too advanced for you at this point. Within an isomultiplet, the mass differences are indeed of the order of a few MeV, but these are not dumb additive combinations of quark mass effects, and enter, instead, in astoundingly intricate ways, controlled by chiral symmetry breaking.

The interaction terms, which pull everything together, entail exchange of u and d quarks which you are probably learning about in your course. WP details exactly how.

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I think there may be some confusion in terminology in the question. The pions belong to the $I=1$ representation of isospin SU(2), while the nucleons belong to the $I=1/2$ representation. They are not different groups. This is directly analogous to a spin $1/2$ particle and a spin $1$ particle being different representations of the rotation SU(2) group.

The mass difference between pions and nucleons is also not a problem. If the up and down quark masses were identential (and if QED were turned off), then isospin would be a perfect symmetry. But the pion mass would still be very different from the nucleon masses. Isospin symmetry just means that the energy is independent of isospin projection $I_3$ (i.e. it is invariant under isospin rotations). It still allows for different isospin magnitudes $I$ to result in different energies.

Flavor SU(3) relates to the inclusion of the strange quark, which is not particularly relevant for pions and nucleons.

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  • $\begingroup$ So are you saying that since in both the cases of nucleons and pions the isospin symmetry is just 'treating up and down quarks as identical', the 'level of approximateness' of the isospin symmetry is the same? And since much of the mass differences of the composite particles come from strong binding energy differences, this does not break the isospin symmetry? $\endgroup$
    – Alex Gower
    Apr 9, 2021 at 22:05
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    $\begingroup$ Essentially, yes. $\endgroup$
    – ragnar
    Apr 9, 2021 at 22:33
  • $\begingroup$ Thank you. Am I right in thinking then that only once you start invoking isospin conservation between particles like Kaons (so that they include strange quarks) and nucleons/pions does the symmetry get worse because now you're treating the strange quark as identical too which is a worse symmetry? (Because, loosely, eventhough you're only conserving an isospin aspect of the flavour group, you're still in the flavour group which is a worse approximate symmetry?) $\endgroup$
    – Alex Gower
    Apr 9, 2021 at 22:43
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    $\begingroup$ Yes, flavor SU(3) will be a worse approximation. However, if you separately conserve SU(2) isospin and strangeness, it will work fine. $\endgroup$
    – ragnar
    Apr 10, 2021 at 2:43
  • $\begingroup$ Why do we call ($u,d$) isospin symmetry but ($u,d,s$) flavor symmetry? Is there any difference between isospin symmetry and flavor symmetry? or are they just different names but no essential difference? Put it in another way, can we call ($u,d$) flavor symmetry and ($u,d,s$) isospin symmetry? $\endgroup$
    – Shen
    Dec 13, 2021 at 11:59

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