# Why is it not accurate to view pions as bound states of quarks?

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This question is inspired by Ron Maimon's answer to the question What does it mean that the neutral pion is a mixture of quarks?

In that answer Ron insisted that pions must be viewed as a sort of collective excitation of vacuum, instead of a bound state of quarks(if I didn't misunderstand the answer):

Saying that pions are made of quarks is like saying that sound is made of atoms. It's true that if there are no atoms, there is no sound, but that's about it. The QCD vacuum is like a condensed matter system, and it has a quark condensate at the pion scale. The eigenstates of motion of the quark condensate define the low-lying excitations of QCD, and the lightest motion of the condensate is moving it's parts chirally against each other. By this, I mean turning the left-handed u/d and right-handed u/d quarks in the condensate by an opposite phase. This would do nothing to the energy if chiral symmetry were exact, that is, if the quarks were massless. This means that you could "move" the vacuum in the chiral direction without any energy cost, and this gives massless "phonons" (Goldstone bosons) for this process, by moving the vacuum over here a little, and not moving the vacuum over there. These phonons carry the same quantum numbers as the isospin triplet u-dbar/symmetric/d-udbar. These phonons are the pions.

I must first confess I don't know exactly how to define bound states in QFT, but whatever it means, I don't get why "being a collective excitation" and "being a bound state" should be mutually exlusive. In QFT context, isn't every particle a collective excitation? But this shouldn't prevent us from saying, for example, that a positronium is a bound state of an electron and a positron(or can we?). The only difference between positronium and pion is that the latter is excited from an (approximately) symmetry-spontaneously-broken vacuum, and I don't see how it can make such a drastic difference on the ontology.

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Part of your question is how to define bound states in QFT.  Just as in QM, a bound state in QFT is a pole of the nonperturbative T-matrix (, i.e., the S-matrix minus 1 stripped from a delta function), and its possible wave functionals are the functionals in the range of the projector linear operator obtained as the residual of the T-matrix at the pole. (A proof is given at the end of this post. For related information see my entry ''Bound states in relativistic quantum field theory'' of Chapter B6: Bound states and applications of my theoretical physics FAQ.)

In particular, elementary particles as conventionally modelled by quantum fields as their elementary collective excitations are bound states iff they can appear as asymptotic states of the theory. Thus electrons and photons are bound states of QED or the standard model, but (because of confinement) quarks in QCD or the standard model are not.

Pions exist as asymptotic states of QCD, hence form a bound state. Since the contribution of the singe-particle fields to their wave functionals vanishes, they are nonelementary (composite) collective excitations. The same holds for all mesons and baryons. In general, an asymptotic $k$-particle composite makes its first appearance in the $k$-particle sector of the T-matrix, i.e., in S-matrix elements with $2k$ external legs.

Positronium is not a bound state of QED but only a resonance (unstable state), as it is represented by a complex pole of the analytic continuation of the T-matrix into the second (unphysical) sheet. The same holds for the pion when regarded as a state of the standard model, since pions decay under the weak force.

In terms of Ron Maimon's parable, made slightly more precise, all particles (whether elementary or composite) are to quantum fields like sound is to the pressure field of air.

Proof of the T-matrix property used above: The Lippman-Schwinger equation for the T-matrix gives $V=(1-VG_0)T=((E-H_0)G_0-VG_0)T=(E-H)G_0T$, hence $G_0T=(E-H)^{-1}V$. Now the spectral theorem implies that if $\psi$ is an eigenvector for a simple eigenvalue $E_0$ of $H$ then the spectral decomposition of $(E-H)^{-1}$ contains a term proportional to $(E-E_0)^{-1}\psi\psi^*$, hence $G_0T$ contains a term proportional to $(E-E_0)^{-1}\psi\psi^*V$. But this means that the residue of $E_0$ is $\psi\psi^*V$. The proof in the degenerate case is similar, but gives a higher rank residue, which has a higher-dimensional eigenspace as range.

answered Mar 8, 2015 by (14,019 points)
edited Mar 9, 2015

I suppose that one has to be careful about the analogy with air insofar as friction ensures that there are no asymptotic phonons, unless there is an external power input that compensates for friction losses.

Thanks and +1.

Just as in QM, a bound state in QFT is a pole of the nonperturbative T-matrix

The definition seems incomplete: how would one obtain the information of the state besides the fact that it corresponds to a pole? The statement

its possible wave functionals are the functionals in the range of the projector obtained as the residual of the T-matrix at the pole.

seems to be related, but it's too terse.

What you call too terse is already the full story. In QM, the T-matrix at energy $E$ contains for every nondegenerate bound state with eigenvalue $E_k$ a multiple of $\psi\psi^*/(E-E_k)$ exhibiting the pole and as residue a multiple of the projector to the eigenspace. For degenerate bound states, the same holds but the projector is higher-dimensional. This persists in QFT, except that instead of wave functions one has wave functionals (in the functional Schroedinger equation picture, which is the picture appropriate here) and  the procedure to obtain the T-matrix is adapted to the field context.

In QM, the T-matrix at energy E contains for every nondegenerate bound state with eigenvalue Ek a multiple of ψψ∗/(E−Ek) exhibiting the pole and as residue a multiple of the projector to the eigenspace.

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