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The quarks of neutral pions don't exist in a pure flavour state, and instead are described as a superposition of up-antiup with down-antidown:

$\frac{u\bar{u}-d\bar{d}}{\sqrt{2}}$

However up and down quarks have different charges. Considering a single quark in isolation, this appears to indicate a superposition of charge states. Such a charge superposition is forbidden for free particles such as electrons/positrons.

  1. How does the standard model reconcile this apparent contradiction? Is this a limitation of the quark model?

  2. Are there any circumstances in which free particles can exist in a superposition of charge states?

Jamie S
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1 Answers1

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You might be confused. Ignoring the sign complication for the moment (having to do with SU(2) conjugate representations), consider the rotation group, in the "addition" of angular momentum constituents/components.

A total spin 0 singlet state composed of two spin 1/2 constituents, $$\frac{1}{\sqrt 2}(|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle),$$ is really a superposition of two spin 0 states with eigenvalues S=0, and $m=0$. Such states may mix and interconvert.

  1. Similarly, you are superposing two middle ($I_3=0$) states with isospin 1 (isotriplets), and $Y=0$, so Q =0. These two states my interconvert into each other, and so they can mix and superpose.

Raising the isospin gives you a Q=1 state, $$ |\bar d u\rangle , $$ for instance. But you may not superpose this on the neutral pion you started with:

  1. As @Hyperon has commented, superselection rules prevent mixed charge superpositions: each charge sector lives in a "different world" so so speak. There are several questions illustrating this on this site. To the extent you are considering representations of conserved charge, you'll never have non-vanishing dot products of states of different charge. E.g., superselection sectors of different fermion number will stay apart, as fermion number cannot change.

  2. An illustration: You'll presumably learn later on that S is violated by a bit (2nd order) by the weak interactions (box diagram), leading to, e.g. strangeness oscillations between $K^0=\bar s d$ and $\bar K^0= \bar s d$, at a frequency ~ 3.5μeV. That is to say, this neutral meson can transition to its antiparticle, so it does not form a leak-proof superselection sector! So you can, and do, form superpositions of indistinct strangeness, e.g., $$ K_L^0=\frac{1}{\sqrt 2} (|\bar s d\rangle -| \bar d s \rangle ) , $$ but distinct charge (still inviolable!). This is a propagating state, whose strangeness flip-flops in time, in collusion with $K_S^0$, its orthogonal combination. A formally analogous phenomenon occurs for neutrinos.

Cosmas Zachos
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  • Thanks for your answer. The part explaining forbidden mixed charge superpositions was very helpful. However I'm still not clear on the meaning of superposing two (apparently) different flavour states up-antiup and down-antidown. I'm aware that for both "pure" states the total flavour and charge is zero, allowing superposition, but wouldn't deep inelastic scattering allow us to "collapse" the system into a single state up-antiup or down-antidown? Forgive me if I'm misunderstanding DIS. – Jamie S Jan 31 '23 at 09:21
  • If by DIS you mean S channel photon exchange, it is precisely this channel that allows mixing/superposition of uubar with ddbar, ssbar, etc. have you learned about ω-φ mixing? – Cosmas Zachos Jan 31 '23 at 12:19
  • What's the relationship between deep inelastic scattering and S channel photon exchange? I've looked for a source mentioning both but failed to find one. – Jamie S Feb 01 '23 at 06:47
  • Tree level photon exchange diagram. Your question is irrelevant to DIS!! – Cosmas Zachos Feb 01 '23 at 08:09