Do up and down quarks obey flavour quantum conservation?

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Main Question or Discussion Point

I was searching for a "upness" over the internet and found nothing but what hints of up and down quarks having quantum number conservation is this isospin thing, which I haven't learnt.

May I ask if there is an "upness" or "downness"?

i.e number of up quarks - number of antiup quarks for upness.

and if there is, then what is the value for upness quantum number... +1 or -1?
 

Answers and Replies

  • #2
tom.stoer
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It is just isospin which is strictly speaking +1/2 and -1/2 and which is conserved in strong interactions. Going to three quarks and including the s quark one indeed talks about strangeness, but for u and d it's isospin.
 
  • #3
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My group theory knowledge is a little rusty, so perhaps you can clarify.

Naively, if I look at the QCD Lagrangian, it would appear that there is a U(1) symmetry corresponding to multiplying the up quark field by a phase that would correspond to conservation of "upness". Thus, I would either expect that "upness" really is conserved--i.e., that the U(1) is a subgroup of larger isospin symmetry, or some combination of isospin and the baryon number U(1). Otherwise, if "upness" is not a conserved quantity, there must be an anomaly that breaks the symmetry of the QCD action (presumably the axial anomaly, but from my limited knowledge of group theory, it doesn't seem like it should be related to the above-mentioned U(1) that I think would correspond to a transformation that acts equally on right and left handed up quarks.)

Or perhaps I'm way off. I haven't thought about this stuff in a long time. Any clarification would be appreciated.
 
  • #4
tom.stoer
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The QCD Lagrangian has a global N flavor symmetry; for N=2 this is just SU(2), not U(1). In addition it has a local color SU(3) symmetry which is not relevant here.

In QCD the SU(2) isospin symmetry is nearly exact; it is broken only by slightly different masses of u- and d-quarks. For high energy processes massless quarks with exact isospin symmetry may be a good approximation.

In el.-weak interactions (which we do not discuss here) the u- and d-quarks couple via W-bosons; this is by no means an anomaly but already present at tree level.
 
  • #5
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The QCD Lagrangian has a global N flavor symmetry; for N=2 this is just SU(2), not U(1). In addition it has a local color SU(3) symmetry which is not relevant here.

In QCD the SU(2) isospin symmetry is nearly exact; it is broken only by slightly different masses of u- and d-quarks. For high energy processes massless quarks with exact isospin symmetry may be a good approximation.
Yes, I am aware that if the mass of the up and down are the same, there is as SU(2) symmetry (actually U(2), or SU(2)xU(1), which is why I mentioned baryon number).

If we ignore the down quark for the moment, I can multiply just the up quark by a phase, and I believe the QCD action is unchanged. (Are you saying this is not correct?) If this is correct (and if the symmetry is not anomolous) then "upness" should be conserved. Of course, I assume it must just be part of a larger symmetry, since no one ever talks about "upness" by itself. From memory, I thought I remembered that there are U(1) subgroups of SU(2), and the natural assumption is that the "upness" U(1) is just part of the isospin SU(2), although, again, my memory of group theory is hazy. Of course, even if this is true, it's more useful to use the full isospin symmetry than to pay attention to upness by itself, but it's still interesting to note the relationships between the symmetries.

Is there something wrong with my reasoning?

*Edit* Yes, I think you are right about the electroweak interactions, which would break any "upness" conservation in QCD. I'm still curious about the case where you just consider the strong interactions, though.
 
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  • #6
bcrowell
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I think it's true that "upness" is conserved in strong interactions. Let upness, U, be the number of up quarks minus the number of anti-ups. Then [itex]U=I_z+(3/2)B[/itex], where Iz is the z-component of isospin and B is the baryon number. Since Iz and B are conserved, so is U.

Probably the reason people prefer to talk about isospin rather than upness is that you also have conservation of I(I+1), which isn't implied by conservation of U.
 
  • #7
tom.stoer
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In QCD all flavors are conserved independently. This is due to the fact that the Lagrangian

[tex]\bar{q}_f \gamma^\mu D_\mu q_f[/tex]

as a sum over all flavors is diagonal in flavor space, that means that the strong interactions mediated by the gluons in [tex]\gamma^\mu D_\mu[/tex] does not mix flavors. Mathematically this is simply

[tex]\bar{q}_f \gamma^\mu D_\mu q_f = \bar{q}_{f} (\gamma^\mu D_\mu)_{ff^\prime} q_{f^\prime} [/tex]
 
  • #8
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Thanks. It's all clear to me now.
 

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