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Weak analog of the 'Strong CP Problem'

  1. Dec 29, 2008 #1
    I understand that there is an issue about the FF-dual term in the strong sector. But I believe I can write an analogous term in the SU(2) part of the electroweak sector. Why don't people worry quite as much about this term (the Weak analog of the 'Strong CP Problem')?
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  3. Dec 29, 2008 #2


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    I remember wondering this very question. There are two answers I have in my mind, neither of which I can back up with the current state of my brain (sorry).

    (1) You can always remove this term by field redefinitions.
    (2) On the off chance (1) is wrong, remember that the electroweak couplings are much smaller than the QCD coupling, so any contributions from this term are invisible in experiment, and there is no "problem." The strong CP problem is b/c the contribution to the neutron EDM is much too big, which is not the case for the analogous electroweak term.

    Well, I'll let someone else back these statements up (or correct them). But there are two quick thoughts.
  4. Dec 29, 2008 #3
    I also thought I had an answer to this question but I now doubt it is correct:

    Recall that one possible resolution of the strong CP problem is that one of the quarks have a vanishing mass parameter (with the up quark being the best candidate). Lattice simulations indicate that not even the up quark is massless, so the strong CP problem isn't resolved.

    As far the weak anolog of the problem is concerned, recall that neutrinos are massless in the Standard Model. I thought this would resolve the weak anolog of the CP problem. However, by the turn of the millennium, neutrinos were found to oscillate, implying them to have a small mass. However, this didn't seem to raise the issue about the weak CP problem, so something else must be at work...
  5. Dec 29, 2008 #4


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    It's a little more subtle than that, I think. Let me try to repeat the argument that lives on the surface of my brain, with the caveat that it's only the surface thoughts I'm talking about here... :wink:

    The reason why the strong CP problem is a "problem" is that you cannot set [itex]\theta_{\rm QCD}=0[/itex] naturally, since the physical phase is actually

    [tex]\theta_{\rm QCD}+{\rm Arg~det}M[/tex]

    so as long as you write down the most general Yukawa couplings (and why wouldn't you?) you will always get the effective vacuum angle to be nonzero, unless the Yukawa's have real determinant. This relationship follows from an instanton calculation and is related to the fact that the axial symmetry is anomalous.

    So far, all facts. My confused memory is saying something like, SU(2)_W vacuum angle does not have this kind of relationship. Recall that QCD has a U(3)xU(3) vectorlike symmetry that is broken by a quark condensate (the chiral Lagrangian story) and the U(1)_A lives inside this symmetry group. EW does not have this kind of symmetry structure, so I'm not sure how it works.

    All still very vague, but I think that's how the argument goes. Again, take all this with a grain of salt. But I don't think the neutrinos are the key. I think is has more to do with how there is only one vacuum angle, and we can always choose it to be the QCD angle by suitable axial field redefinitions.

    Enough rambling. Hope this helps (and doesn't make things worse!).
  6. Dec 30, 2008 #5
    Blechman above is right,

    The weak force has no CP problem, because it explicitly breaks CP. its group is SU(2)_L acting only on left handed particle and right handed antiparticles. This is maximal CP violation. Now it might be that there is also a right handed version of weak force carriers. The mass limit for them are currently that they must be heavier than about 800Gev, so there not very ruled out at all. In the case with both left and right handed carriers we can ask what happens if we swap handness, U(1)_A would exist for particles effected by the weak force(s). In fact it seems to me, that we would need U(1)_A as a local symmetry i.e. as a force, In the independent research forum I've a paper about what it would entail to have U(1)_A as a independent force. It has been looked at by other researchers, but so far many buried in very obscure or old journals.

    See however: <A HREF="http://lanl.arxiv.org/abs/hep-ph/0512324">L.M. [Broken] Slads</A> paper, where is seems to show that U(1)_A is demended as a local symmetry for any theory with both left and right handed weak carrier.
    Last edited by a moderator: Apr 24, 2017 at 9:47 AM
  7. Dec 30, 2008 #6

    Vanadium 50

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    None of that is about CP violation. That's about parity violation.
  8. Dec 30, 2008 #7

    Vanadium 50

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    Blechman is correct about the symmetry structure: in chiral QCD you end up with a leftover U(1) (and it must be a U(1), since you need the generator to be a number, [tex]\theta[/tex] and not a matrix) and that gives you a term in the Lagrangian like [tex]\frac{\theta}{16\pi^2} F^a_{\mu \nu} \tilde{F}^{\mu \nu a}[/tex]. That term is the troublemaker, because it can give rise to a non-zero neutron EDM.

    As an aside, that's the real problem here - not that there is strong CP violation, but that CP violation can give rise to a neutron electric dipole moment, where the experimental constraints are extraordinarily strong.

    If I have an ordinary U(1) theory, this term doesn't contribute: you'll get something like
    [tex]\int d^4x F \tilde{F}[/tex] and since [tex]F \tilde{F} = \vec{E} \cdot \vec{B} =
    \frac{1}{2} \partial_\mu \epsilon^\mu_{\nu \rho \sigma} A^{\nu} F^{\rho \sigma}
    [/tex]: an integral of a total divergence. Now, since dimensionally F falls off faster than 1/r2 and A falls off faster than 1/r, you're left with something that falls off too fast to give you any surface term.

    So, what you need is a non-Abelian theory that has an approximate U(1) built in.

    You get that in chiral QCD. You don't get that in GWS weak theory, because U(1) hypercharge is an exact symmetry, but so is U(1) electromagnetism.

    Now, you do get a neutron EDM from electroweak effects. It's very small, because it can only come in loops, and relies on an intereference from all three families. I would estimate the size of the effect to be of order:

    [tex] \left[ \frac{\alpha}{16 \pi^2} \left( \frac{m^2_u - m^2_d}{M^2_W} \right) \left(\frac{m^2_c - m^2_s}{M^2_W}\right) \left(\frac{m^2_t - m^2_b}{M^2_W}\right) (3.0 \pm 0.3) \times 10^{-5} \right] ea_0 \approx 10^{-31}[/tex]

    which is, in fact, the SM prediction. (In units of e-cm) Experimental limits are about a million times larger.
  9. Dec 30, 2008 #8
    Thanks for the enlightening responses. I'd like to summarize your claim, Vanadium, to see if I understand this properly:

    The chiral limit of QCD is a non-abelian gauge theory with an exact U(1) flavor symmetry. Therefore, I may rotate away the theta term (and hence, it is unobservable). In the real-world QCD, the U(1) flavor symmetry is approximate. Hence, the theta may not be rotated away, as the angle will simply get transferred to the quark mass matrices: the phase is an observable.

    On the other hand, GWS theory is a non-ablelian gauge theory with an exact U(1) of hypercharge. I believe now I should claim that I should be able to rotate away the theta term for weak isospin. However, since the U(1) symmetry is not even anomalous (required for consistency of the GWS theory), doesn't that mean performing a rotation leaves the theta term untouched? So did you mean that I need an approximate global (not local) U(1) symmetry? In that cause we have the anomalous B+L symmetry to remove the theta term; right?

    I believe I'm finally understanding the issue; thanks!
  10. Dec 30, 2008 #9


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    To add to Vanadium-50's comments:

    I just want to emphasize his point that a U(1) has no physical vacuum angle. A slick way to see that goes as follows: Even if you ignore the phase in the mass matrices for a moment, you might think that you can just set [itex]\theta=0[/itex] -- we're theorists, we can do what we want! The problem is that even if you try to do this by "playing G-d," you will just regenerate the theta term from instanton effects in the gauge theory. That is why it is not "natural" to set the vacuum angle to zero.

    However, there are no U(1) instantons, since you cannot embed a circle into a 3-sphere in a nontrivial way. Therefore the abelian vacuum angle has no physical meaning. You really need a nonabelian gauge theory to get a nontrivial vacuum angle. This is just what Vanadium-50 is saying. This leaves the SU(2) and SU(3) theories.

    Vanadium-50: can I ask you to just go through the argument for needing an "approximate U(1)" again? Under an ANOMALOUS U(1) symmetry, [itex]\theta\rightarrow\theta+\alpha[/itex] where [itex]e^{i\alpha}[/itex] is the U(1) phase. Using this explicitly broken symmetry we can set a phase convention where [itex]\theta\rightarrow 0[/itex] but since the mass matrix also rotates under this symmetry we will reintroduce it there, and so the effect does not just go away. What part of this argument goes wrong if we replace SU(3) with SU(2)? U(1)_{Y,EM} are not the anomalous symmetries we are using to remove the vacuum angle...

    As far as it being a "problem" -- well, I agree that IF we saw a sizable neutron EDM then we would say, "Wonderful! QCD predicts that, and we have a great theory!" Unfortunately, we don't see this, and now we are left with asking where the small value of [itex]\theta[/itex] comes from.
  11. Dec 30, 2008 #10


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    you have it backwards: if you have an EXACT symmetry, any "rotation" does not do anything to the parameters, since it is an exact symmetry!! You need the symmetry to be BROKEN, so that the parameters become "spurions" of the broken symmetry and therefore can be rotated away. This is, for example, how the usual flavor symmetry works when diagonalizing the mass matrices in the SM - the U(3)^5 symmetry is explicitly broken by the Yukawa matrices, and so we can use that symmetry to do field redefinitions and therefore remove the off-diagonal masses (up to the CKM matrix).

    BY THE WAY: It is important that the symmetry is broken EXPLICITLY, either by anomalies or spurions, not spontaneously. Remember: if the symmetry is broken spontaneously, it is still an "exact" symmetry, only realized nonlinearly. For this argument to work, you need an explicit breaking.

    again, I'm a little confused by that. I have asked Vanadium-50 to go into more detail. But I bet you are right - it is probably B+L that does it!

    AHAH! Yes, I bet you got it in one! The B+L U(1) can rotate away the EW vacuum angle, and since the mass terms are invariant under this symmetry, it does not come back in the form of a phase like in the QCD case!

    I bet that's the answer. Anyone want to have a go at my guess?
  12. Dec 30, 2008 #11
    Very interesting discussion. Thanks guys.
    I've been re-reading a little since the beginning of the discussion. I think I still need to check in Coleman's "symmetry", but from what I collected so far, I think your bet is correct :smile: Maybe Vanadium has a better reference ?
  13. Dec 30, 2008 #12


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    Coleman is, of course, great! I'm doing all this from home, so I don't have any textbooks readily available to me and have been doing it all from memory (why I've been sure to include all those caveats about possibly being wrong).

    here's an amusing question: we don't REALLY believe that the SM Lagrangian is the full story; we know that the infamous (unique) dimension 5 operator gives Majorana neutrino masses and explicitly breaks L (and therefore B+L). dimension 6 operators also break these symmetries. Does that mean, by my half-brained arguments above, that when including these higher-dimension operators, the electroweak vacuum angle is no longer unphysical? I imagine it is still irrelevant since its contributions to physics are highly suppressed...

    Thoughts, anyone?
    Last edited: Dec 30, 2008
  14. Dec 30, 2008 #13

    Vanadium 50

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    Actually, at the time I wrote this, I thought it was obvious. Now I am not so sure. As a former professor once said, "It's obvious. Um...I think it's obvious. It's either obvious or wrong - I'm not sure which."
  15. Dec 30, 2008 #14


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    yeah, the more obvious it is, the more likely it's wrong! :wink:

    I think I have it though - see my other posts and feel free to rip them apart if you disagree. I don't think it has to do with hypercharge, though. that makes me suspicious...
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