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QCD vacuum and anomaly [Axions]

  1. Nov 6, 2014 #1

    ChrisVer

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    I am studying this paper about the strong CP problem and axions by R.D. Peccei:
    http://arxiv.org/abs/hep-ph/0607268
    At the moment I read through the introduction and I have some questions about some points, so please could you help me clear out some things?. [In general I'll try to feed back this thread with more questions around the topic].

    #1. It says that for massless quarks, and 2 light quarks [for QCD scale] we have an approximate global symmetry [itex]U[2]_V \times U[2]_A [/itex], so a vector and axial current symmetry. However if I look at the massless lagrangian [just the kinetic term] I have:
    [itex] \mathcal{L}= i \bar{q} \gamma^\mu \partial_\mu q [/itex]
    What are the transformations done for the U[1] cases? I think the axial will have something like [itex]\propto \exp[ia \gamma_5][/itex]. In that case I don't see how the [itex] \exp[-ia \gamma_5] \gamma^\mu \exp[ia \gamma_5]= \gamma^\mu[/itex] (it's not even a symmetry for the massless lagrangian)

    #2. In Eq 1: I am having a problem in seeing that the triangle chiral anomaly gives rise to a divergence to the axial current of the form: [itex] \partial_\mu J^\mu_5 = g F^{\mu \nu}_a \tilde{F}_{a \mu \nu}[/itex]. Do you have any source where I can see how does someone extract this result?

    #3. In Eq 7: The gauge transformation of [itex]A[/itex] goes like:
    [itex]A^i \rightarrow \Omega A^i \Omega ^{-1} + \frac{i}{g} \nabla^i \Omega \Omega^{-1}[/itex]
    I think the second term is zero ([itex]\Omega \Omega^{-1}=1[/itex]) or does it mean that [itex](\nabla^i \Omega ) \Omega^{-1}[/itex]. Then, also why does it give to the Omegas a spatial dependence ([itex]\textbf{r}[/itex])?

    #4 Finally for now, under Eq8, I don't understand what it means by the Jacobian of the mapping [itex]S_3 \rightarrow S_3 [/itex] (what is [itex]S_3[/itex]? the symmetry group of order 3!=6?)
     
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  3. Nov 6, 2014 #2
    #1 Be careful with the transformation for [itex]\bar q[/itex]. In fact we have: [itex]q\to e^{ia\gamma_5}q[/itex] and so [itex]q^\dagger\to q^\dagger e^{-ia\gamma_5}[/itex]. Therefore: [itex]\bar q=q^\dagger\gamma_0\to \bar q e^{ia\gamma_5}[/itex]. And so when they switch with the [itex]\gamma_\mu[/itex] they change sign again, thus canceling each other.

    #2 Have you tried to take a look at Peskin? There should be a chapter on QCD axial anomaly. Or also Donoghue's book.

    #3 It's the second thing you wrote: [itex](\nabla^i\Omega)\Omega^{-1}[/itex]. Also, [itex]\Omega(x)[/itex] has a space-time dependence because that's how you define a gauge trasformation, i.e. a symmetry which depends on the space-time point where it si applied.

    #4 I'm not sure but I think that [itex]S_3[/itex] is supposed to be the 3-dimensional sphere.
     
  4. Nov 6, 2014 #3

    ChrisVer

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    Thanks for the reply. Oh my god, I didn't even care about the [itex]\gamma^0[/itex] in (1) .... :s

    True about the gauge transformations, but isn't that supposed to be the case for local symmetry transformations and not global? In fact I'm getting confused of how he changes transformations: first he has U_A(2) xU_V(2), then goes to a U(1)_A that is not a symmetry. Then in the A^i case, he works with an SU(2) QCD (what is that SU(2)?- SU(2)_isospin?)
     
  5. Nov 6, 2014 #4
    Yes, in fact [itex]\Omega(x)[/itex] is supposed to be a local symmetry. In particular, [itex]\Omega\in SU(2)[/itex] since he is saying that he is dealing with SU(2) QCD. This just means to build QCD with an SU(2) group (i.e. 2 colors and using the Pauli matrices instead of the Gell-Mann ones) instead of an SU(3) group. It's just to make everything simpler.
     
  6. Nov 6, 2014 #5

    nrqed

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    I haven't looked in details at the paper but it seems to me that the U_A(2) xU_V(2) is a global symmetry.
    But by SU(2) QCD we mean a gauge SU(2), so it is a local transformation. This SU(2) is not isospin (which is an approximate symmetry of SU(3) QCD). By SU(2) QCD, they mean an hypothetical theory where the SU(3) of QCD is replaced by SU(2) so there would be two colours.
     
  7. Nov 12, 2014 #6

    ChrisVer

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    Also how can someone decide on approximate symmetries? I think this question concerns the experimental verification.
    For example up until today SU(3) for strong interactions [QCD] has been experimentally verified...
    But when you talk about approximate symmetries, like that of isospin, due to [itex]\Lambda_{QCD} \gg m_{u,d}[/itex], how can you experimentally see that? I mean we know that the up and down quarks have masses, and so I don't get the approximation : [itex]m_{u,d}=0[/itex] ...
     
  8. Nov 13, 2014 #7
    The isospin approximation is not [itex]m_{u,d}=0[/itex] but it's the [itex]m_u=m_d[/itex] which is experimentally verified simply looking at the mass difference between proton and neutron being very small.
     
  9. Nov 14, 2014 #8

    ChrisVer

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    Thanks. Back to the topic, extending my #3....

    I don't understand how the transformation [itex] \Omega \rightarrow \exp (2i \pi n)[/itex]

    All I was able to find is that (for SU(2)): [itex]\Omega_j = \exp (- i \pi \frac{x_j \sigma_j}{\sqrt{x^2 + \lambda^2}})[/itex] in the n-th power... as:
    F0000TEK.jpg
     
    Last edited: Nov 14, 2014
  10. Nov 14, 2014 #9
    In my opinion, getting to the actual result in Eq. (9) can be a rather complicated problem (try to take a look at reference [6] cited in the paper). However, Eq. (8) is rather simple to understand. You are performing an SU(2) gauge transformation, [itex]\Omega(x)[/itex]. This simply means that, for each point of your space-time, you are performing a different transformation. Now, it is clear that you want this transformation to be trivial when you are at spatial infinite (you don't want to trasform your field at infinite!). This means that for [itex]\vec{r}\to\infty[/itex] the gauge transformatio [itex]\Omega(t,\vec{r})[/itex] must go to a unity transformation. However, this does not specify the phase with which [itex]\Omega(t,\vec{r})[/itex] goes to unity.
    This is what the paper is saying. Even if all your transformations go to 1 at infinite you can still classify them based on how the go to unity (from which direction if you want).
     
  11. Nov 20, 2014 #10

    ChrisVer

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    One other question...
    http://pdg.lbl.gov/2013/reviews/rpp2013-rev-axions.pdf
    page 3:
    How can this be true for a scalar field? I know this might sound stupid, but the axion as a scalar field it should have a superpartner that is a fermion... (I don't have access in ref 18)
     
  12. Nov 20, 2014 #11
    I think you are right, but it also says:
    This makes me think that in some models axions can also be fermions. I think the important aspect is that they have to carry PQ charge, in order to acquire a non-zero VEV.
     
  13. Nov 20, 2014 #12

    ChrisVer

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    However, if you try to give nonzero vev to a fermion you are going to break the Lorentz invariance...
     
  14. Nov 20, 2014 #13
    Absolutely correct! I have no idea, I'm sorry :D
     
  15. Nov 23, 2014 #14

    ChrisVer

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    This is going to be a very fast question, however I'd like to make it clear for my understanding.
    In the pdg review, they state that the axion is coupled to the nucleons with some dimensionless constants $$C_{p,n}$$ which are related to the nucleon axial vector current matrix elements by the generalized Goldberger-Treiman relation [I'll restrict myself only to the proton]:
    $$C_p =(C_u - \eta) \Delta u + (C_d - \eta z) \Delta d + (C_s - \eta w) \Delta s $$
    where
    $$\eta = (1+z+w)^{-1}$$, $$z= \frac{m_u}{m_d}$$ and $$w=\frac{m_u}{m_s}$$
    and $$\Delta q$$ represent the axial vector current coupling to the proton:
    $$\Delta q S_\mu = <p|~\bar{q} \gamma_\mu \gamma_5 q ~|p>$$
    where $$S_\mu$$ the proton's spin.

    I guess the bracket represents the proton right?

    Can the last equation be rewritten:
    $$\Delta q S_\mu S^\mu= <p|~\bar{q} \gamma_\mu \gamma_5 q ~|p> S^\mu \Rightarrow \Delta q = \frac{<p|~\bar{q} \gamma_\mu \gamma_5 q ~|p> }{S^2} S^{\mu}$$
    So it depends on the spin and thus allows for the axion to couple to the proton's spin?
    I am afraid that the last equation kills dofs of the initial one....
     
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