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Instantons and the QCD vacuum

  1. Jun 27, 2010 #1
    I was wondering about the [tex]G\tilde{G}[/tex] term that can be added to the QCD Lagrangian.
    It seems clear that the analogous QED term can be left out because it is (just as the [tex]G\tilde{G}[/tex] term) a total divergence and thus has no influence on the physics.
    But in the case of a non-abelian gauge theory that doesn´t seem to be the case. So
    a) why is only the QCD case important? The weak interactions would exhibit that as well, wouldn´t they?
    b) I don´t really understand why the term doesn´t drop out just as in QED. When integrating over the [tex]G\tilde{G}[/tex] term, one gets the winding number n - which is simply a constant. Sure, it depends on the type of instanton, so n=n([tex]A_\mu[/tex]) still depends on the gauge fields and therefore does the action S, but if one fixes the intial vacuum at t=-\infty (lets say it has winding number n') and the final vacuum at t=\infty (winding number n''), then all possible gauge fields connecting those two vacua will have winding number n=n'' - n' and so the corresponding term in the action will not depend on [tex]A_\mu[/tex] at all.
    Thus, variating the action with respect to [tex]A_\mu[/tex] will give the same result wether or not I leave the term out.

    I understand that the term is important because it gives us the winding number and therefore there is physics in it, but I don´t understand why it has to stand in the Lagrangian for that...
    (Maybe it is not possible to fix intial and final vacuum? But why?)

    I hope somebody here can help me with that :)

    Thanks in advance,
    Last edited: Jun 27, 2010
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  3. Jun 27, 2010 #2


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    [tex]G^{\mu\nu}\tilde{G}_{\mu\nu} = \partial_\mu K^\mu[/tex]


    [tex]K^\mu = \epsilon^{\mu\nu\alpha\beta}(A^a_\nu\partial_\alpha A^a_\beta-\frac{2}{3}f^{abc}A^a_\nu A^b_\alpha A^c_\beta)[/tex]

    up to factors of 2. It is that A^3 term that is the problem: in QED there is no such term, and there is no contribution to the action. But in nonabelian theories, the term can be nonzero at infinity and so it can contribute, as you say.

    You are absolutely right that there is an SU(2)_W and an SU(3)_C term like this. However, the SU(2) one can always be set to zero in the Standard Model. This is not a trivial point. It is because B+L is anomalous. Anomalous symmetries mean that we can take a phase out of the quark/lepton mass determinant and put it into the coefficient of this operator. It turns out (as a direct calculation will show) that the QCD anomaly vanishes, so B+L can only be used to set the coefficient of the SU(2) term to zero. Since B+L is also a symmetry of the quark and lepton mass matrices, this means that you can safely set the SU(2) term to zero without reintroducing a phase in the mass matrices. Once you do this, you've exausted your phase freedom, and there is no more room to set the QCD phase to zero. You can only do that at the expense of reintroducing it in the quark mass matrix, which is where the "[itex]\bar{\theta}[/itex]" comes from.

    This is PURELY a consequence of the fermion content of the SM, and is not true in general. But there it is.

    I"m not entirely sure I understand the second question. My understanding is that in the generating functional (that is, the "partition function" or Path integral) we must integrate over all field configurations, including ones with nonzero winding number. Such contributions can be shown (by a brutal, direct calculation) to introduce phases [itex]e^{in\theta}[/itex], where [itex]\theta[/itex] is the "vacuum angle" - a free parameter of the theory, and n is the winding number. Then you have to sum over all n. This is reproduced in the path integral if you include your term with coefficient [itex]\theta[/itex].

    I suppose what you are suggesting is that you can always set [itex]\theta=0,\pi[/itex] in which case you wouldn't need this term. However, why should that be true? It is a free parameter, like the CKM phase or the gauge couplings or the fermion masses, and should be kept arbitrary and measured by experiment. Unless you propose an extension to the SM, such as a PQ axion, for instance.

    Anyway, I'm not sure that answers your question, but I hope it at least gets the conversation started.
    Last edited: Jun 27, 2010
  4. Jun 27, 2010 #3
    thank you very much for your nice answer. I really appreciate it :)

    I'm sorry but I still have a few questions to your answer...

    At first:

    Do you probably know where I can find a calculation of this? Maybe a textbook or article or something? Because I think I have understood the main points but I'm not entirely sure :P

    Concerning the second question: Well, that's the part where I have the most problems with so that's probably the reason why I can't formulate my question that nicely...
    I think it has to do with what you said here:

    At first I have three questions which are probably very stupid:
    1) Why does the A^3 term vanish in QED
    2) The reason why the A^3 term can be nonzero at infinity is because in a non-abelian gauge theory, it is sufficient that A approaches [tex]U\partial U^{+}[/tex] for some "gauge transformation" U for the action to be finite, right? But in QED, that would be enough too... I mean, in that case U would be a U(1) transformation and thus A would approach a constant (and thus, [tex]F_{\mu\nu}[/tex] would vanish at infinity). So why are there no instantons in QED?
    Is it because than U would be a map from [tex]S_3[/tex] (sphere at infinity) to U(1) (which I guess is isomorphic to a circle) and the corresponding homotopy group would be trivial (and thus we can assume A approaches 0)?
    3) In textbooks, instantons are always introduced in SU(2) - but then it is assumed that there would also be instantons in SU(3). In SU(2), U is a map from [tex]S_3[/tex] to SU(2) = [tex]S_3[/tex] and the corresponding homotopy group [tex]\pi_3(S_3)[/tex] is not trivial. Therefore, we have instantons.
    But then SU(3) should be isomorphic to [tex]S_4[/tex] and U is a map from [tex]S_3[/tex] to [tex]S_4 [/tex] and the corresponding homotopy group is trivial. So why are there instantons in SU(3) gauge theory?

    As to explain my orignial question:
    If the term [tex]G^{\mu\nu}\tilde{G}_{\mu\nu} = \partial_\mu K^\mu[/tex] where to stand in the Lagrangian, wouldn't the action be S = ... + K(infinity) - K(-infinity)? As you said, because of this A^3 term, K(infinity) - K(-infinity) could be nonzero. But it is still a constant so it doesn't change the stationary points of S? (if n' is the winding number of the vacuum at infinity and n'' the one at -infinity, than K(infinity) - K(-infinity) = n' -n'' = winding number of soliton solution between those two vacua = doesn't depend on A) So why not drop it from the Lagrangian if it doesn't alter the stationary points of S?

    Thank you very much for your help and patience,
    Last edited: Jun 27, 2010
  5. Jun 27, 2010 #4


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    you know, I don't! I had to figure this out on my own. I"m sure someone talks about it somewhere. There is a very good set of lectures by Scott Willenbrock (TASI, 2004 lectures) where he talks about how you can use the flavor symmetry of the standard model to reduce the number of parameters, but I don't think he talks about these terms, unfortunately. Does anyone else out there know of a good reference??

    that one's the easiest: [itex]f^{abc}[/itex] are the structure constants of the gauge group. Abelian gauge groups have vanishing structure constants, so that term simply isn't there.

    well, as I said, the A^3 term is zero in QED anyway, at ALL points in spacetime, not just at infinity. But your reasoning why there are no instantons in QED being because "You cannot lasso a sphere" is correct as well.

    Well, yes, but we live in 4 dimensions, which has S^3 as a boundary, so that's not relelvant. The issue is that SU(3) is a Lie group of rank 2, and therefore it contains 2 SU(2)s. Either of them can have instantons. THIS is why people only concern themselves with SU(2) instantons: higher gauge groups will have instantons through an SU(2) SUBGROUP. That is how QCD can have an instanton.

    You are assuming that the gauge group starts in a specific winding state. Why does that have to be the case? In fact, it could be in any linear combination of winding states, and each state can get its own phase (this is what I was trying to say about the "theta"). Just like in ordinary quantum mechanics: overall phases are irrelevant, but relative phases ARE important. It's exactly the same thing here. In your scenario, the winding number generates an OVERALL phase. But we must sum over ALL possibilities in the path integral, and this means over all winding numbers, each with a phase. This can lead to nontrivial physics, the same way you get a shift in the interference patterns from the Aharanov-Bohm effect.

    Hope that helps!
  6. Jun 27, 2010 #5
    Well, then, if I ever find a good source, I will send it to you :)

    Oh, dammit! That's really embarrassing oO I actually thought that at first but then I somehow convinced myself that it had to be something else... :P

    Ok, so it is really only the A^3 term that is important? Does the [tex]\partial_{\alpha}A_{\beta}^a[/tex] term always vanish at infinity?

    (Quick-and-a-little-off-topic question: Is it really "You cannot lasso a sphere" in that case? I guess that that would rather correspond to the group [tex]\pi_1(S_{2})[/tex] - or more general [tex]\pi_1(S_{n})[/tex] for n greater than 1 - being trivial. But in this case we would have a mapping from [tex]S_3[/tex] to U(1) - corresponding to [tex]\pi_3(S_{1})[/tex] and I hope to god that's trivial oO ... I really should know more about topology...)

    Hm, ok, let's say we are looking at the strong interactions - thus SU(3) - and we have two "gauge transformations" at infinity U and U' both lieing in the same SU(2) subgroup of SU(3). Now lets say - within this specific subgroup - it is not possible to continually transform U into U'. Then we would have two vacua corresponding to U and U' respectively. But are those two vacua really "different"? I mean, they can be transformened into each other in SU(3) - just not in those SU(2) subgroups. So is the separation of those two vacua really a, well, real separation?

    Oh boy, the longer I think about the things, the more I don't understand... there is another question that just popped into my mind as I wrote this:

    What does this "there are many different vacua" actually mean? I mean, a vacuum is a space where there are no fields, i.e. A=0 or a gauge transformation of 0. Lets say we are at infinity in space time. In that case I could suppose there is a region with [tex]A=0[/tex] and another one with [tex]A'=U\partial U^{+}[/tex] and U should not be homotopic to a point. Then it is not possible to change A' continually into A without leaving the vacuum (per definition of U) and those to regions will be separated by a region of higher potential. Thus, it is only logical to say those are "different vacua".

    But this is just at infinity in space time. Lets now say we are at some place in euclidean space time instead. In this case, U is not just some (gauge-) function on [tex]S_3[/tex] but on the whole of four-dimensional euclidian space time. If we continuously continue U to [tex]S_3[/tex] and call the resulting function U', it can be shown that U' can, on [tex]S_3[/tex], be continuously deformed into the identity matrix, i.e. a "point" in SU(2) and therefore the two vacua (the one with A=0 and the one with [tex]A'=U\partial U^{+}[/tex]) both have winding number 0.

    So how can we, in euclidean space, have different vacua? Or does this notion of the theta-vacua - or simply vacua with differing winding numbers - ony make sense at an infinitely far away place in space time?

    Ah, yeah, I think I understand now. If there is a [tex]G\tilde{G}[/tex] term in the Lagrangian, then in the action I will have [tex]S=... + \int \partial K dx=... + \int_{S_3} K dx=...+ \int_{S_3} A^3 dx[/tex]. So this integral will be different for A's with differing winding numbers. For A=0, the integral will vanish, but for other possible A's, it will not. Therefore, it DOES depend on A (and so does S) and one cannot leave the [tex]G\tilde{G}[/tex] term out. Is that right?

    Yes, it does a lot! Thank you very much,
  7. Jun 27, 2010 #6


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    cool! you might also check out Sidney Coleman's instanton chapter in "Aspects of Symmetry" where he really gets into these things, including the things you are worried about below.


    Yeah, I think you're right. I might have been a little too quick with this one. I was conceptually on the right track though... :blushing:

    I've completely lost you! :surprised:

    Look: SU(2) gauge fields admit instanton solutions. These are solutions of finite action in Euclidean spacetime (that's the definition of an instanton). The point is that these field configurations cannot come from perturbation theory (they go like 1/g^2 - you'll never get that from perturbation theory) so you must sum over them explicitly in the path integral.

    I also do not agree with your definition of "vacuum"! The vacuum is the state of lowest energy in your theory; that does not necessarily mean "zero field"! Think of the vacuum of a superconductor (or a ferromagnet for that matter).

    Anyway, take a step back and think clearly about these things. Instantons are already in euclidean space, NOT Minkowski space, so I don't quite understand your logic. And all that is required is that the field be a pure gauge AT INFINITY (to be finite energy). In fact, an instanton can NEVER be a pure gauge everywhere: then it would be zero, as you say! Maybe that's what's confusing you...

    I suppose.... again, I'm not sure I understand. All I was trying to say is that the path integral is a sum over all field configurations. The point is that this sum divides itself into sums of terms with arbitrary winding number n, where each winding number gets a phase [itex]e^{in\theta}[/itex]. You seemed to be saying that if we fix the winding number, nothing happens. But this is exactly like saying that you INSIST that the particle goes through the first slit in the 2-slit experiment. You cannot do that!! You must sum over all winding numbers, and since each contributes a different phase, there is interference. THAT is why you need this term.

    Does that make sense?
  8. Jun 28, 2010 #7


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    I don't know if this helps (especially as I have no good reference yet; it's just what I remember from some papers I studied years ago):

    The extra term in the Lagrangian can be cast into a non-trivial definition of the canonical momentum operators (which is the functional derivative w.r.t. the gauge field acting on the wave functional). This can be understood from the finite dimensional example where a wave function can pick up a phase when a winding number is introduced. So the nontrivial phase and the modified canonical momentum operator in Hamiltonian formulation are equivalent to the extra term in the Lagrangian formulation.
  9. Jun 28, 2010 #8
    Thank you. I'm looking through it right now :) I hope it helps me with some of my questions.

    Yes, that's exactly what's confusing me. Because I read that instantons (with winding number n) couple one vacuum (with winding number n') to another vacuum (with winding number n''= n+n'). But I don't see how a vacuum can have a winding number that is not zero.

    Ah,... the path integrals... . The problems is that they were somehow left out in the lecture I heard :(
    So I don't really understand what they are all about. You are always talking about THE path integral - is there a path integral that describes the whole theory? I always thought that path integrals correspond to specific problems?
    So is this maybe the path integral that sums over all paths that begin in an undefined pure gauge at -infinity and end in another undefined pure gauge at infinity? (undefined meaning that it just has to be SOME pure gauge, not a specific one or with a specific winding number)
    If yes, what does this path integral describe?

    Thank you very much,
  10. Jun 28, 2010 #9


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    I hope it does! Let me know.

    but again, remember that "vacuum" does NOT mean zero field! I think that might be your hitch.

    yes, path integral: that's an important step!

    "THE Path Integral" in QFT is akin to the "partition function" in stat mech. It is defined as:

    [tex]Z=\int\mathcal{D}\phi~ e^{iS[\phi]-i\int d^n x~J(x)\phi(x)}[/tex]

    And correlation functions are given by taking appropriate functional derivatives with respect to J(x). See you favorite QFT textbook for details.

    It is this object that truly defines your theory, NOT the action! Nominally, the action gives you the path integral, but there are subtleties from the boundary conditions! This subtlety is exactly where things like instantons (and the related "anomalous symmetry" comes in!).

    BTW: you do not need the path integral - you can do it other ways. But I thiink I speak for MOST people when I say that this is the most common way to do it these days. And in my humble opinion, the easiest. I'm sure there are those out there who disagree with me, though. :wink:

    Anyway, the path integral gives you your correlation functions (just like a partition function in stat mech) and therefore contains all the interesting physics. When [itex]\phi[/itex] includes the gauge field, you must not only sum over all gauge fields of a fixed instanton number, but you must also sum over all instanton numbers! This is where the interference comes in.
  11. Jun 28, 2010 #10
    Hm, now I am utterly confused... I thought the problem was in the [tex]A^3[/tex] term in [tex]K_\mu[/tex]. But what does this term have to do with the path integral?

    I have understood it as follows according to your explanation, but I don't know in the least if this is right:
    I have my functional

    [tex]Z=\int\mathcal{D}\phi~ e^{iS[\phi]}[/tex]

    in which all information about my theory is contained. Now, if I sum over all paths [tex]\phi[/tex], I will have to sum over all winding numbers n and their corresponding paths [tex]\phi_n[/tex]. So I could probably write sth like this

    [tex]Z=\int\mathcal{D}\phi~ e^{iS[\phi]} = \sum_{n} \int\mathcal{D}\phi_n~ e^{iS[\phi_n]} [/tex]


    [tex] S[\phi_n] = ... + \int d^4 x G\tilde{G} = ... + \int d^4 x \partial_{\mu} K^{\mu} = ... + \int_{S_3} d^3 x K = ... + n [/tex]

    So if I call [tex]S[\phi][/tex] the action minus the [tex]G\tilde{G}[/tex] term, than I will get

    [tex]Z=\sum_{n} e^{in} \int\mathcal{D}\phi_n~ e^{iS[\phi]} [/tex]

    which is different from what I would have obtained if I had left the [tex]G\tilde{G}[/tex] term out of the theory:

    [tex]Z=\sum_{n} \int\mathcal{D}\phi_n~ e^{iS[\phi]} [/tex]

    Am I on the right way?
    But I still don't see a [tex]\theta[/tex] in here...
  12. Jun 28, 2010 #11


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    you have it! The only thing you are missing is the [itex]\theta[/itex] is the coefficient of the topological term in the action:

    [tex]\Delta S = \frac{g^2\theta}{16\pi^2}\int d^4x G_{\mu\nu}\tilde{G}^{\mu\nu}=\theta n[/tex]

    So your phase is [itex]e^{in\theta}[/itex]. Other than that, you're right.
  13. Jun 28, 2010 #12
    Ah, right, I forgot the coefficient... :P
    Oh boy... questions always seem so silly afterwards...

    Thank you very much for your help!
  14. Jun 28, 2010 #13
    Now that my instanton question seems to be solved, I have one more question concerning the [tex]G\tilde{G}[/tex] term:

    Is it always true that if the Noether current of some symmetry (lets call the current [tex]J[/tex]) is not conserved due to quantum corrections, then it adds a term

    [tex] \Delta S = \int d^4 x \partial_{\mu} J^{\mu} [/tex]

    to the action and we therefore should have added (or "should add") the term [tex]\partial_{\mu} J^{\mu} [/tex] (or probably something proportional to this term??) to the Lagrangian?

    If yes, do you know some standard textbooks where it is expained why?

    Thank you very much,
    Last edited: Jun 28, 2010
  15. Jun 28, 2010 #14


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    No, that isn't right. [itex]J^\mu[/itex] in this case would be the axial current [itex]\bar{\psi}\gamma^\mu\gamma^5\psi[/itex] which is NOT what you add!

    I don't know how much QFT you have, so I don't know how to answer this question. Take a look at your favorite QFT textbook to read up on anomalous symmetry. Peskin+Schroeder, or Itzikson+Zuber, or even Zee's (slightly dumbed down) Nutshell book.
  16. Jun 28, 2010 #15
    Ah, no, I meant I add [tex]\partial_{\mu}J^{\mu}[/tex] to the Lagrangian, not [tex]J^{\mu}[/tex].
    In the case of my example, that seems to be the case - thus, the [tex]G\tilde{G}[/tex] term. But I don't know if that is always true or just happens to be right in this case...
  17. Jun 28, 2010 #16

    Ben Niehoff

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    This issue looks resolved already, but I wanted to add:

    Towards the beginning of the thread you mentioned that this term is a topological term that measures a "winding number". Specifically, this is the winding number of a map [itex]\mu: SU(3) \rightarrow S^3_\infty[/itex] which maps the gauge group SU(3) onto the 3-sphere at infinity in the Euclidean QFT. Typically when we look at explicit instanton solutions, we take a simplified view with an SU(2) subgroup of SU(3). SU(2) is topologically a 3-sphere, so we have a map from the 3-sphere to itself with integer winding number.

    The reason a similar term cannot occur in QED is that the gauge group U(1) is a circle, and there are no mappings from the circle into the 3-sphere with nontrivial winding number (because every circle in S^3 can be continuously deformed to a point).

    Edit: I see after reading more that you already talked about that. Whoops. :)
    Last edited: Jun 28, 2010
  18. Jun 28, 2010 #17


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    But you added [itex]\partial_\mu K^\mu[/itex], not [itex]\partial_\mu J^\mu[/itex]! K is not the current!
  19. Jun 28, 2010 #18


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    Always happy to have a confirmation! :smile:
  20. Jun 28, 2010 #19
    Hm, funny, but I think it holds that
    [tex]\partial_{\mu} J_5^{\mu} = -\frac{g^2}{16\pi^2}G\tilde{G}[/tex]
    after quantum corrections. (See f.e. Peskin Schroeder, Srednicki etc.)

    So I do in fact add [tex] -\theta \partial_{\mu} J_5^{\mu} [/tex]

    The question is: why?
    Is that simply a coincidence?
  21. Jun 28, 2010 #20


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    No, hang on! The equation you wrote down is NOT J = K! J and K are two different things that come from different sectors (fermions vs gauge fields). There is an equation relating the DIVERGENCES of the two, so that the new object J-K is conserved, but it is NOT zero!

    You must treat these as DIFFERENT objects that are related through an equation of motion.
  22. Jun 28, 2010 #21
    Does [tex]\partial J = \partial K[/tex] imply that J=K?

    Because I didn't mean that J equals K. I just read that
    [tex]\partial J = - g^2/16\pi^2 G\tilde{G}[/tex]

    And so it seems to me as if [tex]-\theta \partial J[/tex] is then added to the Lagrangian...
  23. Jun 28, 2010 #22


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    Whether or not the symmetry is broken (dynamically, spontaeously, or explicitly) is irrelevant to what goes into the action! You write everything of dimension 4 or less that is invariant under all symmetries (regardless of whether or not they are anomalous). The G-Gdual is one of those allowed operators, and so it must be included. The normal story is that it is a total divergence and therefore can be dropped, but when instantons are allowed then you must keep it. At this point, it has nothing whatsoever to do with anomalous symmetries.

    The anomalous symmetry thing comes in later, when trying to understand the physical effects of this term. For example: if you have a chiral gauge theory, you can absorb the effects of this term into phases that appear in the mass terms of the fermions. But that is neither here nor there when it comes to including this operator in the action.

    Do you see what I'm saying? You do NOT use equations of motion inside the action itself, only when you compute matrix elements. When writing down the action you are not allowed to use equations like dJ = dK. You must treat J and K as totally independent objects, and only when computing physical matrix elements can you set them equal. This is the same idea behind "offshell particles" and whatnot.
  24. Jun 28, 2010 #23
    I think my question has to do with what you said here. I know that everything that can be put in the Lagrangian SHOULD be put in the Lagrangian so that is not the problem. What I mean is:
    After electroweak symmetry breaking it seems there can be a phase in the quark mass matrix (I don't exactly know why because I always thought that the mass matrix could be diagonalized in such a way that it is real... ???). This phase could be rotated away by a chiral transformation.

    At first it seems that that is the end of it because this chiral transformation is a symmetry of the Lagrangian. But it's Noether current [tex]J_5^{\mu}[/tex] is not conserved, suggesting that it was never a symmetry in the first place.
    Is that correct thus far?

    Because now comes my problem: Rotating away the phase in the mass matrix seems to add a term in the Lagrangian that is proportional to [tex]G\tilde{G}[/tex] and I don't know how this comes to be.

    It has to have something to do with the non-vanishing divergence of J (and incidentally it is proportional to said divergence) but I don't see how...

    I distinctly remember something about a Jacobian but I am neither sure about that nor do I know what that Jacobian might or might not do...
  25. Jun 28, 2010 #24


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    OK, now I begin to understand your question. Let's consider it carefully. I'm going to have to take back a few things I said earlier. Here we go:

    Consider a CHIRAL GAUGE THEORY of a single fermion (never mind that this theory is sick, but that won't enter into our calculation). There is a mass term of the form

    [tex]-\mathcal{L}_m = m\bar{\psi}_L \psi_R + m^{*}\bar{\psi}_R \psi_L[/tex]

    where m is a complex number. Now to remove the phase from the complex number, you can do a phase redefinition on, say, the right handed fermions:

    [tex]\psi_R\rightarrow \psi_Re^{i\alpha}[/tex]

    Then if I choose [itex]\alpha[/itex] to cancel the phase of m, all is well and the fermion mass is real.

    But that phase change is a chiral rotation, and this rotation has a Noether current. So the Lagrangian will change as:

    [tex]\mathcal{L}_m\rightarrow \mathcal{L}_m+\alpha\partial_\mu J_R^{\mu}[/tex]

    where the Lagrangian on the RHS now has a REAL mass. This is shown in Ch9 of Peskin and Schroeder, for example.

    Now if the current was truly conserved, the added term would be a total divergence and hence totally irrelevant, and that would be that. BUT if the current had an anomaly, then the divergence would be that G-Gdual, as you say above. So:

    [tex]\mathcal{L}_m\rightarrow \mathcal{L}_m+\alpha\frac{Ag^2}{16\pi^2}G\tilde{G}[/tex]

    where A is some number that comes from the details of the group theory (called the "Anomaly coefficient") But that new term was already in our Lagrangian: it is nothing but the [itex]\theta[/itex] term we had before!! So we find that when we remove the phase of the mass, we correspondingly shift the parameter [itex]\theta[/itex]:

    [tex]\theta\rightarrow \theta+A\alpha[/tex]

    That's the idea.

    I've been very sloppy with factors of 2 and details and whatnot, but hopefully that will help you see what's going on, and you can fill in the gaps.

    Now I can fill in the details about why there is no electroweak theta vacuum in the SM: In that case, we first make the masses of all the particles real. At this point, [itex]\theta_{EW}[/itex] is not zero. NOW we go ahead and do a B+L rotation. Such a rotation does absolutely nothing to the mass terms, since they are invariant under both B and L. HOWEVER, B+L is anomalous, and so it shifts the theta vacuum as I just showed you. So in the end, since [itex]\alpha[/itex] is arbitrary, we can safely set it equal to [itex]-\theta_{EW}/A[/itex] and by the above equation, the EW theta angle can be set to zero! That's why there is no such object in the SM.
  26. Jun 28, 2010 #25
    Thank you very, very much! The whole time, my problem was that I didn't know why the Lagrangian would change in that manner (under such a symmetry transform).

    I will search through chapter 9 and hopefully that will resolve the issue :)
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