Why is the QCD vacuum important in non-abelian gauge theories?

[QuantumCosmo]

Hi,
I was wondering about the $$G\tilde{G}$$ 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 $$G\tilde{G}$$ 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 $$G\tilde{G}$$ term, one gets the winding number n - which is simply a constant. Sure, it depends on the type of instanton, so n=n($$A_\mu$$) 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 $$A_\mu$$ at all.
Thus, variating the action with respect to $$A_\mu$$ 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,
Quantum

 

[blechman]

a) why is only the QCD case important? The weak interactions would exhibit that as well, wouldn´t they?

$$G^{\mu\nu}\tilde{G}_{\mu\nu} = \partial_\mu K^\mu$$

where

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

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 "$\bar{\theta}$" comes from.

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

b) I don´t really understand why the term doesn´t drop out just as in QED. When integrating over the $$G\tilde{G}$$ term, one gets the winding number n - which is simply a constant. Sure, it depends on the type of instanton, so n=n($$A_\mu$$) 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 $$A_\mu$$ at all.
Thus, variating the action with respect to $$A_\mu$$ 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,
Quantum

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 $e^{in\theta}$, where $\theta$ 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 $\theta$.

I suppose what you are suggesting is that you can always set $\theta=0,\pi$ 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.

 

[QuantumCosmo]

Hi,
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:

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 "$\bar{\theta}$" comes from.

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

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:

$$G^{\mu\nu}\tilde{G}_{\mu\nu} = \partial_\mu K^\mu$$

where

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

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.

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 $$U\partial U^{+}$$ 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, $$F_{\mu\nu}$$ would vanish at infinity). So why are there no instantons in QED?
Is it because than U would be a map from $$S_3$$ (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 $$S_3$$ to SU(2) = $$S_3$$ and the corresponding homotopy group $$\pi_3(S_3)$$ is not trivial. Therefore, we have instantons.
But then SU(3) should be isomorphic to $$S_4$$ and U is a map from $$S_3$$ to $$S_4$$ 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 $$G^{\mu\nu}\tilde{G}_{\mu\nu} = \partial_\mu K^\mu$$ 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,
Quantum

 

[blechman]

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

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??

1) Why does the A^3 term vanish in QED

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

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 $$U\partial U^{+}$$ 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, $$F_{\mu\nu}$$ would vanish at infinity). So why are there no instantons in QED?
Is it because than U would be a map from $$S_3$$ (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)?

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.

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 $$S_3$$ to SU(2) = $$S_3$$ and the corresponding homotopy group $$\pi_3(S_3)$$ is not trivial. Therefore, we have instantons.
But then SU(3) should be isomorphic to $$S_4$$ and U is a map from $$S_3$$ to $$S_4$$ and the corresponding homotopy group is trivial. So why are there instantons in SU(3) gauge theory?

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.

As to explain my orignial question:
If the term $$G^{\mu\nu}\tilde{G}_{\mu\nu} = \partial_\mu K^\mu$$ 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?

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!

 

[QuantumCosmo]

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??

Well, then, if I ever find a good source, I will send it to you :)

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

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

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.

Ok, so it is really only the A^3 term that is important? Does the $$\partial_{\alpha}A_{\beta}^a$$ 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 $$\pi_1(S_{2})$$ - or more general $$\pi_1(S_{n})$$ for n greater than 1 - being trivial. But in this case we would have a mapping from $$S_3$$ to U(1) - corresponding to $$\pi_3(S_{1})$$ and I hope to god that's trivial oO ... I really should know more about topology...)

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.

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 let's 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. Let's say we are at infinity in space time. In that case I could suppose there is a region with $$A=0$$ and another one with $$A'=U\partial U^{+}$$ 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. Let's now say we are at some place in euclidean space time instead. In this case, U is not just some (gauge-) function on $$S_3$$ but on the whole of four-dimensional euclidian space time. If we continuously continue U to $$S_3$$ and call the resulting function U', it can be shown that U' can, on $$S_3$$, 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 $$A'=U\partial U^{+}$$) 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?

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.

Ah, yeah, I think I understand now. If there is a $$G\tilde{G}$$ term in the Lagrangian, then in the action I will have $$S=... + \int \partial K dx=... + \int_{S_3} K dx=...+ \int_{S_3} A^3 dx$$. 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 $$G\tilde{G}$$ term out. Is that right?

Hope that helps!

Yes, it does a lot! Thank you very much,
Quantum

 

[blechman]

Well, then, if I ever find a good source, I will send it to you :)

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.

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

yes.

(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 $$\pi_1(S_{2})$$ - or more general $$\pi_1(S_{n})$$ for n greater than 1 - being trivial. But in this case we would have a mapping from $$S_3$$ to U(1) - corresponding to $$\pi_3(S_{1})$$ and I hope to god that's trivial oO ... I really should know more about topology...)

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...

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 let's 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. Let's say we are at infinity in space time. In that case I could suppose there is a region with $$A=0$$ and another one with $$A'=U\partial U^{+}$$ 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. Let's now say we are at some place in euclidean space time instead. In this case, U is not just some (gauge-) function on $$S_3$$ but on the whole of four-dimensional euclidian space time. If we continuously continue U to $$S_3$$ and call the resulting function U', it can be shown that U' can, on $$S_3$$, 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 $$A'=U\partial U^{+}$$) 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?

I've completely lost you! :

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...

Ah, yeah, I think I understand now. If there is a $$G\tilde{G}$$ term in the Lagrangian, then in the action I will have $$S=... + \int \partial K dx=... + \int_{S_3} K dx=...+ \int_{S_3} A^3 dx$$. 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 $$G\tilde{G}$$ term out. Is that right?

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 $e^{in\theta}$. 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?

 

[tom.stoer]

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.

 

[QuantumCosmo]

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.

Thank you. I'm looking through it right now :) I hope it helps me with some of my questions.

I've completely lost you! :
...
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...

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.

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 $e^{in\theta}$. 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.

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,
Quantum

 

[blechman]

Thank you. I'm looking through it right now :) I hope it helps me with some of my questions.

I hope it does! Let me know.

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.

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

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?

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:

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

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.

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 $\phi$ 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.

 

[QuantumCosmo]

Hm, now I am utterly confused... I thought the problem was in the $$A^3$$ term in $$K_\mu$$. 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

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

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

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

with

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

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

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

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

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

Am I on the right way?
But I still don't see a $$\theta$$ in here...

 

[blechman]

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

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

So your phase is $e^{in\theta}$. Other than that, you're right.

 

[QuantumCosmo]

Ah, right, I forgot the coefficient... :P
Oh boy... questions always seem so silly afterwards...

Thank you very much for your help!

 

[QuantumCosmo]

Now that my instanton question seems to be solved, I have one more question concerning the $$G\tilde{G}$$ term:

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

$$\Delta S = \int d^4 x \partial_{\mu} J^{\mu}$$

to the action and we therefore should have added (or "should add") the term $$\partial_{\mu} J^{\mu}$$ (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,
Quantum

 

[blechman]

No, that isn't right. $J^\mu$ in this case would be the axial current $\bar{\psi}\gamma^\mu\gamma^5\psi$ 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.

 

[QuantumCosmo]

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

Ah, no, I meant I add $$\partial_{\mu}J^{\mu}$$ to the Lagrangian, not $$J^{\mu}$$.
In the case of my example, that seems to be the case - thus, the $$G\tilde{G}$$ term. But I don't know if that is always true or just happens to be right in this case...

 

[Ben Niehoff]

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 $\mu: SU(3) \rightarrow S^3_\infty$ 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. :)

 

[blechman]

Ah, no, I meant I add $$\partial_{\mu}J^{\mu}$$ to the Lagrangian, not $$J^{\mu}$$.
In the case of my example, that seems to be the case - thus, the $$G\tilde{G}$$ term. But I don't know if that is always true or just happens to be right in this case...

But you added $\partial_\mu K^\mu$, not $\partial_\mu J^\mu$! K is not the current!

 

[blechman]

Edit: I see after reading more that you already talked about that. Whoops. :)

Always happy to have a confirmation!

 

[QuantumCosmo]

Hm, funny, but I think it holds that
$$\partial_{\mu} J_5^{\mu} = -\frac{g^2}{16\pi^2}G\tilde{G}$$
after quantum corrections. (See f.e. Peskin Schroeder, Srednicki etc.)

So I do in fact add $$-\theta \partial_{\mu} J_5^{\mu}$$

The question is: why?
Is that simply a coincidence?

 

[blechman]

Hm, funny, but I think it holds that
$$\partial_{\mu} J_5^{\mu} = -\frac{g^2}{16\pi^2}G\tilde{G}$$
after quantum corrections. (See f.e. Peskin Schroeder, Srednicki etc.)

So I do in fact add $$-\theta \partial_{\mu} J_5^{\mu}$$

The question is: why?
Is that simply a coincidence?

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.

 

[QuantumCosmo]

Does $$\partial J = \partial K$$ imply that J=K?

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

And so it seems to me as if $$-\theta \partial J$$ is then added to the Lagrangian...

 

[blechman]

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.

 

[QuantumCosmo]

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.

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 $$J_5^{\mu}$$ 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 $$G\tilde{G}$$ 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...

 

[blechman]

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

$$-\mathcal{L}_m = m\bar{\psi}_L \psi_R + m^{*}\bar{\psi}_R \psi_L$$

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:

$$\psi_R\rightarrow \psi_Re^{i\alpha}$$

Then if I choose $\alpha$ 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:

$$\mathcal{L}_m\rightarrow \mathcal{L}_m+\alpha\partial_\mu J_R^{\mu}$$

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:

$$\mathcal{L}_m\rightarrow \mathcal{L}_m+\alpha\frac{Ag^2}{16\pi^2}G\tilde{G}$$

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 $\theta$ term we had before! So we find that when we remove the phase of the mass, we correspondingly shift the parameter $\theta$:

$$\theta\rightarrow \theta+A\alpha$$

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, $\theta_{EW}$ 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 $\alpha$ is arbitrary, we can safely set it equal to $-\theta_{EW}/A$ 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.

 

[QuantumCosmo]

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

$$\mathcal{L}_m\rightarrow \mathcal{L}_m+\alpha\partial_\mu J_R^{\mu}$$

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

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 :)

 

[QuantumCosmo]

Ok, so my questions have been narrowed down to:

1) Why does the Lagrangian change in that manner?
Answer: I haven't fully understood how the calculation works, but this term seems to arise from the change in the measure (in our functional integral) after the transformation and one can find it for example in [Weinberg,Ch 22].

2) Why was there even a phase in the quark mass matrix to begin with? I always thought it was possible to diagonalize it so that all the mass terms are real. Does it have something to do with the possible phase in the CKM matrix?
(Ok, so the question actually is rather: Why CAN there be a phase in the mass matrix?)

 

[blechman]

Ok, so my questions have been narrowed down to:

1) Why does the Lagrangian change in that manner?
Answer: I haven't fully understood how the calculation works, but this term seems to arise from the change in the measure (in our functional integral) after the transformation and one can find it for example in [Weinberg,Ch 22].

I think this is a failure on my part. I don't think I explained this properly. Let me try again, but first, let me answer your second question

2) Why was there even a phase in the quark mass matrix to begin with? I always thought it was possible to diagonalize it so that all the mass terms are real. Does it have something to do with the possible phase in the CKM matrix?
(Ok, so the question actually is rather: Why CAN there be a phase in the mass matrix?)

The mass is an ARBITRARY parameter, and it can be whatever it wants to be. There is no reason for it to be real (for fermions)!

You want to "diagonalize it" - but HOW do you do that?! You have to do what I just did! That is how you make the mass real - you have to absorb the phases into the fields. See Scott Willenbrock's TASI 2004 lecture notes for details, for example.

And yes, this is PRECISELY where the CKM phase comes from.

But anyway, the point is that the phase of the mass is arbitrary, and you have to use chiral rotations to absorb phases. But once you do that, since some of those rotations are anomalous, the phase will not go away completely, but will shift the theta vacuum angle.

Now I have to convince you of that. Stand by for the next post.

 

[blechman]

OK, I've spent the last 1.5 hrs trying to write this up! I've got to move on! But let me at least point you to some references.

Peskin, 19.2, he goes through the functional integral method. There he shows how to get the anomaly term from the Jacobean when you do a field redefinition on your fermion $\psi\rightarrow(1+\alpha\gamma^5)\psi$. This term is proportional to $\alpha$, so that as long as the fermions are massless, you can chose it to precisely cancel the theta term: in a theory with massless fermions, the theta term is an irrelevant phase (a gauge choice, if you wish).

However, if the mass is nonzero and you do this field redefinition to cancel the phase of m, then you have lost the arbitrariness of $\alpha$. You can make the mass real, but only at the expense of moving it into the theta term. Now this term cannot be arbitrarily set to zero, and it IS physical.

I was going to write more, but I found myself getting caught up over details, so I think I'll just leave it at that. You can also look at "Dynamics of the Standard Model" by Donoghue, Golowich and Holstein, Chapters 3.5 and 9.5.

Hope that helps!

 

[QuantumCosmo]

Thank you very much! I realize you put a lot of time into this and I really appreciate it!

So let me summarize (I hope I got it right):
I have the Lagrangian

$$\mathcal{L} = \bar{\psi}m e^{i\theta'}\psi + \theta_{weak} W\tilde{W} + \theta_{strong} G \tilde{G}$$

(where I didn't really care about constant factors... :P)

G is the field strength tensor of the strong interactions and W that of the weak interactions.

Now I perform the chiral rotation

$$\psi \rightarrow e^{-i \frac{\theta'}{2}\gamma_5}\psi$$

which results in

$$\mathcal{L} = \bar{\psi}m \psi + \theta_{weak} W\tilde{W} + (\theta_{strong}-\theta') G \tilde{G}$$

Now I do a B+L U(1) rotation (although I don't know what that is) by some angle $$\alpha$$. This gives me

$$\mathcal{L} = \bar{\psi}m e^{i\theta'}\psi + (\theta_{weak}-\alpha) W\tilde{W} + (\theta_{strong}-\theta') G \tilde{G}$$

Since $$\alpha$$ is arbitrary, I can choose it to be equal to $$\theta_{weak}$$ and thus the weak term vanishes.

... the only question is: how can one know that the first rotation (the axial U(1)) changes the strong $$\theta$$ term and the second (the B+L U(1)) changes the weak $$\theta$$ term?

Is it because U(1)_A is a symmetry of the QCD Lagrangian and the supposed U(1)_{B+L} is a symmetry of the weak Lagrangian?

 

[blechman]

Now I perform the chiral rotation

$$\psi \rightarrow e^{-i \frac{\theta'}{2}\gamma_5}\psi$$

which results in

$$\mathcal{L} = \bar{\psi}m \psi + \theta_{weak} W\tilde{W} + (\theta_{strong}-\theta') G \tilde{G}$$

If the fermion is charged under both QCD and EW, and the anomaly coefficient (those constant factors you're ignoring!) doesn't vanish(!) then BOTH theta_{weak} and theta_{strong} are going to be shifted. But other than that, this is right.

Now I do a B+L U(1) rotation (although I don't know what that is) by some angle $$\alpha$$.

B+L = baryon number + lepton number. The SM lagrangian is (classically) invariant under both of these, but it can be shown that they are both anomalous symmetries. HOWEVER, B-L is NOT anomalous (the anomaly coefficient vanishes). So instead of B and L, we often use B+L and B-L, where the first of these is anomalous, and the other is not.

But the anomaly coefficient for the QCD anomaly vanishes, even for the B+L, but NOT so for the weak. Therefore...

This gives me
$$\mathcal{L} = \bar{\psi}m e^{i\theta'}\psi + (\theta_{weak}-\alpha) W\tilde{W} + (\theta_{strong}-\theta') G \tilde{G}$$

Since $$\alpha$$ is arbitrary, I can choose it to be equal to $$\theta_{weak}$$ and thus the weak term vanishes.

Quod Erat Demonstrandum!

EDIT: There is no phase in the mass term anymore. You have that right above, so I assume you just have a typo here.

... the only question is: how can one know that the first rotation (the axial U(1)) changes the strong $$\theta$$ term and the second (the B+L U(1)) changes the weak $$\theta$$ term?

Is it because U(1)_A is a symmetry of the QCD Lagrangian and the supposed U(1)_{B+L} is a symmetry of the weak Lagrangian?

Again, this depends on the theory at hand. It all depends on which anomaly coefficients vanish and which do not.

To get it all right, you have to make a distinction: I have presented you with BOTH a toy model (that is, one fermion) AND the Standard Model, which is a complicated 3-generation model with quarks and leptons, left and right, each with their own interactions! The B+L argument applies to the STANDARD MODEL, not to any model that happens to come along! If the universe were different, then theta_weak might be a real parameter.

See what I'm saying?

 

[QuantumCosmo]

Ah yeah I think I got it :)

But how does this B+L U(1) symmetry look?
I think U(1)_B is the vector symmetry of QCD:
$$\psi_L \rightarrow e^{i\alpha}\psi_L$$
and
$$\psi_R \rightarrow e^{i\alpha}\psi_R$$

and B is the conserved charge of that symmetry.

But I do not know the symmetries corresponding to L, B+L or B-L...

 

[blechman]

The Standard Model has quarks and leptons. The quarks have baryon number +1/3, and the leptons have lepton number +1, with their antiparticles the negative. This is a U(1) symmetry of the SM Lagrangian.

HOWEVER: both of these symmetries are anomalous: if you compute the triangle diagrams where you insert a B current on one vertex, and W bosons on the other two vertices, that triangle diagram is nonzero. Same with inserting an L current instead of a B current.

But it is also true that these anomalies (the B and the L) are equal. So that if you take the DIFFERENCE of the two, they cancel and that current is not anomalous. That is the B-L current. By a similar analysis, the sum of the two currents is 2x the anomaly of either one of them.

Notice however, that there is NO QCD anomaly: if you replace the W bosons with gluons, the triangle with the B current vanishes on its own. The one for L vanishes trivially, since leptons have no color charge.

So we have the equations:

$$\partial_\mu J_{B+L}^\mu = \frac{Ag^2}{16\pi^2}W\tilde{W}$$
$$\partial_\mu J_{B-L}^\mu =0$$

Notice that THERE ARE NO GLUONS IN THIS EQUATION! Just W bosons.

So by the arguments given above: performing a phase redefinition on all the fermions (quarks AND leptons) in accord with the B+L charge (using Weyl notation where the R-fermions are ANTIfermions; you can translate to Dirac easily by inserting $\gamma^5$s in the appropriate places):

$$q_L\rightarrow e^{i\alpha/3}q_L$$
$$q_R\rightarrow e^{-i\alpha/3}q_R$$
$$l_L\rightarrow e^{i\alpha}l_L$$
$$l_R\rightarrow e^{-i\alpha}l_R$$

will not change the lagrangian sans the theta terms (since it's a symmetry), and it won't change the QCD theta term, since there is no (B+L)-G-G anomaly. But it WILL shift the W-theta term by an amount proportional to $\alpha$. So if you choose $\alpha$ just right, you can cancel the SU(2) theta angle without doing any more damage.

If you try to shift the QCD angle, you cannot use this phase redefinition I wrote down. You have to use another one (such as rotating both qL and qR by the SAME angle rather than opposite angle). Such a redefinition will reintroduce phases into the masses of the quarks, and so THAT is why $\theta_{QCD}$ is physical. There will always be a phase there somewhere.

UNLESS one of the quark masses vanish. In that case, we're golden. Some people believe the up quark might have vanishing mass. It would mean that $\theta_{QCD}$ is unphysical, and therefore solve the strong CP problem; but it also would go against various other results such as lattice calculations. Personally I am inclined to believe the up mass is nonzero, but plenty of perfectly respectable physicists disagree with me.

Anyway, I hope that helps. Does that make sense?

 

[QuantumCosmo]

Anyway, I hope that helps. Does that make sense?

Hm, yeah, I think I might have understood it :)

My whole problem is that my QFT basics aren't that good... and that I often have problems with (I guess) very easy parts, for example those U(1) transformations.

I now suspect (after what you just wrote) that
$$q_L\rightarrow e^{i\alpha/3}q_L$$
$$q_R\rightarrow e^{-i\alpha/3}q_R$$
is a (global) symmetry of the Lagrangian and it's corresponding charge is B.

And
$$l_L\rightarrow e^{i\alpha}l_L$$
$$l_R\rightarrow e^{-i\alpha}l_R$$
is another symmetry of the Lagrangian with charge L.

And if I want the symmetry with charge B+L, I simply apply both of them. Is that correct?

As for the results of applying that symmetry (the change in the weak $$\theta$$ term), I think I have understood it :)

Thanks a million for that!

 

[QuantumCosmo]

Hi, I have found an article that explains why the $$\theta_{weak}$$ term can be eliminated.
It seems, though, this is different from what we discussed here (although I don't really understood what they were doing...)
A. A. Anselm and A. A. Johansen, Nucl. Phys. B407(1993) 652

 

[ansgar]

see my presentation about this topic:

http://www.isv.uu.se/~wouda/axion-beamer-GW.pdf