A pseudoscalar Goldstone boson, $\pi(x)$, is protected by a shift symmetry: it shows up with a derivative in its interaction terms in a Lagrangian. As a pseudoscalar, we may also write it with the usual $i\gamma^5$ interaction. There are thus two ways to encode the interaction:

  1. Shift symmetry manifest: $$\mathcal L = \left(\frac{\partial_\mu \pi}{v}\right)\bar\Psi\gamma^\mu \gamma^5\Psi$$

  2. Pseudoscalar manifest: $$\mathcal L = g \pi \bar\Psi i\gamma^5 \Psi$$

These two are related by the equation of motion, so that $g = 2qm/f$, where $m$ is the fermion mass, $f$ is the order parameter of symmetry breaking, and $q$ is the charge with respect to the broken axial symmetry.

My question is: In the shift-symmetry manifest form, we know that fermion loops do not generate a pseudoscalar mass. However, pseudoscalar manifest form of the interaction looks like a generic pseudoscalar interaction with no symmetry protecting $\pi$ from receiving mass corrections from $\Psi$ loops. So:

  1. How do we see that $\pi$ is protected by a symmetry when we write the interaction in the manifestly pseudoscalar format?

  2. Conversely, I could write any pseudoscalar interaction as $ g \pi \bar\Psi i\gamma^5 \Psi$ Does this mean that I can use the fermion equation of motion to convert any pseudoscalar interaction into one that has a shift symmetry?

Details follow, but the main question is stated above.


Set up: Goldstone interaction with fermions

We show how to convert between the shift-symmetric and pseudoscalar forms of the interaction. For simplicity, assume the case of a global, internal, compact U(1) symmetry that is spontaneously broken by a field $H$ that obtains a vev $\langle H \rangle = f$. Let the theory contain a left-handed fermion $\psi_L$ and a right-handed fermion $\psi_R$. Assume axial U(1) charges such that

$$Q[H] = 2q\\ Q[\psi_L] = q\\ Q[\psi_R] = -q$$

Then we may write out the theory with a Yukawa interaction:

$$\mathcal L_\text{Yuk} = y H^*\bar\psi_L \psi_R + \text{h.c.}$$

We now "pull out the Goldstone fields" from the fields. In order to do this, we transform each field $\Phi \in \{H,\psi_L,\psi_R\}$ by the spontaneously broken symmetry:

$$ \Phi = e^{iq_\Phi \epsilon} \Phi'$$

On the right-hand side, $\Phi'$ is understood to be the field with no Goldstone component. The Goldstone lives in the exponential. For the U(1) case, $\epsilon$ is the transformation parameter, and $q_\Phi$ is the U(1) charge of the $\Phi$.

Then we simply promote the transformation parameter to the Goldstone field, $\epsilon \to \pi(x)/f$. This is a nonlinear transformation to help identify the Goldstone interaction (Sec 19.6 of Weinberg Vol II, or CCWZ II). This gives

$$ \Phi(x) = \exp\left(iq_\Phi \frac{\pi(x)}{f}\right) \Phi'$$

When we do this, $$\mathcal L_\text{int} = y H'^* e^{-2iq} \bar \psi_L' e^{iq} e^{iq} \psi_R' + \text{h.c.} = y H'^*\bar \psi_L' \psi_R' + \text{h.c.} $$

The Goldstone has been completely removed from the Yukawa term and doesn't show up there. This is a consequence of U(1) conservation of the Lagrangian term. Where did the interaction go? We know that the Goldstone must have a derivative interaction, so the natural place to look is the fermion kinetic term.

Writing the kinetic terms with implicit projection operators (alternatively, you can replace $\gamma^\mu$ with $\sigma^\mu$ or $\bar\sigma^\mu$ as appropriate):

$$\mathcal L_\text{kin} = i \bar \psi_L \gamma^\mu \partial_\mu \psi_L + i \bar \psi_R \gamma^\mu \partial_\mu \psi_R $$

Replacing $\psi_{L,R}$ by the fields with the Goldstone pulled out:

$$ \mathcal L_\text{kin} = i \bar \psi_L' e^{-iq \pi(x)/f} \gamma^\mu \partial_\mu \left( e^{iq \pi(x)/f}\psi_L' \right)+ = i \bar \psi_R' e^{iq \pi(x)/f} \gamma^\mu \partial_\mu \left( e^{-iq \pi(x)/f}\psi_R' \right)$$

In addition to the usual kinetic terms, these give terms where the derivative acts on the Goldstone, $\pi(x)$. These are the interaction terms that are our primary focus. For simplicity, let us combine the left- and right handed chiral spinors $\psi_{L,R}'$ into a Dirac spinor, $\Psi = (F',f')^T$ and use the projection operators $\frac{1}{2}(1\pm \gamma^5)$:

$$\mathcal L_\text{int} = i \left(i\frac{q}{f}\partial_\mu \pi \right) \bar\Psi \gamma^\mu \frac{1}{2}\left(1-\gamma^5\right) \Psi + i \left(-i\frac{q}{f}\partial_\mu \pi \right) \bar\Psi \gamma^\mu \frac{1}{2}\left(1+\gamma^5\right) \Psi $$ These terms combine simply into: $$\mathcal L_\text{int} = \frac{q}{f}\left(\partial_\mu \pi\right)\bar\Psi \gamma^\mu \gamma^5 \Psi \ . $$

We thus arrive at the the Goldstone--fermion interaction term in the shift-symmetric manifest form: clearly $\pi$ is invariant under $\pi(x) \to \pi(x) + c$ and so it is protected from quantum corrections that might generate a mass term $m_\pi^2 \pi^2$.

Using the fermion equation of motion

Now we can use the fermion equation of motion to convert this shift-symmetric form of $\mathcal L_\text{int}$ into one that is manifestly pseudoscalar. Recall that the equation of motion in Dirac notation is:

$$i\gamma^\mu\partial_\mu \Psi = m\Psi$$

Armed with this, we may now integrate $\mathcal L_\text{int}$ by parts to shift the derivative from the $\pi$ to the fermion bilinear. We assume that there's no surface term so that integration by parts in the action amounts to a minus sign and moving the derivative in the Lagrangian:

\begin{align} \mathcal L_\text{int} &= \frac{q}{f} \pi \partial_\mu \left(\bar \Psi \gamma^\mu \gamma^5 \Psi \right) \\ & = \frac{q}{f} \pi\left[ (\partial_\mu\bar\Psi)\gamma^\mu\gamma^5 \Psi + \bar\Psi \gamma^\mu \gamma^5 \partial_\mu \Psi \right] \\ & = \frac{q}{f} \pi\left[ (\partial_\mu\Psi)^\dagger \left(\gamma^0\gamma^\mu \gamma^0\right) \gamma^0\gamma^5 \Psi - \bar\Psi \gamma^5 \gamma^\mu \partial_\mu \Psi \right] \\ & = \frac{q}{f} \pi\left[ (\gamma^\mu\partial_\mu\Psi)^\dagger \gamma^0\gamma^5 \Psi - \bar\Psi \gamma^5 \gamma^\mu \partial_\mu \Psi \right] \\ & = \frac{q}{f} \pi\left[ (-im\Psi)^\dagger \gamma^0\gamma^5 \Psi - \bar\Psi \gamma^5 \left(-im\Psi\right) \right] \\ & = \frac{2iqm}{f} \pi \bar\Psi\gamma^5 \Psi \ . \end{align}

This now gives us our manifestly pseudoscalar interaction between the Goldstone $\pi$ and the fermions.

Reiteration of the puzzle

So the puzzle is that:

  • The shift-symmetric and manifestly pseudoscalar forms of the interaction seem perfectly equivalent.

  • However, the pseudoscalar form of the interaction seems perfectly general. One could tune the fermion mass $m$ to be whatever you want by tuning the Yukawa coupling $y$. This, in turn, tunes $g = 2qm/f$ to be any pseudoscalar coupling. Does this mean that any pseudoscalar interaction between massive fermions can be written as a Goldstone interaction?

  • In the manifestly pseudoscalar version of the theory, are loop contributions to the $\pi$ mass manifestly zero? This does not seem to generically be the case. (See, e.g. this discussion based on a problem in Peskin & Schroeder)

  • So: in the case where there really is a spontaneously broken symmetry, there should be a shift symmetry protecting the $\pi$, but how can we see the effect of that shift symmetry when we calculate loops in the pseudoscalar theory?

  • Alternatively, if we took a generic pseudoscalar theory with no shift symmetry (i.e. the pseudoscalar is not a Goldstone), then what prevents me from using the equation of motion to write the interaction in a manifestly shift-symmetric form and waving my hands that there ought to be a shift symmetry?

  • $\begingroup$ Dear @Cosmas Zachos, thank you for noticing this. I have corrected (1) to include a missing $\gamma^5$ so that it is an axial coupling. I have also updated the following discussion to simplify it by making the global symmetry manifestly axial. Please let me know if there are any further unreasonable expectations on the reader. I look forward to further insights. $\endgroup$ Aug 23, 2017 at 0:55
  • $\begingroup$ Thank you for your additional comment, I apologize for the confusion. I will think more carefully if there are ways for me to clarify both for myself and for the reader. I think even without the explicit model (which are the details), the question is simply posed in the first part of the question: given form (1), can you not integrate by parts to get form (2) and hence argue that they are the same? $\endgroup$ Aug 23, 2017 at 1:12
  • $\begingroup$ For simplicity: one may ignore the explicit model altogether and consider the axion, just turn off the Standard Model gauge group so that it becomes a rather simple theory. Does that bring you to equation (1) and (2)? $\endgroup$ Aug 23, 2017 at 1:13
  • 2
    $\begingroup$ The two descriptions are not equivalent as I have explained in the comments to Cosmas Zachos. Th equations of motion you use are only the free ones, and those hold only for the external legs and no, for example, in the fermionic loop that generates the pion mass in the yukawa pseudoscalar theory. The two theories are equivalent only at linear order in pi. You can also calculate an amplitude in the two theories and show that they are different. $\endgroup$
    – TwoBs
    Aug 24, 2017 at 20:49
  • $\begingroup$ As you see the answer has changed since Cosmas had to reintroduce sigma' field. He really didn't have to, it could have just added the non linear pi corrections from solving correct,y the constraints since he was working in the infinite signa' mass limit. Anyway, again, the takeaway message is exactly the opposite that was given: the two theories, yukawa vs gradient, are not equivalent. In particular, the former will generate a pi mass at one loop. $\endgroup$
    – TwoBs
    Aug 25, 2017 at 5:41

2 Answers 2


The two theories, namely the ``gradient model'' $\partial_\mu\pi \bar{\Psi}\gamma^5\gamma^\mu\Psi$ and the Yukawa model $g\pi\bar{\Psi}\gamma^5\Psi$ (both with a massive $\Psi$), are definitely not equivalent. They have different symmetries, spectrum and scattering amplitudes, hence are physically distinct theories. The main mistake that you (the OP) are doing is using the free equations of motion for the fermions, but that's ok only for the external legs and not for the virtual ones that enter e.g. in the one-loop calculation of the $\pi$ mass, or as I'll show below in a scattering amplitude with an intermediate virtual $\psi$ exchanged. (The mistake that Cosmas Zachos was doing in his earlier answer and that in part is still doing in the marginally improved answer is explained in my comments to his answer, I will not repeat it here).

The gradient model is indeed invariant under $\pi\rightarrow \pi+const$ which clearly forbids a mass term for $\pi$. This isn't the case for the Yukawa model where a bare mass is needed to remove the quadratic divergent mass generated by the fermion loops. A physical pole mass is therefore generically non-zero, barring fine-tuning.

More importantly, Goldstone bosons (GB's) aren't just massless particles, they have various special features. For example, soft GBs (that is the limit of vanishing $\pi$-momentum) give vanishing scattering amplitudes (the so-called Adler zero condition). This is realized for the gradient theory but not for the Yukawa theory. Let's see this in more detail looking at a physical scattering amplitude $\pi\Psi\rightarrow \pi\Psi$. For the Yukawa theory one has $$ M^{Yukawa}_{\pi\Psi\rightarrow\pi\Psi}=-g^2 \left[\bar{u}(p_2^\prime)\frac{i(\gamma_\alpha p_1^\alpha+\gamma_\alpha p_2^\alpha-m)}{s-m^2+i\epsilon}u(p_2)+\mbox{crossed diag.}\right] $$ for $\pi(p_1)\Psi(p_2)\rightarrow\pi(p_1^\prime)\Psi(p_2^\prime)$. The $\gamma^5$ have been moved around and simplified with the numerator of the fermion propagator, i.e. $\gamma^5i(\gamma_\alpha p_1^\alpha+\gamma_\alpha p_2^\alpha+m)\gamma^5=-i(\gamma_\alpha p_1^\alpha+\gamma_\alpha p_2^\alpha-m)$. We could have simplified the numerator using $\gamma_\alpha p_2^\alpha u(p_2)=m u(p_2)$, where $m$ is the fermion mass, but it's more convenient this form in the following. There is an s-channel contribution, explicitly displayed, along with a crossed diagram that we do not display explicitly.

(disclaimer: I am doing this calculation by hand on an Ipad, I hope it is not grossly incorrect :-), although factors of 2 and minus signs are most likely off)

This $M^{Yukawa}_{\pi\Psi\rightarrow\pi\Psi}$ doesn't vanish for $p_1\rightarrow 0$ because, even though the numerator goes to zero (namely $\gamma_\alpha p_1^\alpha+\gamma_\alpha p_2^\alpha-m)u(p_2)=\gamma_\alpha p_1^\alpha u(p_2)\rightarrow 0$), so does the denominator at the same rate ($s-m^2=2p_{1\alpha} p_2^\alpha\rightarrow0$; here I am assuming that we have tuned the spectrum to be the same, that is the $\pi$ mass in the Yukawa model has been tuned to zero by hand, otherwise the numerator wouldn't even vanish and the comparison between the two models would make no sense).

On the other hand, for the gradient theory we get $$ M^{gradient}_{\pi\Psi\rightarrow\pi\Psi}=\frac{1}{f^2}\left[\bar{u}(p_2^\prime)\frac{i(\gamma_\alpha p_1^\alpha+\gamma_\alpha p_2^\alpha-m)^3}{s-m^2+i\epsilon}u(p_2)+\mbox{crossed diag.}\right]\rightarrow 0 $$ which is not only different (hence the two theories are physically distinct, period) but it gives a vanishing amplitude in the GB soft limit $p_1\rightarrow 0$ since the numerator can be written as $i\bar{u}(p_2^\prime)(\gamma_\alpha p_1^\alpha+\gamma_\alpha p_2^\alpha-m)^3u(p_2)=i\bar{u}(p_2^\prime)\gamma_\alpha p_1^{\prime\alpha}(\gamma_\beta p_1^\beta+\gamma_\beta p_2^\beta-m)(\gamma_\gamma p_1^\gamma)u(p_2)$, and for momentum conservation $p_1^\prime\rightarrow 0$ too.

The takeaway message is: the two models are distinct physically and mathematically. The gradient theory describes a GB whereas the Yukawa theory describe a scalar with a mass tuned to be zero.

Extra edits I have finally found some times to add a last remark that I mentioned in the comments but it is actually worth to report in the full answer. It is also related to the answer by @Cosmas Zachos.

Having established that the two theories are different, one may wonder how much different and what is the relation between the two, given that the simple use of the equations of motion by the OP was flawed. The answer is very simple: the two theories differ at the non-linear $\pi$-level, starting from the quadratic order. In particular, the claim is that the theory given by $$ \mathcal{L}_{new}=\bar{\Psi}(i\gamma^\alpha \partial_\alpha-m e^{-2i\gamma^5\pi/f})\Psi+\frac{1}{2}(\partial_\mu\pi)^2\qquad f\equiv 2m/g\,, $$ which differs from $\mathcal{L}_{Yukawa}=\bar{\Psi}(i\gamma^\alpha \partial_\alpha-m)\Psi+ig\pi \bar{\Psi}\gamma^5\Psi+\frac{1}{2}(\partial_\mu\pi)^2$ starting from $o(\pi^2)$, is in fact equivalent to the gradient theory $$ \mathcal{L}_{gradient}=\frac{1}{2}(\partial_\mu\pi)^2+\bar{\Psi}(i\gamma^\alpha \partial_\alpha-m)\Psi+\frac{1}{f}\partial_\mu\pi \bar{\Psi}\gamma^5\gamma^\mu\Psi\,. $$ Indeed, it's enough to perform the field redefinition $\Psi\rightarrow e^{i\gamma^5 \pi/f}\Psi$ to move the $\pi$ from the non-derivative term to the gradient coming from the $\Psi$-kinetic term.

As ultimate check, let's see the behavior under the soft limit $p_1\rightarrow 0$. The contributions from two linear-$\pi$ vertex insertions is like in the Yukawa theory, but now there is also a contact term coming from expanding the exponential, $\frac{2m}{f^2}\pi^2\bar{\Psi}\Psi$, i.e. $$ M^{new}_{\pi\Psi\rightarrow\pi\Psi}=M^{Yukawa}_{\pi\Psi\rightarrow\pi\Psi}+i\frac{4m}{f^2}\bar{u}(p_2^\prime)u(p_2)\,. $$ Now, in the soft limit $p_1\rightarrow 0$, we have $p_2^\prime \rightarrow p_2$ hence the Yukawa terms give $$ M^{Yukawa}_{p_1\rightarrow 0}=-g^2 \left[\bar{u}(p_2)\frac{i\gamma_\alpha p_1^\alpha}{2 p_1 p_2}u(p_2)+\mbox{crossed diag.}\right]=-2ig^2 $$ where I have used that $\bar{u}(p)\gamma^\alpha u(p)=2p^\alpha$. On the other hand, the new contact term in from $\mathcal{L}^{new}$ gives
$$ i\frac{4m}{f^2}\bar{u}(p_2)u(p_2)=i\frac{8m^2}{f^2}=2i g^2\,, $$ from $\bar{u}(p)u(p)=2m$ (no sum over the two spin orientations, we are considering definite polarizations). Summing the two contributions we see that they vanish each other out, in agreement with the Adler-zero condition.

But again, the equivalence with the gradient theory is achieved only after modifying the theory at the $o(\pi^2)$ level in the way shown above, which corresponds to render the $\pi$ a GB. The Yukawa coupling alone instead is for non-GB particles.

  • $\begingroup$ Well, I have addressed the Adler zeros presence in the linear sigma model in my own answer. I fear the OP's question was, implicitly, how to see the custodial broken axial symmetry ("shift symmetry") in the "Yukawa model" he never fully specified. By itself, it is not axially invariant, but, in a blink, with suitable choices of couplings, it is the linear sigma model, the mother of all such symmetry broken structures. $\endgroup$ Aug 26, 2017 at 0:45

You are really asking a question about the controlling $U(1)_A$ symmetry of the one-flavor σ-model. You thus need to first display the symmetry you are really probing.

Let's start from the linear σ-model. Schematically (being cavalier with over-all factors...), $$ {\cal L}_{lin}= i\bar{\Psi}\partial \!\! / ~\Psi +\tfrac{1}{2} \partial \pi \cdot \partial \pi +\tfrac{1}{2} \partial \sigma \cdot \partial \sigma + g \bar\Psi (\sigma +i\gamma_5 \pi) \Psi -V(\pi,\sigma), $$ invariant under the U(1)A symmetry, $$ \delta \Psi= \frac{i}{2}\theta \gamma_5 \Psi,\\ \delta \sigma= \theta \pi,\\ \delta \pi= -\theta \sigma, $$ which you must check leaves the kinetic, Yukawa, terms invariant; and dictate it does so on the stable potential; you may choose the latter to be Goldstone's sombrero, etc, but I assume you are familiar with the SSB monkey business, and you know how to carry out the requisite field shifts, etc.

The corresponding (on-shell) conserved current is $$ J_A^{\mu}=-\tfrac{i}{2} \bar\Psi \gamma_5 \gamma^{\mu} \Psi + \pi \partial_\mu \sigma - \sigma \partial_\mu \pi, $$ so $ \partial\cdot J_A=0$.

Now, assuming the potential minimum enforces $\langle \pi\rangle =0$, $\langle \sigma\rangle=-f$ and redefining $\sigma\equiv \sigma ' -f$, s.t. $\langle \sigma' \rangle=0 $, observe this gives the fermion a mass m=gf, and the σ' a mass dependent on the arbitrary curvature of the potential V at its minimum--which you may take to be large, so the scalar σ' may be made arbitrarily massive so as to decouple from the low-energy model.

The result is the associated low energy σ-model, virtually trivial, involving, crucially, a massless goldston π, $$ {\cal L}_{low}= i\bar{\Psi}\partial \!\! / ~\Psi - gf \bar\Psi \Psi +\tfrac{1}{2} \partial \pi \cdot \partial \pi + g \bar\Psi i\gamma_5 \pi \Psi +\sigma' ~ {\mathrm terms}, $$ invariant under $$\delta \Psi= \frac{i}{2}\theta \gamma_5 \Psi, \qquad \delta \pi= \theta f ~~~(-{\small \theta \sigma'}), \qquad ({\small \delta \sigma'= \theta \pi}), $$ so that $$ J_A^\mu=-\tfrac{i}{2} \bar\Psi \gamma_5 \gamma^{\mu} \Psi + f\partial_\mu \pi ~~~(+{\small \pi\partial_\mu \sigma' - \sigma'\partial_\mu \pi}). $$ This hallmark shift in the transformation law for the π identifies it as a Goldstone boson.

(Note added to address concerns of @TwoBs: in fact, a residual term in the variation here is cancelled by the $-\theta \sigma'$ originally left out of $\delta \pi$, and the omitted $g\bar{\Psi} \sigma' \Psi$ in the lagrangian. I am omitting the ultra-heavy σ's here, which are still necessary for the full axial invariance, albeit out of sight here. Nevertheless, the amplitudes of the linear sigma model do, of course, have Adler zeros, as well known; however, the ππσ' vertex in the potential is involved.)

This symmetry enters into and constrains perturbation theory and "protects" the tree-level masslessness of the Goldstone mode π, as an induced mass for it would break it. (In fact, far more transparent than your naturally deprecated derivative coupling model, basically not relevant to the question. See VanderBij & Veltman 1984 for the pitfalls of the infinite-Higgs-mass limit.)

You may then see that comparison with the unrenormalizable gradient model is completely gratuitous, and avoidable. The pseudo scalar Yukawa term by itself has all crucial ingredients of a SSB symmetry in it, and is instantly made axially invariant with an innocuous σ scalar Yukawa, and a potential of your design, renormalizable if you wish, chosen to minimize most visible effects of the σ. The linear σ model (whose non-abelian generalization is the Higgs sector of the standard model) is the prototype that all future variant representations and tweaks are ineluctably based on.

Takeaway: A massless pseudoscalar coupling to a fermion in a parity-preserving Yukawa mode is perforce (or can be naturally promoted to) a goldston of an axial SSB and will remain such in perturbation theory--unless extraneous couplings explicitly spoil this custodial axial symmetry.

Added Ref: The historic 1960 σ-model paper reviewed in most good texts is as good as any to reset one's compass.

  • $\begingroup$ Thank you, Costas. I was following up on your suggested references as you posted this (Itzykson & Zuber, eq 11-194) but your "takeaway" summary clarified everything. I appreciate that time that you put into this question, as I suspect from my interpretation of your language that this exchange caused you some annoyance. I apologize for my lack of clarity and thank you again for your efforts. $\endgroup$ Aug 23, 2017 at 16:00
  • 1
    $\begingroup$ L@CosmasZachos Sorry, but your takeaway message can't be right: just do a loop with the yukawa-like interactions and you will get a mass for $\pi$! while no mass is generated for the GB. The reason is that the matching between the two theories is only at the linear-pi level. Another way to see it in terms of the OP question is that the free equations of motions don't hold inside the loop. $\endgroup$
    – TwoBs
    Aug 24, 2017 at 13:19
  • 1
    $\begingroup$ @CosmasZachos it's not just a detail: the conclusion of this discussion is exactly the opposite of what you have originally written. The yukawa model (alone) and the gradient theory are not equivalent, that would have been a fair takeaway message. They have differ symmetry, spectrum and amplitudes. But the two theories can be made equivalent at low energy by adding a new interaction with a $\sigma^\prime$ field to restore axial symmetry, or adding extra non- linear interactions that make it equivalent up to field redef. But this new theory is physically different than original yukawa's. $\endgroup$
    – TwoBs
    Aug 25, 2017 at 5:23
  • 1
    $\begingroup$ Besides, it was you that wanted to first introduce and then eliminate sigma'. There is nothin wrong with that, just you need to do it properly and solve the constraint, which again would have shown the inequivslnce of the two theories asked by the OP. I see that you reedit slightly your takeaway message, although is silly to say is 'necessarily' followed by its negation 'can be easily promoted' . Again, a more fair answer to the OP is that the Yukawa and the gradient theory are not equivalent. $\endgroup$
    – TwoBs
    Aug 25, 2017 at 5:36
  • 1
    $\begingroup$ Finally, even your, say, 'improved' version of the answer gives a totally wrong message concerning EFT, namely that one should keep somehow the sigma' field back in his mind to get the right answer at low-energy. But that's wrong. One can just integrate it out and work with the pi field alone instead (as you originally wanted to do btw), you just need to write the correct lagrangian that is the one from solving a constraint, which generates extra non-linear pi terms. I strongly advise you to remove your answer all together. If I'll find time a could post a correct one in one or two weeks $\endgroup$
    – TwoBs
    Aug 25, 2017 at 8:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.