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On P&S's QFT book, page 693, the book discussed "Systematics of the Higgs Mechanism". For the kinetic part of Lagrangian, $$ \frac{1}{2}\left(D_\mu \phi_i\right)^2=\frac{1}{2}\left(\partial_\mu \phi_i\right)^2+g A_\mu^a\left(\partial_\mu \phi_i T_{i j}^a \phi_j\right)+\frac{1}{2} g^2 A_\mu^a A^{b \mu}\left(T^a \phi\right)_i\left(T^b \phi\right)_i \tag{20.15}$$ if $\phi_i$ acquire vacuum expectation values of $\left\langle\phi_i\right\rangle=\left(\phi_0\right)_i$, we can obtain the gauge boson mass term through the last term of (20.15). $$\Delta \mathcal{L}=\frac{1}{2} m_{a b}^2 A_\mu^a A^{b \mu} \ \ \ \ m_{a b}^2=g^2\left(T^a \phi_0\right)_i\left(T^b \phi_0\right)_i\tag{20.17 & 18} $$ Also, their have a term after symmetry breaking: $$\Delta \mathcal{L}=g A_\mu^a \partial_\mu \phi_i\left(T^a \phi_0\right)_i . \tag{20.19} $$ (20.19) is related with explaining gauge boson mass. P&S write in page 691 that:

a gauge boson cannot obtain a mass, unless this mass term is associated with a pole in the vacuum polarization amplitude.

Using (20.19) we can derive the Feynman rules for gauge field and goldstone boson scattering, like in (20.20) enter image description here And then we can obtain the vacuum polarization in following form: enter image description here

Here comes my question:

  1. in the paragraph following (20.19), the book said: "the interaction term does not involve all of the components of $\phi$, only those that are parallel to a vector $T^a \phi_0$ for some choice of $T^a$. These vectors represent the infinitesimal rotations of the vacuum; thus the components $\phi_i$ that appear in (20.19) are precisely the Goldstone bosons." Why $\phi_i$ can be Goldstone bosons? I am really troubled for the book's description;

  2. in (20.21), the book give the vacuum polarization form, now there have a pole ($k=0$), and what's the meaning of "transverse" of this diagram? As I remember in chapter 7's description, this vacuum diagram don't have pole at $q=0$;

  3. Also, in eq.(20.20), their have a relative minus sign between $k^{\mu}$ and $k^{\nu}$, I thought the reason is for partial derivate act on in and out particle have a sign difference.

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I don't know if you were able to figure out the answers to your questions by now but I was recently reading this part of P&S, and I would like to give an answer here incase anyone else had these questions while reading the text.

Regarding the first question, there are a few things I want to point out:

  • The vector $T^a\phi_0$ is the infinitesimal transformation of the vacuum, specifically it's the one generated by the generator $T^a$. Making its components explicit we would write $(T^a\phi_0)_i$. This object, if it's a nonzero vector (i.e. if $T^a$ generates a broken symmetry), must then be a null vector of the mass squared matrix $$m^2_{ij}=\frac{\partial{^2V}}{\partial\phi^i\partial\phi^j}\bigm|_{\phi_0}$$ per the discussion in pages 351-352 in P&S (proof of Goldstone's theorem in the book). This means that the corresponding linear combination $\phi_i(T^a\phi_0)_i$ has zero mass and is precisely the Goldstone boson corresponding to the broken symmetry generator $T^a$.

  • Lets diagonalize $m^2_{ij}$ and use the linear combinations of fields associated with the eigenvectors as our new fields instead. Now, the vector $(T^a\phi_0)_i$, being an eigenvector of $m^2_{ij}$, will have only one nonzero component, which then selects the corresponding field component when doing $\phi_i(T^a\phi_0)_i$, and that selected field is precisely the the Goldstone boson corresponding to the broken symmetry generator $T^a$. [I will remark here that if you write down explicitly the three vectors $T^a\phi_0$ ($a=1,2,3$) for the SU(2) example, specifically the one in p.695, you will see that the ones associated with broken generators ($a=1,2$) each have only one nonzero component already so, just as an example, this diagonalization step is already satisfied there.]

  • With the diagonalization and field redefinition complete, it becomes clear that in the Feynman rule for the vertex eq. 20.19, only the index associated with the nonzero component of $(T^a\phi_0)_i$ (for some $T^a$) contributes when we sum over these indices in Feynman diagrams, so these vertices will only connect to the massless propagators of the corresponding Goldstone boson hence we get eq. 20.20 for the Goldstone boson diagram. We can enlarge to all possible values of the indices in the sum in eq. 20.20 (as P&S go on to do) because the rest of the components of $(T^a\phi_0)_i$ are zero anyway.

Regarding the second question:

  • The pole they refer to is from the $k^\mu k^\nu/k^2$ term, it is important for that term to be there so that the contribution to the vacuum polarization (1PI amplitude) is transverse i.e. orthogonal to $k^\mu$ in the sense that $k_\mu (g^{\mu\nu} - k^\mu k^\nu/k^2)=0$. This orthogonality is important for consistency with the Ward identity of the generic non-Abelian theory at hand, it also gives the appropriate form of the contribution to the 1PI amplitude that guarantees a shift in the pole of the exact gauge boson propagator to the nonzero value of $m^2_{ab}$ showing in equation 20.21 (thereby showing that physical mass was indeed acquired by the gauge boson). A similar argument is made in chapter 7 regarding consistency with the Ward identity of spinor QED (see page 245), a similar argument also applies here for our generic non-Abelian gauge theory; it will come with its own Ward identity that dictates a transverse structure for the 1PI vacuum polarization amplitude.

  • To be concrete, let's also see how the gauge boson acquires physical mass, if you compare the form of eq. 20.21 to eq. 7.23 you can identify $\Pi(k^2)=m^2_{ab}/k^2$ and following the calculation in p. 245-246, the exact propagator (2pt. function) for the gauge boson (eq. 7.75) receives a nonzero contribution to its isolated pole as shown below

$$\frac{-ig_{\mu\nu}\delta_{ab}}{k^2(1-\Pi(k^2))}=\frac{-ig_{\mu\nu}\delta_{ab}}{k^2(1-\frac{m^2_{ab}}{k^2})}=\frac{-ig_{\mu\nu}\delta_{ab}}{k^2-m^2_{ab}}$$.

Regarding the third question:

  • I believe you are correct, the relative sign is merely because momentum flow is going out of one vertex while going into the other. The actual vertex Feynman rule is $gk^\mu (T^a\phi_0)_i$ with the convention chosen for e.g. as momentum going into the vertex like they do in eq. 20.10. You immediately get eq.20.20.
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