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I am reading Strominger's lecture notes on the infrared structure of gravity and gauge theory. I am trying to understand subchapter 2.11, where the author focuses on the notions of "Spontaneous Symmetry Breaking, Vacuum Degeneracy, and Goldstone Bosons" for the case of large gauge transformations.

I will explain my understanding of the notions above and I will ask some questions about what these notions mean in the context of large gauge symmetry.

First of all, for the concept of "symmetry breaking", I understand that there are cases in which the Lagrangian (hence the equations of motion) are invariant under a specific transformation, but the vacuum is not invariant by the latter transformation. For instance, taking $\Phi(x)\rightarrow e^{i\theta}\Phi(x)$, where $\Phi(x)$ is a scalar field in a scalar field theory with the Mexican hat potential $$V(\Phi)=-m^2(\Phi^{\star}\Phi)+\lambda (\Phi^{\star}\Phi)^2,$$ and $\theta$ is some parameter, then the equations of motion are invariant under the transformation above, whereas the vacuum state, say $|\rho_0\rangle$, is not invariant under the previously discussed global symmetry transformations. If $Q_{\theta}$ generates those transformations, then one can see that they do not leave the vacuum invariant by acting on them with $e^{iQ_{\theta}}$: $$e^{iQ_{\theta}}|\rho_0\rangle=|\rho_0+\theta\rangle\ne0$$

In addition, Goldstone's theorem states that whenever a continuous global symmetry is spontaneously broken, there exists a massless excitation about the spontaneously broken vacuum. Decomposing $\Phi(x)=|\Phi(x)|e^{i\rho(x)}$ , $\rho$ transforms as $\rho(x)\rightarrow\rho(x)+\theta$. Hence, the Lagrangian can depend on $\rho$ only via the derivative $\partial_{\mu}\rho(x)$; there cannot be any mass term for $\rho$, and it is a massless field. $\rho$ — identified as the field that transforms inhomogeneously under the broken symmetry — is referred to as the Goldstone boson.

Now, selecting the symmetry transformation to be the large gauge transformation, $A_{\mu}(x)\rightarrow A_{\mu}(x)+\partial_{\mu}\varepsilon(x)$, what would the potential term in QED be? Would it be $-j^{\mu}A_{\mu}$, or are we interested in the potential term regarding the matter fields, i.e. $m\bar{\Psi}\Psi+e\bar{\Psi}\gamma^{\mu}\Psi A_{\mu}$? I can not see if (and how) any of the potentials above are similar to the Mexican hat case, so I can not guess a form for the vacuum field for the matter field. But perhaps I do not need to... In addition, do we still have $\Psi^{\alpha}(x)=|\Psi^{\alpha}(x)|e^{i\rho(x)}$, for some spinor index $\alpha=1,2,3,4$, helping us to identify the massless boson $\rho(x)$, just like in the Mexican hat case? And if so, why does the author support that this massless boson is a soft photon?

I feel it would be greatly beneficial for someone to demonstrate what would $\rho(x)$, the vacuum state $\rho_0$ and the charge $Q_{\theta}$ and $\theta$ in the case of an abelian gauge theory.

I have read subsection 2.11 and I am still confused about this. Thanks in advance.

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  • $\begingroup$ Can you say a bit more about the "large gauge transformations"? According to my memory, they refer to gauge transformations that are associated with topological defects such as instantons. Is that what you are referring to? $\endgroup$ Feb 8 at 13:54
  • $\begingroup$ Hi @filippiefanus and thanks for the comment. Large gauge transformations are the usual local gauge transformations that do not vanish at the asymptotic boundary of the spacetime. $\endgroup$
    – schris38
    Feb 8 at 14:22
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    $\begingroup$ Just a small comment on "Lagrangian (hence the equations of motion) are invariant ...". If the Lagrangian is invariant the EoM's sometimes are invariant but usually they are $\textbf{covariant}$. That's just because the EoM's tend to have indices so they don't remain invariant, they transform as a tensor. $\endgroup$ Feb 10 at 22:57

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In the absence of magnetic charges, the value of the gauge field on the boundary is $$ A_\mu |_{{\cal I}^+_-}(x) = \partial_\mu C(x) . $$ Under large gauge transformations, $C(x) \to C(x) + \varepsilon(x)$. $C(x)$ is the analogue of $\rho(x)$.

In the quantum theory, $C(x)$ commutes with itself and the Hamiltonian so it has zero energy. The vacuum Hilbert space is spanned by eigenstates of $C(x)$, $$ C(x) |\theta\rangle = \theta(x) |\theta\rangle. $$ $|\theta\rangle$ is the analogue of $|\rho_0\rangle$

The large gauge charge $Q_{\varepsilon} \sim \int_{{\cal I}^+_-} \varepsilon \star F$ generates large gauge transformations, so its action on the vacuum states are $$ e^{iQ_{\varepsilon}}|\theta\rangle = |\theta + \varepsilon \rangle . $$

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  • $\begingroup$ Hi @Prahar and thanks for tha answer. Can I have one more question though? What would the superselection sectors be in that case? For instance, will a single superselection sector be comprised from states having the same number of soft photons? Could you please clarify this additional point as well? $\endgroup$
    – schris38
    Feb 9 at 8:51
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    $\begingroup$ @schris38 - there isn't a superselection sector (at least in four dimensions). That's what makes this version of SSB different from the standard discussion. Typical finite-energy scattering processes induce a change in the vacuum. This vacuum shift is responsible for memory effects and infrared divergences. Additionally, there is no number of operators in the soft sector. Hence, "number of soft photons" is an ill-defined concept. It's like asking how many oscillators there are in the coherent state of a harmonic oscillator system. $\endgroup$
    – Prahar
    Feb 9 at 9:37
  • $\begingroup$ Okay, I see... Thank you so much @Prahar $\endgroup$
    – schris38
    Feb 9 at 11:52
  • $\begingroup$ Hi @Prahar and sorry for disturbing you again, but I have noticed that your last equation is $Q_{\varepsilon}|\theta\rangle=|\theta+\varepsilon\rangle$, whereas Strominger has $e^{Q_{\varepsilon}}|\theta\rangle=|\theta+\varepsilon\rangle$ in his lecture notes, when your arguments are applied in his case. Is this true? Or am I missunderstanding something? $\endgroup$
    – schris38
    Feb 9 at 15:44
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    $\begingroup$ That was a typo on my part. Strominger is correct. I will edit my answer. $\endgroup$
    – Prahar
    Feb 9 at 16:41

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