✎ Stueckelberg Trick [The Physics Travel Guide]

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====== Stueckelberg Trick ======

<note tip>
Explanations in this section should contain no formulas, but instead colloquial things like you would hear them during a coffee break or at a cocktail party.
</note>
  
<tabbox Student>

<blockquote>we review the implementation of the St\"uckelberg trick in the theory of massive $SU(N)$ YM gauge bosons which, for simplicity, all have the same mass $m$,
\begin{align}
\mathcal{L}&=-\frac{1}{4g^{2}}{\rm tr}F_{\mu\nu}F^{\mu\nu}-\frac{m^{2}}{2g^{2}}{\rm tr} A_{\mu}A^{\mu}\ .\label{MassiveYMforStuckelbergSection}
\end{align}

This lagrangian is not gauge invariant, but gauge invariance can be restored by coupling in new fields, $\pi^{a}(x)$, with $a\in\{1,\ldots,N^2-1\}$, {\it i.e.}, one field for each generator of $SU(N)$.  In order to insert the $\pi^{a}(x)$'s appropriately, one first performs a gauge transformation with $\pi^{a}(x)$ as the gauge parameter,
\begin{align}
A_{\mu}&\longmapsto U^\dagger(\pi)(A_{\mu}+\partial_{\mu})U(\pi)\equiv A'_{\mu}\nn
F_{\mu\nu}&\longmapsto U^\dagger(\pi) F_{\mu\nu}U(\pi)\equiv F'_{\mu\nu}\label{StuckelbergTransformationExample}
\end{align}
where $U(\pi)=e^{\pi^{a}(x)T_{a}}$ is an element of $SU(N)$.  We then define a new lagrangian $\mathcal{L}'$ by taking \eqref{MassiveYMforStuckelbergSection} and replacing $A_{\mu}\mapsto A_{\mu}'$ and $F_{\mu\nu}\mapsto F_{\mu\nu}'$.  That is,
\begin{align}
\mathcal{L}'&=-\frac{1}{4g^{2}}{\rm tr}F'_{\mu\nu}F'^{\mu\nu}-\frac{m^{2}}{2g^{2}}{\rm tr}A'_{\mu}A'^{\nu}\nn
&=\!-\frac{1}{4g^{2}}\!{\rm tr}F_{\mu\nu}F^{\mu\nu}\!-\!\frac{m^{2}}{2g^{2}} {\rm tr}D_{\mu}U(\pi)D^{\mu}U^\dagger(\pi)~, \label{StuckelbergedLagrangianEx}
\end{align}
where $D_{\mu}U(\pi)= \partial_{\mu}U(\pi)+A_{\mu}U(\pi)$ is the gauge covariant derivative of $U(\pi)$.  The lagrangian $\mathcal{L}'$ then enjoys a gauge symmetry under which we simultaneously change
\begin{align}
A_{\mu}&\longmapsto V^\dagger(x)(A_{\mu}+\mathds{1}\partial_{\mu})V(x)\nn
U(\pi)&\longmapsto V^\dagger(x)U(\pi)~,
\end{align}
where $V(x)\in SU(N)$.

The physics of the $\mathcal{L}$ and $\mathcal{L}'$ lagrangians is the same, we have just made the degrees of freedom in~\eqref{MassiveYMforStuckelbergSection} manifest.  In $\mathcal{L}'$, we introduced $N^{2}-1$ new fields, but also restored $N^{2}-1$ gauge symmetries and hence degree of freedom counting is the same for both cases. We can demonstrate the equivalence explicitly by using the gauge symmetry of $\mathcal{L}'$ to go ``unitary gauge" in which we set $U(\pi)\to \mathds{1}$, where the two lagrangians coincide.

The above process and its generalizations are known collectively as the St\"uckelberg trick, which is often a useful tool for elucidating the physics in certain regimes of theories, especially at high energies. See \cite{Burgess:1992gx,Hinterbichler:2011tt,Preskill:1990fr,ArkaniHamed:2002sp} for good discussions of St\"uckelberg fields in various contexts.

<cite>https://arxiv.org/abs/1405.5532</cite></blockquote>
 
<tabbox Researcher>

<note tip>
The motto in this section is: //the higher the level of abstraction, the better//.
</note>

--> Example1#

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--> Example2:#

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</tabbox>

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