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advanced_tools:gauge_symmetry:stueckelberg_trick [2017/11/08 14:33]
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advanced_tools:gauge_symmetry:stueckelberg_trick [2022/06/07 10:04] (current)
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 <tabbox Student> ​ <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 $m$,+<​blockquote>​[W]e review the implementation of the Stückelberg ​trick in the theory of massive $SU(N)$ YM gauge bosons which, for simplicity, all have the same mass $m$,
 \begin{align} \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}+\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} \tag{ 3.22}
 \end{align} \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,+This lagrangian is not gauge invariant, but gauge invariance can be restored by coupling in new fields, $\pi^{a}(x)$,​ with $a\in\{1,​\ldots,​N^2-1\}$,​ 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} \begin{align}
-A_{\mu}&​\longmapsto U^\dagger(\pi)(A_{\mu}+\partial_{\mu})U(\pi)\equiv A'​_{\mu}\nn+A_{\mu}&​\longmapsto U^\dagger(\pi)(A_{\mu}+\partial_{\mu})U(\pi)\equiv A'​_{\mu}\\
 F_{\mu\nu}&​\longmapsto U^\dagger(\pi) F_{\mu\nu}U(\pi)\equiv F'​_{\mu\nu}\label{StuckelbergTransformationExample} F_{\mu\nu}&​\longmapsto U^\dagger(\pi) F_{\mu\nu}U(\pi)\equiv F'​_{\mu\nu}\label{StuckelbergTransformationExample}
 \end{align} \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, 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} \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+\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}\\
 &​=\!-\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} &​=\!-\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} \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 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} \begin{align}
-A_{\mu}&​\longmapsto V^\dagger(x)(A_{\mu}+\mathds{1}\partial_{\mu})V(x)\nn+A_{\mu}&​\longmapsto V^\dagger(x)(A_{\mu}+ {1}\partial_{\mu})V(x) \\
 U(\pi)&​\longmapsto V^\dagger(x)U(\pi)~,​ U(\pi)&​\longmapsto V^\dagger(x)U(\pi)~,​
 \end{align} \end{align}
 where $V(x)\in SU(N)$. 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 physics of the $\mathcal{L}$ and $\mathcal{L}'​$ lagrangians is the same, we have just made the degrees of freedom in Eq. (3.22) ​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 $U(\pi)\to {1}$, 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.+The above process and its generalizations are known collectively as the Stückelberg ​trick, which is often a useful tool for elucidating the physics in certain regimes of theories, especially at high energies. ​
  
 <​cite>​https://​arxiv.org/​abs/​1405.5532</​cite></​blockquote>​ <​cite>​https://​arxiv.org/​abs/​1405.5532</​cite></​blockquote>​
advanced_tools/gauge_symmetry/stueckelberg_trick.1510148037.txt.gz · Last modified: 2017/12/04 08:01 (external edit)