I saw in the text that


and says the operator $X$ has charge $c$ under the generator $Q$. I tried to understand why the coefficient $c$ means the charge. So I used this relation to get the result(using the property of $e^{a}=\lim_{n\rightarrow\infty}(1+\frac{a}{n})^{n}$)

$e^{iQ} X e^{-iQ}=e^{ic}X$.

which seems like the commutator really shows the charge of the operator $X$ as $c$. Is it right way to understand the statement?

  • 1
    $\begingroup$ It is not quite clear what your question is. Could you perhaps try to elaborate your question a bit? Do you want help with the physical interpretation of the math? $\endgroup$ – Codename 47 Mar 1 at 12:41
  • $\begingroup$ Do you perhaps mean $[Q,X] = cX$? $\endgroup$ – Javier Mar 1 at 13:07
  • $\begingroup$ Yes, your final equation is right. It follows from the standard so-called "Hadamard lemma" combinatoric identity. $\endgroup$ – Cosmas Zachos Mar 3 at 1:46

The basic idea is that we transform in proportion to such commutators to produce a symmetry. It may help to discuss an example that's a little more complicated than what you've described.

The infinitesimal BRST transformation $X\mapsto X+\delta X$ is a symmetry of the BRST-quantised Yang-Mills Lagrangian. The BRST operator $Q$ (also called the BRST charge) satisfies $\delta b=\theta [Q,\,b],\,\delta f=\theta\{ Q,\,f\}$ for an infinitesimal constant Grassmann number $\theta$, with arbitrary bosonic field $b$ and fermionic field $f$. The need for anticommutators with fermions comes from $Q$ being fermionic (which is why $\theta$ needs to be a Grassmann number to make $\delta$ bosonic); if $Q$ were bosonic, in which case we wouldn't be discussing the BRST example, it'd be commutators all the way.

We can verify $\delta(XY)=(\delta X)Y+X\delta Y$ for fields $X,\,Y$ of definite exchange symmetry, and $\delta$ commutes with $\partial_\mu$.


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