The charge given by a commutator

I saw in the text that

$$[Q,X]=cX$$

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?

• 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? – Codename 47 Mar 1 at 12:41
• Do you perhaps mean $[Q,X] = cX$? – Javier Mar 1 at 13:07
• Yes, your final equation is right. It follows from the standard so-called "Hadamard lemma" combinatoric identity. – Cosmas Zachos Mar 3 at 1:46

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 $$\partial_\mu$$.