difference of the quark OAM in the two decompositions with an extra minus sign, is given in the following form $$\begin{aligned}
  \bm{L}^g - \bm{L}^{\prime g} \ = \ - \,(\bm{L}^q - \bm{L}^{\prime q})
  \ = \ \int \,\rho^a \,(\bm{x} \times \bm{A}^a_{phys}) \,d^3 x
  \ = \ \int \, \psi^\dagger \,\bm{x} \times 
  \bm{A}_{phys} \, \psi \,\,d^3 x ,\end{aligned}$$ and we call it the [*potential angular momentum*]{} by the following reason. (In the above equation, $\bm{L}^q$ and $\bm{L}^g$ stand for the quark and gluon OAMs in our decomposition, while $\bm{L}^{\prime q}$ and $\bm{L}^{\prime g}$ the quark and gluon OAMs in the decomposition of Chen et al.) That is, the QED correspondent of this term is nothing but the angular momentum carried by the electromagnetic field or potential, which appears in the famous Feynman paradox of classical electrodynamics [@Feynman65]. An arbitrariness of the decomposition arises, because this potential angular momentum term is [*solely*]{} gauge-invariant. This means that one has a freedom to shift this potential OAM term from the gluon OAM part to the quark OAM part in our decomposition, which in fact leads to the quark OAM in the Chen decomposition in such a way that $$\begin{aligned}
 &\,& \bm{L}^q
 \ + \ 