I am learning about the Theory of intense laser-induced molecular dissociation. To set the notations, the radiation field Hamiltonian is for a single charged particule : $$H = \frac{1}{2m}(p-q\hat{A}(r))^2 + V(r) + \hbar \omega (\hat{a}^{\dagger}\hat{a}+1/2)$$ with $\hat{A}$ the vector potential. The length gauge (Electric-field) has been introduced along side the property that, with the long-wavelength approximation :$$\hat{O} p \hat{O}^\dagger = p + q\hat{A} $$ $$\hat{A} = \beta\{\hat{a} + \hat{a}^{\dagger}\}\epsilon$$ with the gauge transformation $\hat{O} = e^{-\frac{i}{\hbar}qr\cdot \hat{A}} = e^{-\frac{i}{\hbar}\beta q r\cdot \epsilon[\hat{a} + \hat{a}^\dagger]}$, $\beta = (\frac{\hbar}{2\epsilon_0 \omega V})^{1/2}$ where $\omega = c|k|$ is the frequency of the mode and $V$ the finite volume leading to a discretization of the laser mode.
But I get, with $p = -i\hbar\nabla_r$ : $$\hat{O} p \hat{O}^\dagger = qr\cdot \nabla_r \hat{A} + q\hat{A} = q\hat{A}$$
Edit, derivation : $$ \begin{align} \hat{O} p \hat{O}^\dagger &= \hat{O} (-i\hbar)\nabla_r \hat{O}^\dagger = \hat{O} (-i\hbar)\nabla_r e^{\frac{i}{\hbar}qr\cdot \hat{A}}\\ &= \hat{O} (-i\hbar) \frac{i}{\hbar}q(\nabla_r r\cdot \hat{A}(r))\hat{O}^\dagger \\ &= q(\hat{A}(r) + r\cdot \nabla_r\hat{A}(r)) \\ &= q\hat{A} \end{align} $$
since the dot product is bilinear