Sign in infinitesimal Unitary transformations When studying unitary transformations in QM, most of the textbooks I used defined the unitary transformation operator as
$$ \hat{U}(\alpha) = e^{-i \alpha \hat{G}} \, \, , $$
where $\alpha$ is the parameter of transformation and $\hat{G}$ is the generator of transformation.
My doubt
Can the definition of unitary transformation be
$$ \hat{U}(\alpha) = e^{+i \alpha \hat{G}} \, \, ? $$
As far as I understand, when taking an infinitesiaml unitary transformation ($\alpha \to 0$)
$$ \hat{U}(\alpha) \approx 1+i\alpha \hat{G} \, \,  $$
and using the fact that $\hat{U}^{\dagger}(\alpha)\hat{U}(\alpha) = \hat{1}$, we can still prove that, to first order,
$$\hat{G}^{\dagger} = \hat{G} \, \, ,$$
as required.
 A: You are just considering the inverse transformation, as said by Frank.
In general you want do describe transformations of the system, like rotations or translations...  These tranformations are generalized in the concept of group, that is a set of elements with certain poroperties, among these properties there is the existence of the inverse element.
Because you want to have the possibility to transform back your system to the same configuration it was in before.
In QM you want to be sure that these transformations of the systems don't affect the probabilities, which are the predictive aspect of the theory. The trasformations that leaves invariant the inner product of the Hilbert space are unitary or antiunitary operators. So you are representing your transformations with unitary operators, but again for every element $g$ of the group there is a $g^{-1}$ such that $gg^{-1}=\mathbb{I}$. So if you are representing the element $g$ by $\hat{U}=e^{i\alpha G}$ then you are also representing the inverse $g^{-1}$ by $\hat{U}^{-1}=e^{-i\alpha G}$, so that you have $\hat{U}\hat{U}^{-1}=\mathbb{I}$. It doesn't matter which one has the minus sing, because it doesn't matter wich one is the inverse element between $g$ and $g^{-1}$, they are respectively the inverse for each other.
