Error in Kac's "Vertex algebra for beginners" proof that a Wightman QFT gives rise to a vertex algebra? Given
$$ 
i[Q_k,\Phi_a(x)]=((x_0^2-x_1^2)\partial_{x_k}-2\eta_k x_k E - 2\Delta_a \eta_k x_k) \Phi_a(x), \quad (1.1.8)
$$
applying a coordinate change $t= x_0-x_1$, $\bar{t}= x_0+x_1$ and defining
$$
Q=-\frac{1}{2} (Q_0+Q_1),
$$
does it imply
$$
i[Q, \Phi_a(t,\bar{t})]= (t^2\partial_t+2\Delta_a t) \Phi_a(t,\bar{t})\;??? \quad (1.2.5c)
$$
Here $k=0$ or $1$, $\eta_0 = 1$, $\eta_1=-1$ and $E= x_0 \partial_{x_0}+x_1 \partial_{x_1}$.
I do not think that formula $(1.2.5c)$ is correct because the coordinate transformation is actually a rotation by $\pi/4$ followed by a dilation by $\sqrt{2}$. So the scalar field $\Phi_a$ should transform as
$$
\Phi_a(t, \bar{t})\to (\sqrt{2})^{-\Delta_a}\Phi_a(x).
$$
BACKGROUND:
Starting from the special conformal covariance for a scalar field
$$
U(0,1,b) \Phi_a(x) U(0,1,b)^{-1} = \varphi^{-\Delta_a} \Phi_a(x^b) \quad\quad (1.1.5)
$$
with
$$
\varphi(b,x)=1+2x \cdot b + |x|^2 |b|^2             \quad (1.1.6)
$$
and dilation covariance
$$
U(\lambda) \Phi_a(x) U(\lambda)^{-1} = \lambda^{\Delta_a}\Phi_a(\lambda x)
$$
with $\Delta_a$ being the conformal weight of the field $\Phi_a$ , Kac then says that formula $(1.1.5)$ implies that the infinitessimal special conformal generators are represented by selfadjoint operators $Q_k$ $(k=0, \ldots, dimension-1)$ on Hilbert space $\mathcal{H}$ such that
$$ 
i[Q_k,\Phi_a(x)]=(|x|^2\partial_{x_k}-2\eta_k x_k E - 2\Delta_a \eta_k x_k) \Phi_a(x), \quad (1.1.8)
$$
where $E=\sum^{dimension-1}_{m=0} x_m \partial_{x_m}$ and $\eta_0=1, \eta_k=-1$ for $k\ge1$. 
Then he focuses to $dimension=2$ and introduces the light cone coordinates $t=x_0-x_1$, $\bar{t}=x_0+x_1$ together with the operator
$$
Q=-\frac{1}{2}(Q_0+Q_1)
$$
to get from $(1.1.8)$
$$
i[Q, \Phi_a(t,\bar{t})]= (t^2\partial_t+2\Delta_a t) \Phi_a(t,\bar{t}). \quad (1.2.5c)
$$
 A: Please note that the light-cone coordinate transformation is not an element of the Poincaré group, nor of the full Conformal group. The easiest way to see this for the conformal group is to consider the transformation properties of the metric tensor. For the Poincaré group note that the proper distance $x_0^2-x_1^2 \rightarrow t\bar{t}$ which is not of the form $x_0^2-x_1^2 \rightarrow {x'}_0^2-{x'}_1^2$.
Therefore the scalar field $\Phi_a(x)$does not transform in a representation of some corresponding group when moving from $(x_0,x_1)$ to $(t,\bar{t})$.
The way I would go about deriving eqn. $(1.2.5c)$ and $(1.2.5d)$ (which are the correct forms of the respective commutators) would be in exact analogy with the steps required to go from eqn. $(1.1.5)$ to eqn. $(1.1.8)$. Namely, one starts with:
\begin{equation}
     U(\gamma)\Phi_a(t,\bar{t})U(\gamma)^{-1} = (ct+d)^{-2\Delta_a}(\bar{c}\bar{t}+\bar{d})^{-2\bar{\Delta}_a}\Phi(\gamma(t,\bar{t})) \tag{1.2.4}
  \end{equation}
Note that exactly as in equation $(1.1.8)$ where $-Q_k$ is the operator corresponding to the special conformal transformation parametrised by the parameter $q_k$, in this case the operator $-Q(-\bar{Q})$ corresponds to special conformal transformation parametrised by $b_+(b_-)$ (ref. eqn. (1.2.3)). 
Therefore the Lie Group elements $U(\gamma)$ are given by:
$$
U(\gamma)= e^{-i(b_+Q+b_-\bar{Q})}
$$
In the following I drop the $\bar{t}$ dependence here since the treatment is exactly analogous, and the left and right moving (holomorphic and anti-holomorphic) components decouple. Now, eqn. $(1.2.4)$ gives:
$$
e^{-ib_+Q}\Phi_a(t)e^{ib_+Q} = (1+b_+t)^{-2\Delta_a}\Phi_a(\frac{t}{1+b_+t}) 
$$
expanding as a series in $b_+$
$$
(1-ib_+Q+\dots)\Phi_a(t)(1+ib_+Q+\dots) = (1-2\Delta_ab_+t+\dots)(\Phi_a(t)-b_+t^2\partial_t\Phi_a(t)+\dots)
$$
Collating terms in powers of $b_+$:
$$
\Phi_a(t)-ib_+[Q,\Phi_a(t)]+\mathcal{O}(b_+^2) = \Phi_a(t)-b_+(t^2\partial_t+2\Delta_at)\Phi_a(t)+\mathcal{O}(b_+^2)
$$
and comparing the coefficients of $b_+$ one gets:
$$
i[Q,\Phi_a(t)] = (t^2\partial_t+2\Delta_at)\Phi_a(t)
$$
Finally, remember that in 2 dimensions, there are in general 2 distinct scaling dimensions ($\Delta_a$ and $\bar{\Delta}_a$) which are not the same as the scaling dimension $\Delta_a$(Let's call it $\Delta_a'$ for clarity) in eqn. $(1.1.5)$. These satisfy $\Delta_a+\bar{\Delta}_a=\Delta'_a$ and only in case of a spinless field have $\Delta_a$ = $\bar{\Delta}_a$.
