Isn't it incorrect for the minimal gauge coupling and related calculations in Prof. Ezawa's book on quantum Hall effect? He is CORRECT. I use $\mathbf{B}=\left(0,0,B_{\perp}\right)$ and he use $\mathbf{B}=\left(0,0,-B_{\perp}\right)$. $B_{\perp}>0$.
Nov.28.2012
Basically I got mad with conventions.
1.Here is the link of the book (second edition): 
http://books.google.fr/books/about/Quantum_Hall_Effects.html?id=p3JpcdbqBPoC
Here is another link for one of his review articles:
http://iopscience.iop.org/0034-4885/72/8/086502
2.I am not happy with the negative sign of the commutator $\left[X,Y\right]=-il_{B}^{2}$. Here is my calculation:
$\left[X,Y\right]=\left(\left[x,p_{x}\right]+\left[-p_{y},y\right]\right)/eB+\left[-P_{y},P_{x}\right]/e^{2}B^{2}=il_{B}^{2}$
3.In my calculation, I used some conventions different from Prof.Ezawa's book and article.
Here is my convention:
$X:=x-P_{y}/eB\;;\; Y:=y+P_{x}/eB$
However, Prof.Ezawa use this convention:
$X:=x+P_{y}/eB\;;\; Y:=y-P_{x}/eB$
To be prepared for being driven mad, please compare them carefully.
4.I think he must be wrong somewhere, for example, in his book (2nd.ed), (10.2.5) and in his article (2.15)
$\left[P_{x},P_{y}\right]=i\hbar^{2}/l_{B}^{2}$
but you know we use minimal coupling $\mathbf{p}\rightarrow\mathbf{p}-\frac{q}{c}\mathbf{A}$ in this problem, $q=-e,e>0,{\mathbf{p}+\frac{e}{c}\mathbf{A}}$ for electrons, as Prof.Ezawa suggested in his book (10.2.3) and his article (2.12). Under this convention, I calculated $\left[P_{x},P_{y}\right]$ as following:
$\left[P_{x},P_{y}\right]=-i\hbar e\left(\left[\partial_{x},A_{y}\right]+\left[A_{x},\partial_{y}\right]\right)/c=-i\hbar eB/c=-i\hbar^{2}/l_{B}^{2}$
Oh my god here is a negative sign.
5.To summarize, I think if we take Prof.Ezawa's convention and apply his result for $\left[P_{x},P_{y}\right]=i\hbar^{2}/l_{B}^{2}$ during the calculation of $\left[X,Y\right]$, we will get his result. But his result for $\left[P_{x},P_{y}\right]=i\hbar^{2}/l_{B}^{2}$ seems not correct.
6.Someone save my day...

 A: I have done this calculation some time ago. My convention was:
$$ X = x - \frac{P_y}{m \omega_c}\quad  Y = y + \frac{P_x}{m \omega_c} $$
and
$$P_i = p_i +\frac{e}{c} A_i$$
And the magnetic field is $B = \nabla \wedge A = B_z \hat{z}$. Note that in particular: $X = x-\frac{1}{m\omega_c}(p_i + \frac{e}{c}A_i)$. My notes say that this gives:
$$ [X, Y]  = i l_B^2\qquad \text{and}\qquad [P_x,P_y] = -\frac{i}{l_B^2}$$
If you now take his convention, you essentially flip the magnetic field, $\vec{B} \rightarrow -\vec{B}$. This replaces:
$$ X = x + \frac{P_y}{m \omega_c}\quad  Y = y - \frac{P_x}{m \omega_c} $$
but you still have $P_i = p_i +\frac{e}{c} A_i$ -- that stays the same. Therefore we have:  $X = x+\frac{1}{m\omega_c}(p_i + \frac{e}{c}A_i)$ (!!!! compare this to the other convention), and to compute the commutator we get:
$$\begin{align}[X,Y] &= \left[x+\frac{1}{m\omega_c}(p_y + \frac{e}{c}A_y),y-\frac{1}{m\omega_c}(p_x + \frac{e}{c}A_x)\right] \\
&= (-[x,p_x] + [p_y,y])/m\omega_c + \frac{e}{c (m\omega_c)^2}(-[p_y,A_x]+[A_y,p_x])
\end{align}$$
Now, $[x,p_x] = i$, as always. The other commutator depends on the orientation of the magnetic field:
$$-[p_y,A_x]+[A_y,p_x] = i ([\nabla_y, A_x] - [\nabla_x, A_y]) = -i(\nabla\wedge A)_z = iB_z$$, and so you get
$$[X,Y] = -\frac{2i}{m\omega_c} + \frac{e}{c (m\omega_c)^2}  iB_z = -i l_B^2$$
Long story short: your derivation of $[X,Y]$ does not apply to his conventions. Your conve
Final note: If you switch conventions, you essentially replace $B_z \rightarrow -B_z$, so the magnetic length and cyclotron frequency also switch sign, $l_B^2\rightarrow -l_B^2$ and $\omega_c \rightarrow - \omega_c$. So you see that both commutators  (involving $X$ and $Y$ and $P_x$ and $P_y$) pick up a minus sign, because they both involve $l_B^2$. 
