Commutation of operators in quantum theory I have always written the commutation rules of quantum theory as ,
$[q,p] = i\hbar\delta _{ij}$ 
But seems that some people write this as,
$[q^i,v_j]= \frac{i\hbar}{M}\delta^i _{j}$ 
(..this is often done in the context of taking the Galilian group limit of the Poincare group…though I am not sure which aspect of it does it emphasize-- the non-relativstic aspect or the non-quantum aspect?..) 


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*But somehow dimensionally the second form doesn't look okay.
Am I missing something?


In the same strain, it seems that the operators for "finite boost by $v$ " and is done by the operator $exp(\frac{iK.v}{\hbar})$ and the "finite translation by $q$" is effected by the operator $exp(\frac{iMv.q}{\hbar})$. (..where $v$, $q$ and $K$ are all $3-vectors$..)


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*I would like to know how the  above is rationalized. 
To ask again - is the above taking just the non-relativistic limit or is it also a non-quantum limit ? 

 A: There is nothing wrong about the dimensional analysis. $\hbar$   has the dimension of $x\cdot p$, so $[x,p]=i\hbar$   is the most standard commutator of quantum mechanics.
Now, the velocity is $v=p/m$, which is just another way of writing the usual simple definition of the momentum, $p=mv$, so the right commutator $[x,v]$ will obviously include $i\hbar/m$, too. It's the same thing divided by $m$.
The exponentials are just two examples of the most standard way to get a finite transformation from the (infinitesimal, Hermitian) generator $G$. The finite transformation is always
$$ \lim_{N\to\infty} \left( 1 + \frac GN \right)^{\phi N}  = \exp(i\phi G) $$
where $\phi$ is the finite amount of the transformation. For Galilean boosts, the generator is $G=(\sum q_i M_i)/M_{\rm total}$ - the center of mass (your statement that the generator is $v$ is just incorrect). For translations, the generator is $G=p$. 
At any rate, getting finite elements of a Lie group from the generator of a Lie algebra - by an exponential - is the first thing that an undergrad learns when he hears both about Lie groups and Lie algebras, and if you haven't understood this point before, it just proves that every single question that you have previously posted on this server was totally inappropriate because you are missing at least three critical layers of knowledge that are pre-requisites for the topics discussed in all your previous questions (about advanced topics such as supermultiplets in exotic supersymmetric theories).
