# 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?..)

• 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 $3-vectors$..)

• 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 ?
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I suspect that (at least a part of) the question (v1) is caused by a combination of (1) putting $\hbar=1$ and (2) misreading formulas in this answer (v3) physics.stackexchange.com/q/11839/2451 For instance, the second formula should have been $[q^i, v_j]= \frac{i\hbar}{M}\delta^i_j$. – Qmechanic Jul 16 '11 at 12:24
'Notation' tag might be apt here – qftme Jul 16 '11 at 16:39
@Qmechanic Edited the question. Interesting that you located that reference. I happened to have some recent conversations with Arun (the OP of that reference of yours). There were a lot of things said in that discussion which is far from clear. – user6818 Jul 16 '11 at 20:50
I thought of $[q^i,v_j]=\frac{i\hbar}{M}\delta_j^i$ as being the regular commutation relation, except each side is 'divided' by $M$. Intuitively at least it made sense to me. And the Galilean group limit of Poincare group is the non-relativistic limit. About your other point, $v$, $q$, and $K$ are indeed vectors, but $v$ is a c-number while $q$ and $K$ are vector operators. You can just resolve the dot product into a sum of terms, and then separate the exponentials since the different components of $K$ and $q$ commute. Then each factor is well-defined. – Arun Nanduri Jul 16 '11 at 21:14
With the new update (v2), the first bullet is completely solved because $p_j=Mv_j$, where $M$ is a c-number mass, i.e. $M$ commutes with everything. – Qmechanic Jul 16 '11 at 21:15

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 $G=p$.
The issue with the dimensions was solved by the comment of Qmechanic. What I wasn't sure of is how one can justify substituting $mv$ for $p$. The general commutation relations are defined for $p$ being a generalized momentum. I am not sure how to reconcile these. – user6818 Jul 17 '11 at 13:55