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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
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'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

1 Answer 1

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).

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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
    
Also in reference to your insensitive barbed comments towards the end about my previous questions - its not my prerogative to explain to anyone about the circumstances from which those previous questions had come. –  user6818 Jul 17 '11 at 13:57
    
I don't see any problem about discovering blind spots in one's understanding. It would be extremely egoistic to think that a perfectly linear education is the only right way. One's educational trajectory is often determined by various immediate needs rather than conscious choices…but then you don't seen to have an understanding of the varied difficult social circumstances in which people have to pursue science. –  user6818 Jul 17 '11 at 13:58

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