1
$\begingroup$

In Dirac's 'Principles of Quantum Mechanics' the definition of quantum Poisson bracket is $uv-vu=i\hbar[u,v]$ with $[q_r,p_s]=\delta_{rs}$. But in lots of other books the definition is $uv-vu=[u,v]$ with $[q_r,p_s]=i\hbar\delta_{rs}$. Now it certainly appears to me as a matter of convention but my question is is there any reason to use one convention over another that might have any advantage or is this purely a matter of choice? Most textbooks I read use the second convention so is Dirac's convention outdated for some reason?

$\endgroup$
1
3
$\begingroup$

It is important to distinguish between a Poisson bracket and a quantum mechanical commutator. A Poisson bracket is a classical concept, defined for two phase space functions $u$ and $v$ in $n$ coordinates by $$[u,v] = \sum_{k=1}^n \left( \frac{\partial u}{\partial q_k}\frac{\partial v}{\partial p_k} - \frac{\partial u}{\partial p_k} \frac{\partial v}{\partial q_k} \right),$$ which for $u = q_i$ and $v = p_j$ immediately gives $\delta_{ij}$. In quantum mechanics you get the latter result (in the form of a scaled identity operator) when you divide the commutator of position and momentum operators by $i\hbar$, but in more complicated situations the correspondence is broken in the order of $\hbar$ (the commutator can be perceived as a "deformation" of the P.B., converging to it when $\hbar \to 0$).

It might have been correct to call the $i\hbar$-adjusted commutator a Poisson bracket in Dirac's time but this is outdated terminology, and with this, there is no reason to trying to make it equal either. In modern literature and communication you should use the modern definition of the $[]$ bracket to be implicitly understood, and that is exclusively $$[u,v] = uv - vu.$$

$\endgroup$
2
$\begingroup$

Yep, it is actually just a matter of convention. I don't think that choosing the first or the second one would change nothing but the convenience in the computations. As far as I know, the second definition of the commutator is most widely used and should coincide also with the definition in group theory... so as to avoid mistakes and confusion.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.