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In this paper, I saw the following statement, which suggests that the value of $\hbar$ is not a fixed constant: enter image description here

I am given to believe that the value of $\hbar$ is Planck's constant divided by $2 \pi$. Is this not correct, or is the $\hbar$ here different from that encountered in the context of, say, the momentum operator?

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    $\begingroup$ Taking limits that are in a naive way insensible is one of the many tools of mathematical physics. It doesn't mean anything fundamental about $\hbar$. I recall a seminar from grad school where we were shown a trick that worked if you assumed the number of colors in the strong force $n_c$ was greater than 16 while you did the work and then took the limit as $n_c \to 3$ at the end. Don't get worked up about it. $\endgroup$ – dmckee Feb 8 '17 at 3:00
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    $\begingroup$ The limit hbar to zero is short hand for the semi-classical limit of quantum mechanics. Indeed, hbar is a constant, and one does not have the freedom to tune it. However, to be more precise usually requires a careful statement of the actual physical question and will typically involve looking at states involving large quantum numbers. $\endgroup$ – user2309840 Feb 8 '17 at 3:00
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    $\begingroup$ have a look at this and similar arxiv.org/pdf/quant-ph/9504016.pdf $\endgroup$ – anna v Feb 8 '17 at 9:07
  • $\begingroup$ @user2309840 Thank you very much for your helpful comment. Kindly post it as an answer so that I may accept it. $\endgroup$ – NewDogOldTricks Feb 16 '17 at 23:46
  • $\begingroup$ Overlap with 61569. $\endgroup$ – Cosmas Zachos Feb 20 '17 at 1:01
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The limit $\hbar \to 0$ is short hand for the semi-classical limit of quantum mechanics. Indeed, $\hbar$ is a constant, and one does not have the freedom to tune it. However, to be more precise usually requires a careful statement of the actual physical question and will typically involve looking at states involving large quantum numbers.

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