-4

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?

GnomeSort
  • 345
  • 1
    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. – dmckee --- ex-moderator kitten Feb 08 '17 at 03:00
  • 5
    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. – user2309840 Feb 08 '17 at 03:00
  • 1
    have a look at this and similar https://arxiv.org/pdf/quant-ph/9504016.pdf – anna v Feb 08 '17 at 09:07
  • @user2309840 Thank you very much for your helpful comment. Kindly post it as an answer so that I may accept it. – GnomeSort Feb 16 '17 at 23:46
  • Overlap with 61569. – Cosmas Zachos Feb 20 '17 at 01:01

1 Answers1

4

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.

user2309840
  • 1,460