In this post: Dimensionless Constants in Physics there is a discussion about dimensionful vs. dimensionless constants in physics. In the context of this discussion, I'm wondering about the significance of setting $\hbar=1$ in calculations, specifically, in calculations where we are expanding the classical action around the classical solution to the equations of motion + quantum fluctuations. If I understand correctly, then, the quantum fluctuations are expanded in powers of $\hbar$, which measures just how much quantum effects you're taking into account in your approximation.

I was wondering then, since $\hbar$ is dimensionful, then it must not be a fundamental measure of the amount of quantum effect in a calculation, is there another perspective on this? I have seen an ad-hoc coupling constant introduced, but I also remember specifically reading in books that $\hbar$ is an answer to the question "Is the action big?", a question which has no meaning in classical physics (e.g. in S. Coleman's report on instantons).

  • $\begingroup$ Yes, $\hbar$ is exactly the typical scale of the contribution of quantum effects to any quantity with the units of action. $\endgroup$ Aug 11, 2014 at 13:49
  • $\begingroup$ But then $\hbar$ does have a fundamental meaning, despite the linked post which says it shouldn't as it is dimensionful, also despite the usual convention of setting it equal to 1, where we actually physically need it to be small (assuming classical physics is the dominant effect in the world and quantum mechanics is always a perturbative correction). $\endgroup$
    – PPR
    Aug 11, 2014 at 14:00

1 Answer 1


Planck's constant $\hbar$ is the scale of quantum mechanics' contributions to all quantities with the units of action – e.g. the action itself, angular momentum, product of position and momentum, the ratio of energy and frequency, and so on. Quantum mechanics dictates rules that specify not only the "order of magnitude estimate" but the precise values of everything that can be computed.

Whenever quantum mechanics affects the laws governing a physical system, $\hbar$ is bound to appear in the equations. Such equations therefore simplify if we set $\hbar=1$, and this is always possible because there are a priori independent units for the mass, time, and distance that may be wisely chosen to set $\hbar=1$.

Setting $\hbar=1$ doesn't make Planck's constant less fundamental. On the contrary, it makes it very fundamental. It makes it as fundamental... as the number one. It is the unique nonzero number that squares to itself.

We say that $\hbar$ is small but this statement is only true if 1) we keep $\hbar$ dimensionful i.e. if we do not set it to one or anything else (because one isn't too small), and, at the same moment, 2) if we compare the value of $\hbar$ to the values of quantities with the same units that appear in the macroscopic world well enough described by classical physics.

Because of the clarification in the last sentence, a more natural and accurate statement would be that the size of $\hbar$ is "normal", of order one (like the number one itself) while the angular momenta of macroscopic objects, the actions computed for histories of classical objects, and so on are "very large". But we usually say the former – the statement really doesn't mean anything else than the statement that the numerical value of $\hbar$ in usual units originally chosen to describe the macroscopic world is tiny, much smaller than one, e.g. $\hbar=1.054\times 10^{-34}\,{\rm J}/{\rm s}$. There is clearly no contradiction anywhere as long as one avoids giving these statements additional, wrong meanings.

The numerical value of $\hbar$ is small in all units designed to "easily" describe the macroscopic phenomena, those are well enough described by classical physics. But the value of $\hbar$ is of order one – or directly one – in units that are designed to describe Nature at the fundamental, i.e. quantum and atomic, level. Because quantum physics picks a preferred scale of the action, $\hbar$, it also breaks the "scale invariance" of the action in classical physics. But note – and this is just another way to say the very same thing – that this "scale invariance" only holds in classical physics i.e. if the actions are much greater than $\hbar$.


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