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I am going through Groenewold's theorem and in his book: On The Principles of Elementary Quantum Mechanics, page 8, eq. 1.30:

$$[\mathbf{p}, \mathbf{q}]=1\left(\text { i.e. } \mathbf{p q}-\mathbf{q} \mathbf{p}=\frac{\hbar}{i}\right),\tag{1.30} $$

and he wrote:

The classical quantities $a(p,q)$ can be regarded as approximations to the quantum Operators $\mathbf{a}$ for $\lim \hbar \rightarrow 0$.

How did he assume that $\frac{\hbar}{i}=1$? And if $\hbar$ (as we have learned it) is a constant and it is precisly equal to $6.5821 × 10^{-16} eV s$, how can we say that it goes to zero?

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    $\begingroup$ Just for the title, Planck's constant has recently been defined to be exactly $h= 6.62607015×10^{-34} $Js, so it is now a constant even if it could hypothetically have varied before. Fortunately $\hbar$ remains $\frac{h}{2\pi}$ $\endgroup$
    – Henry
    Commented May 28, 2020 at 22:41
  • $\begingroup$ We can imagine (and calculate) alternate universes with decreasing values of hbar. $\endgroup$
    – Criticizing Israel not allowed
    Commented May 30, 2020 at 20:02

3 Answers 3

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How did he assume that $\frac{\hbar}{i}=1$?

He didn't. Check the definition he gives of the commutator in equation (1.02).

And if $\hbar$ (as we have learned it) is a constant how can we say that it goes to zero?

I think the point here is to say: if $\hbar \rightarrow 0$ we recover the classical mechanics (CS), therefore if in nature $\hbar = 0$ we wouldn't have QM only CS. And classical mechanics is a limit of QM and this is fundamental since we see that classical mechanics works. Moreover, it tells us that since $\hbar \neq 0$ but it's small we see QM only at small scales.

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    $\begingroup$ We can talk about small and big of dimensionful numbers with no problem. It's small in the "units of classical physics". It is small with respect to the action on a classical trajectory. $\endgroup$
    – TheoPhy
    Commented May 29, 2020 at 14:22
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    $\begingroup$ A semiclassical way to look at it is that $\hbar$ much smaller than the angular momentum we run into in the macroscopic world. A spinning bicycle wheel has an angular momentum of something like $L \sim 10^{34} \hbar$ $\endgroup$
    – WaterMolecule
    Commented May 29, 2020 at 15:20
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    $\begingroup$ @ludmicseveric You can deduce 1000 s is greater than 1 s because you can implicitly divide 1 s / 1 s and get 1 without dimension and divide 1000 s / 1 s and get 1000 without dimension again and because 1000 > 1 then $t_0 < t_1$. In this example it’s pretty quick and trivial but it’s not that trivial for $\hbar$. WaterMolecule brought up a good point here I believe. $\endgroup$
    – Mithridates the Great
    Commented May 29, 2020 at 19:36
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    $\begingroup$ More generally we can define Planck units for length, mass, time etc in terms of ℏ and other universal constants. These form a natural set of units which don't have any arbitrary anthropocentric basis. When we compare, say, metres and seconds to the Planck length and Planck time, we can say that ℏ is indeed small, in the sense that the scale at which quantum mechanics is visible is much smaller than the scale of our everyday mundane existence $\endgroup$
    – Mark Allen
    Commented May 29, 2020 at 22:09
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    $\begingroup$ @AloneProgrammer Difficulty and impossibility are two different things. Avoidance should only be recommended for people who find such things difficult, no more no less. For everyone else there is no difficulty. Dimensionful comparison is not only routine, but is pivotal to theoretical physics. Take for example one quantity of theoretical interest: the shear viscosity to entropy density ratio; this is a dimensionful quantity (Ks) that is compared to several values in this paper: arxiv.org/pdf/1108.0677.pdf (pp. 4ff.). The literature can be searched for many more such examples. $\endgroup$
    – Zorawar
    Commented Jun 1, 2020 at 20:06
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  1. Groenewold is working in the framework of deformation quantization, where the (reduced) Planck constant $\hbar$ is treated as a formal parameter that doesn't have to be the actual physical value $\sim 10^{-34}{\rm Js}$.

  2. Eq. (1.30) is explained by an unconventional normalization of the commutator $$ [{\bf a},{\bf b}]~:=~\frac{i}{\hbar}({\bf ab}-{\bf ba}). \tag{1.02} $$

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For a slightly different perspective, in natural units one can set $\hbar = 1$. That is, in natural units we agree to measure action in units of $\hbar$ (instead of, say, $\rm J\cdot s$). Seen this way, it makes no more sense to send $\hbar$ to $0$ than it does to send $1 \, \rm J \cdot s$ to $0$. Put differently, sending $\hbar$ to $0$ is like sending $1 \rm m$ to $0$ by writing it as $1 \times 10^{-9} \,\rm Gm$. Such a change can't actually affect the physics of the system.

To recover the notion of sending $\hbar$ to $0$ in natural units, we consider the natural scales of the system under consideration. For example, the classical limit of the quantum harmonic oscillator is achieved when $E \gg \hbar \omega_0 $, i.e. when the energy of the system is much greater than the spacing between energy eigenvalues. So while it doesn't make sense to send $\hbar$ to $0$ from the natural units perspective, it does make sense to send $\frac{\hbar \omega_0}{E}$ to $0$.

As Qmechanic alluded to, there is also the deformation quantization perspective, where quantum effects are treated perturbatively in a parameter suggestively (but perhaps misleadingly for the uninitiated) written as $\hbar$. To be more precise, $\hbar$ plays the role usually denoted by $x$ in the Taylor expansion of the quantum mechanical commutator in terms of the Poisson bracket associated with the classical system. In this case, when $\hbar$ goes to $0$, we really do recover the classical situation, essentially by construction. I should say that I'm not very knowledgeable about deformation quantization, so hopefully someone else can expand on what I've said here and correct any mistakes I might have made.

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    $\begingroup$ I don't agree with the first paragraph. Sending $\hbar$ to 0 is the same as defining an universe without QM. Writing $\hbar = 1$ has no meaning, without the units (it's not a pure number). Physicists are very sloppy on notation. They should write $\hbar = 1 \, \mathrm{u}$ instead, where $\mathrm{u}$ is just a special action unit, as arbitrary as any other one (like $\mathrm{J \cdot s}$). In that unit system, the value of $\hbar$ is just 1. But we still could decide to define $\hbar$ as a "variable" and make it goes to 0, even in the same unit system. $\endgroup$
    – Cham
    Commented May 28, 2020 at 17:39
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    $\begingroup$ In other words: you take an arbitrary object on your desk, of mass $M = 3,218 \, \mathrm{kg}$. Then you define it as an unit, so $M = 1 \, \mathrm{M}$, which is trivial. Obviously, once you made that choice, you're stuck with it and can't apply any kind of limit to it ! Doing $M \rightarrow 0$ is equivalent to changing the choice of standard (or choice of object on your desk). $\endgroup$
    – Cham
    Commented May 28, 2020 at 17:45
  • $\begingroup$ I pretty much agree with your comment. An assumption that is getting snuck in here is that there really is a meaningful quantum of action worthy of the name $\hbar$. But once we convinced of that fact and agree to use it to measure action, it no longer makes sense to talk about sending it to $0$ any more than it makes sense to talk about sending the mass of the object of your desk to $0$. It's a physical parameter of the universe under examination. If we consider the object on your desk interacting with another mass, say $m$, then we can talk about what happens when $M/m \to 0$ $\endgroup$
    – Charles Hudgins
    Commented May 28, 2020 at 17:56
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    $\begingroup$ I agree. However, many physicists (especially theorists) are writing non-senses like $\hbar = c = 1$ for exemple. $\hbar$ and $c$ don't have the same dimensions. Even if we separate them: $\hbar = 1$ and $c = 1$, is non-sense! They should write $\hbar = 1 \, \mathrm{\hbar}$ and $c = 1 \, \mathrm{c}$, which is trivial (and do make sense). Once we adopt that convention, we are stuck with those values and can't vary the "constants". Physical constant are like 1D vectors: a component (value) and a base (unit). We may fix the base and vary the component. To me, this is what the limit means. $\endgroup$
    – Cham
    Commented May 28, 2020 at 18:04
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    $\begingroup$ I'm being warned to avoid extended discussion, but this is a point worth following up on. There might be some sense to writing $\hbar = c = 1$ if you think there isn't really a physical distinction between the quantities they represent. In the case of GR, everything is ultimately written in terms of $m$ (or some unit of distance) because the GR perspective is that everything (at the classical level) really is geometry. Time isn't a different kind of quantity from distance. They both describe the geometry of the system once you take a 4d perspective. $\endgroup$
    – Charles Hudgins
    Commented May 28, 2020 at 18:10

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