Is Planck's Constant Really a Constant?

Question

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?

Answer 1

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.

Answer 2

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 $$ [{\bf a},{\bf b}]~:=~\frac{i}{\hbar}({\bf ab}-{\bf ba}). \tag{1.02} $$

Answer 3

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.