Why is Planck's constant the same for all particles?

Question

This question came to me while reading Where does de Broglie wavelength $\lambda=h/p$ for massive particles come from? This question has a nice answer that explains that wave number has be proportional to momentum because of Lorentz symmetry. The proportionality constant is Planck's constant $h$.

As I understand it $h$ can be thought of as the charge under translation transformation,

$$U(d) \, f(x) \equiv f(x+d) \cong 1 + \frac{df}{dx} d \, .$$

$U(d)$ is the time translation operator which is defined as $U(d) = \text{e}^{i P/\hbar}$. This definition leads to

$$ P = -i \hbar \frac{\text{d}}{\text{d}x} \, ,$$

from which $\lambda p = h$ can be deduced.

My question is the following: How come there is only one proportionality constant for all particles? Why is is not like with the relation in between spin and magnetic moment where there is a particle dependent gyromagnetic ratio?

Answer 1

We know from translation symmetry that the expectation value of the wavevector operator is constant -- that is, $$ \langle \mathbf{k} \rangle = \langle \mathbf{k}_1 + \mathbf{k}_2 + \ldots \rangle = \text{const.} $$

In other words, wavevector is a conserved quantity. If the constant of proportionality between $\mathbf{k}$ and momentum were different for different particles, then momentum would not be conserved.

Here's how I think of these things: the wavevector of a particle is the really fundamental quantity. Momentum is simply defined to be proportional to the wavevector because it can be shown using Schrodinger's equation that if a particle of mass $m$ is moving as a wavepacket with an average wavevector $\mathbf{k}$, then $m\mathbf{v} = \hbar\mathbf{k}$, where $\mathbf{v}$ is the particle's group velocity.

Answer 2

I discussed this with a colleague yesterday and I think that I get it.

It has to do with the normalisation of the group generators (momentum operators). Within each representation of the translation symmetry we are free to normalise the momentum operator as we like. Then it is natural to choose this normalisation such that all representations have the same pre-factor.

The momentum operator is chosen in each representation in such a way that $h$ is always the same.

It would be nice if someone could comment on this. Especially if I am wrong.

Answer 3

I think a much more natural way to come to $\hbar$ is the one used by Dirac in his principles of quantum mechanics. In it you start by stating that you want a Poisson Bracket which has the same algebraic structure as that of classical mechanics. Then, since quantum operators don't commute (unlike what happens in classical physics) you quickly find out that if $A$ and $B$ are operators, and we denote the Poisson Brackets with {}, we have $$\{A,B\}=i\hbar[A,B]$$where $[A,B]=AB-BA$. In it $\hbar$ is just a constant (to be determined experimentally) but it is quite obvious that is independent of the particle under study since it is only a relationship of operators.