# The constants of physics not constant

1. Dec 14, 2007

### Jim Kata

I'm in a hurry so I'm going to say this quick.

For angular momentum you have

$$[J_i ,J_j ] = i\hbar \varepsilon _{ijk} J_k$$

and for isospin you have

$$[t_i,t_j] = ig\varepsilon_{ijk}t_k$$

now, $$\hbar$$ is viewed as a constant, but $$g$$ is allowed to change it's value due to renormalization group flow. It seems to me that $$\hbar$$ can not be a constant then. Where is my reasoning going wrong.

2. Dec 20, 2007

### blechman

Quick answer: the g (and the $\hbar$) cancels out.

With this choice of normalization for your generators, the isospin group elements are of the form:

$$e^{-i\vec{t}\cdot\vec{\theta}/g}$$

So that in the end, the coupling cancels out. This is as it should be, since the group theory doesn't care what the physical coupling is! To see things more clearly, it is better to use a different normalization, where $\vec{t}\rightarrow g\vec{t}$. Then the g cancels out of your commutator right away, and the group element makes no reference to g at all.

3. Dec 20, 2007

### blechman

Another way to say what I said above is that you usually want your generators to be dimensionless. That is why, for example, we often use the Pauli Matrices and not the actual angular momentum matrices. That way, the group generators are all dimensionless, and when you need the physical angular momentum/isospin/etc, THEN you can add the necessary factors of g and $\hbar$.