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We often mix up 2 different spins. I want to understand how they are related to each other. In many QFT textbooks, we use the natural units $\hbar=c=1$. So, $\hbar/2$ is written as $1/2$.

Representation theory spin: This tells us how the field transforms under the Poincaré group actions. It is often equal to the rank of the tensor; for example, electromagnetism $A^\mu$ has spin 1 and gravity $g_{\mu\nu}$ has spin 2. But this is not always true. For example, if we have a $p$-brane, it will couple to a spin 1 field that is $A_{\mu_1 \mu_2 \dots \mu_{p+1}}$, like Kalb–Ramond field for $p=1$ is not spin 2.

Spin angular momentum: This is the internal angular momentum of a particle/field. This has units of angular momentum. In the classical limit $\hbar\to0$, this will become zero. Of course, $\hbar$ is a constant that cannot be changed by $\hbar\to0$; I mean, all angular momentum scales in our system are much larger than $\hbar$.

So, classical electromagnetism is an example where the Representation theory spin is $1$, but the Spin angular momentum is $1\hbar=0$. But are there any quantum examples where these 2 spins are different? Like for example, even though classical physics violates the spin-statistics relations due to Maxwell Boltzmann distribution, it is a trivial counter-example. Non-trivial examples of Spin–statistics relation violation are like Faddeev–Popov ghost etc. Similarly, is there a nontrivial quantum example where these 2 spins are different? Why are they generally the same?

Qmechanic
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K. Sreeman Reddy
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1 Answers1

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The use of Planck's constant in relation to spin is at best misleading. Although one often finds expressions such as, for example $$ [J_x,J_y]=i\hbar J_z , $$ the appearance $\hbar$ in the expression is misleading. One can make a trivial redefinition $J_n\rightarrow \hbar J_n$ and cancel $\hbar^2$ from both sides of the equation to get $$ [J_x,J_y]=i J_z , $$ without having set $\hbar=1$. The second equation still describes the same physics as the first equation. They make the same predictions. It then follows that when one can perform such a transformation and the physics stays the same, then the transformation has no physical significance. Hence, $\hbar$ did not play any physical significant role in the first equation.

Spin is related to internal symmetries. Classical theories also have internal symmetries. These symmetries may sometimes be described by non-abelian Lie groups whose generators don't commute. An example of such a case is the polarization of classical optical fields. For a paraxial classical optical field, all the states of polarization are represented by the points on the Poincaré sphere, which is directly related to the spin half representation of SU(2). Therefore, the relationship for polarization of classical optical field is governed by the second equation above and not by the result that one would get by setting $\hbar=0$ in the first equation.

flippiefanus
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