What is the exact value of the spin of a particle with a 'spin' of 'one'? In units of Nms (Newton-meter-second)?

And does a boson really have a spin of exactly one, or has that been 'normalized'?

Lastly, if a spin-1/2 particle has to fully rotate twice to return to its original 'position' or 'value', does that mean a spin-2 particle only has to rotate 180-degrees to return to its original 'position' (meaning spin-up or spin-down, I presume?)


The total (orbital + spin) angular momentum $J$ of a particle is measured with reference to a fundamental unit of angular momentum given by the reduced Planck constant $\hbar$ as
$J = \sqrt{j (j + 1)} \hbar$
$j$ total angular momentum quantum number
$j = \vert l \pm s \vert$
$l$ orbital angular momentum quantum number
$s$ spin angular momentum quantum number
$\hbar = 1.0546 × 10^{−34} N m s$ SI units

A particle with zero orbital angular momentum and spin quantum number $s = 1$ has a spin angular momentum $S$ given by
$S = \sqrt{s (s + 1)} \hbar = \sqrt{2} \hbar$

A boson is a particle with integer spin, however not necessarily one. For instance a scalar has spin zero, a photon has spin one, a graviton has spin two. To complete the description a fermion is a particle with half-integer spin. For example an electron has spin $\frac{1}{2}$.

A spin-2 particle has a rotational symmetry of 180 degrees.


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