# Precise value of spin in Nms?

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?)

## 1 Answer

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
$$J = \sqrt{j (j + 1)} \hbar$$
where:
$$j$$ 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

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

2)
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 $$\frac{1}{2}$$.

3)
A spin-2 particle has a rotational symmetry of 180 degrees.