# 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?)

On Wikipedia, it gives a very brief description of how spin is calculated, in Nms, then says but, 'this is not the full computation of this value'. No further explanation or computation of spin in Nms is given in the rest of the article... Not that I've been able to find...

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.

• If you've just given the full computation of spin in Nms, that should be easy enough to be on Wikipedia in full... Can you or someone else edit Wikipedia so that it doesn't just say, 'this is not the full computation of this value' in the opening section? – Kurt Hikes Sep 18 at 21:21
• So, the spin of a particle with s=.5 is the square root of .75 times the reduced Planck constant? According to your equation? – Kurt Hikes Sep 22 at 7:11
• @Kurt Hikes. Yes, that is the spin angular momentum $S$. – Michele Grosso Sep 22 at 12:02