We can determine the orbital angular momentum by $mvr$, how is the intrinsic spin angular momentum of an electron?

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    $\begingroup$ Did you read this article en.wikipedia.org/wiki/Spin_(physics) or any material related to the early history of spin, as an idea to explain experimental results? $\endgroup$ – user108787 Dec 1 '16 at 16:17
  • $\begingroup$ You should first take it as an experimental fact. Have a look at the Stern-Gerlach experiment. Later on, Dirac arrived at the spin in a natural way combining relativity with QM. $\endgroup$ – FGSUZ Feb 6 '19 at 14:09

Spin is a complicated matter that requires a bit of sophisticated mathematics to understand fully. I'll do my best to give a brief overview. The idea of spin is intimately related to the idea of group representations. The Hilbert space of every particle carries a (projective) representation of the rotation group $SO(3)$. (Or if you prefer, a standard representation of the double cover SU(2) which is what you will usually see in these discussions. I find it more intuitive to think about the actually 3 dimensional rotation symmetry.) From pure mathematics relating nothing to spin or angular momentum, we find that the representations of this group are defined by either an integer, or a half-integer. In physical language, this means roughly that the angular momentum of a quantum-mechanical particle, either spin or orbital, is either an integer or half-integer multiple of $\hbar$.

The representations of the group $SO(3)$ are the linear operators on the space of states that we have come to call the "angular momentum operators". One particular representation gives us the operators for spin $1/2$ particles $$ S_x = \frac{\hbar}{2} \sigma_x, \quad S_y = \frac{\hbar}{2} \sigma_y, \quad S_z = \frac{\hbar}{2} \sigma_z, \quad $$ where the $\sigma_i$ are the Pauli spin matrices. These operators act on a state, denoted $|{s,m_s}\rangle$ and "return" the spin of the particle. I.e, $$ S_z |{s,m_s}\rangle = \hbar m_s |{s,m_s}\rangle $$ In the case of spin 1/2 particles, $m_s = \pm1/2$. Hence, we find that the spin of an electron for example is $\pm \frac{\hbar}{2}$, or as you wrote it, $\frac{h}{4\pi}$. I would be happy to elaborate more if you are interested in seeing more of the mathematics behind this explanation.


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