0
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.