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Why is electron spin quantized? I've seen the derivation for the Hydrogen atom's energy levels, but my professor jumped to electrons having spin 1/2 or -1/2 as experimental. Why do electrons obey the same quantization rules for angular momentum as the Hydrogen atom does? Why must the two states be one apart?

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Are you asking why angular momentum is quantised, or why the electron has a angular momentum of 1/2? –  John Rennie Aug 13 '13 at 8:27
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I am going to give you, not a rigourous explanation, but a feeling.

The discretization of angular momentum is a consequence of the quantum nature of particles.

Imagine a classical rotating particle, the variation of action could be written :

$\Delta S = J \Delta \theta$

where $J$ is the angular momentum, and $\theta$ an angle.

At this point, there is no reason why $J$ needs to have discrete values.

Now, turn to quantum mechanics. The main point is that, in quantum mechanics, $ \large \frac{S}{\hbar}$ is a phase. For instance, if we want to calculate the transition amplitude (propagator), we have :

$$<x't'|xt> = \int D \Phi ~e^{i\large \frac{\Delta S(\Phi)}{\hbar}}$$ where the sum is done for all the paths $\Phi$, with $\Phi(t) =x$ and $\Phi(t') =x'$

It is very clear, in this expression, that $ \large \frac{S}{\hbar}$ is really a phase.

Now, we may write, formally, $e^{i\large \frac{\Delta S}{\hbar}}$ as, for your question, as $e^{i\large \frac{ J \Delta \theta}{\hbar}}$

But, because $\theta$ is an angle, it is identified to $\theta + 2 \pi$, and if we want that $e^{i\large \frac{ J \Delta \theta}{\hbar}}$ does not change, we need :

$$\frac{J (2 \pi)}{\hbar} = 2 \pi n$$ That is :

$$J = n \hbar$$

Now, if you want to be more rigourous, one may say, that, in quantum mechanics, we are going to search unitary representations of groups. For instance, we may look at the group of rotations $SO(3)$. This is a compact group (because of the identification $\theta$ with $\theta + 2 \pi$), so we can find finite dimensional representations. To find all representations, we have to consider in fact $SU(2)$. The representations of $SU(2)$ are labelled by a semi-integer or an integer : $0,1/2,1,3/2,2$, etc... In a representation $s$, you have $2s+1$ states, labelled by $-s, -s+1,....,s-1,s$. For the representation $s= 1/2$, you have 2 states labelled $-1/2, +1/2$

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