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?
[Physics] Quantization of Electron Spin
angular momentumdiscretequantum mechanicsquantum-spinrepresentation-theory
Related Solutions
The orbital wavefunctions of the hydrogen atom, which obey the eigenvalue equation $$ \left[-\frac{1}{2\mu}\nabla^2-\frac{e^2}{r^2}\right]\psi_{nlm}=E_{nl}\psi_{nlm}, $$ are functions of the separation vector $\mathbf r$ which points from the proton towards the electron. This is a standard trick in the two-body problem and it is done in both the classical and the quantum versions to factor away the motion of the bigger body (which is close to the centre of mass) and leave an effective one-body problem which is easier to treat.
This means that the orbital angular momentum, with total angular momentum number $l$, is in fact the combined angular momentum of the electron and the proton about their centre of mass. In essence, the proton partakes in part of the orbital motion and takes out some of the angular momentum from the electron. (Note, though, that this is classical language and it explicitly does not hold for the hydrogen atom, where the angular momentum of the motion is essentially indivisible.)
This raises an apparent paradox, which is resolved through the fact that the separation vector obeys dynamics through the reduced mass $\mu=1/\left(\tfrac{1}{m_e}+\tfrac{1}{m_p}\right)\approx\left(1-O\left({m_e\over m_p}\right)\right)m_e\lesssim m_e$ of the system, and this is slightly smaller than the electron mass. This slightly enlarges the orbital radius of the electron (since the Bohr radius is inversely proportional to the mass). The velocity stays constant (at $\alpha c$), which means that the angular momentum $L\sim \mu r v$ stays constant as well.
That said, the proton does have spin angular momentum of its own, but this couples weakly to the electronic motion. This coupling is via the same spin-orbit couplings as the electron, but its much higher moment of inertia means that the relevant energies are much smaller, as are the corresponding hyperfine splittings in the spectrum.
The selection rule $\Delta S=0$ is an approximation, nothing more, and in suitable circumstances it can easily break. One prominent example of this is the hydrogen 21cm line.
Electromagnetic atomic and molecular transitions are arranged into a series in order of multipolarity, which describes the atomic operators that enact the interaction hamiltonian, with higher multipolarities scaling down in coupling strength as powers of $a/\lambda$, i.e. the ratio between the system's size and the radiation's wavelength, which is generally small. Thus, you get
- as the leading-order term, electric dipole (E1) transitions;
- weaker by a factor of ${\sim}a/\lambda$,
- magnetic dipole (M1), and
- electric quadrupole (E2) transitions;
- still weaker by another factor of ${\sim}a/\lambda$,
- magnetic quadrupole (M2), and
- electric octupole (E3) transitions;
- and so on.
Generally, the no-spin-flips rule holds only for electric dipole transitions, for which the interaction hamiltonian is the electric dipole operator $\hat{\mathbf d}$, which does not couple sectors with different spin (unless you have strong spin-orbit coupling).
However, it is perfectly possible to have magnetic dipole transitions between states of spin that differ by $\Delta S=1$. Here the coupling is weaker (so you will need a higher intensity, or longer pulse times, to excite them), and thus typically the linewidth is smaller (so you will require a sharper laser), also giving a longer decay lifetime, but those are things that make the transition harder to observe, not impossible.
Best Answer
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 $-1/2, +1/2$