Why does the particle always have the same spin magnitude $\pm \frac{1}{2}\hbar$?

In classical mechanics we can split a vector into components but in quantum mechanics due to uncertainty principle we can't. So how does the $\hat z$ component of spin always have the same magnitude (there are infinite possible orientations between $\hat z$ axis and a magnetic dipole)? It is like the particle “knows” what direction we will measure it and always give us a result the same magnitude? Whatever coordinate system we choose if we measure along 1 axis then the magnitude will be the same. How is it possible?

  • $\begingroup$ This is actually a deep question; the uncertainty relation has nothing to do with it but rather it is an experimental “fact” that the measurement apparatus determines the direction of the projection. $\endgroup$ – ZeroTheHero Aug 16 '19 at 2:10

This has nothing to do with that particular experiment, and everything to do with the nature of electrons themselves.

The spinning of the electron - i.e. its possession of a nonzero rotational angular momentum - is an inherent property. It is part of what makes an electron what it is, and is as immutable and essential to its identity as such as its charge is. Yes, it may seem "weird" to you that an object can possess an angular momentum that is immutable in this sense, but we haven't yet found a better model for our world that doesn't involve this.

Personally, I think it's fun. Everything else in the universe spins - asteroids, stars, planets, galaxies. Granted, these angular momenta supplies are not immutable, as they can be spun up and down. So why should that not also be the case on the scale of elementary particles? And all the more, that we can't alter it?

However, there's also another aspect to your question here, and that's regarding the specific value taken. In fact, the intrinsic angular momentum of the electron is not equal in magnitude to $\frac{1}{2}\hbar$, but instead $\frac{\sqrt{3}}{2} \hbar$. But any time you know one component of the angular momentum, it always is $\frac{1}{2} \hbar$, and you wonder how that can be.

The reason is this: as you've surmised, the other two components in such a situation are subject to "uncertainty" - more formally, it means they have nontrivial Shannon entropy, meaning the information in them has been reduced. If you had gotten $\frac{\sqrt{3}}{2} \hbar$ for any component, then that would imply that, given this is the magnitude of the spin angular momentum, thanks to the laws of vector algebra you would know the other components would have to be zero (remember that the component of a vector along itself is its mangnitude): that is, you would have perfect information, about all three components - a contradiction in that the laws of physics do not permit that perfect information to exist. Hence you must never get this value, and always something else. In particular, given that the magnitude of the full vector is fixed at $\sqrt{3}{2} \hbar$, the components have to be uncertain "in just the right way" as for that to always be certain, and the way that those uncertainties cancel each other out ends up as having any fully-known component to be $\pm\frac{1}{2} \hbar$.

  • $\begingroup$ Thanks for the nice explanation. $\endgroup$ – Antonios Sarikas Aug 16 '19 at 19:58

When we hear the word 'spin,' we can't help but imagine something like the earth rotating about its axis. But electrons are dimensionless point particles, and so imagining an electron doing that is bound to cause some confusion. Myself, I prefer calling this thing 'intrinsic angular momentum.' Informative, but maybe not very catchy.

So if electrons don't spin in space, what gives them this additional angular momentum? We can't say for sure, but physicists much clever than me hypothesise that electrons 'spin' in the same way that electrons orbit around other particles like protons. So this means that all the quantum laws that apply to angular momentum should apply to spin too. (At least, according to the hypothesis)

So back to your question of why we can only observe spin states of $\hbar /2 $ and $ -\hbar / 2 $. The short answer is that from these postulates, we can derive math tools called spin operators, which represent the act of observing our system. But we can't observe any value because the math itself just won't let us. In technical jargon one would say "The eigenvalues of the spin operators represent the observables," (Again, this is a fancy way of saying The math won't let us and we're too darn big to go down there and see what's really happening!)

So I've waffled on for a bit here, and chances are you're probably not satisfied with this answer. I don't blame you! This is a really subtle and difficult thing that requires a lot of practice and patience. I recommend Griffith's Intro to Quantum Mechanics. Sectiiion 4.4 covers everything one could hope to know about spin. Good luck!

  • $\begingroup$ Thanks for the answer. So can we say that this specific value of spin of the electrons is the consequence of the Maths we use to describe them which are consistent with the experimental data? $\endgroup$ – Antonios Sarikas Aug 16 '19 at 19:57
  • $\begingroup$ Yes that's correct. There are many instances like this in QM where the maths seems to impose constraints on our physical reality. It's quite surprising! $\endgroup$ – Visipi Aug 17 '19 at 3:37
  • $\begingroup$ I think you're putting it upside down. The mathematical description was designed to give the answer that matches the experiment (see Pauli equation; Dirac was also looking to reproduce this result when designing his relativistic equation), not the other way around. Heavy holes in cubic semiconductors like silicon have spin $\frac32\hbar$, for example, and it's not a problem, since the math describing them (the effective Hamiltonians) is designed to make this possible. $\endgroup$ – Ruslan Dec 21 '19 at 20:29

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