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I am unclear about what is meant by spin quantum number being an inherent number. I understand the spin quantum number can either be +1/2 or -1/2.

So, we can think of the spin quantum number like a arrow pointing upward or downward, and it always have the same magnitude.

What I am unclear about is this: Can there be a transition from spin up state to spin down state, or vice versa?

If a such a rotation/transition can occur, does that necessarily mean that it is the only magnitude of the spin quantum number that is intrinsic, and not the direction of it?

Qmechanic
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kolmogorov
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    There are two “spin quantum numbers”: $s$ (always 1/2) and $m_s$ (either 1/2 or -1/2). Thus $s$ is “inherent” but $m_s$ is not. – Ghoster Feb 02 '24 at 06:36
  • Not necessarily, if you have two entangled 1/2 spin particles, not even the total spin is well defined, as it can either be zero or one, depending on chance, or whatever you want to name it. – Pato Galmarini Feb 02 '24 at 07:13
  • I don't think it's a good idea to think of spin as an arrow, even though we denote it that way. If a system with spin (or better, an ensemble of spins) is exposed to a magnetic field it will show a "precession"-like dynamic. A "spinor" is a different class of "object" that changes according to a different set of symmetry rules than a vector or pseudo-vector like angular momentum does. We have several answers about the properties of spinors like https://physics.stackexchange.com/questions/582612/rotation-of-a-spinor. – FlatterMann Feb 02 '24 at 07:36

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It depends on what you mean by intrinsic. It's all linear algebra.

The generic spin state for the electron is $$ |\psi\rangle= \cos\theta|\uparrow\rangle +e^{i\phi}\sin\theta |\downarrow\rangle $$ and acting on it by the three 2$\times$2 matrix operators $\vec s$ yields $$ s_z|\psi\rangle = {\hbar\over 2} (\cos\theta|\uparrow\rangle -e^{i\phi}\sin\theta |\downarrow\rangle),\\ s_x|\psi\rangle = {\hbar\over 2} (\cos\theta|\downarrow\rangle +e^{i\phi}\sin\theta |\uparrow\rangle),\\ s_y|\psi\rangle = {i\hbar\over 2} (\cos\theta|\downarrow\rangle -e^{i\phi}\sin\theta |\uparrow\rangle),\\ \vec s\cdot \vec s |\psi\rangle = {\hbar^2} {1\over 2}\left ( {1\over 2}+1\right )|\psi\rangle. $$

So, e.g., starting from the state with θ=0, you get eigenvalues for $s_z$ and $\vec s\cdot \vec s$, but rotated states for the other two operators, $s_x$ and $s_y$. See WP.

For your title question, start from θ=π/2.

Cosmas Zachos
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