0

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
  • 220,844

1 Answers1

1

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
  • 67,623