0
$\begingroup$

I have had a true/false question in a practice exam stating:

For a spin 3/2 system (S=3/2), there are only four spin eigenstates.

which is true. (solutions)

I do not understand how one can determine how many eigenstates exist for a given spin system.

All I know is a s=1/2 system has two eigenstates.

$\endgroup$

closed as off-topic by Thomas Fritsch, Jon Custer, ZeroTheHero, GiorgioP, tpg2114 Jul 5 at 23:49

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Thomas Fritsch, Jon Custer, ZeroTheHero, GiorgioP, tpg2114
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ A general formula for a number of the spin eigenstates is $2S+1$ $\endgroup$ – Gec Jul 5 at 7:36
0
$\begingroup$

The states may take any value between $S$ and $-S$ in steps of 1. I.e. for $S = \tfrac{3}{2}$ valid states would be $\lbrace\tfrac{3}{2},\tfrac{1}{2},-\tfrac{1}{2},-\tfrac{3}{2}\rbrace$, for $S = \tfrac{1}{2}$ we get $\lbrace\tfrac{1}{2},-\tfrac{1}{2}\rbrace$ and for $S = 2$ we get the five states $\lbrace2,1,0,-1,-2\rbrace$.

Notice that then nececarrily follows that $S$ may take only full or half integer-values.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.