# Two spin 1/2 partcles and spin operators

## Main Question or Discussion Point

Dear forumers,

I was thinking about how the Sz operator "couples" (has non zero matrix elements) states with the same expectation values for the projection of spin on the z-axis (duh! α and β are its eigenvectors), and how Sx and Sy couple different states (once again, duh!). I was also thinking that this would hold for a two-particle system, i.e. that triplet states would couple to singlet states. I sat down and wrote down the expressions for the operators in a two-particle space (4-by-4 matrices) in both the "usual" basis (αα, αβ, βα, ββ) and the basis consisting of the singlet state+triplet states. I found that the singlet state does not "couple" to any other state through Sx or Sy, which if found to be counterintuitive. Do you see an inherent flaw in my reasoning? Do you think my conclusions are correct?

For reference, a lot of the expressions for the operators are found here:

http://electron6.phys.utk.edu/qm1/modules/m10/twospin.htm

Thank you in advance and all the best :)

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DrClaude
Mentor
I was thinking about how the Sz operator "couples" (has non zero matrix elements) states with the same expectation values for the projection of spin on the z-axis
I donated understand what that means. Eigenstates of $S_z$ are, by definition, not "coupled" by $S_z$.

I found that the singlet state does not "couple" to any other state through Sx or Sy, which if found to be counterintuitive.
The singlet and triplet states are eigenstates of the total spin operators $S^2$ and (usually) $S_z$. Since $S^2$ and $S_x$ (or $S_y$) commute, there can be no coupling between states of different total spin, only between the different $M_S$ components.