- #1
davidbenari
- 466
- 18
I'm confused about the statement that if operators commute then eigenstates are shared.
My main confusion is this one:
##L^2## commutes with ##L_i##. Then these two share eigenstates. But ##L_x## and ##L_y## do not commute, so they don't share eigenstates. Isn't this violating some type of transitivity? Namely, if ##[L^2,L_i]=0## then exactly what accounts for ##L_x## and ##L_y## not sharing eigenstates (considering they share eigenstates with ##L^2##)?
A mathematical explanation of this would be much appreciated.
If my confusion is not well understood please ask for clarification.
Many thanks.
My main confusion is this one:
##L^2## commutes with ##L_i##. Then these two share eigenstates. But ##L_x## and ##L_y## do not commute, so they don't share eigenstates. Isn't this violating some type of transitivity? Namely, if ##[L^2,L_i]=0## then exactly what accounts for ##L_x## and ##L_y## not sharing eigenstates (considering they share eigenstates with ##L^2##)?
A mathematical explanation of this would be much appreciated.
If my confusion is not well understood please ask for clarification.
Many thanks.