Does $\langle A^2B^2\rangle = \langle B^2A^2\rangle$ (i.e. the operators $A$ and $B$ commute) somehow imply
$\langle A^2\rangle = \langle B^2\rangle$? If so, why?
To add some context, my operators are the $x$ and $y$ components of the angular momentum operator. I require the one to imply the other for a proof I am attempting!
The answer is no. You can try with $A,B$ numbers (i.e., scalar multiplies of the identity). If $A=1$, $B=2$, then $\langle A^2B^2\rangle = \langle B^2A^2\rangle=4$, while $\langle A^2\rangle=1$, $\langle B^2\rangle = 4$.