I have a Hamiltonian of the Hydrogen atom: $H=H_0+H_1+H_2$ , when: $H_0 $ is the hamiltonian from central force and from electron momentum , $H_1$ is the relativistic kinetic fixing, and $H_2$ is the Darwin element.

We are in the ground state, and it given to us that this state have degeneracy of 2 so in $|n,l,m_l,m_s>$ base, we are in the state of $|1,0,0,\pm 1/2 >$

in this state I need to find the expectation value of the operator: $A=XYS_xS_y+XZS_xS_z+YZS_yS_z$ when :$S_i$ represent spin in $i$ coordinate

$<1,0,0,\pm 1/2|A|1,0,0,\pm 1/2>=?$

I think it's need to be zero , but I can`t proof why? it's somehow related with the fact that the state have some symmetry in the angular momentum $l$. I don't get it...


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