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In the explanation of diamagnetism, the atoms do not have a total magnetic moment $\bf{m}$. This magnetic moment comes from the sum of all magnetic moments of electrons, that have an orbital $\bf{m_{o}}$ and a spin component $\bf{m_{s}}$.

My question is: can the total angular momentum $\bf{L}$ of an atom (which is the sum of all angular momenta of electrons, that have an an orbital $\bf{L_{o}}$ and a spin component $\bf{L_{s}}$) be non-zero even if $\bf{m}$ is zero (as it should be in diamagnetism)? That is, can I have

$$\bf{m}=0 \,\,\, \mathrm{and} \,\,\,\, \bf{L}\neq 0$$ ?

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  • $\begingroup$ You're unlikely to be able to have both of those vectors have well-defined values simultaneously. If you only care about one component, though, it's a lot easier. $\endgroup$ – Emilio Pisanty Jan 8 '17 at 11:09

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