I read one Phys.SE question similar to mine, in Total angular momentum in a full shell but the question was so confusing and vague. The answer, though, was helpful for me to understand a part of my question. It's said that the total orbital and spin angular momentum for a closed shell is zero. I understand, as explained in the link above, that paired electrons in a closed shell have zero net spin. That's because each pair has one up electron and one down. But I can't realize why this holds for total orbital angular momentum too.
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The total angular momentum of a closed shell is zero because for fixed $l$, we have the possible states labeled by eigenvalues of $L_i$ as $m_{l,i} = l,\dots,0,\dots,-l$ in integer steps. The sum over all $m_l$ inside a shell is always zero, so total angular momentum of a shell is zero.
This is just the generalization of the argument with "up/down" for spin, which is the case $l = \frac{1}{2}$.
answered Jul 23 '15 at 10:50
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$\begingroup$ Hi, I have a question about your answer; for any $\ell$ the sum over the $m_{\ell}$ is always zero. So if we consider hydrogen for simplicity and $\ell = 3$ then $\sum_{-3}^{m_{\ell}=3}(\ell=3)=-3+-2+-1+0+1+2+3=0$. Clearly, this cannot be the case and I am not understanding part of your answer. I would just like to see an example where the total orbital angular momentum is non-zero in the case of an unfilled shell. $\endgroup$ – BLAZE Jul 27 '19 at 4:08
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1$\begingroup$ @BLAZE Of course, to get the total angular momentum of an unfilled shell, you only sum those $m_l$ that correspond to a filled state. $\endgroup$ – ACuriousMind♦ Jul 27 '19 at 9:05
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1$\begingroup$ This shows that the z-projection $m_L$ of total orbital angular momentum (coming from the operator $L_z = \sum_{i} l_{z,i})$ is zero. But to show that total orbital angular momentum is zero, it is also necessary to show that the magnitude $\mathbf{L}^2$ is zero. $\endgroup$ – Quantum Jun 13 '20 at 15:32
