Angular Momenta and Magnetic Dipole Moments — Question 9

Question 9

A complete magnetic multiplet cancels its vector moment

A model subshell contains one electron in every $(m_l,m_s)$ state for fixed l, with $m_l=-l,\ldots,l$ and $m_s=\pm1/2$. Use $\mu_z=-\mu_B(m_l+2m_s)$. Treat the shell as filled; exclusion will be developed separately.

1. Sum the orbital and spin z moments.

2. For $l=1$, calculate $\sum_i\mu_{zi}^2$ and explain why it differs from the square of the total z moment.

Question 9 - Solution

A complete magnetic multiplet cancels its vector moment

(a) Method and calculation

$$\sum_i\mu_{Lz,i}=-2\mu_B\sum_{m=-l}^lm=\boxed{0}.$$$$\sum_i\mu_{sz,i}=-2\mu_B(2l+1)(1/2-1/2)=\boxed{0}.$$

(b) Interpretation and further calculation

$$l=1:\quad\mu_{zi}/\mu_B=2,1,0,0,-1,-2\Rightarrow\boxed{\sum_i\mu_{zi}^2=10\mu_B^2}.$$$$\boxed{(\sum_i\mu_{zi})^2=0};\quad 2\sum_{i<j}\mu_{zi}\mu_{zj}=-10\mu_B^2.$$

Physical check

Cancellation concerns the total vector, including cross terms when squared. A sum of individual squared components does not measure the net moment.