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\begin{document}

In a classical current loop, \( \frac{eRv}{2c}( \)\( \widehat{n}\times \overrightarrow{B})=\overrightarrow{\mu }\times \vec B = \)torque.

Work required to go from \( \theta =0 \)to \( \theta _{1} \)is \( W=-\overrightarrow{\mu }*\vec B  \)

classically \( \overrightarrow{\mu }=\frac{e}{2mc}\vec L  \)

QM 

\( \overrightarrow{\mu }=g\frac{e}{2mc}\vec S  \) where g is a proportionality constant (=1 for L).

\( |\vec S |=\frac{\hbar }{2}\Rightarrow  \)\( \overrightarrow{\mu }=g\frac{e\hbar }{4mc}\widehat{S} \) 

\( \frac{e\hbar }{2mc}=\mu _{B} \) =bohr magneton = \( 5.8*10^{-11}MeV/Tesla=9.3*10^{-21}erg/gauss \)

classical electron radius is \( \frac{e^{2}}{R_{c}}=m_{e}c^{2}\Rightarrow R_{c}\simeq 3fm \)

g=2 from stern gerlach experiment. Actually, \( \frac{g}{2}=1.00115965219 \). This can be calculated
to almost the same accuracy, and it agrees.

QM derivation:

Hamiltoian in external classical uniform B field:

\( H\rightarrow H_{0}-\frac{e}{2mc}\sum _{i}(p_{i}A(r_{i})+A(r_{i})p_{i})+\sum _{i}\frac{e^{2}A^{2}(r_{i})}{2mc^{2}} \)

since \( \{p_{i},A_{i}\}=B_{i}L_{i} \), this is \( H\rightarrow H_{0}-\frac{e}{2mc}B_{j}\sum _{i}L_{ij}+\frac{e^{2}}{2mc^{2}}\frac{1}{2}B^{2}\sum _{i}r^{2}_{\bot i} \)

= \( H\rightarrow H_{0}-\mu _{j}*B_{j}+O(B^{2}) \)

\( u_{j}=\frac{e}{2mc}\sum _{i}L_{i}=-\mu _{B}\frac{\sum _{i}L_{j}}{\hbar } \) where \( \mu _{B}=\frac{|e|\hbar }{2mc} \).

Ratio between 2nd and 3rd terms is \( 10^{-5}*Btesla \)

\end{document}
