D.73 Or­bital mo­tion in a mag­netic field

This note de­rives the en­ergy of a charged par­ti­cle in an ex­ter­nal mag­netic field. The field is as­sumed con­stant.

Ac­cord­ing to chap­ter 13.1, the Hamil­ton­ian is

$$H = \frac{1}{2m}\left({\skew 4\widehat{\skew{-.5}\vec p}}- q\skew3\vec A\right)^2 + V$$

where $m$ and $q$ are the mass and charge of the par­ti­cle and the vec­tor po­ten­tial $\skew3\vec A$ is re­lated to the mag­netic field $\skew2\vec{\cal B}$ by $\skew2\vec{\cal B}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\nabla$ $\times$ $\skew3\vec A$. The po­ten­tial en­ergy $V$ is of no par­tic­u­lar in­ter­est in this note. The first term is, and it can be mul­ti­plied out as:

$$H = \frac{1}{2m}{\skew 4\widehat{\skew{-.5}\vec p}}^{ 2} ... ...} \right) + \frac{q^2}{2m} \left(\skew3\vec A \right)^2 + V$$

The mid­dle two terms in the right hand side are the changes in the Hamil­ton­ian due to the mag­netic field; they will be de­noted as:

$$H_{{\cal B}L} \equiv - \frac{q}{2m}\left({\skew 4\widehat{\... ...{\cal B}D} \equiv \frac{q^2}{2m} \left(\skew3\vec A \right)^2$$

Now to sim­plify the analy­sis, align the $z$-​axis with $\skew2\vec{\cal B}$ so that $\skew2\vec{\cal B}$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${{\hat k}}{\cal B}_z$. Then an ap­pro­pri­ate vec­tor po­ten­tial $\skew3\vec A$ is

$$\skew3\vec A= -{\hat\imath}{\textstyle\frac{1}{2}} y {\cal B}_z + {\hat\jmath}{\textstyle\frac{1}{2}} x {\cal B}_z.$$

The vec­tor po­ten­tial is not unique, but a check shows that in­deed $\nabla$ $\times$ $\skew3\vec A$ $\vphantom0\raisebox{1.5pt}{$=$}$ ${{\hat k}}{\cal B}_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\skew2\vec{\cal B}$ for the one above. Also, the canon­i­cal mo­men­tum is

$${\skew 4\widehat{\skew{-.5}\vec p}}= \frac{\hbar}{{\rm i}}\... ...} + {\hat k}\frac{\hbar}{{\rm i}}\frac{\partial}{\partial z}$$

There­fore, in the term $H_{{{\cal B}}L}$ above,

$$H_{{\cal B}L} = - \frac{q}{2m} ({\skew 4\widehat{\skew{-.5}... ...rtial}{\partial x} \right) = - \frac{q}{2m} {\cal B}_z \L _z$$

the lat­ter equal­ity be­ing true be­cause of the de­f­i­n­i­tion of an­gu­lar mo­men­tum as ${\skew0\vec r}$ $\times$ ${\skew 4\widehat{\skew{-.5}\vec p}}$. Be­cause the $z$-​axis was aligned with $\skew2\vec{\cal B}$, ${\cal B}_z\L _z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\skew2\vec{\cal B}\cdot{\skew 4\widehat{\vec L}}$, so, fi­nally,

$$H_{{\cal B}L} = - \frac{q}{2m} \skew2\vec{\cal B}\cdot{\skew 4\widehat{\vec L}}.$$

Sim­i­larly, in the part $H_{{{\cal B}}D}$ of the Hamil­ton­ian, sub­sti­tu­tion of the ex­pres­sion for $\skew3\vec A$ pro­duces

$$\frac{q^2}{2m} \left(\skew3\vec A\right)^2 = \frac{q^2}{8m} {\cal B}_z^2 \left(x^2+y^2\right),$$

or writ­ing it so that it is in­de­pen­dent of how the $z$-​axis is aligned,

$$H_{{\cal B}D} = \frac{q^2}{8m} \left(\skew2\vec{\cal B}\times{\skew0\vec r}\right)^2$$