13.4 Par­ti­cles in Mag­netic Fields

Maxwell’s equa­tions are fun, but back to real quan­tum me­chan­ics. The se­ri­ous ques­tion in this sec­tion is how a mag­netic field $\skew2\vec{\cal B}$ af­fects a quan­tum sys­tem, like say an elec­tron in an hy­dro­gen atom.

Well, if the Hamil­ton­ian (13.2) for a charged par­ti­cle is writ­ten out and cleaned up, {D.73}, it is seen that a con­stant mag­netic field adds two terms. The most im­por­tant of the two is

H_{{\cal B}L} = - \frac{q}{2m} \skew2\vec{\cal B}\cdot{\skew 4\widehat{\vec L}}
$} %
\end{displaymath} (13.35)

where $q$ is the charge of the par­ti­cle, $m$ its mass, $\skew2\vec{\cal B}$ the ex­ter­nal mag­netic field, as­sumed to be con­stant on the scale of the atom, and ${\skew 4\widehat{\vec L}}$ is the or­bital an­gu­lar mo­men­tum of the par­ti­cle.

In terms of clas­si­cal physics, this can be un­der­stood as fol­lows: a par­ti­cle with an­gu­lar mo­men­tum $\vec{L}$ can be pic­tured to be cir­cling around the axis through $\vec{L}$. Now ac­cord­ing to Maxwell’s equa­tions, a charged par­ti­cle go­ing around in a cir­cle acts as a lit­tle elec­tro­mag­net. Think of a ver­sion of fig­ure 13.6 us­ing a cir­cu­lar path. And a lit­tle mag­net wants to align it­self with an am­bi­ent mag­netic field, just like a mag­netic com­pass nee­dle aligns it­self with the mag­netic field of earth.

In elec­tro­mag­net­ics, the ef­fec­tive mag­netic strength of a cir­cling charged par­ti­cle is de­scribed by the so called or­bital “mag­netic di­pole mo­ment” $\vec\mu_L$, de­fined as

\vec\mu_L \equiv \frac{q}{2m} \vec L. %
\end{displaymath} (13.36)

In terms of this mag­netic di­pole mo­ment, the en­ergy is
H_{{\cal B}L} = - \vec \mu_L \cdot \skew2\vec{\cal B}.
\end{displaymath} (13.37)

which is the low­est when the mag­netic di­pole mo­ment is in the same di­rec­tion as the mag­netic field.

The scalar part of the mag­netic di­pole mo­ment, to wit,

\gamma_L = \frac{q}{2m} %
\end{displaymath} (13.38)

is called the “gy­ro­mag­netic ra­tio.” But since in quan­tum me­chan­ics the or­bital an­gu­lar mo­men­tum comes in chunks of size $\hbar$, and the par­ti­cle is usu­ally an elec­tron with charge $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom{0}\raisebox{1.5pt}{$-$}$$e$, much of the time you will find in­stead the “Bohr mag­ne­ton”
\mu_{\rm B}= \frac{e\hbar}{2m_{\rm e}}
\approx \mbox{9.274~10$\POW9,{-24}$\ J/T}
$} %
\end{displaymath} (13.39)

used. Here T stands for Tesla, the kg/C-s unit of mag­netic field strength.

Please, all of this is se­ri­ous; this is not a story made up by this book to put physi­cists in a bad light. Note that the orig­i­nal for­mula had four vari­ables in it: $q$, $m$, $\skew2\vec{\cal B}$, and ${\skew 4\widehat{\vec L}}$, and the three new names they want you to re­mem­ber are less than that.

The big ques­tion now is: since elec­trons have spin, build-in an­gu­lar mo­men­tum, do they still act like lit­tle mag­nets even if not go­ing around in a cir­cle? The an­swer is yes; there is an ad­di­tional term in the Hamil­ton­ian due to spin. As­ton­ish­ingly, the en­ergy in­volved pops out of Dirac's rel­a­tivis­tic de­scrip­tion of the elec­tron, {D.74}. The en­ergy that an elec­tron picks up in a mag­netic field due to its in­her­ent spin is:

H_{{\cal B}S} = - g_e \frac{q}{2m_{\r...
...skew 6\widehat{\vec S}}
\qquad g_e \approx 2 \quad q=-e
$} %
\end{displaymath} (13.40)

(This sec­tion uses again $S$ to in­di­cate spin an­gu­lar mo­men­tum.) The con­stant $g$ is called the “$g$-​fac­tor”. Since its value is 2, elec­tron spin pro­duces twice the mag­netic di­pole strength as the same amount of or­bital an­gu­lar mo­men­tum would. That is called the “mag­netic spin anom­aly,” [52, p. 222].

It should be noted that re­ally the $g$-​fac­tor of an elec­tron is about 0.1% larger than 2 be­cause of the quan­ti­za­tion of the elec­tro­mag­netic field ig­nored in the Dirac equa­tion. The quan­tized elec­tro­mag­netic field, whose par­ti­cle is the pho­ton, has quan­tum un­cer­tainty. You can think of it qual­i­ta­tively as vir­tual pho­tons pop­ping up and dis­ap­pear­ing con­tin­u­ously ac­cord­ing to the en­ergy-time un­cer­tainty $\Delta{E}\Delta{t}$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $\hbar$, al­low­ing par­ti­cles with en­ergy $\Delta{E}$ to ap­pear as long as they don't stay around longer than a very brief time $\Delta{t}$.Quan­tum elec­tro­dy­nam­ics” says that to a bet­ter ap­prox­i­ma­tion $g$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ $2+\alpha$$\raisebox{.5pt}{$/$}$$\pi$ where $\alpha$ $\vphantom0\raisebox{1.5pt}{$=$}$ $e^2$$\raisebox{.5pt}{$/$}$$4\pi\epsilon_0{\hbar}c$ $\vphantom0\raisebox{1.1pt}{$\approx$}$ 1/137 is called the fine struc­ture con­stant. This cor­rec­tion to $g$, due to the pos­si­ble in­ter­ac­tion of the elec­tron with a vir­tual pho­ton, [19, p. 116], is called the “anom­alous mag­netic mo­ment,” [25, p. 273]. (The fact that physi­cists have not yet de­fined po­ten­tial de­vi­a­tions from the quan­tum elec­tro­dy­nam­ics value to be “mag­netic spin anom­aly anom­alous mag­netic mo­ment anom­alies” is an anom­aly.) The pre­dic­tion of the $g$-​fac­tor of the elec­tron is a test for the ac­cu­racy of quan­tum elec­tro­dy­nam­ics, and so this $g$-​fac­tor has been mea­sured to ex­quis­ite pre­ci­sion. At the time of writ­ing, (2008), the ex­per­i­men­tal value is 2.002 319 304 362, to that many cor­rect dig­its. Quan­tum elec­tro­dy­nam­ics has man­aged to get things right to more than ten dig­its by in­clud­ing more and more, in­creas­ingly com­plex in­ter­ac­tions with vir­tual pho­tons and vir­tual elec­tron/positron pairs, [19], one of the great­est achieve­ments of twen­ti­eth cen­tury physics.

You might think that the above for­mula for the en­ergy of an elec­tron in a mag­netic field should also ap­ply to pro­tons and neu­trons, since they too are spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$ par­ti­cles. How­ever, this turns out to be un­true. Pro­tons and neu­trons are not el­e­men­tary par­ti­cles, but con­sist of three “quarks.” Still, for both elec­tron and pro­ton spin the gy­ro­mag­netic ra­tio can be writ­ten as

\gamma_S=g\frac{q}{2m} %
\end{displaymath} (13.41)

but while the $g$-​fac­tor of the elec­tron is 2, the mea­sured one for the pro­ton is 5.59.

Do note that due to the much larger mass of the pro­ton, its ac­tual mag­netic di­pole mo­ment is much less than that of an elec­tron de­spite its larger $g$-​fac­tor. Still, un­der the right cir­cum­stances, like in nu­clear mag­netic res­o­nance, the mag­netic di­pole mo­ment of the pro­ton is cru­cial de­spite its rel­a­tive small size.

For the neu­tron, the charge is zero, but the mag­netic mo­ment is not, which would make its $g$-​fac­tor in­fi­nite! The prob­lem is that the quarks that make up the neu­tron do have charge, and so the neu­tron can in­ter­act with a mag­netic field even though its net charge is zero. When the pro­ton mass and charge are ar­bi­trar­ily used in the for­mula, the neu­tron's $g$-​fac­tor is -3.83. More gen­er­ally, nu­clear mag­netic mo­ments are ex­pressed in terms of the “nu­clear mag­ne­ton”

\mu_{\rm N}=\frac{e\hbar}{2m_{\rm p}}
\approx \mbox{5.050\,78~10$\POW9,{-27}$\ J/T}
$} %
\end{displaymath} (13.42)

that is based on pro­ton charge and mass. There­fore nu­clear $g$-​fac­tors are sim­ply twice the nu­clear mag­netic mo­ment in mag­ne­tons. (Need­less to say, some au­thors leave out the fac­tor 2 for that ad­di­tional touch of con­fu­sion.)

At the start of this sub­sec­tion, it was noted that the Hamil­ton­ian for a charged par­ti­cle has an­other term. So, how about it? It is called the “dia­mag­netic con­tri­bu­tion,” and it is given by

H_{{\cal B}D} = \frac{q^2}{8m}\left(\...
...l B}\times {\skew 2\widehat{\skew{-1}\vec r}}\,\right)^2
$} %
\end{displaymath} (13.43)

Note that a sys­tem, like an atom, min­i­mizes this con­tri­bu­tion by stay­ing away from mag­netic fields: it is pos­i­tive and pro­por­tional to ${\cal B}^2$.

The dia­mag­netic con­tri­bu­tion can usu­ally be ig­nored if there is net or­bital or spin an­gu­lar mo­men­tum. To see why, con­sider the fol­low­ing nu­mer­i­cal val­ues:

\mu_{\rm B}= \frac{e\hbar}{2m_{\rm e}} \approx \mbox{5.788~...
...a_0^2}{8m_{\rm e}} = \mbox{6.156\,5~10$\POW9,{-11}$\ eV/T$^2$}

The first num­ber gives the mag­netic di­pole en­ergy, for a quan­tum of an­gu­lar mo­men­tum, per Tesla, while the sec­ond num­ber gives the dia­mag­netic en­ergy, for a Bohr-ra­dius spread around the mag­netic axis, per square Tesla.

It fol­lows that it takes about a mil­lion Tesla for the dia­mag­netic en­ergy to be­come com­pa­ra­ble to the di­pole one. Now at the time of this writ­ing, (2008), the world record mag­net that can op­er­ate con­tin­u­ously is right here at the Florida State Uni­ver­sity. It pro­duces a field of 45 Tesla, tak­ing in 33 MW of elec­tric­ity and 4 000 gal­lons of cool­ing wa­ter per minute. The world record mag­net that can pro­duce even stronger brief mag­netic pulses is also here, and it pro­duces 90 Tesla, go­ing on 100. (Still stronger mag­netic fields are pos­si­ble if you al­low the mag­net to blow it­self to smithereens dur­ing the frac­tion of a sec­ond that it op­er­ates, but that is so messy.) Ob­vi­ously, these num­bers are way be­low a mil­lion Tesla. Also note that since atom en­er­gies are in elec­tron volts or more, none of these fields are go­ing to blow an atom apart.