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.74}, it is seen that a constant magnetic field adds two terms. The most important of the two is

\begin{displaymath}
\fbox{$\displaystyle
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

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

In terms of this magnetic dipole moment, the energy is
\begin{displaymath}
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,

\begin{displaymath}
\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 $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$e$, much of the time you will find instead the “Bohr magneton”
\begin{displaymath}
\fbox{$\displaystyle
\mu_{\rm B}= \frac{e\hbar}{2m_{\rm e}}
\approx 9.274\;10^{-24}\mbox{ 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.75}. The energy that an electron picks up in a magnetic field due to its inherent spin is:

\begin{displaymath}
\fbox{$\displaystyle
H_{{\cal B}S} = - g_e \frac{q}{2m_{...
...w 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,” [51, 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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$ 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

\begin{displaymath}
\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”

\begin{displaymath}
\fbox{$\displaystyle
\mu_{\rm N}=\frac{e\hbar}{2m_{\rm p}}
\approx 5.050{,}78\;10^{-27}\mbox{ 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

\begin{displaymath}
\fbox{$\displaystyle
H_{{\cal B}D} = \frac{q^2}{8m}\left...
...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:

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

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.