Subsections


10.7 Additional Points

This section mentions a couple of additional very basic issues in the quantum mechanics of solids.


10.7.1 About ferromagnetism

Magnetism in all its myriad forms and complexity is far beyond the scope of this book. But there is one very important fundamental quantum mechanics issue associated with ferromagnetism that has not yet been introduced.

Ferromagnetism is the plain variety of magnetism, like in refrigerator magnets. Ferromagnetic solids like iron are of great engineering interest. They can significantly increase a magnetic field and can stay permanently magnetized even in the absence of a field. The fundamental quantum mechanics issue has to do with why they produce magnetic fields in the first place.

The source of the ferromagnetic field is the electrons. Electrons have spin, and just like a classical charged particle that is spinning around in a circle produces a magnetic field, so do electrons act as little magnets. A free iron atom has 26 electrons, each with spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. But two of these electrons are in the 1s states, the K shell, where they combine into a singlet state with zero net spin which produces no magnetic field. Nor do the two 2s electrons and the six 2p electrons in the L shell, and the two 3s electrons and six 3p electrons in the M shell and the two 4s electrons in the N shell produce net spin. All of that lack of net spin is a result of the Pauli exclusion principle, which says that if electrons want to go two at a time into the lowest available energy states, they must do it as singlet spin states. And these filled subshells produce no net orbital angular momentum either, having just as many positive as negative orbital momentum states filled in whatever way you look at it.

However, iron has a final six electrons in 3d states, and the 3d states can accommodate ten electrons, five for each spin direction. So only two out of the six electrons need to enter the same spatial state as a zero spin singlet. The other four electrons can each go into their private spatial state. And the electrons do want to do so, since by going into different spatial states, they can stay farther away from each other, minimizing their mutual Coulomb repulsion energy.

According to the simplistic model of noninteracting electrons that was used to describe atoms in chapter 5.9, these last four electrons can then have equal or opposite spin, whatever they like. But that is wrong. The four electrons interact through their Coulomb repulsion, and it turns out that they achieve the smallest energy when their spatial wave function is antisymmetric under particle exchange.

(This is just the opposite of the conclusion for the hydrogen molecule, where the symmetric spatial wave function had the lowest energy. The difference is that for the hydrogen molecule, the dominant effect is the reduction of the kinetic energy that the symmetric state achieves, while for the single-atom states, the dominant effect is the reduction in electron to electron Coulomb repulsion that the antisymmetric wave function achieves. In the antisymmetric spatial wave function, the electrons stay further apart on average.)

If the spatial wave function of the four electrons takes care of the antisymmetrization requirement, then their spin state cannot change under particle exchange; they all must have the same spin. This is known as “Hund’s first rule:” electron interaction makes the net spin as big as the exclusion principle allows. The four unpaired 3d electrons in iron minimize their Coulomb energy at the price of having to align all four of their spins. Which means their spin magnetic moments add up rather than cancel each other. {D.56}.

Hund’s second rule says that the electrons will next maximize their orbital angular momentum as much as is still possible. And according to Hund’s third rule, this orbital angular momentum will add to the spin angular momentum since the ten 3d states are more than half full. It turns out that iron’s 3d electrons have the same amount of orbital angular momentum as spin, however, orbital angular momentum is only about half as effective at creating a magnetic dipole.

In addition, the magnetic properties of orbital angular momentum are readily messed up when atoms are brought together in a solid, and more so for transition metals like iron than for the lanthanoid series, whose unfilled 4f states are buried much deeper inside the atoms. In most of the common ferromagnets, the orbital contribution is negligible small, though in some rare earths there is an appreciable orbital contribution.

Guessing just the right amounts of net spin angular momentum, net orbital angular momentum, and net combined angular momentum for an atom can be tricky. So, in an effort make quantum mechanics as readily accessible as possible, physicists provide the data in an intuitive hieroglyph. For example

\begin{displaymath}
\strut^5\!D_4
\end{displaymath}

gives the angular momentum of the iron atom. The 5 indicates that the spin angular momentum is 2. To arrive at 5, the physicists multiply by 2, since spin can be half integer and it is believed that many people doing quantum mechanics have difficulty with fractions. Next 1 is added to keep people from cheating and mentally dividing by 2 - you must subtract 1 first. (Another quick way of getting the actual spin: write down all possible values for the spin in increasing order, and then count until the fifth value. Start counting from 1, of course, because counting from 0 is so computer science.) The $D$ intimates that the orbital angular momentum is 2. To arrive at $D$, physicists write down the intuitive sequence of letters $S,P,D,F,G,H,I,K,\ldots$ and then count, starting from zero, to the orbital angular momentum. Unlike for spin, here it is not the count, but the object being counted that is listed in the hieroglyph; unfortunately the object being counted is letters, not angular momentum. Physicists assume that after having practiced counting spin states and letters, your memory is refreshed about fractions, and the combined angular momentum is simply listed by value, 4 for iron. Listing spin and combined angular momentum in two different formats achieves that the class won’t notice the error if the physics professor misstates the spin or combined angular momentum for an atom with zero orbital momentum.

On to the solid. The atoms act as little magnets because of their four aligned electron spins and net orbital angular momentum, but why would different atoms want to align their magnetic poles in the same direction in a solid? If they don’t, there is not going to be any macroscopically significant magnetic field. The logical reason for the electron spins of different atoms to align would seem to be that it minimizes the magnetic energy. However, if the numbers are examined, any such aligning force is far too small to survive random heat motion at normal temperatures.

The primary reason is without doubt again the same weird quantum mechanics as for the single atom. Nature does not care about magnetic alignment or not; it is squirming to minimize its Coulomb energy under the massive constraints of the antisymmetrization requirement. By aligning electron spins globally, it achieves that electrons can stay farther apart spatially. {N.22}.

It is a fairly small effect; among the pure elements, it really only works under normal operating temperatures for cobalt and its immediate neighbors in the periodic table, iron and nickel. And alignment is normally not achieved throughout a bulk solid, but only in microscopic zones, with different zones having different alignment. But any electrical engineer will tell you it is a very important effect anyway. For one since the zones can be manipulated with a magnetic field.

And it clarifies that nature does not necessarily select singlet states of opposite spin to minimize the energy, despite what the hydrogen molecule and helium atom might suggest. Much of the time, aligned spins are preferred.


10.7.2 X-ray diffraction

You may wonder how so much is known about the crystal structure of solids in view of the fact that the atoms are much too small to be seen with visible light. In addition, because of the fact that the energy levels get smeared out into bands, like in figure 10.11, solids do not have those tell-tale line spectra that are so useful for analyzing atoms and molecules.

To be precise, while the energy levels of the outer electrons of the atoms get smeared out, those of the inner electrons do not do so significantly, and these do produce line spectra. But since the energy levels of the inner electrons are very high, transitions involving inner electrons do not produce visible light, but X-rays.

There is a very powerful other technique for studying the crystal structure of atoms, however, and it also involves X-rays. In this technique, called X-ray diffraction, an X-ray is trained on a crystal from various angles, and the way the crystal scatters the X-ray is determined.

There is no quantum mechanics needed to describe how this works, but a brief description may be of value anyway. If you want to work in nano-technology, you will inevitably run up against experimental work, and X-ray diffraction is a key technique. Having some idea of how it works and what it can do can be useful.

First a very basic understanding is needed of what is an X-ray. An X-ray is a propagating wave of electromagnetic radiation just like a beam of visible light. The only difference between them is that an X-ray is much more energetic. Whether it is light or an X-ray, an electromagnetic wave is physically a combination of electric and magnetic fields that propagate in a given direction with the speed of light.

Figure 10.27: Depiction of an electromagnetic ray.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...}
\put(-195,25){\makebox(0,0)[r]{${\cal E}$}}
\end{picture}
\end{figure}

Figure 10.27 gives a sketch of how the strength of the electric field varies along the propagation direction of a simple monochromatic wave; the magnetic field is similar, but 90 degrees out of phase. Above that, a sketch is given how such rays will be visualized in this subsection: the positive maxima will be indicated by encircled plus signs, and the negative minima by encircled minus signs. Both these maxima and minima propagate along the line with the speed of light; the picture is just a snapshot at an arbitrary time.

The distance between two successive maxima is called the wave length $\lambda$. If the wave length is in the narrow range from about 4,000 to 7,000 Å, it is visible light. But such a wave length is much too large to distinguish atoms, since atom sizes are in the order of a few Å. Electromagnetic waves with the required wave lengths of a few Å fall in what is called the X-ray range.

The wave number $\kappa$ is the reciprocal of the wave length within a normalization factor $2\pi$: $\kappa$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2\pi$$\raisebox{.5pt}{$/$}$$\lambda$. The wave number vector $\vec\kappa$ has the magnitude of the wave number $\kappa$ and points in the direction of propagation of the wave.

Figure 10.28: Law of reflection in elastic scattering from a plane.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...''$}}
\put(-40,5){\makebox(0,0)[b]{$\theta$}}
\end{picture}
\end{figure}

Next consider a plane of atoms in a crystal, and imagine that it forms a perfectly flat mirror, as in figure 10.28. No, there are no physical examples of flat atoms known to science. But just imagine there would be, OK? Now shine an X-ray from the left onto this crystal layer and examine the diffracted wave that comes back from it. Assume Huygens’ principle that the scattered rays come off in all directions, and that the scattering is elastic, meaning that the energy, hence wave length, stays the same.

Under those conditions, a detector A, placed at a position to catch the rays scattered to the same angle as the angle $\theta$ of the incident beam, will observe a strong signal. All the maxima in the electric field of the rays arrive at detector A at the same time, reinforcing each other. They march in lock-step. So a strong positive signal will exist at detector A at their arrival. Similarly, the minima march in lock-step, arriving at $A$ at the same time and producing a strong signal, now negative. Detector A will record a strong, fluctuating, electric field.

Detector B, at a position where the angle of reflection is unequal to the angle of incidence, receives similar rays, but both positive and negative values of the electric field arrive at B at the same time, killing each other off. So detector B will not see an observable signal. That is the law of reflection: there is only a detectable diffracted wave at a position where the angle of reflection equals the angle of incidence. (Those angles are usually measured from the normal to the surface instead of from the surface itself, but not in Bragg diffraction.)

For visible light, this is actually a quite reasonable analysis of a mirror, since an atom-size surface roughness is negligible compared to the wave length of visible light. For X-rays, it is not so hot, partly because a layer of atoms is not flat on the scale of the wave length of the X-ray. But worse, a single layer of atoms does not reflect an X-ray by any appreciable amount. That is the entire point of medical X-rays; they can penetrate millions of layers of atoms to show what is below. A single layer is nothing to them.

Figure 10.29: Scattering from multiple planes of atoms.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...(0,-1){7}}
\put(-33,9){\makebox(0,0)[r]{$d$}}
\end{picture}
\end{figure}

For X-rays to be diffracted in an appreciable amount, it must be done by many parallel layers of atoms, not just one, as in figure 10.29. The layers must furthermore have a very specific spacing $d$ for the maxima and minima from different layers to arrive at the detector at the same time. Note that the angular position of the detector is already determined by the law of reflection, in order to get whatever little there can be gotten from each plane separately. (Also note that whatever variations in phase there are in the signals arriving at the detector in figure 10.29 are artifacts: for graphical reasons the detector is much closer to the specimen than it should be. The spacing between planes should be on the order of Å, while the detector should be a macroscopic distance away from the specimen.)

The spacing between planes needed to get a decent combined signal strength at the detector is known to satisfy the Bragg law:

\begin{displaymath}
\fbox{$\displaystyle
2d\sin\theta = n\lambda
$}
\end{displaymath} (10.16)

where $n$ is a natural number. A derivation will be given below. One immediate consequence is that to get X-ray diffraction, the wave length $\lambda$ of the X-ray cannot be more than twice the spacing between the planes of atoms. That requires wave lengths no longer than of the order of Ångstroms. Visible light does not qualify.

The above story is, of course, not very satisfactory. For one, layers of atoms are not flat planes on the scale of the required X-ray wave lengths. And how come that in one direction the atoms have continuous positions and in another discrete? Furthermore, it is not obvious what to make of the results. Observing a refracted X-ray at some angular location may suggest that there is some reflecting plane in the crystal at an angle deducible from the law of reflection, but many different planes of atoms exist in a crystal. If a large number of measurements are done, typically by surrounding the specimen by detectors and rotating it while shining an X-ray on it, how is the crystal structure to be deduced from that overwhelming amount of information?

Clearly, a mathematical analysis is needed, and actually it is not very complicated. First a mathematical expression is needed for the signal along the ray; it can be taken to be a complex exponential

\begin{displaymath}
e^{{\rm i}\kappa(s-ct)},
\end{displaymath}

where $s$ is the distance traveled along the ray from a suitable chosen starting position, $t$ the time, and $c$ the speed of light. The real part of the exponential can be taken as the electric field, with a suitable constant, and the imaginary part as the magnetic field, with another constant. The only important point here is that if there is a difference in travel distance $\Delta{s}$ between two rays, their signals at the detector will be out of phase by a factor $e^{{\rm i}\kappa\Delta{s}}$. Unless this factor is one, which requires $\kappa\Delta{s}$ to be zero or a whole multiple of $2\pi$, there will be at least some cancelation of signals at the detector.

Figure 10.30: Difference in travel distance when scattered from P rather than O.
\begin{figure}
\centering
\setlength{\unitlength}{1pt}
\begin{picture}(...
...\put(145,78){\makebox(0,0)[b]{$\vec\kappa\,'$}}
\end{picture}
\end{figure}

So, how much is the phase factor $e^{{\rm i}\kappa\Delta{s}}$? Figure 10.30 shows one ray that is scattered at a chosen reference point O in the crystal, and another ray that is scattered at another point P. The position vector of P relative to origin O is ${\skew0\vec r}$. Now the difference in travel distance for the second ray to reach P versus the first one to reach O is given by the component of vector ${\skew0\vec r}$ in the direction of the incoming wave vector $\vec\kappa$. This component can be found as a dot product with the unit vector in the direction of $\vec\kappa$:

\begin{displaymath}
\Delta s_1 = {\skew0\vec r}\cdot\frac{\vec\kappa}{\kappa}
...
...i}\kappa\Delta s_1}=e^{{\rm i}\vec\kappa\cdot{\skew0\vec r}}.
\end{displaymath}

The difference in travel distance for the second ray to reach the detector from point P versus the first from O is similarly given as

\begin{displaymath}
\Delta s_2 = - {\skew0\vec r}\cdot\frac{\vec\kappa'}{\kapp...
...}\kappa\Delta s_2}=e^{-{\rm i}\vec\kappa'\cdot{\skew0\vec r}}
\end{displaymath}

assuming that the detector is sufficiently far away from the crystal that the rays can be assumed to travel to the detector in parallel.

The net result is then that the phase factor with which the ray from P arrives at the detector compared to the ray from O is

\begin{displaymath}
e^{{\rm i}(\vec\kappa-\vec\kappa')\cdot{\skew0\vec r}}.
\end{displaymath}

This result may be used to check the law of reflection and Bragg's law above.

First of all, for the law of reflection of figure 10.28, the positions of the scattering points P vary continuously through the horizontal plane. That means that the phase factor of the rays received at the detector will normally also vary continuously from positive to negative back to positive etcetera, leading to large-scale cancelation of the net signal. The one exception is when $\vec\kappa-\vec\kappa'$ happens to be normal to the reflecting plane, since a dot product with a normal vector is always zero. For $\vec\kappa-\vec\kappa'$ to be normal to the plane, its horizontal component must be zero, meaning that the horizontal components of $\vec\kappa$ and $\vec\kappa'$ must be equal, and for that to be true, their angles with the horizontal plane must be equal, since the vectors have the same length. So the law of reflection is obtained.

Next for Bragg’s law of figure 10.29, the issue is the phase difference between successive crystal planes. So the vector ${\skew0\vec r}$ in this case can be assumed to point from one crystal plane to the next. Since from the law of reflection, it is already known that $\vec\kappa-\vec\kappa'$ is normal to the planes, the only component of ${\skew0\vec r}$ of importance is the vertical one, and that is the crystal plane spacing $d$. It must be multiplied by the vertical component of $\vec\kappa-\vec\kappa'$, (its only component), which is according to basic trig is equal to $\vphantom0\raisebox{1.5pt}{$-$}$$2\kappa\sin\theta$. The phase factor between successive planes is therefore $e^{-{\rm i}{d}2\kappa\sin\theta}$. The argument of the exponential is obviously negative, and then the only possibility for the phase factor to be one is if the argument is a whole multiple $n$ times $\vphantom0\raisebox{1.5pt}{$-$}$${\rm i}2\pi$. So for signals from different crystal planes to arrive at the detector in phase,

\begin{displaymath}
d 2\kappa \sin\theta = n 2 \pi.
\end{displaymath}

Substitute $\kappa$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2\pi$$\raisebox{.5pt}{$/$}$$\lambda$ and you have Bragg’s law.

Now how about diffraction from a real crystal? Well, assume that every location in the crystal elastically scatters the incoming wave by a small amount that is proportional to the electron density $n$ at that point. (This $n$ not to be confused with the $n$ in Bragg’s law.) Then the total signal $D$ received by the detector can be written as

\begin{displaymath}
D = C
\int_{{\rm all\ }{\skew0\vec r}} n({\skew0\vec r})...
...ppa-\vec\kappa')\cdot{\skew0\vec r}}{\,\rm d}^3{\skew0\vec r}
\end{displaymath}

where $C$ is some constant. Now the electron density is periodic on crystal lattice scale, so according to section 10.3.10 it can be written as a Fourier series, giving the signal as

\begin{displaymath}
D = C \sum_{{\rm all\ }{\vec k}_{\vec n}}
\int_{{\rm all...
...ppa-\vec\kappa')\cdot{\skew0\vec r}}{\,\rm d}^3{\skew0\vec r}
\end{displaymath}

where the ${\vec k}_{\vec n}$ wave number vectors form the reciprocal lattice and the numbers $n_{{\vec k}_{\vec n}}$ are constants. Because the volume integration above extends over countless lattice cells, there will be massive cancelation of signal unless the exponential is constant, which requires that the factor multiplying the position coordinate is zero:
\begin{displaymath}
\fbox{$\displaystyle
{\vec k}_{\vec n}= \vec\kappa'-\vec\kappa
$}
\end{displaymath} (10.17)

So the changes in the x-ray wave number vector $\vec\kappa$ for which there is a detectable signal tell you the reciprocal lattice vectors. (Or at least the ones for which $n_{{\vec k}_{\vec n}}$ is not zero because of some symmetry.) After you infer the reciprocal lattice vectors it is easy to figure out the primitive vectors of the physical crystal you are analyzing. Furthermore, the relative strength of the received signal tells you the magnitude of the Fourier coefficient $n_{{\vec k}_{\vec n}}$ of the electron density. Obviously, all of this is very specific and powerful information, far above trying to make some sense out of mere collections of flat planes and their spacings.

One interesting additional issue has to do with what incoming wave vectors $\vec\kappa$ are diffracted, regardless of where the diffracted wave ends up. To answer it, just eliminate $\vec\kappa'$ from the above equation by finding its square and noting that $\vec\kappa'\cdot\vec\kappa'$ is $\kappa^2$ since the magnitude of the wave number does not change in elastic scattering. It produces

\begin{displaymath}
\fbox{$\displaystyle
\vec\kappa\cdot{\vec k}_{\vec n}=
...
...style\frac{1}{2}}{\vec k}_{\vec n}\cdot{\vec k}_{\vec n}
$}
\end{displaymath} (10.18)

For this equation to be satisfied, the X-ray wave number vector $\vec\kappa$ must be in the Bragg plane between $-{\vec k}_{\vec n}$ and the origin. For example, for a simple cubic crystal, $\vec\kappa$ must be in one of the Bragg planes shown in cross section in figure 10.24. One general consequence is that the wave number vector $\kappa$ must at least be long enough to reach the surface of the first Brillouin zone for any Bragg diffraction to occur. That determines the maximum wave length of usable X-rays according to $\lambda$ $\vphantom0\raisebox{1.5pt}{$=$}$ $2\pi$$\raisebox{.5pt}{$/$}$$\kappa$. You may recall that the Bragg planes are also the surfaces of the Brillouin zone segments and the surfaces along which the electron energy states develop discontinuities if there is a lattice potential. They sure get around.

Historically, Bragg diffraction was important to show that particles are indeed associated with wave functions, as de Broglie had surmised. When Davisson and Germer bombarded a crystal with a beam of single-momentum electrons, they observed Bragg diffraction just like for electromagnetic waves. Assuming for simplicity that the momentum of the electrons is in the $z$-​direction and that uncertainty in momentum can be ignored, the eigenfunctions of the momentum operator ${\widehat p}_z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\hbar\partial$$\raisebox{.5pt}{$/$}$${\rm i}\partial{z}$ are proportional to $e^{{\rm i}\kappa{z}}$, where $\hbar\kappa$ is the $z$-​momentum eigenvalue. From the known momentum of the electrons, Davisson and Germer could compute the wave number $\kappa$ and verify that the electrons suffered Bragg diffraction according to that wave number. (The value of $\hbar$ was already known from Planck's blackbody spectrum, and from the Planck-Einstein relation that the energy of the photons of electromagnetic radiation equals $\hbar\omega$ with $\omega$ the angular frequency.)