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 $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. 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. {A.34}.

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

$$\strut^5\!D_4$$

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.

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.

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.

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:

$$\fbox{$\displaystyle 2d\sin\theta = n\lambda $} %$$ (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

$$e^{{\rm i}\kappa(s-ct)},$$

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.

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 $\vec\kappa$:

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

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

$$\Delta s_2 = - {\skew0\vec r}\cdot\frac{\vec\kappa'}{\kappa... ...i}\kappa\Delta s_2}=e^{-{\rm i}\vec\kappa'\cdot{\skew0\vec r}}$$

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

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

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 $\vphantom{0}\raisebox{1.5pt}{$-$}$$2\kappa\sin\theta$. The phase fac­tor be­tween suc­ces­sive planes is there­fore $e^{-{\rm i}{d}2\kappa\sin\theta}$. The ar­gu­ment of the ex­po­nen­tial is ob­vi­ously neg­a­tive, and then the only pos­si­bil­ity for the phase fac­tor to be one is if the ar­gu­ment is a whole mul­ti­ple $n$ times $\vphantom{0}\raisebox{1.5pt}{$-$}$${\rm i}2\pi$. So for sig­nals from dif­fer­ent crys­tal planes to ar­rive at the de­tec­tor in phase,

$$d 2\kappa \sin\theta = n 2 \pi.$$

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

$$D = C \int_{{\rm all }{\skew0\vec r}} n({\skew0\vec r}) e... ...appa-\vec\kappa')\cdot{\skew0\vec r}}{ \rm d}^3{\skew0\vec r}$$

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

$$D = C \sum_{{\rm all }{\vec k}_{\vec n}} \int_{{\rm all ... ...appa-\vec\kappa')\cdot{\skew0\vec r}}{ \rm d}^3{\skew0\vec r}$$

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:

$$\fbox{$\displaystyle {\vec k}_{\vec n}= \vec\kappa'-\vec\kappa $}$$ (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

$$\fbox{$\displaystyle \vec\kappa\cdot{\vec k}_{\vec n}= - ... ...xtstyle\frac{1}{2}}{\vec k}_{\vec n}\cdot{\vec k}_{\vec n} $}$$ (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.)