Manyofthepropertiesofelectronbandscanbeseenthroughafairlysimple,exactlysolv- ablemodel,knownastheKronig–Penneymodel,whichisanextensionofthesquare-well
11 1.2 TheKronig–PenneyModel
U0
2 1 3
a –b 0
.
. .. .. .. .. ..
±Fig.1.9 ThepotentialenergyoftheKronig–Penneymodel.
structuresweexaminedinSection1.1.1.Weimagineanelectroninaninfinite,perfectly periodic,one-dimensionalstructure,asshowninFigure1.9.
Asinthestandardsquare-wellmodel,weguesstheformofthesolutionasfollows:
ψ1(x) = A1eiKx+ B1e−iKx, 0 < x < a
ψ2(x) = A2eκx+ B2e−κx, −b < x < 0. (1.2.1) Weonlyneedtoworryaboutthesetworegions,becausetherestofthestructureisidentical tothese.
Becauseeverycellisidentical,thereisnoreasonforthewavefunctionofaneigenstate inonecelltohavegreatermagnitudethaninanyothercell.Therefore,itisreasonableto expectthatthesolutionmustbethesameineverycellexceptpossiblyforanoverallphase factor. The phase factor will in general bea function of the cell position, but constant withinanygivencell.Furthermore,thephaseshiftfromonecelltothenextshouldbethe same,sincethereisnowaytotellanytwoadjacentcellsfromanotherpair.Wetherefore setthe phase factorequalto eikX,where X is the cell positionand k is aconstant.This implies
ψ3(x) = ψ2(x − a − b)eik(a+b), a < x < a + b (1.2.2) andtheboundaryconditions
ψ1(0) = ψ2(0), ∂ψ1
∂x
±±±±
x=0= ∂ψ2
∂x
±±±±
x=0
ψ1(a) = ψ3(a), ∂ψ1
∂x
±±±
±x=a= ∂ψ3
∂x
±±±
±x=a. (1.2.3) Pluggingthedefinitionsof ψ1, ψ2,and ψ3intotheseequationsgivesthematrixequation
⎛
⎜⎜
⎜⎝
1 1 −1 −1
iK −iK −κ κ
eiKa e−iKa −eik(a+b)e−κb −eik(a+b)eκb iKeiKa −iKe−iKa −κeik(a+b)e−κb κeik(a+b)eκb
⎞
⎟⎟
⎟⎠
⎛
⎜⎜
⎝ A1 B1
A2
B2
⎞
⎟⎟
⎠ = 0.
(1.2.4) Settingthedeterminantofthismatrixtozerogivestheequation
(κ2− K2)
2κK sinh(κb)sin(Ka) + cosh(κb)cos(Ka) = cos(k(a + b)). (1.2.5)
The Schrửdinger equation in the well region gives the energy E = ±2K2/2m, and the Schrửdingerequationinthebarrierregiongives U0− E = ±2κ2/2m.Since κ and K both dependon E,Equation(1.2.5)canbesolvedfor E foranygivenvalueof k.
Wecangreatlysimplifythemodelbytakingthelimit b → 0, U0→∞,insuchaway thattheproduct U0b remainsconstant,whichimpliesthat κ2b isaconstantindependentof E,and κb → 0.Then(1.2.5)reducesto
κ2b
2K sin(Ka) + cos(Ka) = cos(ka). (1.2.6) Foranygiven k,wecansolvethisequationnumericallyfor K,whichgivesustheenergy E = ±2K2/2m. Since(1.2.6)depends onlyoncoska,thesolutions for ka ± 2π willbe indistinguishable from the solution at ka.We therefore need only find the solutions of (1.2.6) forarangeof k values such that ka variesby2π.Foreach value of k, thereare manysolutionsfor K.
Figure1.10shows E vs k foraKronig–Penneymodel,for k from −π/a to π/a.Two generalfeaturesstandout.Thefirstisthattherearegapsintheelectronenergy E,which occurwhen |(κ2b/2K)sin(Ka)+ cos(Ka)| > 1;inotherwords,forsomevaluesofthetotal energy E = ±2K2/2m thereisnocorrespondingvalueof k.Thesegapsappearatvaluesof k thataremultiplesof π/a.Asecondfeatureisthatthefirstderivativeof E withrespectto k vanishesatthesepoints.
E
N=3
0 ka π
–π
gap band N=1N=2
N=4
±Fig.1.10 Energyvsk fortheKronig–Penneymodel,for κ2b = 4.Differentsolutionsof(1.2.6)forthesamek arelabeledby integersN.
13 1.2 TheKronig–PenneyModel
Therange −π < ka < π iscalledthe Brillouin zoneofaperiodicstructure.1 Inthe caseoftheKronig–Penneymodel,itisfairlysimple,sinceweareconsideringonlyaone- dimensionalsystem.AswewillseeinSection1.9,inthreedimensionstheBrillouinzone becomesmorecomplicated.
Exercise1.2.1 Verifythealgebraleadingto(1.2.5)and(1.2.6).Mathematicacanbevery helpfulinsimplifyingthealgebra.
Exercise1.2.2 Findthezero-pointenergy,thatis, E = ±2K2/2m,at k = 0,using(1.2.6) inthe limit b → 0 and U0 → ∞ and U0b finite but small.To dothis, usethe approximationsfor sinKa à Ka andcos Ka à 1 − 12(Ka)2,assuming Ka isvery smallat k = 0.
Exercise1.2.3 Findthefirstgap energyat ka = π using(1.2.6) inthelimit b → 0and U0b issmall.Youshouldwriteapproximationsforsin Ka andcos Ka near Ka = π, that is, Ka à π + (±K)a, where ±K is small.You should find that you get an equationintermsof K thatisfactorizableintotwotermsthatcanequalzero.The differencebetweentheenergies E = ±2K2/2m forthesetwosolutionsfor K isthe energygap.
Doyourzero-pointenergyandgapenergyvanishinthelimit U0b → 0?
In (1.2.2),we introduced thephase factor eikX for different cells,where X is the cell position X = na,inthelimit b → 0,and n issomeinteger.Wecanthinkofthis phase factorasaplanewavethatmodulatesthesingle-cellwavefunction;inotherwords,
ψ(x) = ψcell(x mod X)eikX, (1.2.7) where ψcell is the same forall cells and “x mod X” gives the positionwithin each cell relativetothecelllocation X = na.WecanthereforeviewFigure1.10asthedispersion relationforthisoverallplanewave.
We can get a feel for whythe gaps appear at the points ka = nπ in the dispersion relationifwethinkofthephysicaleffectoftherepeatedcellsontheplanewave.Thesetof interfaceswithspacing a makeupa Bragg reflector.ABraggreflectorisalargenumber ofequallyspaced, partiallyreflecting, identicalobjects.As illustratedin Figure1.11, if thedistancebetweentheobjectsis a,thentheround-tripdistancebetweentwoobjectsis 2a.Therefore,thereflectionsofatravelingwavewithwavelength2a/n willallbeinphase andaddconstructively.Evenifonlyasmallamountofthewaveisreflectedfromanygiven object,awavewithwavelength2a/n willbeperfectlyreflectedinaninfinitesystem.The incidentwaveplusthereflectedwavetravelingintheoppositedirectionmakeastanding wave.Astandingwavehasgroupvelocityofzero;inotherwords, vg(k) = (1/±)∂ω/∂k = 0.Since E = ±ω forelectrons,thisimplies ∂E/∂k = 0.(Wewillreturntodiscussgroup velocityinSection3.3.)
Formally,wecanseethatthebandsmusthave ∂E/∂k = 0at thesymmetrypointsby implicitdifferentiationof(1.2.6).Setting U0= ±2κ2/2m inthelimit U0ả E,wehave
1 It is sometimes called the “first” Brillouin zone, because Brillouin came up with a series of zones based on the symmetry of a system. (See Section 1.9.3.) It is typically called simply the Brillouin zone, however.
(a) 2a
(b)
(c)
a
±Fig.1.11 ReflectionofawavefromaBraggreflector.(a)Twowavefronts(solidlines)ofawavewithwavelength2a approacha setofpartiallyreflectingplanes;thefirstwavefrontemitsareflectedwavefront(dashedline)movingintheopposite direction.(b)Afterthewavefrontshavemovedadistance a,thefirstwavefrontemitsanotherreflectedwavefront.
(c)Aftertheyhavemovedanotherdistance a,thesecondwavefrontemitsareflectedwavefrontalso.Thereflected wavesfromthefirstandsecondwavefrontsareinphase.Allofthepartialreflectionsthereforeaddupconstructively forwaveswithwavelength2(a).
mU0ba
±2
cos(Ka)
Ka dK −mU0ba
±2
sin(Ka)
(Ka)2 dK − sin(Ka)dK =− sin(ka)dk, (1.2.8) or
F(K)dK =− sin(ka)dk. (1.2.9)
Using dE = (±2K/m)dK,wethenfindthatwhen ka = nπ,then
∂ E
∂k = ∂E
∂K
∂K
∂k =−±2K m
sin ka
F(K(k)). (1.2.10)
Sincesin ka = 0 when ka = nπ, ∂E/∂k goes to zero at thesame points. (One might wonderif F(K)cangotozero,butif F(K) = 0,thensin Ka andcos Ka mustbothhavethe samesign,inwhichcasefortheleftsideof(1.2.6), |(κ2b/2K)sin(Ka) + cos(Ka)| > 1,so thatthereisnoreal k whichcorrespondstothiscase.)
We can alsosee whythe zone boundary is aboundary byrealizing that the solution at ka = ±π corresponds tothe single-wellwave functionmultiplied byaphase factor of eika = −1 fromone cellto thenext. Thisis themaximumpossible phasedifference betweencells.Increasingthephaseangle ka beyond π isjustthesameasstartingatphase angle ka =−π andmakingthephaseanglelessnegative.
15 1.2 TheKronig–PenneyModel
Inthecaseofweakcoupling(large U0b),theenergybandsoftheKronig–Penneymodel correspondtothesingle-wellquantizedstates,withenergyproportionalto N2.Toseethis, wecantakethelimit U0b →∞,inwhichcase(1.2.6)becomes
U0b
K sin Ka = 0, (1.2.11)
which implies Ka = πN or λ = 2a/N. This is just what we expect from the discus- sionof Section 1.1 –in the limit of very weak coupling, that is, U0b → ∞, we have the single-cellstates, whilewhen the coupling is increased, the gaps shrink, and these states are smeared out into bands. In the limit U0b → 0, we are left with K = k and E = ±2k2/2m, whichis, not surprisingly,just the energy ofa free particleinvac- uum, since there are no barriers left. Figure 1.12 shows the energy dispersion of the Kronig–Penneymodelwiththehigherbands plottedinadjacentzones.Thisis calledan extended zone plot of the energy dispersion, while Figure 1.10 is called the reduced zone plot. As one may expect from Figure 1.12, when the barriers between the cells become small,thedispersionof theKronig–Penney modelapproaches thefree-electron dispersion.
Inthis Kronig–Penney model,the energybands goupinenergy forever,to E = ∞, because we assumed U0 = ∞, which makes each cell a squarewell withsingle-state energies proportional to N2. This is not true for bands arising from atomic orbitals,as innormalsolids–theboundstateenergiesofatomsareproportionalto −1/N2,not N2. Thismeansthereis amaximum energyfor thebands ina solidformedfrom atomic or molecularbound states. Above thisenergy, there willbea continuum ofstateswith no
3π 2π
π 0
zero-point energy
gap
ka E = h2m2K2
E=h2m2k2
±Fig.1.12 Heavylines:extended-zoneplotfortheKronig–Penneymodel,for κ2b = 4.Grayline:theenergyofafreeelectron.
energy gaps. Thismaximumenergy is notnecessarily the sameas theenergy of afree electroninvacuum,however.Theenergyofamotionlessfreeelectron,aninfinitedistance awayfromthesolid,relativetotheenergybands,dependsonthepropertiesofthesurface ofthematerialaswellasthebandenergies.
Exercise1.2.4 UseMathematicatoplot Re k asafunctionof E = ±2K2/2m usingEqua- tion1.2.6.Assume thatyou haveasetof unitssuch that ±2/2m = 1,set a = 1, and choose various values of U0b from 0.1 to 3. This plot is just the Kronig–
Penneyreduced zone diagram turned onits side.Plot the first three bands. How dothegaps dependonyour value of U0b? Thenplot aclose-upofthe firstband and band gap along with the free-electron dispersion, k = √
E, on the same graph.