Chng 6: in trng v t trng

Chng 6: in trng v t trng in trng v t trng nm trong s my d c gi tr nht ca mt h in t. Mt tc dng r rng ca in trng l iu khin dng qua mt vt dn. Ta nghin cu s dn nh vo s xuyn hm trong chng 5 v s kho st trng hp ngc li i vi s lan truyn t do ca cc in t b tn x yu bi cc tp hoc cc fonon trong cc chng sau. iu ngc nhin hn l thng tin c ch hoc cc ng dng thc t c th thu c bng cch tc dng mt in trng ln mt vt liu cch in. Mt v d ca iu l s thay i hp th quang gn mt bin vng gay ra bi mt in trng mnh. l hiu ng Franz Keldysh m ta s nghin cu trong phn 6.2.1. iu ny tr nn thm ch c ch hn khi cc in t v l trng b giam cm trong mt h lng t v c dng lm mt b bin iu quang t. Mt t trng c cc nh hng ng ch ln mt h thp chiu. V d nh mt trng thi lin tc ca mt 2DEG tch thnh mt h gin on ca cc hm gi l cc mc Landau. iu ny c phn nh trong dn dc nh l hiu ng Shublikov de Haas m n cho mt du hiu r rt ca dng iu 2 chiu. Hiu ng Hall l mt cng c dng rng ri trong cc cht bn dn v vic kt hp vi cc mc Landau trong mt 2DEG cho hiu ng Hall

Chng 6: in trng v t trng lng t nguyn, trong dn Hall l bi s chnh xc ca By gi, n c dng nh mt chun c bn. Cc gi tr l ca hiu ng Hall lng t ho tm c trong cc mu vi nhiu lead v c th hiu c khi s dng hnh thc lun i vi s vn chuyn kt hp nghin cu trong phn 5.7.2. Nhng thay i xa hn xut hin trong mt h chun 1 chiu v s gim mt t c t ra. Cc mu c linh ng cao nht cn ch ra mt tnh cht khc gi l hiu ng Hall lng t phn s.6.1. Phng trnh Schrodinger vi in trng v t trng Mt in trng hoc mt t trng thng phi c a vo trong c hc lng t thng qua th v hng v th vect K hiu c dng cho in trng trnh nhm ln vi nng lng E v ln ca c o bng tesla (T) trong h SI. Cc trng c rt ra t cc th nh cc h thc C mt s t do rt ln trong vic la chn cc th, c bit l iu ny c mo t nh mt s la chn gauge.

6.1. Phng trnh Schrodinger vi in trng v t trng Trc ht xt mt in trng tnh iu ny c m t ph bin nht bi mt th tnh in v hng Thm ch n c mt yu t ty v ta c th thm bt k mt hng s no vo m khng nh hng n trng. N c ngha l th tuyt i (hoc th nng) khng c ngha g v cc kt qu vt l ch ph thuc vo nhng s khc nhau. Mc d n l s la chn ph bin nht, thay th n ta c th rt ra in trng t mt th vect y c s t do nhiu hn v ta c th thm bt k hm no vo khng ph thuc vo thi gian m khng lm thay i Ta cng c th dng mt s kt hp ca th v hng v th vect. Khng c la chn no l l tng v cc nghim ca phng trnh Schrodinger ph thuc mnh vo vic la chn gauge. in trng l nh nhau ti mi im trong khng gian v n s l tt nu cc th phn nh tnh cht ny nhng th v hng th khng. Th vect tha nhn bt bin khng gian nhng l mt hm ca thi gian m n c ngha l s khng c cc trng thi dng. Trong thc t, ngi ta la chn th n gin ho bi ton nhiu nht c th nh ta s thy trong phn tip theo. Vic la chn gauge c vai tr ln hn na i vi mt t trng. Xt mt

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6.1. Phng trnh Schrodinger vi in trng v t trng trng c ln B dc theo hng z. Rota cho Hai la chn r rng l hoc Chng c gi l gauge Landau v c u im l ch cn mt thnh phn ca th vect m n thng lm n gin ho cc tnh ton. Nhc im l ch chng chn ra mt hng c bit trong mt phng xy mn khng phi l ng hng. Mt la chn khc l gauge i xng iu ny bo ton tnh ng hng ca mt phng ngang i vi nhng cc hm sng phc tp hn. Bt k hm no vi rota bng khng u c th thm vo m khng lm nh hng n gi tr ca Do i vi mi hm nn vic thm vo s khng lm thay i Tuy nhin, n s lm thay i v ta s to ra mt s thay i b cho th v hng gi cho khng i. Nh vy, mt php bin i gauge khng lm thay i trng l

V d nh tha mn php bin i gauge. N cho ta mi lin h gia th v hng v th vect khi m t mt in trng u. C mt s la

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6.1. Phng trnh Schrodinger vi in trng v t trng chn c bit ca gauge m chng c ch n gin ha nghin cu in t trng ph thuc vo thi gian nhng ta khng cn chng. By gi ta c cc th. Phng trnh Schrodinger i vi mt ht c in tch q trong mt in t trng l

Mt c tnh quan trng l by gi c 2 xung lng trong phng trnh. Xung lng c gi l xung lng chnh tc m n c thay bng ton t Xung lng th hai l N l biu thc tham gia vo ng nng (bnh phng ca xung lng trn 2m) v c gi l xung lng c hay xung lng ng. Cng mt s phn bit c to ra trong c hc c in. Mt c tnh khc khng c trong c hc c in v c sinh ra trong c hc lng t do phng trnh Schrodinger (6.3) cha cc th khc vi cc trng. N c ngha l cc in t c th b nh hng bi cc th thm ch trong cc vng m khng c cc trng. Hiu ng Aharonov Bohm l mt h qu ca iu ny v s c m t trong phn 6.4.9. Biu thc ca mt dng thay i khi c mt mt th vect. Phng trnh

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6.1. Phng trnh Schrodinger vi in trng v t trng (1.32) tr thnh

S hng b sung c gi l dng nghch t. Ngun gc ca n tr nn r rng hn mt t nu dng c vit li ging (1.42) nh sau

iu ny chng t rng ton t vn tc thc ra l vi xung lng c khc vi xung lng chnh tc.6.2. in trng u Dng iu c in ca mt in tch q trong in trng l n gin: n gia tc u ti vn tc v chuyn ng ca n vung gc vi khng b nh hng. Bc tranh trong c hc lng t khng d hiu nh vy ch yu l do ta cn dng cc th thay cho trng. Xt mt in tch q = -e trong mt in trng F u dc theo z. Do , th nng l m n tng theo z nu F > 0. V th ch ph thuc vo z,

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6.2. in trng u phng trnh Schrodinger c th c tch ra theo cch thng thng (phn 4.5) v ta c th tp trung vo bi ton 1 chiu. Th l khng i theo thi gian v do , ta tm cc trng thi dng ca phng trnh Schrodinger

iu ny l d dng v ta cng gii n i vi h th tam gic phn 4.4. Cc hm sng cha chun ho vi nng lng l

trong cc thang chiu di v nng lng l

iu ny l d dng hn so vi h tam gic do in t c th chuyn ng trong ton b khng gian v nng lng c th ly mi gi tr. Mt t hm sng c v th trn hnh 6.1. Cc trng thi khc nhau n

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6.2. in trng u Hnh 6.1. (a) Th nng eFz, 3 hm sng v cc nng lng i vi ccin t trong GaAs trong mt in trng u l (b)Mt trng thi nh x ti z = 0m n ch ra cc c tnh tng ngvi cc hm sng nh th no.

gin trt v pha trc trong khng gian v trt ln v nng lng trong khi gi cng mt dng hm. Chng i xuyn hm vo trong th i vi v chng gim nhanh hn khi th tng ln. Dng chnh xc rt ra t dng tim cn ca hm Airy khi (xem (A5.2)). Cc hm sng dao ng i vi Chng nhp nh nhanh hn khi tr nn m hn v ng nng tng ln. Bin ca chng gim ti cng mt thi im bo ton dng. Dng hm c th tm c t dng iu tim cn (A5.3). Cc hm ny l cc sng ng khc vi sng truyn. D dng thy ti sao n

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6.2. in trng u cn phi nh vy: mt in t vi nng lng khng i chuyn ng theo hng +z cho ti khi n va chm vi th ti Khi , n phn x hon ton v quay tr li dc theo hng z. S giao thoa gia 2 sng c cng cng to nn mt sng ng. iu ny mu thun vi bc tranh ca chng ta l in t c gia tc u trong mt in trng v mt quan im khc c tm ra bng cch dng mt th vect thay cho th v hng. Mt nhn xt lin quan n vic la chn th v hng l mt hng s c th thm vo th tnh in m khng nh hng n in trng nhng n lm dch chuyn cc nng lng ca tt c cc trng thi mt cch chc chn. Nh vy, nng lng tuyt i khng c ngha m ch c nhng s khc nhau v nng lng mi ng vai tr quan trng. D nhin, iu ny l quen thuc nhng vic t do la chn trong mt th vect dn ti mt xung lng ty m n b n hn nhiu. 6.2.1. Mt trng thi Mt trng thi xut hin lm nhng vic l trong mt in trng. Ch cc nng lng dng l c php trc khi trng tc dng v

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6.2. in trng u trong trng hp 1 chiu (xem (1.89)). Tuy nhin, tt c cc nng lng t n l c php ngay khi trng tc dng v t bn cht ca cc trng thi ring r rng l mt trng thi l hng s. y c iu g sai tri? Vn l ch h l mt bt bin i vi php dch chuyn tnh tin trc khi trng tc dng v do , ta c th o mt trng thi ti bt k im no v tm c cng mt cu tr li. R rng l iu ny khng ng vi th tuyn tnh m n ly cc gi tr t n nhng trng l yu. Tuy nhin, nu ta xem xt mt im c bit, ta bit rng a s cc hm sng c vi ch cc ui do xuyn hm ti cc nng lng thp hn. S phn chia gia trng thi truyn v trng thi xuyn hm s xy ra ti cc nng lng khc nhau ti nhng ch khc nhau v do , c tnh ny mt i nu ta ly trung bnh mt trng thi qua ton b h. Nghim l cn tp trung ln mt im v dng mt trng thi nh x. N c nh ngha trong phng trnh (!.102) nh sau Tng ly theo mi trng thi ring c nh s l k. Ta khng chun ho

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6.2. in trng u hm sng m ta hiu chnh tha s trc trong kt qu cui cng. Cc hm sng trong (6.7) c k hiu bi nng lng ca chng v khng c s lng t ging nh s sng i vi cc sng phng. Cng c mt phm vi lin tc ca khng ging nh h ca cc gi tr gin on ca k c gi thit trong (6.9). Do , tng c thay bng tch phn (mt ln na li b qua hng s). Nh vy, mt trng thi trong mt in trng trong trng hp 1 chiu l

Tch phn l tm thng v ta thu c

trong C l tha s trc cha bit. Lu rng nh ta k vng ch ph thuc vo E eFz m n l s khc nhau gia nng lng ton phn v th nng ti im quan st. V mt c in, n n gin l ng nng. c nh C, ta c th k vng rng mt trng thi ti cc ng nng ln s tr nn tng t vi mt trng thi i vi cc in t t do khi khng

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6.2. in trng u c trng

(6.11) v dng tim cn (A5.3) cho

Tha s trc c s ph thuc chnh xc vo nng lng nhng bnh phng ca cosin lun lun dao ng gia 0 v 1. Vic ly gi tr trung bnh ca n l 1/ 2 v lm bng cc tha s trc c nh C v cui cng ta c

i vi mt trng thi nh x ca h 1 chiu trong mt in trng. N c v th trn hnh 6.2(a). im k d cn bc 2 nghch o ti y vng c tm ra vi mt ui gim theo hm m ti nhng nng lng m. iu ny tng ng vi ui ca cc hm sng m chng i xuyn hm vo trong vng cm c in c ng nng m. Mt trng thi dao ng i vi ng nng dng do n ph thuc vo mt hm sng m n cha cc nt

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6.2. in trng u Hnh 6.2. Mt trng thi nh x i vi cc in t trong GaAs trong mt in trng l nh mt hm ca ngnng nh x Ccng cong mnh l cc kt qu i vi cc in t t do. Cc n v can (E, z) l trong trng hp d chiu. t sng ng. S tng ng c ch ra trn hnh 6.1. Mt trng thi c th c m rng cho cc h 3 chiu nh ta lm i vi siu mng (xem (5.92)). Kt qu l

Tch phn c cho trong (A5.6). N c v th vi kt qu tng ng

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6.2. in trng u i vi trng hp 2 chiu trn hnh 6.2. Cc c tnh chung l ging nh trong trng hp 1 chiu nhng cc dao ng c lm sch bi s cun v dng iu cn bc 2 nm di l r rng. Mt trng thi c th o c bng s hp th quang. Cc thay i sinh ra bi in trng c gi l hiu ng Franz Keldysh v n c minh ha trn hnh 6.3. Mt s chuyn tip quang ti tn s ch c th xy ra nu 2 trng thi c th c tm thy vi s tch v vi iu kin l cc trng thi xen ph trong khng gian. Thng thng s hp th trong mt bn dn l khng th xy ra nu do khng c cc trng thi sn c. Tuy nhin, trong mt in trng c cc trng thi trong c CB v VB ti tt c cc nng lng. S xen ph ca chng trong khng gian ph thuc vo s khc nhau trong cc nng lng ca chng. N l mnh nu vHnh 6.3. Hiu ng Franz Keldysh trn s hpth gia cc vng. Cc trng thi ch ra trong ccVB v CB c tch ra bi nhngxen ph do ui i xuyn hm vo trong khevng.

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6.2. in trng u iu ny gim nhanh vi Nh vy, bin hp th t c mt ui xuyn hm vo trong khe cm trc cng nh cu trc nh vo nhng thay i ca cc hm sng nh ch ra trn hnh 6.2. N thng c ni rng bin hp th b dch xung di bi in trng nhng t hnh v r rng l iu ny thc ra khng phi nh vy: bin c m rng khc vi dch chuyn. 6.2.2. in trng t mt th vect S m t c hc lng t i vi mt in t chuyn ng trong mt th v hng c mt s c im khng mong mun v d nh trng thi ring l mt sng ng m n khng phi l mt sng gia tc nh ta mong mun v s ng nht khng gian b ph v. Nhng vn ny khng xut hin nu dng mt th vect thay th nhng cc kh khn khc li thay th chng. Phng trnh Schrodinger i vi mt in t trong trng hp 3 chiu c dng Ton t Hamilton l mt hm ca thi gian v do khng c cc trng thi

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6.2. in trng u dng. Tuy nhin, th khng ph thuc vo v tr v do , ta c th th mt hm sng phng cho phn khng gian ca hm sng, ngha l

Ton t xung lng (chnh tc) n gin cho tc dng ln hm sng ny v sng phng mt i. Do , (6.16) tr thnh

N l mt phng trnh vi phn bc nht v c nghim l

D dng gii thch mt s c im ca n. Phn ch thi gian c th c xem nh s tng qut ho dng thng thng cho trng hp khi nng lng l mt hm ca thi gian. Xung lng xut hin trong khng phi l mt hng s nhng thay i nh l N chng t s gia tc u vi mt lc l m n chnh l ci ta k vng i vi in

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6.2. in trng u tch e trong trng Mt cng ng u qua ton b khng gian v n phn nh tnh ng u ca in trng. Mt khc, ta c th k vng rng s gia tc s c phn nh trong phn khng gian ca qua mt vect sng thay i nhng y l hng s. N nh vo 2 xung lng cp trong phn 6.1: nng lng ph thuc vo xung lng c m n thay i do s gia tc ca in trng nhng hm sng khng gian ph thuc vo xung lng chnh tc m n l hng s do th l khng i khp khng gian. Khi ny sinh cc kh khn ca vic gii thch ging nh trn, n l quan trng tnh ton v mt vt l cc i lng c th quan st thy nh mt dng. Biu thc bin dng (6.5) m n cha th vect cho

N l hng s qua ton b khng gian nh ta k vng v tng tuyn tnh theo thi gian phn nh s gia tc u. C l bc tranh vi mt th vect gn hn vi quan im c in trong vic ch ra rng mt ht c gia tc u v mi im trong khng gian l tng ng. Ta cn ri b cch m t thng thng ca chng ta theo cc trng

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6.2. in trng u thi dng t c iu m n c ngha l vic nh ngha cc i lng nh mt trng thi l kh hn nhiu. iu tt hn l kt hp 2 bc tranh v ri b s ph thuc vo gauge dng biu din in trng nhng n i hi cc hm Green. 6.2.3. Vng hp trong mt in trng Trc khi chuyn qua t trng, ta s xem xt ngn gn v nh hng ca in trng ln mt in t trong mt vng hp trong mt tinh th. iu ny cung cp mt phi cnh mi ln cc kt qu nh dao ng Bloch trong phn 2.2. Mt s khc bit quan trng xut hin khi so snh chng vi cc khc bit i vi cc in t t do. Tm quan trng ca cc vng hp l ch ta s b qua s xuyn hm Zener gia cc vng v do cc khe cn phi rng. Ta dng php gn ng cosin i vi mt vng vi b rng W, ngha l

Theo phn 3.10, k s c thay bng xy dng hm Hamilton hiu dng. Nh vy, phng trnh Schrodinger i vi hm bao ca mt in t 1 chiu trong mt in trng l

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6.2. in trng u

Trc tin xt mt th vect Phng trnh Schrodinger l ging vi phng trnh Schrodinger i vi cc in t t do (xem (6.16)) ngoi tr dng ca v dng ca nghim ging vi (6.18) vi mt tch phn qua Mt s khc bit quan trng ny sinh t bn cht tun hon ca m n lm cho hm sng tun hon theo thi gian. Chu k l theo k m n tr thnh theo thi gian. N ging nh iu ta tm thy trc y i vi cc dao ng Bloch (xem (2.14)). Thc t l n l mt bin minh cho nghin cu ng lc hc ca cc in t trong mt vng trong phn 2.2, trong khng nh rng in trng v t trng iu khin xung lng tinh th. T phng trnh Schrodinger (6.19) suy ra rng mt in trng khng i lm cho xung lng tinh th k tng tuyn tnh min l trng c biu din bi mt th vect v b qua s lin kt vi cc vng khc. Cc kt qu s khc i nu dng mt th v hng mc d tnh ton phc tp

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6.2. in trng u hn mt cht. C nh v y vng u nghing nh eFz nh trn hnh 2.7. Mt in t c nng lng khng i do b gii hn ti mt vng khng gian hu hn mc d hm sng ca n gim trong cc khe vng ging nh cc in t t do trn hnh 6.1. Cc hm sng c cng dng hm v trt dc theo z v E. S khc bit l ch cc in t t do c th trt lin tc trong khi cc hm sng trong tinh th ch c th di chuyn theo cc bi s ca hng s mng a. Nh vy, cc nng lng to thnh mt thang Stark gin on vi khong cch gia cc nc thang l eFa. iu ny tng t nh cc dao ng Bloch trong mt th vect. Khong tch ra ph thuc vo hng s mng nhng khng ph thuc vo b rng vng ban u. Mt trng thi nh x c th tnh c v cc c tnh ti cc bin li b che m ging nh trn hnh 6.2(a). Hm sng b gii hn ti mt vi nguyn t khi trng tng v cc vng b nghing nhiu hn. Trong cc trng rt mnh, s thay i nng lng gia cc v tr ln cn vt qu b rng ban u ca vng, ngha l eFa > W v cc hm sng tr thnh cc hm sng nh x Stark trn cc v tr n. Cc nng lng ca cc nguyn t ln cn by gi khc nhau n mc s xuyn hm

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6.2. in trng u gia chng to thnh vng gn nh bin mt. N c ghi nhn v mt quang hc trong siu mng theo cng mt cch nh hu ng Franz Keldysh. Bn cht ca s dn dc theo in trng cng thay i mnh. Thng thng ta ngh rng s tn x cn tr s vn chuyn ging nh cc trng yu. Tuy nhin, trong s nh x Stark, mt in t ch c th chuyn ng dc theo siu mng bng cch nhy gia cc trng thi nh x v n i hi s pht ra nng lng. Do , s vn chuyn c thc y bi tn x khng n hi. Bc tranh ny trong khng gian thc tng t nh lp lun trong phn 2.2 ch tn x cn gy hn lon cc dao ng Bloch v cho php s vn chuyn. Mt hn ch ln l tnh ton ny b qua lin kt vi cc vng cao hn. Xt siu mng trn hnh 6.4. Khi khng c trng, cc mc trong cc h ln cn lin st c xp thng hng v s xuyn hm gia chng to thnh cc vng. Chng c tch ra bi cc khe rng nu cc ro l rng cho s xuyn hm yu. Mt in trng ko cc mc ra khi s xp thng hng v lm gim s xuyn hm m n gii hn hm sng ti mt s hu hn ging nh tho lun trc y. Tuy nhin, c th xp thng hng mc nng lng

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6.2. in trng u thp nht trong mt h vi mc th hai trong h lin st vi mt gi tr thch hp ca in trng. 2 mc ny khi s lin kt mnh to ra mt s cng hng mc d chng lm tch cc vng trong cc trng yu. By gi c mt thang ca cc trng thi lin kt nh trn hnh 6.4(b) m n c dng lm c s ca laze nhiu tng lng t gia cc vng con. 6.3. Tenx dn in v tenx in tr sut Trc khi bt u m t c hc lng t i vi cc in t trong t trng, iu quan trng l cn xem xt phn b ca in trng v dng in trong t trng. Xt mt h 2 chiu trong mt phng xy vi mt t trng dc theoHnh 6.4. nh hng ca mt intrng mnh ln mt siu mng. (a)S xuyn hm gia cc mc nh nhautrong cc h lin k to thnh cc vngtch ra rng cn bng. (b) Mt intrng mnh c th gy ra s xp thnghng cng hng ca cc mc khc nhau trong cc h lin k.

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6.3. Tenx dn in v tenx in tr sut z. Mt dng in v in trng lin h vi nhau bi khi khng c t trng , trong l dn in. N l mt v hng nu h l ng nht v ng hng. Mt t trng sinh ra hiu ng Hall. l mt in trng vung gc vi dng in v do , dng in v in trng khng cn song song vi nhau. V hng cn phi c thay bng tenx dn in (mt ma trn 2 x 2) m n c xc nh bi hay

N cho mt dng in nh l mt phn ng vi mt in trng. H thc ngc lin quan ti tenx in tr sut bi Cc phn t cho ca l nh nhau v l cc hm chn ca in trng B trong lc cc phn t khng cho l nh nhau v l cc hm l ca B. Ta cng bit rng tenx in tr sut l nghch o ca tenx dn in. Do , chng c th vit thnh

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6.3. Tenx dn in v tenx in tr sut trong cc du l thch hp cho cc in t. C 2 ma trn l cho khi khng c t trng v ta tm c kt qu quen thuc l By gi ta tc dng mt t trng yu ln mt thanh Hall mang mt dng in (hnh 6.5(a)). Ta dng m hnh Drude (2.15) trong dn in khi khng c t trng l trong n l s ht ti trong mt n v din tch, m l khi lng ca cc ht ti v l thi gian hi phc ca chng. Dng in cn c duy tr theo cng mt hng gia mu v do trong bc tranh c in, lc Lorentz do t trng cn phi c cn bng vi in trng (Hall) ngang. Nh vy, Du xut pht t du ca cc ht ti. Lu rng ta ang p t mt dng in v tnhHnh 6.5. Cc mu dng chung o dn in ca cc bn dn: (a) thanh Hall,(b) mu van der Pauw v (c) a Corbino. Cc vng ti l cc tip xc o th hiu hoc dng in v cc vng sng lcc vng hot ng ca mu.

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6.3. Tenx dn in v tenx in tr sut in trng do n sinh ra. N cho cc phn t ca tenx in tr sut

trong v tn s xiclotron c a vo. in tr sut dc khng b nh hng trong php gn ng ny. Hng s Hall qui c l v cho mt ht ti vi du in tch ca chng. iu ny mt phn gii thch tm quan trng cng ngh ca hiu ng Hall. Mt c im khc l cc kch thc ca mu khng tham gia vo khi hng s Hall c biu din theo th hiu v dng in thay cho in trng v mt dng in: i vi thanh Hall, trong l b rng mu m n loi b. Nghch o ca cho dn in

Trong trng hp ny, cc phn t cho b nh hng bi t trng. Lu rng trong l linh ng. H thc ny c v c ch nhng gi tr ca n b hn ch do m hnh Drude ch cha mt

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6.3. Tenx dn in v tenx in tr sut thi gian hi phc n gin. C th nh ngha cc thi gian hi phc khc nhau m gi tr ca chng c th l khc nhau mt bc ln trong mt 2DEG. Mt s knh ca cc ht ti thng c ng gp vo dng in. C c nhng in t v l trng trong mt cht bn dn khi ring. C th c hn mt vng con b chim gi trong mt kh in t chun 2 chiu hoc c th c mt knh song song khng mong mun ca cc in t trong mt cu trc bin iu pha tp . Tt c cc knh tng ng vi cng in trng v do , mt dng in tng cng tm c bng cch thm vo cc tenx dn in trc khi ly nghch o thu c tenx in tr sut tng cng. in tr sut dc cho s ph thuc bc 2 vo t trng m n c ly lm mt du hiu ca cc knh dn song song. Xt 2 knh song song ca cc in t vi cc mt v cc linh ng Ta nh ngha mt mt hiu dng v mt linh ng hiu dng Khi , trong cc trng yu m i vi c 2 knh,

26

6.3. Tenx dn in v tenx in tr sut

Trong cc trng rt mnh m i vi c 2 knh, ta thu li kt qu n gin mc d iu ny c th b che y bi cc nh hng c hc lng t. Cc kt qu ny l c ch trong thc t phn gii cc knh song song. Dng iu gy ngc nhin xy ra trong cc t trng mnh m c th t c iu kin iu ny c bit quan trng trong hiu ng Hall lng t. Trong gii hn ny,

By gi cc thnh phn dc ca v l t l thun thay v t l nghch vi nhau. Cc thnh phn dc ca c dn in v in tr sut c th trit tiu cng nhau trong gii hn ny. iu ny khc hn tnh cht khi khng c t trng. Dng in v in trng vung gc vi nhau nu v do , dng in chy dc theo cc ng th khc vi dng in chy dc

27

6.3. Tenx dn in v tenx in tr sut theo cc ng trng. iu ny lm thay i mnh phn b ca dng in v th bn trong thanh hnh ch nht nh ch ra trn hnh 6.6. xa cc u, dng in cn phi chy dc theo chiu di ca mu. iu ny l ging nh hng ca in trng khi khng c tc dng ca t trng v c mt in trng u dc theo mu. S phn b dng in ng u tip tc cho n cc tip xc ti u ny hoc u kia. iu ny thay i trong mt trng mnh m v dng in

Hnh 6.6. in trng, dng in v cc ng th bntrong mt mu hnh ch nht di vi cc tip xc quami mt u. (a) Khi khng c in trng, dng inl ng u khp mu v chy dc theo in trng.(b) Trong mt t trng mnh m dng in chy dc theo cc ng th.

28

6.3. Tenx dn in v tenx in tr sut chy dc theo cc ng th. Nh vy, in trng chy qua b rng ca mu v n cho mt th hiu Hall gia cc mt di v khng c trng dc. iu ny khng th duy tr gn cc u thanh v cc tip xc cng l cc ng th. Phn b ca dng in v trng b mo trong lc duy tr trc giao ln nhau sao cho tt c dng i vo v dng i ra ti 2 gc i din ca mu m chng ch l nhng ni m c th xy ra mt s bit th dc theo bin (hnh 6.6(b)). Tip xc bn tri ti cng mt th ging nh nh mu trong khi tip xc bn phi c gi cht ti y v do , chng o th hiu Hall bt k dng hnh hc no ca 2 my d. Mc d iu ny l mt gii php ng n v ton hc, r rng l n cn phi ph v gn cc xoy m , dng in i vo v dng in i ra do khng c c mt dng in v in trng u l v hn. Tuy nhin, cc th nghim ch ra rng ton b cng sut b tiu tn gn im ny. Ta s quay tr li phn b dng in khi nghin cu hiu ng Hall lng t phn 6.6. 6.4. T trng u Tnh cht c hc lng t ca cc in t t do trong in trng khng qu xa vi tnh cht mong mun v phng din c in mc d n b che m

29

6.4. T trng u bi s ph thuc vo gauge. nh hng ca t trng l mnh hn c bit l ln mt h 2 chiu. V mt c in, mt t trng khng c nh hng ln chuyn ng song song vi trng m n c ly dc theo trc z nh thng thng. Cc in t thc hin cc qu o trn trong mt phng vung gc vi ti tn s gc khng i

c gi l tn s xiclotron. Bn knh qu o c gi l bn knh xiclotron v c cho bi

trong v l ln (khng i) ca vn tc v E l ng nng. Mt c im quan trng l chu k chuyn ng khng ph thuc vo nng lng nhng bin t l vi bnh phng nng lng. iu ny ging vi dao ng t iu ho (phn 4.3) v kt qu c hc lng t phn nh iu ny. 6.4.1. Li gii trong gauge Landau i s n gin nht l trong gauge Landau m ch c mt thnh phn ca th vect. Ly Phng trnh Schrodinger (6.3) tr thnh

30

6.4. T trng u

T trng sinh ra 2 s hng: mt th t parabol m n c xu hng by cc hm sng theo x v mt o hm bc nht m n lin kt x v y v lm nh li lc Lorentz. S hng th hai ny l o. iu ny l quan trng v n ph v bt bin o thi gian (phn 5.3) da trn c s l hm Hamilton l thc. Ta cng cn phi o ngc t trng nu cn duy tr cng mt tnh cht trong s c mt ca n. Nh vy, nu l mt nghim ca phng trnh Schrodinger vi t trng B, bt bin o thi thi gian i hi cng cn phi l mt nghim. Th V(z) khng c mt i vi cc in t t do hoc n c th l mt h th by cc in t theo chiu ny to ra mt 2DEG. Th ny c thm vo sao cho chuyn ng dc theo z c th c a ra khi phng trnh v khng b nhiu lon bi t trng. Do , ta s b i z v gii phng trnh 2 chiu. Hm sng cui cng s c nhn vi hm z v cn thm vo nng lng tng ng.

31

6.4. T trng u Th vect khng ph thuc vo y v do , hm sng cn phi l mt tch ca mt sng phng ca y vi mt hm cha bit ca x, ngha lu(x)exp(iky). Khi thay n vo phng trnh Schrodinger (6.28) ta thy rng sng phng trit tiu c 2 v v iu chng t rng n l mt d on chnh xc. T ta c mt phng trnh ch i vi x nh sau

l phng trnh Schrodinger i vi mt dao ng t iu ho 1 chiu (phn 4.3) ging nh d on. Lu rng hm sng theo y khng cho ng gp thng thng vo nng lng v phi do s hng ny b st nhp vo trong th nng. Tn s xiclotron xut hin ging nh trong trng hp c in v s khc bit duy nht so vi nghin cu trc y ca chng ta v dao ng t l ch nh ca th parabol b dch chuyn ti Thang chiu di (xem (4.27)) c gi l chiu di t v c cho bi

32

6.4. T trng u N ch ph thuc vo trng m khng ph thuc vo khi lng ca ht. Gi tr thng thng l khi B = 1 T. By gi ta c th vit cc nng lng v hm sng i vi chuyn ng trong mt phng xy nh sau (khng ph thuc vo k), (6.32)

trong n = 1,2,3,... v l nhng a thc Hermite. Cc hm sng khng c chun ho nhng cc tha s c cho trong (4.35). Hm sng thp nht n gin l mt hm Gauss m phn b xc sut ca n c mt lch chun l Nghim ny c mt s tnh cht bt thng. Th nht l nng lng ch ph thuc vo n m khng ph thuc vo k. Cc trng thi c cng n nhng c k khc nhau l cc trng thi suy bin. Mt trng thi bin i t hng s i vi 2DEG thnh mt chui ca cc hm gi l cc mc Landau ti cc nng lng c cho bi (6.82).

33

6.4. T trng u Mt c im khc l cc hm sng to ra cc di song song theo hng y c tch u ra dc theo trc x, ngha l dc theo nhng ng ca N l mt s phc tp do khng c g c bit v bt k hng no trong mt phng xy v s phn bit gia x v y hon ton l do gauge c la chn cho th vect. Ngi ta cng c th chn mt gauge m cc di song song vi trc x hoc ging nh trong phn tip theo trong i xng trn c phc hi. im quan trng l ch tt c cc trng thi trong mt mc Landau u l suy bin (c cng nng lng) v do , bt k hn hp no ca chng cng l mt nghim ca phng trnh Schrodinger vi cng nng lng. Bng cch trn chng li vi nhau, ta c th bin i cc hm sng t mt h ny sang mt h khc. l mt cch th xem xt mt bin i gauge m ta s nghin cu chi tit hn phn 6.4.8. Thng khng c cc khe gia cc di hm sng. Cho mu l mt hnh ch nht vi cc kch thc Cc iu kin tun hon theo hng y (khng phi lun lun bo m trong mt t trng) cho iu kin thng thng trong j l mt s nguyn . S tch gia cc gi tr ln cn ca l S m rng ca mi

34

6.4. T trng u mt hm sng c t bng v do , t s ca tch trn m rng l By gi trong trng B = 1T v do r rng l trong bt k trng no trong cc mc Landau b phn tch v mt s ln cc hm sng ph ln nhau trong khng gian. Mc d t thang ca trng thi thp nht trong th parabol t t trng, cc trng thi cng cao m rng cng xa. Ta c th xc nh n m khng cn cc hm sng bng cch dng kt qu l cc gi tr k vng ca ng nng v th nng l bng nhau i vi mt dao ng t iu ho v mi mt gi tr bng 1/ 2 nng lng ton phn. Hm th nng l trong c o t cc tiu. Do ,

iu ny ph hp vi cong thc c in (6.27) i vi bn knh xiclotron ln ti tha s 2 m n khng gy ngc nhin bn cht khc ca cc trng thi. Sng phng gy ra n tng l cc trng thi mang dng ph thuc vo k v do ph thuc vo v tr Tuy nhin, ta cn phi thn trng v biu thc (6.5) i vi dng cha th vect cng nh o hm ca hm sng.

35

6.4. T trng u Vic thay th ch ra rng 2 s hng trit tiu v trng thi khng mang dng ton phn. Tuy nhin, c mt dng tun hon (dng chy theo vng trn) m n sinh ra mmen t. iu ny lin tng n kt qu c in, trong t trng lm cho cc in t quay theo qu o nhng khng mang dng ton phn. Mt cch khc chng minh cc in t khng mang dng ton phn l lu rng vn tc nhm ph thuc vo N trit tiu do s suy bin nhng l do ny khng cht ch do n b qua s dch chuyn dc theo x khi k thay i. Trc khi xem xt cc h qu ca cc mc Landau trong mt trng thi, ta s kho st ngn gn li gii khi dng mt th vect khc gn hn vi kt qu c in. 6.4.2. Nghim trong gauge i xng C th tm c mt nghim i xng hn nu ta chn gauge i xng biu din t trng. By gi cc ng ca to thnh cc vng trn gn gc v cc ta cc tr em li thun li cho i xng ny vi Ta vit li phng trnh Schrodinger

36

6.4. T trng u trong ta cc (phn 4.7) nh sau

Trong trng hp ny, c bt bin quay thay cho bt bin tnh tin dc theo y ging nh trong gauge Landau v ta c th ly phn bn knh ca hm sng l Vic thay th dn n mt phng trnh i vi hm bn knh v(r) nh sau

l phng trnh i vi mt dao ng t iu ho 2 chiu (phn 4.7) vi mt s hng b sung v nng lng v phi. Nghim ca phng trnh ny dn ti cc nng lng v hm sng sau (khng c s chun ho)

Cc hm l nhng a thc Laguerre lin kt.

37

6.4. T trng u Cc hm sng c cng dng tng qut ging nh trong gauge Landau vi s suy gim Gauss b bin iu bi mt a thc v chng li chy dc theo cc ng ca Th li tm y cc trng thi vi ln ra xa gc v do , chng chim gi cc vng hnh vnh khuyn thay cho cc a. Dng ca nng lng kh l v n tng theo l i vi l > 0 nhng n khng ph thuc vo l i vi l < 0. Tuy nhin, cc gi tr cho php ca cng nh th v cc mc li suy bin cao. Cc hm sng l n gin hn trong gauge Landau v n thng c s dng trong thc t tr cc bi ton nh cc chm lng t trong i xng quay l quan trng. Thc ra l gauge i xng khng i xng nh ngi ta c th thch do tt c cc trng thi u lun chuyn gn cng mt im trong khng gian khng ging nh trng hp c in m , cc qu o c th bt k ch no. Gauge ny bo ton i xng quay ca khng gian nhng gy hao tn i xng tnh tin ca n. 6.4.3. Spin ca in t Cho n by gi, spin ca in t b b qua ngoi tr thc t l n lm gp i s trng thi sn c. iu ny khng c chp nhn trong t trng do

38

6.4. T trng u c mt mmen t gn vi spin. Spin c th ln hoc xung v do , mmen song song hoc phn song song vi trng. Cc cu hnh ny c cc nng lng khc nhau nh vo mmen trong l manheton Bohr. Tha s 1/ 2 xut hin do spin mang xung lng gc khng ging nh cc bi s nguyn ca i vi chuyn ng theo qu o. Tha s g gn vi 2 i vi spin ca cc in t t do tri vi gi tr 1 i vi chuyn ng theo qu o v vic d on gi tr chnh xc ca n l mt thnh cng ln ca in ng lc hc lng t. Tuy nhin, g c th ly cc gi tr rt khc nhau trong mt cht rn do cu trc vng c bit l lin kt spin qu o. N l m i vi GaAs khi (g = - 0,44) nhng n l dng i vi v gi tr nm gia chng trong mt d cu trc. Nhng phc tp hn na bao gm s tng cng trao i m n c th lm tng g mt cch ng k nh mt hm ca t trng v mt ht ti v xut pht t s tng tc ln nhau ca chng. Thng th cc mc Landau cha c 2 trng thi spin trong cc trng nh, tch mt phn thnh hai khi trng tng v cc mc gn vi 2 spin tr thnh

39

6.4. T trng u hon ton tch ri ti cc trng cao nht (hnh 6.10). Trong Si c mt s tch b sung ca cc mc Landau m n sinh ra do cc in t trong 2DEG chim 2 thung lng [001] nhng n khng c mt trong GaAs. 6.4.4. Cc mc Landau C th iu quan trng nht ca nghim phng trnh Schrodinger trong t trng l ph nng lng. Mt trng thi lin tc ca cc in t t do c thay th bi mt chain ca cc hm Mi mt trong cc mc Landau ny cha mt s ln cc trng thi suy bin. By gi ta s kho st cc h qu ca ph ny. Trc tin, ta cn bit s trng thi trong mi mt mc. Gi thit rng h l mt hnh ch nht c cc kch thc v dng gauge Landau. Cc iu kin bin tun hon dc theo y cho iu kin thng thng trong j l mt s nguyn. Gii hn t xy ra do cc hm sng tp trung vo i hi phi trong mu c ngha l

40

6.4. T trng u Nh vy, s trng thi cho php trong mi mt mc Landau ng vi mt n v din tch l Thng qui c khng bao hm h s 2 i vi spin do cc spin ln v xung khng phi l suy bin trong t trng. iu ny c th c vit theo nhiu cch minh ha. Lng t ca t thng c nh ngha bi (mc d h/ 2e thng dng trong siu dn). S trng thi ng vi mt n v din tch khi c th vit thnh hoc trong mu c din tch A vit thnh trong l tng t thng tc dng. Khi , c mt trng thi (ng vi mt spin) trong mi mt mc Landau i vi mi mt lng t t thng i qua mu. Di y, mi mt trng thi trong mt mc Landau chim mt din tch h/ (eB) = Mt h thc khc rt ra t vic a vo nng lng xiclotron m n cho

Nh vy, mi mt mc Landau (tnh n c 2 spin) cha cng mt s trng thi ging nh vng 2 chiu gc trn mt khong l V cng l s tch gia cc mc Landau, n ch ra mt kt qu quan trng l t trng

41

6.4. T trng u khng lm thay i mt trng thi khi ly trung bnh qua nhiu mc Landau. Mi mt khi c b rng c chp li vo trong mt hm r nt ch trong mt h l tng trong , cc in t khng bao gi b tn x bi cc in t khc hoc bi cc tp cht, cc phonon v cc tng t. iu thc t hn l gi thit rng mt in t thng sng trong mt thi gian hu hn gia cc s kin tn x. Khi , nng lng ch c th c nh ngha trong phm vi chnh xc ging nh i vi trng thi chun lin kt trong mt cu trc xuyn hm cng hng. Cc mc Landau i hi mt b rng m n c th c nh ngha mt cch chnh xc ging nh lch chun hoc b rng y ti na cc i. Hnh dng chnh xc ca cc mc Landau trong cc h thc cn l vn tranh ci. Cc gi thuyt chung l mt bin dng Gauss hoc Lorentz. Trong trng hp ny hoc trng hp khc, ta khng k vng nhn thy cc thay i mnh trong mt trng thi tr khi s tch ca cc mc Landau vt qu b rng ca chng: (hnh 6.7(b) v (c)). iu ny cng c th c vit thnh m n c ngha l mt in t cn phi sng trong t nht mt qu o hon chnh trong t trng trc khi mt trng thi b tch ra.Bin

42

6.4. T trng u ny tch cc nh hng bn c in nhn thy cc t trng nh ra khi cc nh hng c hc lng t nh vo cc mc Landau cc t trng ln. Thi gian xut hin trong cc biu thc ny gi l thi gian sng ht n hay thi gian sng lng t. N khng ging nh thi gian sng vn chuyn xut hin trong linh ng S khc nhau l ch tt c cc va chm u c ng gp nh nhau vo nhng cc ng gp vo ph thuc vo s thay i hng xy ra. Khi t trng tng ln t khng, s tch gia cc mc Landau tng v do Hnh 6.7. Mt trng thi trong ttrng khi b qua s tch spin. (a)Cctrng thi trong mi mt khongb p vo trong mt mc Landau dnghm (b) Cc mc Landau c brng khc khng trong mt bc tranh thc t hn v ph nhau nu (c) Cc mc tr nn r rng khi

43

6.4. T trng u s trng thi gi mi mt mc cng tng nh ch ra trn hnh 6.8. Hu ht cc th nghim c thc hin vi mt in t c gi khng i v do s mc Landau b chim gi cn phi thay i. S mc Landau b chim hay h s lp y c th c m t theo nhiu cch

nh ngha ny tnh 2 spin nh cc mc tch ri m n l chung nhng khng ph qut. Ni chung, h s lm y khng phi l mt s nguyn v ti nhit khng, c n mc Landau y trong n l s nguyn ln nht di vHnh 6.8. S chim gi ca cc mcLandau trong t trng khi b qua stch spin.N ch ra mc Fermi chuynng nh th no gi mt int khng i. Cc trng l trong t s 2:3:4 v cho v 2.

44

6.4. T trng u v mc nh s b lm y mt phn. Vic tng B hn na lm cho cc mc Landau tng v nng lng v s trng thi trong mi mt mc Landau tng. Do , mt t in t hn chim mc nh. Mc nh tr nn trng khi ti trng

trong chnh xc c n mc Landau y . Sau , n gim i 1 v mc tip theo bt u lm trng. iu ny c minh ho trn hnh 6.8 i vi 8/3 v 2 trong s tch spin b b qua sao cho mi mt mc Landau cha in t. Cui cng th tt c cc in t nm trn mc Landau thp nht khi v n c gi l gii hn lng t t. N l mt mc tch spin v do , tt c cc in t c cc spin ca chng nm theo cng mt hng. iu ny ngc vi mt kh in t khng trong t trng m trong , s spin ln v s spin xung l bng nhau. Mc Fermi chuyn ng vi mt trng thi gi s in t khng i. Ti cc trng m chnh xc l mt s nguyn n (hoc mt

45

6.4. T trng u s nguyn chn khi khng c s tch spin), nm trong vng trng gi cc mc Landau v bng gi tr ca n trc khi trng tc dng. N chuyn ng ra xa gi tr ny sm nh khi trng thay i. Gi s rng t trng nm trong mt khong sao cho , ngha l C n mc Landau lm y hon ton vi mc n + 1 lm y mt phn cho mt in t chnh xc. Nh vy, mc Fermi nm trn mc n + 1 v ti bn trong phm vi b rng S tng B lm cho tng tuyn tnh cho n khi trng t ti m ti im , mc n + 1 tr nn trng v mc Fermi ri tr li mc Landau di. Chuyn ng ca c minh ho trn hnh 6.9 i vi mt h l tng vi cc mc Landau dng hm khi b qua s tch spin. iu ny c th o c bng thc nghim nh mt th tip xc gia mt 2DEG v mt cng kim loi trn b mt ca mt d cu trc. Cc bc nhy r nt c lm trn nu cc mc c m rng hoc nhit tng. R rng l ta cn quan st thy cc mc Landau c phn gii tt. Mt kt lun quan trng l kh a mc Fermi vo trong cc khe gia cc

46

6.4. T trng uHnh 6.9. S thay i ca mc Fermi nh mt hm ca t trng i vi mt 2DEGtrong GaAs vi meV trc khi tcdng trng. S tch spin c b qua. Qut ca cc ng mnh ch ra cc mc Landautrong khi ng m gin on l

mc Landau. Mt trng thi (hnh 6.7(c)) c v th theo E nh mt bin s c lp v d dng hnh dung rng c th c nh v ty . Vic gi khng i gn cht vo cc mc Landau v iu cn tr vic a vo trong mt khe. V tr ca c nh hng nh tnh ln tnh cht in t ca mt 2DEG. Nu nm trong mt mc Landau, mt trng thi ti mc Fermi l cao v mt s thay i nh ca nng lng gy ra mt s thay i ln v mt . H c gi l h chu nn. Tnh cht ngc li xy ra nu nm trong mt khe. By gi, mt s thay i nh ca nng lng khng c nh hng

47

6.4. T trng u g c n mt ca 2DEG do mt trng thi ti mc Fermi bng khng v h c gi l h khng chu nn. S phn bit ny l quan trng trong cc l thuyt v hiu ng Hall lng t. 6.4.5. Hiu ng Shubnikov de Haas Hnh 6.8 chng t rng mt trng thi ti mc Fermi thay i nh mt hm ca B. N gim ti 0 khi (gi thit rng c cc khe r rt gia 2 mc Landau) v c cc cc i khi iu ny c phn nh trong nhiu i lng c th quan st thy. i lng o c n gin nht trong s l in tr dc v mt php o in hnh c ch ra trn hnh 6.10. in tr sut l khng i nhng t trng nh nhng pht trin cc dao ng vi cc im khng ti nhng t trng ln. l hiu ng Shubnikov de Haas. N l mt du hiu ni bt v tnh cht c hc lng t ca cc in t trong t trng. Cc cc tiu xy ra ti cc t trng khi (xem (6.42)). iu ny l do dn dc xy ra ti mc Fermi (phn 5.4) v do mt i khi mt trng thi tin ti 0 khi Trong nhng iu kin , v do cng trit tiu. Nh vy, mt th ca n theo cc gi tr o c ca

48

6.4. T trng uHnh 6.10. in tr sut Hall dc v ngang ca mt 2DEG c mt nh mt hm cat trng. Cc php o c thc hin tiT = 1,13 K. S lng vo ch ra chiacho n v lng t ca dn nhmt hm ca h s lp y cho mt ng thng c dc l i qua gc. Gin qut ny l mt cch o chun i vi mt v cng cho mt cch chng minh r rng l h l 2 chiu. Mt la chn khc l dn in c cu trc tun hon khi xt nh mt hm ca 1/ B v n c th c lm r bng cch tnh ph cng sut ca n. Mt nh ln sinh ra do hiu ng Shubnikov de Haas v v tr ca n cho iu ny cng c th chng minh s chim gi ca cc vng con in cao hn t chuyn ng theo z. Mt in t s l khc v gy ra mt chu k r rt theo 1/B. Phn tch khc ca cc dao ng Shubnikov de Haas cho nhit v thi

49

6.4. T trng u gian sng ca 2DEG. Ci sau i khi c biu din nh l mt linh ng lng t Mu ch ra c H s 20 gia chng nhn mnh tm quan trng ca vic dng thi gian sng chnh xc. Tin trin ca s tch spin l r rng trn hnh 6.10. Cc mc Landau b suy bin kp m n cha c 2 spin i vi B < 2T. C mt nh kp ti gn 2 T khi cc tiu tch spin ti bt u xut hin v gim gn ti 0 khi gn 3,5 T. C mt cc tiu rt mnh gia 10 v 11 T khi Tt c cc in t nm mc Landau thp nht. N l gii hn lng t t i vi cc trng ln. 6.4.6. Hiu ng Hall lng t nguyn Hiu ng Hall l cch o chun i vi nng ht ti trong mt bn dn v ta nghin cu n i vi cc t trng nh. H.6.10 ch ra tnh cht ng ch ca n trong cc trng ln. in tr ngang ti cc trng nh nh k vng (lu rng b i dng hnh hc). Ti cc h s lp y nguyn m n cho N c cng mt

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6.4. T trng u dng ca dn lng t ho ging nh tm c trong h chun 1 chiu (phn 5.7.1). N ng khi mc Fermi nm gia cc mc Landau nhng ta thy ngay rng iu kin ny t c cc kh. Do , l gii ny khng gip chng ta gii thch c im quan trng nht ca cc kt qu thc nghim l c nhng on bng m rng ti cc gi tr tng ng ca in tr Hall ph hp vi cc cc tiu trong in tr dc. l hiu ng Hall lng t nguyn v s lng t ho dng nh l chnh xc trong phm vi chnh xc thc nghim. Mt l thuyt c a ra trong phn 6.6. S lng vo trong hnh 6.6 ch ra cc i lng khng th nguyn c v th theo V mt c in, n l mt ng thng c dc n v. Mc d n ng v trung bnh, hiu ng Hall lng t dn ti cc on bng khi l nguyn. Cc s nguyn l tng ng vi cc mc tch spin m cc on bng ca n mt dn nhanh hn khi trng gim i. Cc i lng khc ph thuc vo mt trng thi ti mc Fermi cng ch ra tnh cht dao ng trong mt t trng. Mt v d l cm t trong cc dao ng c gi l hiu ng de Haas van Alphen. l my d cc k quan trng i vi b mt Fermi trong cc kim loi. Mt v d khc l

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6.4. T trng u ng gp ca in t vo nhit dung. C hai u kh o trong cc h thp chiu do chng b che lp bi cc ng gp t cht nn. Trong trng hp 3 chiu, ta cn bao hm chuyn ng t do dc theo z song song vi Cc mc Landau tr thnh cc y ca cc vng con 1 chiu vi s phn k trong mt trng thi thay v cc hm Cc hiu ng tng t c quan st thy khi t trng thay i v n dn ti nhng thay i ca mt trng thi ti mc Fermi nhng chng km gy n tng hn nhiu. Trong phn 6.5 ta s thy nhng thay i khi cc in t b giam cm tip tr thnh h chun 1 chiu. 6.4.7. in trng v t trng ngang Chuyn ng c in ca mt ht tch in trong in trng v t trng (ngang) vung gc c 2 thnh phn. C mt chuyn ng ko theo n nh vi vn tc ti cc gc vung ti c 2 trng vi cc qu o xiclotron chng chp c tn s Mt in t thot khi s dng s thc hin mt xicloit

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6.4. T trng u l i vi dc theo z v dc theo x m n lm khp tt nht vi cc tnh ton ca chng ta trong gauge Landau. Vn tc ko theo dc theo y. Ht lc u c gia tc dc theo x bi in trng nhng chuyn ng ny b b cong bi t trng cho mt s ko theo ton phn dc theo y cc gc vung ti c 2 trng. iu ny c ch ra trong v d (iii) ca hnh 6.11 m n c quay ti mt s nh hng thun li hn. Cc qu o khc bt u t cc iu kin ban u khc. Bi ton cng n gin trong c hc lng t khi xy dng nh vo cc kt qu trc ca chng ta. Trng thm th nng eFx vo phng trnh Schrodinger (6.28). Chuyn ng dc theo z li c th c ngt lin kt v Hnh 6.11. Chuyn ng c in ca mt ht tchin trong in trng v t trng ngang vi vung gc vi trang giy v cc khong thi gianbng nhau gia cc k hiu. Cc ng cong tng ng vi cc vn tc v nng lng ban ukhc nhau vi (iii) ch ra xicloit i vi mt ht lcu ng yn.

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6.4. T trng u loi b v c th nhn c mt sng phng i vi y. Phng trnh Schrodinger thu c i vi u(x) l ging vi (6.30) nhng vi th nng b sung. N l tuyn tnh v do n gin l n dch chuyn nh ca th parabol cho

nh parabol b dch chuyn ti

H hm sng (6.33) b dch cng bi in trng v s dch chuyn ca parabol lm thay i cc nng lng ti

S suy bin i vi k by gi b ph v: c mt ng gp t th tnh in ph thuc vo s dch chuyn ca hm sng v mt ng nng b sung

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6.4. T trng u t vn tc ko theo. N ging vi kt qu c in ngoi tr iu l nng lng ca cc dao ng xiclotron b lng t ho. S dch ca parabol bi in trng lm cho cc trng thi c c mt vn tc dc theo y. li l hiu ng Hall. Mt dng t in t ng vi mt n v din tch l Do , do khng c dng dc theo x v m n l h s Hall c in. Lu rng tt c cc in t u c ng gp vo dng ch khng ch cc in t gn mc Fermi nh trong dn dc. 6.4.8. Bt bin gauge i vi c in trng v t trng, ta thy rng c th tm c cc nghim rt khc nhau i vi cng iu kin vt l bng cch la chn cc gauge khc nhau. l cc th vect v th v hng khc nhau. Ta rt ra dng tng qut i vi mt php bin i gauge ca cc th trong (6.2). Ta thm vo mt s thay i tng ng ca hm sng i vi ht c in tch q Ton t trong phng trnh Schrodinger em li mt s hng b sung

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6.4. T trng u m n loi b s thay i trong trong lc o hm thi gian sinh ra m n loi b s thay i trong Nh vy, hm sng mi l mt nghim ca phng trnh Schrodinger vi cc th bin i. S loi b ny l r rng hn nu th v hng c ly theo v phi vit phng trnh Schrodinger di dng

Trc y, ta thy rng v tri c th c gii thch nh l ng nng trong mt trng u v v phi l hiu ca nng lng ton phn v th nng. Do , nhm ny xut hin mt cch t nhin. Bt bin gauge i hi rng cc kt qu vt l khng b nh hng bi mt php bin i gauge nh vy. Mt khng b nh hng do n ph thuc vo m t n h s pha trit tiu do php bin i gauge. Dng cng khng thay i do nhng thay i hy b gia hai phn ca Cc mc nng lng trong mt t trng cng gi khng i nh xem xt trong phn 6.4.2. Mt khc, trong phn 6.2 ta thy rng khng th nh ngha cc mc nng lng trong mt in trng vi mt th vect v do , cc mc nng lng trong mt th v hng ph thuc vo la chn ring ca

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6.4. T trng u gauge v khng th t chng c th c quan st thy. lm v d ca php bin i gauge n gin, xt mt tnh in trng u. N c m t bi hoc Hm chuyn mt th vect thun tu thnh mt th v hng thun tu. Nh vy, ta c th nhn cc hm sng (6.18) trong mt th vect vi mt h s pha v chng tr thnh cc nghim ca phng trnh Schrodinger vi mt th v hng. Chng khng phi l cc tr ring do chng cng m t cc hm gia tc khc vi cc sng ng v c gi l cc hm Houston. Bt bin gauge c th xut hin nh l mt ci g ca s mt mt. thng l trng hp trong thc t m trong d dng thc hin cc php gn ng ch ng trong mt gauge ring v dn ti cc kt qu c hi. l mt bc tranh khng ng v n che lp ngha su sc gn vi i xng gauge l t do thc hin cc php bin i trong cc trng v cc hm sng m chng cn gi nguyn tnh cht vt l. V d nh s c mt ca in tch q trong hm m khi bin i hm sng dn ti s bo ton in tch. Trong vt l ht c bn, dng ca cc l thuyt nh sc ng hc lng t (QCD)

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6.4. T trng u c thit lp bi i hi l chng phi l bt bin i vi cc php bin i hm sng m chng l nhng s tng qut ho ca (6.48). iu ny c ngha l cc trng tng t vi cc th vect v th v hng cn phi c a vo lin kt vi cc ht. Khi xem xt tr li cc bn dn, hiu ng Hall lng t ho lc u c gii thch bng cch dng bt bin gauge. N khng thch hp i su hn na vo bt bin gauge y nhng n gin l cn lu rng n c xem nh mt i xng quan trng ca t nhin v khng n gin nh mt s phin toi. 6.4.9. Hiu ng Aharonov Bohm Hm Hamilton cha cc th thay cho chnh cc trng v iu thng c xem nh mt s kh chu. Tuy nhin, n c mt h qu ng kinh ngc l c th xy ra mt iu l cc in t b nh hng bi mt th vect cho d chng khng bao gi cm nhn c chnh t trng. iu ny dn n hiu ng Aharonov Bohm. Xt mt giao thoa k nh trn hnh 6.12(a). Cc in t i vo t mt ng dn sng bn tri v cha thnh 2 sng c bin bng nhau m mi sng i gn mt nhnh vng. Chng ti hp, giao thoa vi nhau bn phi v thot ra qua ng dn sng khc.

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6.4. T trng uHnh 6.12. (a) S b tr ca mt giao thoa k chng minh hiu ng Aharonov-Bohm vi mt xolenoit nh mang t thng khp kn gia 2 nhnh. (b) S truyn qua gn mt phn chm nh mt hm ca t trng m n ch ra cc dao dng do hiu ng Aharonov-Bohm. By gi gi s rng t trng c sinh ra trong mt xolenoit nh t hon ton bn trong vng sao cho khng c trng i qua cc ng dn sng. Mc d bn ngoi xolenoit, c mt th vect chy trong cc vng xung quanh ln ca n ti bn knh r c cho bi nh l Stokes

trong tch phn th nht l tch phn theo vng kn, cc tch phn th hai v th ba l cc tch phn theo din tch v l t thng bn trong xolenoit. Nh vy,

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6.4. T trng u Tip theo xt nh hng ca ln in t chuyn ng trn pha ny hoc pha kia ca xolenoit. Nng lng ca n v do xung lng c ca n l khng b nh hng nhng xung lng ny c cho bi v n l xut hin trong pha ca hm sng.Trong nhnh nh, v ngc hng v do tng ln v pha ca in t thay i nhanh hn. v cng hng trn y v do , gim i. Th vect do sinh ra mt s thay i pha gia 2 ng i cho d cc in t khng qua t trng. S thay i pha nh hng n s giao thoa chuyn tip bn phi v do nh hng n h s truyn qua ca dn thit b. S chnh lch pha c cho bi

Tch phn cui l tch phn theo vng. Nh vy, trong l lng t t thng. S giao thoa do l tun hon trong s cc lng t t thng i qua vng. N tng khi l mt bi s ca v gim na ng. l hiu ng Aharonov- Bohm. Trong cc ng dn sng 1 chiu l tng, iu ny lm gim

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6.4. T trng u h s truyn qua v dn ti khng m n cho Cc thit b thc xa vi thit b l tng nhng ch ra s biu iu r rt ca dn nh ch ra trn hnh 6.12(b). Cc th nghim nh th khng phi l mt s kim tra nghim ngt i vi hiu ng Aharonov Bohm do t trng xuyn su vo cc nhnh ca giao thoa k ch khng ch din tch b khp kn bi chng. N dn ti cu trc b sung ti cc t trng ln hn nhng t thng khp kn chi phi ti cc trng nh. Hiu ng Aharonov Bohm c xc nhn bi thc nghim c tin hnh trong cc knh hin vi in t trong nhng iu kin nghim ngt hn.6.5. T trng trong mt knh hp Tn ti nhng nh hng khc khi tc dng t trng vo cc in t trong mt knh hp hoc dy lng t. Ta bt u vi mt 2DEG trong mt phng xy v khi giam cm cc in t theo phng x v chng chuyn ng t do dc theo y. Phng trnh Schrodinger (6.30) trong gauge Landau t c th giam cm b sung V(x) tr thnh

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6.5. T trng trong mt knh hp vi Th t c th c vit li thnh m n gim ti ng nng thng thng i vi chuyn ng dc theo y khi B = 0. Bi ton ny c th gii c mt cch gii tch i vi mt th giam cm parabol nhng mt h vung gc su v hn ch ra ngha vt l r rng hn. Cho V(x) = 0 i vi vi mt th v hn bn ngoi. i vi trng khng, cc nng lng l u tie xt k = 0 v trong trng hp ny, th ca t trng l mt parabol xoay quanh gc. Mt t trng yu (hnh 6.13(a)) lm tng nng lng ca trng thi v lm mo hm sng mt t nhng cc thnh cn cho s giam cm

Hnh 6.13. Th nng v trng thi ringthp nht trong t trng i vi mtin t vi s sng k trong mt dy c thnh cng bao quanh vi b rng0,1 m trong GaAs.

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6.5. T trng trong mt knh hp chnh. S tng trng lm tng th ca t trng cho n khi n by cc in t mnh hn. By gi hm sng gim gn ti khng trong parabol trc khi n gp thnh (hnh 6.13(b)). Trong gii hn ny, nng lng gn vi gi tr ca n i vi mt h 2 chiu khng giam cm trong t trng

By gi tng k m n dch chuyn nh parabol ti N thm vo nng lng khi B = 0. Ti cc trng ln khi cc in t b giam cm ti k = 0 bi th ca t trng thay cho cc tng cng, mt s dch parabol dc theo x n gin lm dch chuyn hm sng v lm cho nng lng ca n khng b nh hng. iu ny ging nh trong mc Landau ca mt h khng b giam cm v gi cho n khi hm sng va chm vi thnh ca dy. i vi k ln, nh parabol t hon ton nm ngoi dy ging nh trn hnh 6.13(c). By gi in t trong h th tam gic th, nng lng ca n tng v s suy bin mt i. Cc trng thi p ln cnh ca dy vi nng lng cao hn so vi cc trng thi gia v chng c gi l cc trng thi bin. Chng c vai tr rt quan trng i vi s vn chuyn trong mt t trng. Mt cch minh ho

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6.5. T trng trong mt knh hp s xut hin ca chng l v th nng lng ca chng theo v tr trung bnh ca mt hm sng N c ch ra trn hnh 6.14. Ti cc t trng ln, nng lng gn nh khng i khi nm trong dy nhng tng nhanh khi n gn ti bin. Trong s trng hp, ch cc trng thi mc Fermi mi l cc trng thi bin. Dng iu c in ca cc in t gn bin ca mt dy c thnh cng bao quanh c ch ra trn hnh 6.14(c). Cc in t su bn trong dy c cc qu o xiclotron trn khng c s ko theo ton phn. Tuy nhin gn bin, Hnh 6.14. Cc nng lngca cc in t trong mt dy thnh cng trong t trng (a) B = 2T v (b)B = 5T c v th i vi tm dnsng r rng, ch 3 vng thp nht c ch ra i vi B = 2T. Cc chm biu din mt s cc trng thi b chim v ng t nt l mcFermi. (c) Cc qi o nhy c in dc theo cnh dy.

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6.5. T trng trong mt knh hp cc qu o b gin on khi in t va chm vi bin. Kt qu l mt qu o nhy m n nhy dc theo bin. in t do c mt vn tc ko theo ton phn m n ln hn i vi cc qu o gn bin hn. Cc trng thi trn cc bin i din chuyn ng theo cc hng ngc li. N c th c gii thch theo lc Lorentz theo quan im c in v th giam cm tc ng ging nh mt in trng theo hng ngc li trn hai bin. N cng l mt h qu ca s ph v bt bin o thi gian trong t trng. Khi B = 0, i xng ny i hi rng cc trng thi i v pha trc v cc trng thi i v pha sau chim cng mt vng khng gian ging nh trong tip xc im lng t (phn 5.7.1). Cc trng thi c th tch ra khi N trit s tn x gia chng v l mt thnh phn khc ca hiu ng Hall lng t. Mt th giam cm parabol trong mt dy c th c gii chnh xc v l do r rng l th ca t trng cng l th parabol. Cc nng lng l

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6.5. T trng trong mt knh hp T trng lm tng cc mc nng lng ti k = 0 nhng lm bt parabol i vi mi mt vng con v iu lm tng mt trng thi v lm gim nng lng ca cc trng thi vi k ln. V mt nh tnh, hin tng vt l tng t nh i vi dy c thnh cng bao quanh. Ti cc trng ln, cc trng thi b giam cm bi th ca t trng vi cc nng lng ging nh trong mt 2DEG khng b giam cm. Do , ta hi vng thy hiu ng Shubnikov de Haas thng thng trong t trng c l vi mt s bin dng no nh vo cc trng thi bin. Tuy nhin, tnh cht s khc cc trng nh do ta chuyn t s giam cm bi t trng ti s giam cm bi mt dy dc theo x. Ni ring, s dao ng t c cho bi s vng con b chim gi khi B = 0 khc vi mt chui v hn i vi mt 2DEG l tng m khng c s tn x. iu ny c th c ch ra bi mt th qut ca n theo (nghch o ca t trng ti cc cc tiu ca Mt v d c ch ra trn hnh 6.15. Mu l mt dy b giam cm bi th hiu dch m Dy c m rng ti th hiu dch m nh hn v n cho mt ng thng nh k vng i vi mt h 2 chiu khng giam cm (xem (6.42)). cong tng ti cc trng nh ging nh

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6.5. T trng trong mt knh hpHnh 6.15. Gin qut ch ra s lm gim chim gi t trong mt dy to cng ti cc gi tr khc nhau ca th hiu dch iu khin b rng dy. dy c ch to hp hn v cc vng con in tr nn quan trng. Qu trnh ny c gi l s gim chim gi t ca cc vng con vi s thay i t in trng ti t trng v bn cht khi trng tng. 6.5.1. Chm lng t v siu mng Cc l lun tng t c th c s dng tm nh hng ca t trng ln cc in t trong chm lng t (QD) trong chng b giam cm theo ton b 3 chiu. Thng th cc in t b giam cht theo hng z (hng nui) nhng lan truyn xa hn trong mt phng xy. Hu ht cc QD c hnh trn

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6.5. T trng trong mt knh hp nhiu hoc t v do thun tin, ngi ta dng gauge i xng v cc kt qu ca phn 6.4.2. Ta li c th them th giam cm vo phng trnh Schrodinger (6.36). Trong trng hp ca mt thnh cng, cc nng lng ca cc trng thi vi l = 0 tng ln bi th t trng parabol v cui cng tr thnh cc mc Landau. Xung lng gc l tng t nh s sng k trong dy v th li tm p cc in t p vo bin ngoi ca QD c tr gip bi s hng m n lm gim nng lng i vi l m. Nh vy, cc trng thi bin tng theo mt hng u tin. Trng hp ca th giam cm parabol li c th c gii chnh xc. N l mt kt qu bit t lu v cc mc nng lng l

i vi n = 1,2,3,... v l = T trng c 2 nh hng ln ph. Th nht l tt c cc trng thi tng bi cng h s do th t thm vo s giam cm. S hng th hai l nng lng Zeeman ca mt lng cc trong trng. i vi cc in t lun chuyn trong QD v tc ng ging nh mt dng lun chuyn sinh ra mt lng cc t trong l manheton

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6.5. T trng trong mt knh hp Bohr. Nng lng ca lng cc ny trong t trng l m n gii thch s hng th hai trong (6.54). N ph v s suy bin ti B = 0 trong mc th n ca mt dao ng t iu ho gi n trng thi (phn 4.7.2). Ph c minh ha trn hnh 6.16 i vi meV. Cc th nghim chng minh cu trc ny. Cui cng, dng nh l chuyn ng trong t trng l n gin. Hnh 6.17 ch ra ph ng ch ca mt h tun hon l mt tinh th 2 chiu trong mt t trng mnh. Cc nng lng khi khng c trng c cho bi php gn ng lin kt cht nh l l phin bn 2 chiu ca m hnh n gin(xem (2.9)) m ta s dng nghin cu ng

Hnh 6.16. Cc mc nng lng trong t trngca mt chm GaAs vi th giam cm parabolm n cho meV.

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6.5. T trng trong mt knh hpHnh 6.17. Bm Hofstadter: cc mc nng lng ca mt tinh th 2 chiu trongt trng. Cc im c v th i vi cc nng lng cho php nh mt hm ca (phn phn s ca s lng t t thm nhp vo mi mt mng nv). Vng chim gi khi khng cmt t trng.

lc hc ca cc in t trong trng hp 1 chiu. Tt c cc nng lng trong khong l c php v n to thnh trc ngang ca hnh v. Trc thng ng l t trng o theo n v trong l t thng qua mt mng n v ca tinh th. Thang o l v cc chm en nh du cc nng lng c php i vi mt t trng. Cu trc c gi l bm Hofstadter. Phn bn tri thp hn ch ra y vng trong n l mt parabol kiu in t i vi cc t trng nh. Dy

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6.5. T trng trong mt knh hp thng thng ca cc mc Landau c th nhn thy l ra t gc v cu trc tng t i vi cc l trng xut hin nh vng. Cu trc tr nn phc tp hn xa im nh th nhng n xa s mt trt t. Cc khe v cc vng ln xut hin ti cc gi tr c bit ca trng nh vi cc khe nh hn to cc gi tr t s phc tp hn. C l c im ni bt nht l s t tng t ca cu trc: cu trc thang o ln c lp li cc thang o ngy cng nh trong hnh v. l mt v d kinh in ca mt fractal. Mi mt mc Landau pht trin vo trong mt cu trc kiu cy m n kt thc trn mt vng ti mt trng c bit m t cc h mi ca cc mc Landau pht ra. Mt c im khng tt ca kt qu p ny l ln ca t trng i hi. Mt tinh th thng thng c a < 1 nm m n i hi B > 1000T a mt phn quan trng ca mt lng t thng qua mi mt mng. Mt siu mng nhn to cho c hi tt hn vi Khi , c s m rng ca cc mc nng lng ging nh trong hiu ng Shubnikov de Haas. N i hi mt th tun hon tng i mnh m n dng nh l kh xa ci c th t c trong cc cu trc hin nay m khng cn ph hng

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6.5. T trng trong mt knh hp qung ng t do trung bnh. Tnh hung ny tng t vi tnh hung i vi quan st cc dao ng Bloch trong mt in trng. S thiu ht l khng ln v ta c th hi vng nhn thy xc nhn thc nghim i vi ph ng ch ny trong tng lai.6.6. Hiu ng Hall lng t Cc php o thc nghim i vi hiu ng Hall lng t (nguyn) c ch ra trn hnh 6.10. Cc c im chnh l cc on bng trong in tr Hall ti trong lc ng thi in tr dc gn nh trit tiu. Hng s cu trc tinh t c th rt ra t hiu ng Hall lng t ging nh c v c l cc i lng c nh r. Gi tr ph hp vi cc php o khi dng cc k thut rt khc nhau cng nh cc tnh ton trong in ng lc hc lng t ti chnh xc l Hiu ng Hall lng t by gi c a vo nh mt tiu chun in tr vi nh ngha in tr sut dc trong cc on bng Hall thp hn so vi trong bt k vt liu no khc vi cc cht siu dn vi cc gi tr thp ti .

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6.6. Hiu ng Hall lng t Cc kt qu ny khng ph thuc vo vt liu v c chng minh trong cc thit b c ch to t Si, GaAs v cc cht bn dn khc. Chng cng chnh xc hn nhiu so vi nhiu php gn ng truyn thng c to ra trong l thuyt cu trc in t v vn chuyn nh php gn ng khi lng hiu dng v do , mt l thuyt tt cn phi vt xa nhng s n gin ho ny. Mt s l lun c xut gii thch hiu ng Hall lng t m n thay i t cc l thuyt trn c s bt bin gauge ti bc tranh trn c s cc trng thi bin m by gi ta s xem xt. N xy dng da trn mt l thuyt c t trc v s vn chuyn kt hp trong cc mu vi nhiu lead (phn 5.7.2). Xt thanh Hall trong mt t trng mnh nh ch ra trn hnh 6.18(a). Dng i gia cc my d 1 v 2 trong lc cc my d khc l cc my d th hiu v dch chuyn khng c dng. Gi s rng tn ti t trng sao cho mc Fermi nm gia cc mc Landau gia dy ging nh trong hnh 6.14(b). Trong trng hp ny, ch cc trng thi mc Fermi l cc trng thi bin. C N trng thi ny c cho bi s mc Landau b chim tm. Cc tp hp ca cc trng thi bin trn cc pha i din ca dy c tch ra v

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6.6. Hiu ng Hall lng tHnh 6.18. Mt thanh Hall trong mt t trng mnh mn ch ra s lan truyn ca cc trng thi bin. Mt thhiu dch m trn tip xc 1 bm cc in t ph vo trong N trng thi bin m chng ri b n (ch 2 trongs c v ra). Cc in t i ra qua my d dng khc (2). chuyn ng theo cc hng ngc nhau nh ch ra trn hnh 6.18(a). Cc trng thi bin ny ng vai tr mt phn ca cc mode xem xt trong phn 5.7. S nhn dng ny ca cc trng thi mang dng l iu ct yu. By gi tc dng mt th hiu dch m ti tip xc 1 sao cho dng qui c i t my d 2 ti 1. N nng mc Fermi ca cc in t ri b tip xc 1 mt khong l v do bm mt s in t ph vo trong cc trng thi bin m chng ri b tip xc ny nh ch ra bi cc dng ti trn hnh 6.18(a). Gi thit rng cc trng thi bin ny chy dc theo bin trn ca mu m khng b tn x v i vo my d th hiu 3. iu ny khng c php di chuyn dng ton phn v do , mc Fermi ca n cn phi tng bm mt dng nh nhau vo trong cc trng thi bin m chng ri b n.

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6.6. Hiu ng Hall lng t iu ny i hi Tng t, Nh vy, tt c cc tip xc dc theo bin nh dn ti cng mt th . Cng vy, tt c cc tip xc trn y cng mt th ging nh cc my d dng khc, do cc trng thi bin ca chng (c ch ra bng cc ng xm nht) khng mang dng ph. Mt dng c bm vo trong mi mt trong cc trng thi bin trn nh. N khc vi gi tr dng t trc bi h s 2 do n thch hp hn nghin cu cc spin ln v xung mt cch tch ri khi cc mc Landau c th b tch ra. Nh vy, dng ton phn v in tr Hall l mt gi tr lng t ho. in tr dc bng Nh vy, ta chng minh hiu ng Hall lng t. L lun c th c a ra mt cch chnh tc hn khi s dng phng trnh Landauer Buttiker. By gi ta xt k hn mt s ln cc gi thuyt trong bc tranh ny. iu c bn nht l cc trng thi bin khng b tn x. Tn x v pha trc xy ra nu mt in t b tn x t trng thi bin ny ti trng thi bin khc lan truyn dc theo cng mt bin. N khng c nh hng ln dng ton phn do in t cn i theo cng mt hng v h s truyn qua (ly tng theo tt

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6.6. Hiu ng Hall lng t c cc trng thi bin) do khng b nh hng. Thc nghim pht hin thy rng tn x nh th c th l rt yu vi qung ng t do trung bnh c hng chc m. S tn x vo trong cc trng thi bin i theo hng ngc li tuy nhin s ph hy s lng t ho v cn phi trnh gp. Cc trng thi bin nh vy trn pha khc ca mu v s tn x s cc k yu vi iu kin l mc Fermi gia cc mc Landau gia mu. Nh vy, s lng t ho y l mnh khng ging vi s lng t ho ca mt tip xc im khi khng c t trng (phn 5.7.1). L lun nghe hp dn ny c mt vn nghim trng. Trong phn 6.4.4 ta thy rng mc Fermi t iu chnh nm trong mc Landau i vi hu ht cc gi tr ca t trng trong khi v tnh hung vi mt dy m rng l tng t. Nh vy, l lun trc y ch ng i vi cc khong rt hp ca B v khng gii thch b rng ca cc on bng trn in tr Hall. Thnh phn ch yu khc c th gy ngc nhin l s mt trt t nh vo mt th hn n t cc tp cht hoc cc khuyt tt ti giao din. N lm cho nhiu trng thi trong cc mc Landau tr thnh b nh x m n c ngha l chng b gii hn ti mt vng nh ca mu.

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6.6. Hiu ng Hall lng t C mt s c ch dn n s nh x ph thuc vo bn cht ca th hn n. Mt th hn n phm vi hp l th thay i nhanh trong khng gian trn thang o v n dn ti s nh x Anderson c minh h trn hnh 6.19(a) i vi mt trng thi. Tm ca mi mt mc Landau cha cc trng thi m rng m chng lan truyn khp mu ging nh gi thit t trc nhng cc trng thi trong cc ui b nh x ging nh trong cc trng thi lin kt v khng c vai tr trong s dn. By gi ta i hi iu kin nghim ngt l mc Fermi nm gia cc mc Landau i vi hiu ng Hall lng t v ch c iu l n nm trong cc trng thi nh x. N gii thch s tn ti ca cc on bng nhng lm cho ngun gc ca s lng tHnh 6.19. (a) Mt trng thi ca mt mc Landau trong mt h hn n m n ch ra mt vng ca cc trng thi m rng tm ca mi mt mc vi cc trng thi nh x gia.(b) Cc trng thi bin nhx trong mt th thay i chm vi mt g bn tri v mt thung lng bn phi.

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6.6. Hiu ng Hall lng t ho b n hn do dn Hall ly gi tr mong mun nu tt c cc in t c ng gp trong khi by gi ta cho rng cc in t nh x khng tham gia. Cch gii quyt l cc trng thi m rng b tr cc trng thi nh x bng cch mang dng nhiu hn so vi chng c trong mt h sch. Cc tnh ton chi tit xc nhn tnh cht gy ngc nhin ny v ch ra rng cc in t c tng tc gn cc th tn x tng vn tc trung bnh ca chng. Tn ti bc tranh khc ca s hn n khi th hn n thay i chm trong khng gian. Trong trng hp ny, cc mc Landau b ko ln v xung v nng lng gi trn th hn n ti mi mt im. Trong trng hp ny, hu ht cc cc trng thi bin nm trong cc vng xung quanh cc g hoc cc thung lng trong th nh ch ra trn hnh 6.19(b). Cc trng thi nh vy nh x c hiu qu v khng th dn. Rt t trng thi i qua ton b mu ni vi cc my d v cc chuyn tip gia cc on bng Hall trong xy ra khi cc trng thi nh th nm mc Fermi. Cc l lun ch ra rng cc on bng Hall tt nht c nhn thy khng phi trong cc mu sch nht m l cc mu vi s hn n va phi. Ngi

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6.6. Hiu ng Hall lng t ta pht hin thy rng gn nh l ti u i vi GaAs v i vi Si. Cc chuyn tip khi ch c th c khong 5% b rng ca cc on bng. l mt c trng thch hp i vi o lng hc. Vic kho st k lng hn pht hin thy rng c nhiu bi ton hn b che du. V d nh th hiu Hall c phn b nh th no qua thit b ? Bin dng ca th m ta rt ra i vi thanh Hall trn hnh 6.18 b ngoi trng ging vi bin dng m ta rt ra trn phng din c in trong phn 6.3 t tenx dn. Tuy nhin, dng chy u qua mu trong bc tranh c in (hnh 6.6(b)) trong lc ta cho rng n c mang tt c bi cc trng thi bin. Hn na, bc tranh n gin ca cc trng thi bin khng tnh ng cc vng chu nn v vng khng chu nn (phn 6.4.4). Mt cu hi khc l dng Hall c b iu khin bi mt chnh lch th ho hc hoc th tnh in hay khng? Ta gi thit rng cc trng thi bin c iu khin bi cc th ho hc ca cc tip xc trong khi dng b iu khin bi mt in trng trong trng hp c in. Ta tnh dng Hall nh vo in trng v t trng ngang trong phn 6.4.7 v n li l in

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6.6. Hiu ng Hall lng t trng iu khin dng. dn Hall ti cc dng nh m phn ng l tuyn tnh khng ph thuc vo ranh gii ca th in ho gia cc thnh phn tnh in v ho hc ca n. Nhiu c c im ca hiu ng Hall lng t ang ch mt s hiu bit y . 6.6.1. Trng thi bin v ro Ta thy rng hiu ng Hall lng t nguyn c th c m t theo cc trng thi bin m chng tc ng nh cc mode nh r trong cch ni xuyn hm. Nhiu thc nghim c tin hnh trn cc trng thi ny v c phn tch theo phng trnh Landauer Buttiker (5.7). Mt v d c cho bi mu phc tp hn mt cht nh trn hnh 6.20. N c mt ro gia lm tng nng lng ca tt c cc trng thi sao cho ch M trng thi bin c th lan truyn khc vi N trng thi trong phn cn li ca mu. Gi thit

Hnh 6.20. Mt thanh Hall vi mt ro xuyn qua gia m n ch truyn M trong s N trng thi bin.

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6.6. Hiu ng Hall lng t rng khng c s tn x gia cc trng thi bin. Trong s N trng thi i ra khi my d 3, M trng thi i qua ro ti my d 4 trong lc N M trng thi b phn x v i vo my d 5 thay th. Nh vy, N M trng thi i vo my d 5 mang dng ph nh vo th hiu dch trong lc M trng thi khc m chng bt ngun t my d th khng phi nh vy. My d 5 l mt my d th hiu v do dng ny cn phi c lm cn bng bi cc trng thi ri khi 5 v i vo 1. Mt th hiu do pht trin nhng ln ca n nh hn nh ch ra bi mu xm trung bnh ca cc ng ri khi tip xc. Tip xc chia phn cc in t trong s cc trng thi bin v do n khng b ng v ton b cho d dng ton phn khng b nh hng. tip xc 4 cng xy ra s tng t. Ta c th vit cc h s truyn qua v cc phng trnh i vi cc dng khi s dng (5.108). Ging nh trong thanh Hall n gin, ta hi vng rng v v c th kim tra iu ny. Rt t cc h s truyn qua l khc khng Chng tun theo qui tc cng dng v ct (5.107). Ta cn bit in tr Hall in tr 4 u v in tr 2 u

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6.6. Hiu ng Hall lng t u tin dng i vo my d 3 c cho bi (5.108). N tr thnh v do , Nh vy, ta xc nhn lp lun trc y l cc my d dc theo nh ca thanh Hall dn ti cng mt th hiu nh my d dng bn tri (khng c ro). Tip theo, tnh in tr 4 u gia cc my d 4 v 3 c k hiu l trong cp ch s th nht cho cc my d dng v cp ch s th hai cho cc my d th hiu. Cc phng trnh i vi v l

Loi tr v tnh n cho

in tr dc l khc khng nh k vng t vic lng ro vo. N quay li khng nu b ro v M = N.

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6.6. Hiu ng Hall lng t Kt qu khc c th c rt ra t cng thc 4 my d chung (5.119). V d in tr hn hp

Tng v do , n ph hp vi cc kt qu c trc. 6.6.2. Hiu ng Hall lng t phn s Hnh 6.21 ch ra mt php o hiu ng Hall lng t phn s.

Hnh 6.21. in tr sut dc vin tr sut ngang ca mt 2DEG c linh ng cao ti 150 mK mn ch ra hiu ng Hall lng t phn s. H s lp y c ch ra v crt gn bi mt h s l 2,5 ti cc trng ln.

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6.6. Hiu ng Hall lng t Khng ging vi hiu ng Hall lng t nguyn i hi s mt trt t cho b rng ca cc on bng, hiu ng Hall lng t phn s ch quan st thy nhit thp trong cc mu vi linh ng rt cao. Cc c trng li l cc on bng trong gn vi cc cc tiu trong Hiu ng phn s xut hin ti cc h s lp y trong p, q l cc s nguyn vi q l. c bit l cc c tnh mnh xy ra khi Ngun gc vt l ca hiu ng Hall lng t phn s l rt khc vi ngun gc vt l ca hiu ng Hall lng t nguyn. Tt c cc in t l trong mc Landau thp nht khi v theo l thuyt trc y ca chng ta, tt c cc in t u c cng ng nng v cc spin ca chng nh hng song song. Hin tng vt l m ta b qua by gi c vai tr quan trng v nh thng thng, ta b qua s y Coulomb gia cc in t. Cc in t c th t sp xp trong mt cu hnh a thch c bit lm cc tiu nng lng ny ti cc gi tr ca m n gii thch mt s on quan st thy. Cc trng thi c lin quan mnh ny c cc tnh cht ng ch . Cc kch thch ca chng c th mang mt in tch phn s v d nh Nghin cu gn y tp trung vo xem xt vt l ca cc mc Landau lm

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6.6. Hiu ng Hall lng t y mt na. Khi quay tr li hnh 6.21, ta thy rng c v o theo hng ny hoc hng khc t ging nh cng cc i lng o t B = 0. N c gii thch theo cc fermion hn hp. l cc in t vi 2 ng thng lng gn vo. Mi mt ng thng lng ging nh mt xolenoit v cng b mang thng lng Cc fermion hn hp chuyn ng trong trng hiu dng m n trit tiu khi L thuyt v thc nghim ang c pht trin nhanh chng. Cui cng, mt hin tng khc xut hin ti cc h s lp y chng trong in tr dc tng nhanh chng. iu ny c tin tng l do s nh x ca cc in t khng phi bi s mt trt t m do s y Coulomb ca chng. N dn ti s to thnh mt tinh th Wigner trong cc in t trong mt mng thay v lang thang khp 2DEG. Mt ti 2DEG kt tinh khi B = 0 l cc thp v trong thc t, cc tnh cht b chi phi bi s hn n. Mt t trng mnh lm cho s kt tinh c a thch hn do nh cp trc y, ng nng b dp tt khi tt c cc in t chim mc Landau thp nht v tng tc in t - in t chi phi.

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6.6. Hiu ng Hall lng t Tho lun thng thng tp trung vo cc in t nhng nhiu thc nghim c tin hnh vi cc l trng. Chng c khi lng hiu dng cao hn m n lm gim ng nng v do lm tng vai tr ca tng tc in t - in t. Cc h khc c quan tm hin nay l cc lp xp cht ca cc in t v/ hoc cc l trng trong ngi ta quan st thy cc hiu ng mi do tng tc gia cc lp.

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Bi tp chng 66.1. Chng minh rng cc biu thc (6.4) v (6.5) i vi dng thch hp vi phng trnh lin tc bng cch lp li s bin i ca phn 1.4 vi phng trnh Schrodinger (6.3).6.2. Xc nh hiu ng Franz Keldysh ln s hp th gia cc VB v CB trong GaAs. Khi lng xut hin trong mt trng thi trong trng hp ny l khi lng hiu dng quang (phn 1.3.1) c cho bi trong c th l khi lng hiu dng i vi cc l trng nng v nh. Cn mt in trng c ln bng bao nhiu c mt ui c ngha trong bin hp th (v d nh 0,1 eV)?6.3. Cn mt thang o no ca in trng sinh ra s nh x Stark trong mt siu mng thng thng? Dng v d t phn 5.6. N c thc t khng?6.4. o hoc c phi l cch thng thng o dn v hiu ng Hall trong cc cht bn dn? Cc dng hnh hc thng thng c ch ra trn hnh 6.5.6.5. Xt mt thanh Hall c chiu di v b rng trong c dng chy qua v o cc th hiu v Xc nh v khi dng kt qu

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Bi tp chng 6 (6.22) t m hnh Drude trong t trng. Mt 2DEG c ch to thnh mt thanh Hall vi mm v mm. N cho vi Tm nng v linh ng ca 2DEG. Vic s dng cc cng thc trng thp c ng khng? 6.6. Rt ra phng trnh (6.24) i vi hiu ng Hall vi 2 knh ca cc in t trong cc trng yu v chng minh rng trong cc trng mnh. Cng chng minh rng c mt s hng bnh phng theo B ti cc trng yu. Cc kt qu ny thay i nh th no i vi mt knh ca cc in t song song vi mt knh ca cc l trng? Gi s rng knh 2 c mt gp 10 ln nhng c linh ng ch bng 1/100 so vi knh 1. C d dng ghi nhn c knh 2 bng cch dng hiu ng Hall hay khng? Tnh hung ny thng xy ra trong cc lp bin iu pha tp.6.7. Hy gii thch khi dng h thc gia v lm th no c th lm trit tiu ng thi cc thnh phn dc ca c dn v in tr sut trong cc t

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Bi tp chng 6 trng mnh?6.8. Ci g xy ra i vi s phn b dng v in trng trong mt vng Corbino (hnh 6.5(c)) khi tng t trng cho ?6.9. Chng minh rng mt dng ca mt in t trong t trng khi dng gauge Landau l

trong l hm sng dao ng t iu ho th n xoay quanh Chng minh rng khng c dng ton phn mc d sng chy trong hm sng (6.33) nhng c mt dng lun chuyn.6.10. Chng minh rng cc hm sng (khng chun ho) ca mc Landau thp nht trong gauge i xng c th c vit dng gn trong (khng hon ton l nh ngha thng thng theo cc s phc). Lu rng i vi mi C s ny c chng minh l c bit c ch trong nghin cu hiu ng Hall lng t phn s.6.11. Ci g xy ra i vi cc nng lng ca cc mc Landau bao hm s tch

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Bi tp chng 6 spin nu cc in t trong mt 2DEG c g = 2 nh th chng l t do6.12. Xc nhn mt in t i vi mu trn hnh 6.10 khi dng c hiu ng Hall ti cc trng yu v cc cc tiu ca hiu ng Shubnikov de Haas (gin qut) ti cc trng mnh. Cng rt ra dng b iu khin qua mu t cc on bng trong th hiu Hall.6.13. Dng s liu trn hnh 6.10 c tnh b rng ca cc mc Landau v thi gian sng n ht. Lm th no so snh n vi thi gian sng vn chuyn t linh ng?6.14. Chng minh rng khng c cc h qu vt l ca cc th vect v th v hng khng i do chng khng sinh ra trng. Chng lm g i vi nng lng, xung lng v cc hm sng ca cc in t t do?6.15. Ci g l ng kinh hiu dng ca phn chm dng trong th nghim Aharonov Bohm hnh 6.12?6.16. Xc nh cc mc nng lng ca mt dy vi s by parabol (phng trnh (6.53)) v v tr ca tm dn sng. N s l mt m hnh tt i vi cc dy hp vi mt t in t. Mt gi tr th l meV v do phm vi

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Bi tp chng 6 no ca trng c bao hm? By gi c 2 tn s v Ci no i vo hm sng v ci no i vo nng lng nh mt hm ca tm dn sng?6.17. c tnh b rng ca dy hp nht c ch ra trong gin qut trn hnh 6.15.6.18. Rt ra cc nng lng ca chm parabol trong t trng (phng trnh (6.54)).6.19. Gi s rng c 12 in t trong chm parabol m cc mc nng lng ca n c v th trn hnh 6.16. Lm th no nng lng ca trng thi lp y cao nht thay i nh mt hm ca t trng nu gi thit rng cc trng thi thp nht lun lun b chim gi? S tch spin c th c b qua v do mi mt mc l suy bin kp.6.20. Chng minh rng in tr 2 my d ca thanh Hall vi mt ro (phn 6.6.1) l ch ph thuc vo s trng thi truyn qua ro v in tr Hall cho tng s trng thi bin. 6.21. Rt ra phng trnh (6.59) i vi thanh Hall vi mt ro khi dng cc

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Bi tp chng 6 cng thc 4 my d Landauer Buttiker. N cn c p dng cho cc lead 1, 2, 4 ch tr 5 do ta bit rng cc lead 3 v 6 l khng l th. Cc h s truyn qua c cho bi (6.55) da trn t Chng minh rng nh thc v t

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d./2he.ArfBrFr.Ar)1.6.(,/ArotAcurlBdtAdgradFrrrrr=--=ff..rFrr-=f.Fr.FrAr.tFArr-=Ar.BrArfFr.BrtRFrr.=c)2.6.(/,dtdgradAAcffc-+rrFrBr()().2/1)0,,(2/1RBxyArrr=-=.ByAx-=BxAy=.//yAxABxzy-=cgradc0=crotgrad.Aqpr)r-.-hip)r()()[](){}()()[])3.6.(/,,,,2/12dttRditRtRqtRAqpmrhrrrr)ryyf=+-mFq/rFr()mAqp/r)r-()()())4.6.(,2,2**--=tRAmqimqtRJrrhrryyyyy())5.6.(2,**-+-=yyyymAqpmAqpqtRJr)rr)rrreFzq=fFr())7.6(,/,00-=-=eeeefeFzAizeFzAize()())6.6.(2222zzeFzdzdmeyy=+-h())8.6.(2,203/1203/120eFzmeFmeFz===hhex()eFz/e>.51-=MVmF()./eFze

e-()()2/12//1)(-=mEEnhp()zEnFD,)(1())11.6(,,02)(1--=eeFzECAizEnFD()()---)10.6.(,02)(1eedeedEeFzAizEnFDe())14.6(22,020)(1--=eeeFzEAimzEnFDh())13.6.(432cos~,2/3020)(1---peepeFzEeFzECzEnFD())12.6.(21~,1eFzEmzEnFED-hp[][]{}())15.6.(/,)()(22022'030203)(3eepeeeepeFzEssAissAimmdeFzAimmnEFD--=-=--=-hhdnmeV--1.eFzE-=e15-MVm()zEnF,)(gEE

DgE>21nnneff+=1>>twc())24.6.(,221122221222221222211mmmmmmmmmnnnnnnnnneffeff++=++=,LTss>>LTss>>cR)27.6(,2eBmEvRcc==w)26.6(meBc=wBrBr).0,,0(BxA=r()Bt--,*y()Bt,y()())28.6.()(212222222RERzVzeBxyixmrrhhhyy=+-+-+-)31.6.(eBmlcBhh==w)./(eBkxkh-=meBc/=w)30.6).(()(21222222xuxueBkxmdxdmcew=++-hhd.2/BlnH()())33.6(,exp2exp),(2201ikylxxlxxHyxBBknnk----fcnknweh-=21nmlB26()()./2//22yBykLleBLxpp==Dhkx(),/2jLkyp=.yxLL.Ar.kx()()()())34.6.(/,)2/1(2/12222cnncmxxmweew=D=D()ikyexpxDnmlB26./2/yBBkLllxp=DBl()(),2/122xmcDwBlBklx