28Academic Research in Educational Sciences VOLUME 2 | ISSUE 10 | 2021
ISSN: 2181-1385
Scientific Journal Impact Factor (SJIF) 2021: 5.723 Directory Indexing of International Research Journals-CiteFactor 2020-21: 0.89
DOI: 10.24412/2181-1385-2021-10-527-538
BIR ZARRACHALI SHREDINGER OPERATORI XOS QIYMATI UCHUN
ASSIMPTOTIK FORMULALAR
Yodgor Safarali o'g'li Xurramov
O'MU Jizzax filiali assistenti yxurramov@mail.ru
ANNOTATSIYA
Ushbu ishda bir o'lchamli panjarada tashqi maydon ta'siri bilan kontakt ta'sirlashuvchi bir kvant zarracha harakatini tavsiflovchi bir zarrachali gamiltonianga mos bir zarrachali diskret Shredinger operatori hM ning zarrachalar ta'sirlashuv energiyasi ^ > 0 ga bog'liq chapda yoki o'ngda yagona xos qiymatga ega ekanligi isbotlangan.
Ushbu xos qiymat z(u), ^ > 1 ning aniq ko'rinishi topilgan. O'zaro ta'sir parametri ^ > 1 ga monoton va uzluksiz bog'liqligi ko'rsatilib, moduli yetarlicha kichik ^ > 1 larda ushbu xos qiymatlar uchun yaqinlashuvchi yoyilmalar olingan .
Kalit so'zlar: panjara, operator, xos qiymat, spektr, muhim spektr, yoyilma.
ASSYMPTOTIC FORMULAS FOR EIGENVALUE OF ONE PARTICLE
SCHRODINGER OPERATOR
ABSTRACT
In this study, one particle interaction energy of one particle discrete Schrodinger operator hM corresponding to a quantum particle gamiltonian describing the motion of a quantum particle in contact with an external field effect in a one-dimensional grid depends on ^ > 0 left or right proved to have a eigenvalue.
This eigenvalue z(u), ^ > 1a clear view of was found. The monotonic and continuous dependence of the interaction parameter on ^ > 1 is shown, and approximate distributions are obtained for these eigenvalues at sufficiently small modulus ^ > 1.
Keywords: lattince, operator, eigenvalue, spectrum, essential spectrum
KIRISH
Tashqi maydon ta'siridagi bir kvant zarracha harakatini tavsiflovchi bir zarrachali gamiltonianlar hamda qisqa masofada ta'sirlashuvchi ikkita bir xil va har
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Academic Research in Educational Sciences VOLUME 2 | ISSUE 10 | 2021
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xil zarrachalar sistemasi gamiltonianlariga mos ikki zarrachali diskret Shredinger operatorlarining spektral xossalari Faria, Corolli, S.Albeverio, S.Laqayevlarning ishlarda o'rganilgan.
Ikki zarrachali diskret Shredinger operatori /i^ ( 0 ) bir zarrachali gamiltoniani ga unitar ekvivalent va shuning uchun bir zarrachali Shredinger operatorlarining spektral xossalarini o'rganish panjaradagi ko'p zarrachali Shredinger operatorlar spektral nazariyasida muhim o'rin tutadi.
ADABIYOTLAR TAHLILI VA METODOLOGIYA
Ushbu [S.Lakaev, I.Bozorov] ishda uch o'lchamli panjarada potensialli maydonda tashqi maydon ta'sirida kontakt va bir qadamda ta'sirlashuvga mos bir zarrachali gamiltonian qaralgan. Unga mos diskret Shredinger operatori xos qiymatlar sonining zarrachaning ta'sir energiyalari ß > 0 va X > 0 larga bog'liqligi to'la o'rganilgan.
B.Saymon va M.Klausning ishida uzluksiz Shredinger operatori xos qiymatlari uchun kichik larda da yaqinlashuvchi qator, da esa kichik
X > 0 larda asimptotik yoyilma olingan.
2. Bir zarrachali Shredinger operatorinig spektri
T = (—7T,7t] bir o'lchamli tor. L2 (T ) kvadrati bilan integrallanuvchi funksiyalar Gilbert fazosi. Lc2 ( T) c L 2 ( T ) toq funksiyalar qism fazosi bo'lsin.
Lc2 ( T ) da aniqlangan quyidagi operatorni qaraymiz:
hß = h0+ \w,
Bu yerda / 0 operator £ ( t) funksiyaga kopaytirish operatori, ya'ni
(/i0/)(t) = £(t)/(t); e(t) = 1 — cos2t;
ta'sir operatori (qo'zg'atish operatori) v operator esa integral operator bo'lib,
da quyidagicha aniqlangan:
1 f (v/) (t) =—5 in 2 t I 5 in 2 q ■/ (q) d 2n J
T
ß > 0 — zarrachaning tashqi muhit bilan ta'sir energiyasi.
Lemma 1. Yuqorida aniqlangan operator Lc2 ( T ) da o^z-o^ziga qo^shma, chegaralangan operator boladi.
Academic Research in Educational Sciences VOLUME 2 | ISSUE 10 | 2021
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Isbot. Hilbert fazosida oz-oziga qoshma operator tarifiga kora H operator V/,g e Huchun (4/, g) = (/Mg) shartni qanoatlantirishi kerak. (■,■) — skalyar kopaytma.
Oz-oziga qoshma operator xossasiga kora h0 va v operatorlarning har birining oz-oziga qoshma operator ekanligini ko'rsatish yetarli.
(h0/,^) = J(1 - cos2q)/(q)g ( q== J/(q)(1 - cos2q^ ( q=
T T
= J /(q)( 1 - CO 5 2 q )g (q ^q = (/, ho^);
T
(v/,g) = J(—sin2q J sin2t/(t)dt) ■g (q )dq =
T
T
= J/(i)(2^ 5in2 t J 5 i7i 2 q ■ g (q) dq)dt = (/, vg).
TT
Chegaralanganligi. Operator chegaralanganlik ta rifidan ^ operator uchun 3M > 0 soni mavjud bolib V/ e D(4) c H element uchun quyidagi
N(^/)N < MN/N;
tengsizlik o'rinli bo'lsa, ^ operator chegaralangan deyiladi.
2
T
||(V)(0||2 = J (l-cos2t)/(t)+^sm2tJsm2qr/(qr)dqr
T
<2 Jl(l - cos2t)/(t)|2dt + + 2 J J sm2q/(q)dq
T
T
T
dt <
dt
< 8 J|/(t)|2dt + 2 J ^ J sm2q/(q)dq
T
T
dt =
= 8 J|/(t)|2dt + ^2 J sm2q/(q)dq
TT
Koshi - Bunyakovskiy tengsizligiga ko'ra:
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z
Jsin2qf (q)dq < Jsin22qdq • J\f (q)\2dq
T TT
= 1J( 1~co s 2 q) d q^ J\f iq)\2d q =
T T
Demak, \\(hßf)(t)\\ < V 8 + n\\f\\. Lemma isbotlandi.A
v — chekli o' lchamli (bir o' lchamli) operator bo' lganidan uning kompakt ekanligiga kelamiz.
hß operator o'z-o'ziga qo'shma bo'lganligi uchun uning qoldiq spektri bosh toplam bolib, spektri haqiqiy sonlar toplamining qismidan iborat, ya'ni
a(hß) a R.
Quyidagi belgilashlarni kiritamiz:
£min = min s(q) = 0; £max = max e(q) = 2;
qET qET
Muhim spektr turg'unligi haqidagi Veyl teoremasiga ko'ra. hM operatorning muhim spektri oess(hM) berilgan ^ > 0 parametrlardan bog'liq emas va h0 operatorning spektri bilan ustma-ust tushadi. Shunday qilib
cress( hß) = Gess(h0) = G (h0^ = [£min> £max] = [0' 2]
tenglik o'rinli.
NATIJALAR VA MUHOKAMA
Teorema 1; ß> 1 bolsin h^ muhim spektrdan o^ngda yagona xos qiymatga ega va u quyidagi ko^rinishga ega
o -1)2
z(ji) = 2 +—--;
2/i
Bu xos qiymatga mos xos funksiyaning ko^rinishi.
2[i2Csin2q
^ [L2 + 2ßcos2q + 1 dan iborat.
Natija. Quyidagi assimptotikformulalar o^rinli
z(/i) = 2+i(/i-l)2 + o(/i-l)2 6
Z M-2 + oß).
Kll,
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3. (z) funksiyaning xossalari.
C orqali kompleks sonlar tekisligini belgilaymiz, har bir ^>0 va z£C\[0,2] da aniqlangan
2
u f sin22q
A^(z) = l+5iJic;?)-zd9; (l)
T
funksiyani aniqlaymiz.
Bitiruv malakaviy ishini natijasini isbotlashda muhim o'rin tutadigan quyidagi lemmalarni isbotlaymiz.
Lemma 2. z 6 C\[0,2] soni operatorning xos qiymati bo'lishi uchun
AM (z) = 0. tenglikning bajarilishi zarur va yetarli.
Isbot; Zarurligi. z 6 C\[0,2] soni operatorning xos qiymati bo'lisin. U
holda
U f
(e(q) — z)/(q) =--sin2q I sin2t/(t)dt (2)
2n J
T
tenglama nolmas 0 ^ / (q) 6 LC2(T) yechimga ega. Agar
J 5 in2 t/ (t) d t = C ( 3 )
T
deb belgilash kiritsak. ( C ^ 0 )
\i
(s(q) -z)f(q) = - — sinlqC;
In
tenglikka kelamiz. Bundan / ( q ) ni topamiz
fiC sinlq = (4)
Ushbu (4) tenglik bilan aniqlangan / ( q) funksiyani (3) belgilashga keltirib qoyamiz va quyidagi
ßC f sin22q
= -^L [
2 7T J
(e(q) - z)
T
tenglikka ega bolamiz, ya'ni:
dq
fiC f sin22q C + — J , , ,--d q = 0 .
2n) (£((?)-z) H
T
(1) belgilashga ko' ra
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C (z) = 0.
C ^ 0 ekanligidan
(z) = 0
tenglikka kelamiz.
Yetarliligi. AM (z) = 0 bo'lsin. U holda zeC \ [ 0 , 2 ] soni operatorning xos
qiymati bo'lishini ko'rsatamiz.
fiC sinlq fiC sinlq
^^ 2n (s(q) — z) 2n (1 — cos2q — z) ' funksiyani qurib olamiz. Dastlab / ( q) 6 L2( T) bo'lishini ko'rsatamiz.
ißC sin2q fßC\ [ / sin2q \
2 7T (1 —cos2 q — z) d q ( 2 7T/ J ( ( 1 —co 5 2 q — z)/ d q;
T T
sijilo
z 6 C \ [ 0 , 2 ] ekanligidan 1 — co s 2 q — z ^ 0 va —-— — uzluksiz
funksiya. Bundan uning chegaralangan ekanligiga kelamiz. Demak, / ( q) 6 L 2( T).
s in2 q — toq funksiya, ( 1 — c o s 2 q — z ) —juft funksiya. Bundan / ( q) — toq funksiya. Demak,
Endi bu funksiya z soni uchun (1) tenglamaning yechimi bo'lishini ko'rsatamiz.
/ /¿C sin2q \ V 27T (1 — cos2q — z))
/i f / fiC sin2t \
= —_sin2 qj ^ t(—____jdt.
T
Ushbu tenglikdan
/i f sin22t
s in 2 q H--s in 2 q I 7-rd t = 0
^ 2tc J (1 - cos2t - z)
T
ga kelamiz, ya'ni:
/ /¿f : s in 2 q ( 1 + — J —-——— d t ) = 0 ;
\ ip (1) belgilashga ko' ra
Aß (z)sin2q = 0; (z) = 0.
bundan (2) tenglikning o'rinli ekanligiga kelamiz. Lemma isbot bo'ldi.
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Lemma 2.3. A^ (z) funksiya z 6 ( 2 ; oo ) da monoton o^suvchi bolib
lim A„ (z) = 1;
z—>oo r
limA„ (z) = 1 - /¿.
z-> 2 ^
Isbot; (1) formula bilan aniqlangan AM (z) funksiyadan z boyicha hosila olamiz.
dq > 0.
д AM(z) _ Г sin22q д z _/iJ(£ (o) —z) 2
T
Demak, A^ (z) funksiya monoton o'suvchi.
sin22q e(q) - z
funksiya da uzluksiz va
sin22q lim ———-= 0;
z->co £(qr) — z
ekanligidan, integral belgisi ostida limitga o'tish haqidagi teoremaga ko'ra:
lim Д „ (z) = 1;
z-> oo r
ga kelamiz.
5 in 2 x + с о s2 x = 1 ayniyatga ko'ra:
(ß f sin22q \ [if sin22q 1 H--I -do ) = = 1 H--1 im I -do =
2nj£(q)-z 4j 2nz^2j£(q)-z 4
/i Г sin22q \i f sin22q = 1 + —— I 1 i m z г-d o = = 1 - — I --— d o =
2л J z^2£{q) — z 2л J 1 + cos2q
T T
,2-
ß f 1 — cos 2q ß f = 1--I -d o = = 1--I ( 1 — с о 5 2 o) d o = 1 — / .
2л J 1 + cos2q 4 2л J 4 4 ^
T T
Lemma isbotlandi. ▲
Yuqoridagi lemmalardan kelib chiqadiki. 0 < / < 1 bo lsa
AM (z) = 0
tenglama yechimga ega emas. // > 1 da faqat bitta yechimga ega boladi. // = 1 bo'lsin.
(z) = 0;
tenglama z = 2 yechimga ega. z = 2 6 cress( /iM). Bu son operatorning xos qiymat bolishini tekshiramiz.
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(hßf) (t) = 2f (t); tenglamani qaraymiz. Bu tenglamadan
1 f
( 1 + cos21) f (t) = —5 in21 I sin2q f ( q) dq;
2n J
T
ga kelamiz.
I
sin2qf{q) dq = C;
T
belgilash olamiz (C ^ 0) . Bundan
C sin2t
№ = —
2tc 1 + cos2t' f ( t) ± 0 , endi f ( t) e 4 (T) ni tekshiramiz. 4n2 f 0 f sin22t r 1 — cos2t
—J f2(t)dt = J (TTC02Wdt = J TTC0S2Tdt
TT T
On n
f 1 - cos2t f 1 - cos2t f 1 - cos2t
= I ü-d t + I --—d t = = 2 I --—d t >
J 1 + cos2t J 1 + cos2t J 1 + cos2t
— 71 0 0
( b u x o sm as int e g r a I b o" I i b t = — d a m axsus I i k k a e g a.)
l+s l+s s s
f 1 - cos2t f dt f dy f dy
>2 I --— d t>C I --=cI-^ = 2
J 1 + cos2t J tc\ J y2 J y2
s s ( t~2) -s 0
Ushbu munosabatning o ng qismida turgan integral uzoqlashuvchi. Demak f (t) £ L2(T ) . z = 2 xos qiymat emas. Asosiy natijalar isboti
2.1Teoremaning isboti.
Dastlab,
I
sin22q
dq
s(q) - z
T
integralni hisoblaymiz .
isin22q f sin22q f sin22q -dq = I ----dq = I --—dq;
£{q) — z J 1 — cos2q — z J A — cos2q
T T T
bu yerda
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A = 1 — z < — 1;
Trigonometriyaning asosiy ayniyatidan foydalanib integralni ikki qismga ajratamiz.
isin22q i 1 — cos22q C dq i cos22q
A — co 5 2 q d q J A — co 5 2 q d q J A — co5 2 q J A — co 5 2 q d q
TT TT
= /i + /2; (5)
Birinchi ^ integralni hisoblaymiz:
dq
J
cos2q'
T
Juft funksiyadan simmetrik oraliq boyicha olingan integral xossasidan foydalanib quyidagi tenglikka kelamiz:
0 7T 7T
/ = f dq = i dq i dq _ _ i_dq
1 JA — cos2q JA — cos2q JA — cos2q J A —
cos2q J A — cos2q J A — cos2q J A — cos2q
T -7T 0 0
Quyidagicha almashtirish olamiz
g2iq = £
bundan quyidagilar hosil boladi:
c o 5 2 q = f + 1)
df = 2ie2iqdq; df
dq = 2Tt;
Topilganlarni integralga qoyamiz.
IT
dq f df 2 f df
f dq f df 2 f
1 = 2J A — co 52 q = 2 J 2 £f(A —f+i))==7 J 2Af — f2 — l
o |£|=i2 £ f(A 2( f + f ))> |f| =1
- 2 f - - 2 [
= —7 J f2 —2A f+ l = = —7 J (f — fi )
bu yerda
| | | |
f 1 = A — Va 2 — l ; f 2 = A + V A2 — l ;
Korish mumkinki
|fl|>l|f2|<l .
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T |<f|=1
Koshining integral formulasi yordamida integralni hisoblaymiz.
f dq __ 2 i df _ 2 1
1 J A — c o5 2 q £ J (f — fL^ ^ ' 7
_ 27r
= — VA2—'1;
Demak,
2n
^^VA*—l . (6)
Endi /2 integralni hisoblaymiz.
icos22q
T
Integral ostidagi kasr ifoda suratiga A 2 ni qo'shib ayiramiz va integralni
quyidagi ikkita qismga ajratamiz.
cos22q f A2 — cos22q — A2
/ = _ f cos 2q ^ = i
2 J A — cos2q J A — cos2q
T T
J J
dq
dq
== 2nA - A2It;
cos2q
T T
Demak,
r co
= — J AT—
cos22q 2nAc
/2 = — J --— d q = 2 7tA +
co52 q VA2 — l'
T
Lemma 2.4. ga ko'ra z 6 C \ [ 0 , 2 ] soni operatorning xos qiymati bo'lishi uchun A ß (z) = 0 tenglamani qanoatlantirishi zarur va yetarli. Faraz qilamiz z 6 C \ [ 0 , 2 ] xos qiymat bo'lsin. U holda
2-
ß f sin 2q
A"(z) = l+27JicqT—^d q = 0 ;
T
ga ega bo'lamiz. (5), (6) va (7) tengliklarga ko'ra.
u 2n 2nA2
-+27tA+ , ) = 0;
2 7 v VA2—l VA2—l
Ushbu tenglik
A2-l l + / M a + ) = 0
1 VA2—1/
ga ekvivalent. Bundan Google Scholar Scientific Library of Uzbekistan
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1 + ß (A + ^A2 - 1) = 0 ;
ß^A2 - 1 = -1- ßA; (A < — l,ß > 1 bo^lganidan — 1 — ßA > 0) Tenglamaning ikkala tomonini kvadratga kotaramiz.
ß2A2 - ß2 = 1 + 2ßA + ß2A2. Oxshash hadlarni qisqartirib soddalashtiramiz, natijada quyidagi A noma'lumga nisbatan chiziqli tenglamani hosil qilamiz:
1 + 2 ßA + ß2 = 0;
A = 1 — z ekanligidan, quyidagi tenglikka kelamiz:
2 ßz = 1 + 2ß + ß2;
1 — 2ß + ß2
z = 2 +---;
2ß
yoki
(1 - ßf z = ^ ^ 2ß ' W
Bu xos qiymatga mos xos funksiyani aniqlaymiz. ( 3 ) ga kora
ßCsin2t
f^ (1 — cos2q — z) Demak, (8) ni hisobga olsak bu xos qiymatga mos xos funksiya:
2ß2Csin2q ^^ ß2 + 2ßcos2q + 1" Ko'rinishda bo'lib bunda C = const ^ 0 . Teorema isbotlandi. ▲
REFERENCES
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5. L.A.Lyusternik, V.I.Sobolev: Elements of functional analysis. "Nau-ka". Moscow 1965
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6. N.I.Akhiyezer, I.M.Glazman: Operator theory in Hilbert spaces. "Nau-ka". Moscow 1972
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