Ikki o‘lchamli olmos panjaradagi shredinger
![Ma’lumki, σ(A) ∥ A∥ radiusli doira ichida saqlanar edi. O’z-o’ziga qo’shma
operatorlar uchun esa bu baholash yanada aniqroq.
Teorema 1.3.3
σ(A)⊂[m ,M ] . Shuningdek, m , M ∈ σ ( A )
.
Natija 1.3.1. Har qanday chegaralangan o’z-o’ziga qo’shma operatorning
spektri bo’sh emas.
Teorema 1.3.4 A
o’z-o’ziga qo’shma operator bo’lsin. λ
soni A
operator
uchun xos qiymat bo’lishi uchun R ( A − λI ) ≠ H
bo’lishi zarur va yetarli .
Natija 1.3.2.
λ xos qiymatga mos keluvchi xos funksiyalar fazosi R(A− λI ) ning
ortogonal to’ldiruvchisidan iborat.
O’z-o’ziga qo’shma operatorning spektrini quyidagicha tavsiflash ham
mumkin: agar
R(A− λI )≠R(A− λI ) bo’lsa, λ soni A operatornng uzluksiz spektriga
tegishli bo’ladi va agar
R(A− λI )≠H bo’lsa, λ soni A
operatorning nuqtali spektriga
tegishlidir.
Teorema 1.3.5 Faraz qilamiz, A − H
Hilbert fazosidagi o’z-o’ziga qo’shma
operator bo’lsin. U holda
A qoldiq spektrga ega emas.
H
Hilbert fazosi, A ∈ L ( H ) − ¿
o’z-o’ziga qo’shma operator bo’lsin.
Teorema 1.3.6 Kompakt operatorning nolmas
z xos qiymatiga mos
keluvchi
Xλ xos fazosi chekli o’lchamli.
Teorema 1.3.7 Istalgan
δ>0 son uchun kompakt operator xos
qiymatlarining moduli δ
dan katta bo’lganlari soni chekli.
Bu teoremadan shuni xulosa qilamizki, kompakt operatorning xos
qiymatlarini moduli bo’yicha kamayish tartibida joylashtirish mumkin.
Natija 1.3.1 Kompakt operatorning xos qiymatlari to’plami noldan farqli
limitik nuqtaga ega emas.](https://docx.uz/documents/5de49115-0aae-4a3f-a5e1-7f4c9d180edc/page_14.png?v=1)
![BOB 2 . Olmos panjaradagi diskrit Shredinger operatorining
spektri.
2.1 Olmos panjaradagi diskr et Shredinger operatorining koordinata va
implus ta`sviri
.
Quyidagi to’plamni kiritamiz:
A
2 ={ v ( n ) : v ( n) = n
1 v
1 + n
2 v
2 n = ( n
1 ; n
2 ) , n ϵ Z 2 }
,
bu yerda
v1= (−1;0;1)v2=(0;−1;1) .
2.1.1 ta’rif.
A2 to’plamga 2 o’lchamli olmos panjara deyiladi ( qarang [2]).
Quyidagi to’plamni kiritamiz:
Ω = A
2 ∪
( p + A
2 ) , p = 1
3 ( − 1 ; − 1 ; 2 )
.
l2(Ω)
- orqali Ω da kvadrati bilan jamlanuvchi ^ f( n) = ( ^ f
1 ( n) ,^ f
2 ( n ) )
funksiyalar
juftligini belgilaymiz. Bu fazo Hilbert fazosi bo‘lib, skalyar ko’paytma
quydagicha aniqlangan
(^f,^g)= ∑vϵA2
3^f1(n)^g1(n)+ ∑vϵ(p+A2)
3^f2(n)^g2(n).
T =
( − π ; π ] .
L2(2)(T2)−¿
T 2
da aniqlangan kvadrati bilan integrallanuvchi
f(x)=(f1(x),f2(x))
funksiyalar juftligining Hilbert fazosi bo`lsin. Bu yerda
skalyar ko’paytma quydagicha aniqlangan
( f , g ) = ( f
1 , g
1 ) + ( f
2 , g
2 )](https://docx.uz/documents/5de49115-0aae-4a3f-a5e1-7f4c9d180edc/page_17.png?v=1)
![Bunda ( f
i , g
i ) =
∫
T 2 f
i ( x ) g
i ( x ) dx , i = 1,2.
Quydagi F : l
2
( Ω ) → L
2 ( 2 ) (
T 2 )
unitar operatorni kiritamiz:
F=(
F 0
0 F)
, (F ^f)(x)= √3
2π ∑n∈Z2 ei(x,s)^f(s).
Bu operator teskarisi F − 1
: L
2 ( 2 )
(
T 2 )
→ l
2 ( Ω )
quydagicha aniqlanadi:
F − 1
=
( F − 1
0
0 F − 1 ) , ( F − 1
f )( s) = √ 3
2 π ∫
T 2 e − i ( s , x )
f ( x ) dx .
bu yerda
( s , x ) = s
1 x
1 + s
2 x
2 .
Olmos panjaradagi diskrit Shredinger operatori
^ H
ushbu l2(Ω) fazod a
chegaralangan o‘z-o‘ziga qo‘shma operator sifatida quyidagicha aniqlanadi:
^
H = − 3 ( ∆
2 + 1 ) + ^ Q
Bunda
(−3(∆2+1)^f)(v)=((V1^f2)(n);(V2^f1)(n))
Bu yerda
(V1^f2)(n)= ^f2(n)+^f2(n− e1)+^f2(n−e2)
(V2^f1)(n)= ^f1(n)+^f1(n−e1)+^f1(n−e2)
e1,e2,nϵΩ n=(n1;n2), e1=(1;0),e2=(0;1) .
^Q - Ω
da aniqlangan zarrachalarning o‘zaro ta’sir potensiali bo‘lib, ular
quyidagi formulalar bilan aniqlanadi.
(
^ Q f ) ( n ) = (
^ Q
1 ( n ) 0
0
^ Q
2 ( n ) )(
^ f
1 ( n )
^
f
2 ( n ) ) = (
^ Q
1 ( n ) ^ f
1 ( n )
^
Q
2 ( n ) ^ f
2 ( n ) )
bunda
∑n∈A2
|^Q1(n)|<∞ , ∑n∈(p+A2)
|^Q2(n)|<∞.
^
H
operatorni koordinata ko‘rinishidan impuls tasvirga o‘tish F
almashtirishilari yordamida amalga oshiriladi [2]](https://docx.uz/documents/5de49115-0aae-4a3f-a5e1-7f4c9d180edc/page_18.png?v=1)
![H = F^ H F − 1
= F ( − 3 ( ∆
2 + 1 )) F − 1
+ F ^ Q F − 1
.
H
operator o lmos panjaradagi diskrit Shredinger operatorining impuls tasviri
bo’lib, u quydagicha aniqlanadi [2]
H = H0+Q
, (1)
bu yerda :
H 0 va Q 2×2 matritsa uchun matritsa operatorlari bo’lib, L2(2)(T2) da
quyidagicha aniqlanadi
(H ¿¿0 f)(x)=(
0 E(x)
E(x) 0 )(
f1(x)
f2(x))=(
E(x)f2(x)
E(x)f1(x))¿
,
( Qf )( x ) = ( Q
1 0
0 Q
2 )( f
1
( x )
f
2
( x )) = ¿
,
Bunda, E
( x ) − ¿
2 o`zgaruvchili kompleks qiymatli funksiya
E
( x ) = 1
3 ( 1 + e i x
1
+ e i x
2 )
, Qi− L2(T2) da aniqlangan integral operator
( Q ¿ ¿ i f ¿ ¿ i )
( x ) =
∫
T 2 Q
i ( x − t ) f
i ( t) dt . i = 1 , 2 , ¿ ¿
Qi(∙)− ¿
T2 da aniqlangan haqiqiy qiymatli biror uzluksiz, juft funksiya.
Lemma 2.1.1. H
operator L
2 ( 2 )
(
T 2 )
fazoni L
2 ( 2 ) (
T 2 )
fazoga o`tkazadi, ya`ni
H = H
0 + Q
: L
2 ( 2 )
(
T 2 )
→ L
2 ( 2 ) (
T 2 )
.
Isbot. Biz H = H
0 + Q
: L
2 ( 2 )
(
T 2 )
→ L
2 ( 2 ) (
T 2 )
. Ekanligini tekshirishdan avval Q :
L2(2)(T2)→ L2(2)(T2)
va H
0 :
L
2 2 (
T 2 )
→ L
2 2 (
T 2 )
ekanligini alohida- alohida ko`rsatib
o`tamiz.
Avval
Q : L2(2)(T2)→ L2(2)(T2) . Ekanligini ko`rsatamiz: Bizga ma`lumki bu
yerda
L2(2)(T2)−¿ orqali T2 da kvadrati bilan integrallanuvchi funksiyalar
juftligini belgilagan edik. Demak biz](https://docx.uz/documents/5de49115-0aae-4a3f-a5e1-7f4c9d180edc/page_19.png?v=1)
![2.2 Olmos panjaradagi diskret Shredinger operatorning muhim spektri
Ushbu paragrafda L2(2)(T2)−¿ Hilbert fazosida (1) ko’rinishda aniqlangan
Olmos panjaradagi diskret Shredinger operatori
H = H0+Q ning muhim
spektrini o`rganamiz.
Bizga ma`lumki biror
λ∈C uchun A− λI operator teskarilanuvchan bo’lsa, u
holda
λ soni A
operatorning regulyar nuqtasi deb atalar edi va A
operatorning barcha regulyar nuqtalari to’plami ρ
( A )
kabi belgilagan edik .
Ta`rif 2.2.1
σ(A)=C ¿(A) to’plam A
operatorning spektri deb ataladi.
Ta`rif 2.2.2 λ ∈ σ
( A )
son yakkalangan, A operatorning chekli karrali xos
qiymatlari to’plami diskret spektr deb ataladi va σ
disc ( A )
deb belgilanadi.
Ta`rif 2.2.3 σ
ess
( A ) = σ ( A ) ¿
disc ( A ) ,
A muhim spektr deb ataladi .
Ushbu paragrafning asosiy teoremasi quyidagidan iborat
Teorema 2.2.1 σ
( H
0 ) = [ − 1 ; 1 ] .
Isbot . Bizga ma`lumki [Reed Simon],
H 0 2×2 matritsa operatorining
spektri quyidagi formula bilan aniqlanadi
σ
( H
0 ) = ¿ x ∈ T 2
σ ( H
0 ( x ) ) .
(*)
Bunda
H 0(x)−¿ har bir fiksirlangan x∈T2 da 2×2 sonli matritsa bo`ladi,
ya’ni
H
0 ( x ) =
( 0 E ( x )
E ( x ) 0 ) , E ( x ) = 1
3 ( 1 + e i x
1
+ e i x
2 )
.
Shuning uchun
H 0(x) ning spektri xos qiymatlaridan iborat bo’ladi, ya`ni har
bir tayinlangan
x∈T2 larda det | H
0 − λI | = 0
tenglamaning ildizlaridan iboratdir.
Bu tenglamani tuzamiz:](https://docx.uz/documents/5de49115-0aae-4a3f-a5e1-7f4c9d180edc/page_28.png?v=1)
![(H ¿¿0− λI )(x)=(
0 E(x)
E(x) 0 )−(
λ 0
0 λ)=(
− λ E(x)
E(x) − λ)¿.
det |H 0− λI|=|
− λ E(x)
E(x) − λ|= 0⟾ λ2− E(x)∙E(x)= 0⟾
λ 2
=
| E ( x ) | 2
⟾ λ
1,2 = ± | E ( x )| , x ∈ T 2
.
Bu yerda
|
E ( x )| 2
= E ( x ) E ( x ) = 1
3 ( 1 + e i x
1
+ e i x
2 ) 1
3 ( 1 + e − i x
1
+ e − i x
2 )
= 1
9 ( 3 + e i x
1
+ e − i x
1
+ e i x
2
+ e − i x
2
+ + e i ( x
1 − x
2 )
+ e − i ( x
1 − x
2 ))
= ¿
¿1
9¿
.
Shunday qilib,
σ(H 0(x))={xos qiymatlari }={−|E(x)|;|E(x)|}
.
Demak, (*) ga ko’ra
σ(H0)=¿
¿x∈T2σ(H 0(x))=¿x∈T2{−|E(x)|;|E(x)|}
¿ − Ran
{| E ( x )|} Ran {| E ( x )|}
.
Endi
¿ E
( x ) ∨ ¿ max
x ∈ T 2 1
9 ¿ ¿
va
min
x ∈ T 2
| E ( x )| = 0
ekanligidan, ushbu
− Ran {|E(x)|}=[−1;0]
va Ran {|E(x)|}=[0;1]
tengliklarni hosil qilamiz. Shunday qilib,
Demak σ
( H
0 ) = [ − 1 ; 1 ]
.](https://docx.uz/documents/5de49115-0aae-4a3f-a5e1-7f4c9d180edc/page_29.png?v=1)
![Teorema isbotlandi.
Lemma 2.2.1 Q :L2(2)(T2)→ L2(2)(T2) kompakt operator .
Isbot.
Q :L2(2)(T2)→ L2(2)(T2). operatorni ko`rinishi quydagicha edi:
( Qf )( x ) = ¿
Bunda,
( Q ¿ ¿ i f ¿ ¿ i )
( x ) =
∫
T 2 Q
i ( x − t ) f
i ( t) dt . i = 1 , 2 , μ
1 , μ
2 > 0 ¿ ¿
Biz
Q operatorni kompaktligini ko`rsatishimiz uchun har bir i ∈ { 1 , 2 }
da
Qi:L2(T2)→ L2(T2)
operatorni kompakt ekanligini ko`rsatamiz. Yuqoridagi 1.5.4
teoremadan ma`lumki, ( Q ¿ ¿ i f ¿ ¿ i )
( x ) =
∫
T 2 Q
i ( x − t ) f
i ( t) dt . ¿ ¿
operator kompakt
bo`lishi uchun
∫
T2 ∫
T2 ¿Qi(x−t)¿2dtdx <∞ bo`lishi zarur va yetarli edi.
Shartga ko`ra
Qi(∙)− ¿ ikkala o`zgaruvchi buyicha ham T2 da aniqlangan biror
uzluksiz funksiya.
Bundan ∫
T 2
∫
T 2 ¿ Q
i
( x − t ) ¿ 2
dtdx
integral mavjud va chekli. Demak
Q
i kampakt , ya`ni Q
kampakt operator.
Teorema 2.2.2 σ
ess
( H ) = σ ( H
0 ) = [ − 1,1 ]
.
Isbot. Muhum spektr turg`unligi haqidagi Veyl teoremasi ga ko`ra
H =Q+H 0
operatorning muhim spektri Q kompakt qo`zg`lishda o`zgarmaydi va
H 0
operator spektri bilan ustma-ust tushadi.
Q kampakt operator. Bu yerdan esa xulosa σess(H )=σ(H 0) .
Teorema isbotlandi.](https://docx.uz/documents/5de49115-0aae-4a3f-a5e1-7f4c9d180edc/page_30.png?v=1)
![(−
| E ( x )| 2
+ | E ( x )| 2
− z 2
|
E ( x )| 2
− z 2 μ
1 C
1 − z μ
2 E
( x )
|
E ( x )| 2
− z 2 C
2
− z μ
1
|
E ( x )| 2
− z 2 C
1 + −
| E ( x )| 2
+ | E ( x )| 2
− z 2
|
E ( x )| 2
− z 2 μ
2 C
2
) = ¿
(
− z2
|E(x)|2− z2μ1C1− zμ2E(x)
|E(x)|2− z2C2
− zμ1
|E(x)|2− z2C1− z2
|E(x)|2− z2μ2C2)
= ¿
z
(
− z
|E(x)|2− z2μ1C1− μ2E(x)
|E(x)|2− z2C2
− μ1
|E(x)|2− z2C1− z
|E(x)|2− z2μ2C2)
= z(
ψ1
ψ2)
Yoki bundan
Hψ = z(
ψ1
ψ2)
tenglikka ega bo’lamiz. Demak, ψ =
( ψ
1 , ψ
2 ) H
operatorning xos funksiyasi va z
soni
H
operatorning xos qiymati ekan.
Teorema isbot bo’ldi.
Isbotlangan teoremaga asosan, aytish mumkinki
H operatorning muhim spektri
[−1,1 ]
dan iborat, R ¿ − 1,1 ¿
da esa deskrit spektri mavjud.
Xulosa
Ushbu malakaviy bitiruv ishi Shredinger operatori, uning xossalari va
spektrini, muhim spektri va deskrit spektrlarini o’rganishga bag`ishlangan.
Malakaviy bitiruv ishida operator va uning xossalari, chiziqli, o’z-o’ziga
qo’shma va unitar operatorlarga misollar keltirilgan. Shredinger operatorining](https://docx.uz/documents/5de49115-0aae-4a3f-a5e1-7f4c9d180edc/page_35.png?v=1)
Ikki o‘lchamli olmos panjaradagi shredinger