chuong 4 ch.iv- theory of representation

C HC LNG T L THUYT BIU DIN

CHNG IV

L THUYT BIU DIN

1. Biu din

Khi nim biu din c dng trong vt l hc vi nhiu ngha, nhng

tu chung, u l din t mt th g tru tng thng qua mt th g khc c th. Ring trong gio trnh ny ta dng khi nim biu din vi ba ngha. Phn u, l cch thc thay vect trng thi tru tng thnh mt th c th l hm sng. Tip theo, l cch thc din t c th cc nh lut ca c hc lng t, v cui cng l cch din t khc nhau ca qu trnh din tin theo thi gian ca h vt l. Trong chng V, ta s dng khi nim ny, nh mt cch thc thay nhm bng nhm cc php bin i trong mt khng gian vect no . Lm nh vy, phn t ca nhm tru tng, c thay bng cc ton t, v do , thnh nhng ma trn vi cc php tnh ton c th.

T i s tuyn tnh ta bit rng, mi khng gian vect, tr khng gian khng, u c c s. T nh l B xung h cha y , cng suy ra rng, khng gian c v s c s. Khi mt h c s c la chn, mi vect u c th khai trin mt cch duy nht theo n v cc h s khai trin l thnh phn ca vect trong h c s chn.

Vic thay th vect bng thnh phn ca n, cho php ta thc hin cc php tnh tru tng ca vect thng qua nhng php tnh cng, nhn quen thuc ca cc s trn thnh phn ca chng.

Khng gian Hilbert cc trng thi ca mt h vt l cng lm thnh khng gian vect, mi vect cng l mt i lng tru tng. Bng cch chn c s, mi vect trng thi cng c thay bng cc thnh phn vi nhng php ton quen thuc.

Do cc ton t vt l u l Hermitian cho nn chng lun c th cho ho c. iu ny c chng minh d dng cho khng gian hu hn chiu bng phng php quy np hu hn bc. Trong khng gian Hilbert v hn chiu, tnh kh cho ca cc ton t Hermitian vn c gi thit l ng.

Do tnh cht trn, c s ca khng gian Hilbert cc trng thi, s c chn bng cch ly h cc vect ring ca h y cc ton t vt l no . Vic la chn h hm ring ca nhng ton t vt l no, c gi l vic chn biu din.

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Nu h c s l cc vect ring ca ton t vt l lA , ta ni rng, ta c lAbiu din. Ni ring, khi lA l ton t to , q biu din cn c gi l biu din to . Khi lA l ton t xung lng, p biu din cn c gi l biu din xung lng. Tng t nh vy, ta s c biu din nng lng, biu din s ht,.

Thnh phn ca vect trng thi trong mt biu din, c gi l hm sng trong biu din tng ng. Nu s thnh phn l v hn khng m c, chng c coi l mt hm lin tc i vi gi tr ring.

1.1. Biu din ta

To , s c k hiu tng qut bng ch . N c th c nhiu hay t

thnh phn, tu thuc vo s chiu khng gian cu hnh ca h vt l: q

( ) ( ) ( ) ( )1 2 1 2 1 1 2 2, ,... , , q q q dq dq dq q q q q q q = = = " " Nu l khng gian Euclid ba chiu, thay cho , ta dng . ,q dq ,r dVGTrong c hc lng t, to c hai vai tr khc nhau, mt l tham s, hai

l, bin ng lc c bn. Ni chung, nu xut hin trong hm sng, q ( ),q t , n ng vai tr ca tham s. Cn nu xut hin trong cc biu thc din t ng lc ca h, nh trong hm th chng hn,

q( )U U q= , n s c vai tr ca

bin ng lc. Trong mt s trng hp, din t hm sng, phi cn n hai bin to , mt ch tham s, v mt ch bin ng lc. Khi , bin ng lc s c ghi di dng ch s. V d, hm sng m t trng thi c v tr xc nh , s c dng l 0q (0 ,q q t ) . Tuy nhin, d vit, ta thng k hiu bin ng lc ny di dng bin s: ( ) ( )

0 0, ;q q q t q = 4.1

Hai vai tr khc nhau ca to khng ng vai tr g quan trng trong c hc c in. Tuy nhin, trong c hc lng t, n li c nhng ngha nht nh. Khi c coi l tham s, ta c th thc hin vic i bi ton ang xt tr nn n gin. Khi coi l bin ng lc, khi sang c hc lng t, n s c chuyn t s sang c q s.

Tuy nhin, vic lng t ho, tc l thay t c s sang s, li khng giao hon vi php bin i to .

q

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Ta hy ly mt v d1, l, h mt ht t do khi lng . Lagrangian ca h, trong h to Descartes, s c dng:

m

( )= + + 2 2 22mL x y z Nu thc hin php bin i sang h ta tr: = =cos , sin , =x r y r z z

Lagrangian s tr thnh:

= + + 2 2 2 2( )2mL r r z

Nu coi cc to mi ( , , )r z l bin ng lc, xung lng lin hp vi chng s l:

= = = = = =

2, , r zL L

Lp mr p mr p mz

r z

Khi , Hamiltonian s c dng:

=

= = + + 2 221,2,3

1 1( )2i i r zi

H pq L p pm r

2p

Lng t ho bng cch coi to v xung lng tr l ton t, v d, theo cch thc thy trong cc chng trc, l /i ip p i = = iq :

= =, , r zp i p i p ir z

=

ta s thu c ton t Hamilton:

= + +

=2 2 2 22 2 2 2

12

Hm r r z

Biu thc ny ca ton t Hamilton khng c thc nghim xc nhn. Ngc li, nu thc hin vic lng t ho trc:

( ) l= + + = = G = 2 22 2 22 2m pL x y z H Hm m2 sau mi thc hin php bin i to :

=

sincosx r r

1 Xem A. Messiah, C hc lng t, (ting nga), trang 76-77, Tp I, Moskva, Nauka, 1978.

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= +

cossiny r r

= z z ngha l:

2 2 2

2 2 2sin sin(cos )(cos )

x y z r r r r

+ + = + 2 2 2

2 2 2 2cos cos 1 1(sin )(sin )

r r r r z r r r r

+ + + = + + 2

2z+

Khi , ton t Hamilton s c dng:

= + + +

=2 2 2 22 2 2

1 12

Hm r r r r z2

4.2

Biu thc ny ca ton t Hamilton c thc nghim xc nhn l ng. Mt kh nng khc cng c th dn n nhng biu thc khng ng n

khi lng t ho. V d, Hamiltonian ca mt ht t do chuyn ng mt chiu c dng:

2

1 2pHm

= trong , p l xung lng ca ht. Biu thc ny cn c th vit:

21 1 1

2H pqp

m q q=

Trong c hc c in, hai biu thc ny l mt. Nu coi l bin ng lc, khi lng t ho, hai biu thc c in ging nhau li cho hai ton t khc nhau:

q

= = =2 2

1 1 2

2 2

2p dH Hm mdq

2 2

2 2 2 21 1 1 1 1 1

2 2d d dH pqp H q

m dq dq mq q q q = = =

=4dq q

+

q

trong , biu thc th hai khng c thc nghim xc nhn. Mt trng hp khc, trong , biu thc c in ca mt i lng vt l c cha tch to v xung lng. V d nh:

A p= 4.3Khi lng t ho, ton t tng ng khng l mt trong cc biu thc sau y:

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l l m l l l1 2 3, , =A pq A q p A q p = =

Biu thc th ba khng ng, chng t rng, trong trng hp ny, to khng phi l tham s. Hai biu thc u khng ng, bi v, chng u khng Hermitian. Ta phi chn lA di dng:

l l ( ) l1A pq q p = + trong , c la chn sao cho, lA l Hermitian. Kt qu l 1/ 2 = v:

l l l2

A pq q p = + 4.4T cc v d trn, ta thy rng, bo m tnh ng n ca biu thc

ton t, mi bin i, trong to c vai tr tham s, u phi thc hin sau khi lng t ho. Thm vo na, ton t thu c phi Hermitian ho.

By gi ta tm c s ca ton t to . Gi s , v cc hm ring ca n, tng ng vi to l q q 0q( )0, ;q t q .

0q q = 4.5

Trong trng hp tng qut, do khng c iu kin g rng buc i vi , cho nn ph ca n s lin tc. h hm ring c th ng vai tr l h c

s trc chun ca khng gian Hilbert cc hm sng, chng phi tho mn: 0q

- iu kin trc chun ( ) ( ) ( )0 0 0, ,q q q q dq q q0 = - iu kin y ( ) ( ) ( )0 0 0, ,q q q q dq q q = Cc h thc ny s c m bo, nu: ( ) ( )0 0,q q q q = 4.6Nh vy: Trong biu din, h c s ca khng gian Hilbert cc vect trng thi chnh l hm delta Dirac.

q Xt mt vect trng thi bt k . N s c khai trin di dng: ( ) ( ) ( )0 0 0q q q dq q = = 4.7trong , h s khai trin chnh l ( )q = . iu ny chng t rng, ( )q va c vai tr ca vect trng thi, va c vai tr ca hm sng trong biu din to .

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Hm phn b xc sut cho to q trong trng thi ( )q s l: ( ) ( )2q q 4.9

V do : 2* *q dq qdq q dq = =

Ngha l, ( ) ( )q q q q = 4.10

Nh vy: Trong biu din to , ton t to vn l ta c in.

Trong trng hp khng gian vect ba chiu thng thng: ( ) ( )r r r r =G G G G 4.11

1. 2. Ton t xung lng

Trong biu din to , ton t xung lng khng th l mt tha s nhn,

bi v nh th, xung lng khng lm thay i trng thi ca h, to v xung lng xc nh c ng thi. iu ny tri vi nguyn l bt nh.

tm dng ca ton t xung lng, ta s dng nguyn l tng ng. Trong c hc lng t, s bo ton ca xung lng lin quan n tnh

ng nht ca khng gian, ngha l, tnh bt bin ca phng trnh Schroedinger i vi ton t tnh tin khng gian. n gin, ta hy xt trng hp h mt ht t do. Khi , khng gian i vi ht s l dng nht. Php tnh ti n: n trong khng gia

= = +GG G G Gar r T r r Ga s phi lm bt bin phng trnh Schroedinger. Gi l ton t tnh tin tc ng ln hm sng cm sinh bi php tnh tin aTG trong khng gian:

l l ( ) l ( ) ll ( )a aH r H r a H r = + = G GG G G G 4.12Nh vy, ton t tnh tin giao hon vi ton t Hamilton.

Do php tnh tin lm thay i v tr cu hnh khng gian, cho nn, n s dn n thay i hm sng. :

= +G G G( ) ( ) ( )r r r Ga 4.13V v, hm sng l kh vi v hn, ta c th khai trin:

( ) ( ) ( ) ( = + = G G G G1 ! nnr r a an )Gr

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i vi php tnh tin cc vi (vect tnh tin c di cc nh): ( ) + G1 a 4.14Suy ra, ton t tnh tin l 1 a = + G . Do ton t n v giao hon vi lH , suy ra ton t cng giao hon vi lH . Nh vy, n l ton t ca mt i lng bo ton. Theo nguyn l tng ng, i lng bo ton ny phi t l vi xung lng:

= Gk p tn h s , ta chuyn v gii hn c in, khi hm sng c dng k

= =exp( / )iS . Vi bin thin nh, ta c: = =G G= =

i ip S p 4.15

Cho nn, . Kt qu l, xung lng c in s tng ng vi ton t = =k i = G =p i . Mt cch tng qut: Trong biu din to , thnh phn ca ton t vect xung lng s bng nhn vi ton t o hm theo bin ta lin hp chnh tc: / i=

l ( ) ( )qp q qi q =

= 4.16

Tnh Hermitian ca ton t xung lng c th kim tra mt cch trc tip. V d, vi mt ht t do vi hm sng (3.1), ta c:

= = = = = =G GG G*

*,p dV p dV dVi r i r

= =

= GG ,dV pi r 4.17Cn i vi trng hp h lin kt, hm sng tin ti khng khi x ,

cho nn, bng cch tch phn tng phn, ta c:

( ) = = = = =G G G G*

, =p dV dV dVi r i r i r

= =

= = GG G*

,dV dV pi r i r

4.18 Ta hy tm hm ring ca ton t xung lng trong biu din to .

Gi h hm ring ny l l ( )p q , ngha l:

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l ( ) ( ) ( )pp pqp q p qi q = =

= 4.19

Phng trnh ny cho nghim l:

( ) expp iq C pq = = 4.20Do cng khng c hn ch c bit no i vi hm sng, cho nn, ph

ca xung lng l lin tc. iu kin chun ho s c cho bng hm delta Dirac. Nu b qua mt tha s pha, h s chun ho s c tnh t iu kin:

( ) ( ) ( ) ( )2 2exp 2 niC q p p dq C p p p = = == p v kt qu l, n s bng , trong , l chiu khng gian cu hnh ca h vt l ang xt. Nh vy:

( ) / 21/ 2 n= n( ) ( ) / 2

1 exp2

p niq pq

= == 4.21Trong trng hp ba chiu, phng trnh:

( ) ( ) ( ) = =G GG G G G G=p p p r i r p r s cho nghim trc chun l:

( ) ( ) =

G G G== 3/ 21 exp .

2p

ir p Gr

H hm ring ca ton t xung lng cng tho mn iu kin y . Thc vy, s dng biu din ca hm Dirac, ta c:

( ) ( ) ( ) ( ) ( )* 1 exp

2p p n

iq q dp p q q dp q q = = ==

1. 3. Biu din xung lng

Nu chn h vect (4.21) lm c s cho khng gian Hilbert cc trng thi,

ta s c biu din hay biu din xung lng. p Trong biu din xung lng, nu hm sng ( ) (

0 0,p q q )p = m t

trng thi c xung lng xc nh 0p , trong biu thc khai trin: ( ) ( ) ( )0 0, , pq p C p p q dp = 79


h s s l: ( ) ( )0 0,C p p p p= iu ny l hin nhin trong trng hp hu hn chiu:

( )k ki ie = Nu ( )q l hm sng bt k trong biu din to , khai trin n theo

c s xung lng: ( ) ( ) ( )pq p q dp = 4.22 h s khai trin ( )p :

( ) ( ) ( ) ( ) ( )*

/ 21 exp

2p n

ip q q dq pq q dq = = == 4.23

chnh l hm sng trong biu din xung lng. T cng thc ca , suy ra, n l nh Fourier ca ( )q . Nh vy:

Hm sng trong biu din xung lng, chnh l nh Fourier ca hm sng trong biu din to .

Ta hy tm dng ca cc ton t ng lc trong biu din xung lng. Xut pht t h thc: l ( ) ( ) ( ) ( ) ( ) ( )p pp q q p q dp p p q di q i q

= = = = = p

Mt khc: l ( ) l ( ) ( )pp q p p q d = p

So snh hai biu thc ta c th kt lun rng: Trong biu din xung lng, ton t vect xung lng vn ch l xung lng c in:

l ( ) ( )p p p p = 4.24i vi ton t to , t h thc:

( ) ( ) ( ) ( )pq q q q q p q dp = = = ( ) ( ) ( ) ( )p pp q dp p q dpi p i p

= = = =

4.25

Mt khc: ( ) ( ) ( )pq q q p q d = p

So snh hai biu thc, ta c:

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Trong biu din xung lng, thnh phn ca ton t vect to bng nhn vi o hm theo bin xung lng lin hp chnh tc: / i=

l ( ) ( )pq p pi p =

= 4.26

Trong trng hp mt ht, khng gian l ba chi u, khi :

( ) ( ) ( )p pr p pi i p = = G

G= =G G G G 4.27Tm li, trong biu din to , hm sng l mt hm, trong , to v

thi gian l nhng tham bin s. Ton t to s l tha s nhn, ton t xung lng bng ( )i = nhn vi ton t o hm theo bin to lin hp chnh tc.

Ngc li, trong biu din xung lng, hm sng l nh Fourier ca hm sng trong biu din to , trong , xung lng v thi gian l cc tham bin s. Ton t xung lng ch l tha s nhn, ton t to bng nhn vi ton t o theo bin xung lng lin hp chnh tc.

(i=) V nguyn tc, trng thi ca h khng ph thuc vo vic ta chn biu din to hay biu din xung lng. Vic chuyn t biu din ny sang biu din khc, c thc hin nh mt ma trn i c s U unitary. i vi khng gian hu hn chiu, thnh phn trn mt ct ca ma trn U chnh l thnh phn ca c s mi i vi h c s c. Trong biu din to , vect c s ca biu din xung lng l ( )p q , trong khi vect c s c l ( )0q q , cho nn, trong khai trin:

( ) ( )0p pq C q q dp = 4.28h s khai trin:

( ) ( ) ( )0 0p p pC q q q dq = = q 4.29chnh l mt ct ca ma trn i c s U . Nh vy, theo (4.21):

( ) ( ) ( )/ 21, exp

2 nU q p pq

i = =

= 4.30

Ngc li, nu chuyn t h c s xung lng sang h c s to , ta phi c ma trn chuyn l:

( ) ( ) ( ) ( )1

/ 2

1, , exp2 n

U p q U p q pqi

+ = = =

= 4.31

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Khi thc hin php bin i U ln cc vect c s, thnh phn ca vect trng thi ( )q , s bin i nghch bin, tc l, bng ma trn nghch o ca ma trn i c s:

( ) ( ) ( ) ( ) ( ) ( )1

/ 2

1, exp2 n

ip U p q q pq q d = = == q 4.32

Cn thnh phn ca vect ( )p s bin i ng bin, tc l bin i bng ma trn i c s:

( ) ( ) ( ) ( ) ( ) ( )/ 21, exp

2 niq U q p p pq p dp

= = == 4.33iu ny ngha l, ( )q l nh Fourier ngc ca ( )p .

Gi tr trung bnh ca cc i lng vt l khng ph thuc vo vic la chn biu din. Thc vy, nu chn xung lng v chn biu din to , ta s c:

( ) ( ) = p q p q dq 4.34Mt khc, cc do hm sng ( )q l nh Fourier ngc ca ( )p , cho nn:

( ) ( ) = q p q dq ( ) ( ) ( )

( ) = ==

1 exp2 n

i q p p p p p dpdpdq =

( ) ( ) ( ) ( ) ( )* *p p p p p dpdp p p p dp = = 4.35 Biu thc tch phn cui cng, chnh l gi tr trung bnh ca xung lng trong biu din xung lng. Tng t, nu chn to , ta c:

( ) ( ) = = *q q q q dq ( ) ( ) ( ) ( )

= == *1 exp

2 ni =p p q q p p dqdpdp

( ) ( ) ( ) ( ) = =

==*1 exp

2 ni =p p q p p dqdpdp

i p

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( ) ( )( ) ( ) =

===

*1 exp2 n

p i =p q p p dqdpdpi p

( ) ( ) ( ) ( )* *pp dp pi p i p =

= = p dp 4.36Biu thc cui cng chnh l gi tr trung bnh ca to trong biu din xung lng.

1. 4. Biu din nng lng Khi mt h vt l l ng kn, khng c tng tc vi bn ngoi, hoc c

tng tc, nhng trng ngoi khng ph thuc tng minh vo thi gian, nng lng ca h s bo ton. Khi , cc trng thi ring ca ton t Hamilton s c gi l cc trng thi dng, v nu ph nng lng l khng suy bin, chn h hm ring cua ton t Hamilton lm c s, ta s c biu din nng lng.

Trong biu din to , hm ring ( ) ,n q t ca ton t Hamilton tng ng vi nng lng , s chnh l nghim ca phng trnh Schroedinger khng ph thuc vo thi gian:

nE

= ( , ) ( , )n n nH q t E q t 4.37T phng trnh Schroedinger ph thuc thi gian:

= = n

nH i t, suy ra:

= =n

n ni Et 4.38

Phng trnh ny s cho nghim:

( ) = =, ( )expn niq t q E tn 4.39

trong , ( ) n q cng l hm ring ca ton t Hamilton: l ( ) ( )n n nH q E q = 4.40

v chng l hm sng ca ton t Hamilton trong biu din to . Nu tng ng vi mt gi tr ring ta c nhiu hm ring nE ,ni

c lp tuyn tnh, mc nng lng c gi l suy bin v 1,2,...,i = s s c gi l bc suy bin. Cc vect ny c th chn lm c s cho mt khng gian con bt bin s chiu ca ton t Hamilton. Ni chung, cc vect ny khng

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trc giao, tuy nhin, thng thng chng l h hu hn vect, cho nn lun lun c th trc chun ho c bng phng php Gram-Schmidt. V vy, h hm ring ca ton t Hamilton, cng c gi thit l mt h trc chun. Nu ph nng lng l suy bin, ch mt mnh nng lng khng xc nh duy nht trng thi ca h. xc nh duy nht, ta cn tm thm nhng i lng vt l khc m ton t ca chng giao hon vi ton t Hamilton. Khi , vect trng thi s l vect ring chung ca h cc ton t . Ph nng lng c th l lin tc, gin on hoc mt phn gin on mt phn lin tc.

i vi ph gin on, h hm ring c chun ho theo k hiu Kronecker:

( ) ( )m mq q dq mn = 4.41i vi ph lin tc, h hm ring c chun ho bng hm delta Dirac:

( ) ( ) ( )E Eq q dq E E = 4.42Nu h hm ring ca ton t Hamilton l y , n cng c th chn

lm c s ca khng gian trng thi. Khi ta c E biu din hay biu din nng lng.

Gi s, vect trng thi c din t trong biu din to bng hm sng ( )q v ( ){ }E q l h hm ring ca ton t Hamilton trong biu din .

Khai trin hm sng cho theo h hm ring ca ton t Hamilton: ( ) ( )E Eq C q dE = 4.43Tp hp h s: ( ) ( ) ( )*E EC C E q q dq = 4.44s l hm sng trong biu din nng lng. Gi tr trung bnh ca nng lng s l:

( )l ( ) ( ) ( )* *E E E EE q H q dq C C EdEdE q q dq = = = ( ) 2*E E EC C EdEdE E E C EdE = = 4.45

Nh vy: E EHC EC=

( ) ( ) 2E C E = 4.46ngha l, trong biu din nng lng, tc dng ca ton t Hamilton ln hm sng ch l nhn hm sng y vi nng lng, v bnh phng modul ca hm sng cng chnh l mt xc sut trong biu din nng lng.

ECEC

84


Nu ph nng lng l gin on, mt s cng thc tch phn s tr thnh tng v hn: ( ) ( ) ( ) ( )*, n n n n

nq C q C q q dq = =

2

n nn

E C = E 4.47 Vic chuyn t biu din to sang biu din nng lng cng c thc hin bng ton t unitary. Khi : ( ) ( ) ( ) ( )0 0 0, ,E q U q E q q dq U q E = = 4.48Nh vy, hm ring ca ton t Hamilton trong biu din to d chnh l cc ct ca ma trn i c s: ( ) ( ), EU q E q= 4.49Cng thc: ( ) ( ) ( )*EC E q q dq = 4.50c hiu nh l: Hm sng trong biu din nng lng l nh Fourier ca hm sng trong biu din to . Khc vi to v xung lng, biu thc ca Hamiltonian s khc nhau trong cc trng hp khc nhau, cho nn, dng ca cc ton t to v xung lng trong biu din nng lng cng khc nhau trong tng trng hp.

V d, vic xc nh ton t q trong biu din nng lng tng ng vi vic xc nh h s ( ),E E trong khai trin sau y:

( ),E EqC E E C dE = trong , c th, ta chn trng hp ph lin tc. Khi , s dng tnh cht trc chun ca hm sng , suy ra: EC

( )* *,E E E EC qC dE E E C C dE dE = = ( ) ( ) ( ), ,E E E E dE E E = =

Nh vy, ton t q c xc nh thng qua tch v hng ca vi EC EqC . iu ny c ngha l ton t q c xc nh thng qua ma trn ca n. Ta s kho st chi tit vn ny trong v d ca mc 3.

2. Cc bc tranh m t c hc lng t

Trong c hc lng t phi tng i tnh, thi gian, khc vi to , tuy cng l mt tham s xc nh vect trng thi, cc i lng vt l, nhng n li

85


khng phi l mt bin ng lc. V vy, nh hng ca thi gian phi c xem xt theo kha cnh khc hn.

Ta chp quan im cho rng, thi gian tin trin theo quy lut tng dn mt cch lin tc, do , vect trng thi v cc i lng vt l ph thuc thi gian cng s bin i theo. V nh vy, mt trong nhng vn c bn ca vt l hc, l xc nh quy lut bin thin theo thi gian ca chng. Cc quy lut ny c gi l cc phng trnh ng lc hc ca h.

Trong c hc c in, ng lc hc ca mt h vt l c din t bng nhiu cch. Mi cch din t c in, theo nguyn l tng ng, s phi c mt cch din t lng t. Cc cch din t ny, c th c hiu nh cc cch biu din khc nhau ca c hc c in.

2. 1. Cc biu din c in v lng t tng ng

1. Biu din Newton. Mt trong nhng cch din t c in quen thuc nht, l h phng

trnh Newton. Ta hy h mt ht khi lng trong trng th bo ton lm v d. Cc phng trnh Newton, khi , c dng:

m

dr pdt m

=G G

dp F Udt

= = G G

4.52

Cch din t ny c chuyn sang c hc lng t bng cch coi cc bin ng lc thnh ton t v cc phng trnh c in c din t bng cc phng trnh ton t (hoc nh l Ehrenfest):

ldr pdt m

=G G

l l ld p F Udt

= = G G

4.53

2. Biu din Hamilton. Ta cng c th din t h vt l thng qua mt hm ph thuc vo ta

v xung lng, gi l hm Hamilton hay Hamiltonian. Khi cc phng trnh ng lc s l h phng trnh Hamilton:

{ },i i i ii k k k kdx p H x H x H xdt m p x x p = = 4.54a

86


{ },i i i ii k kk k

dp U H p H p H pdt x p x x p

= = 4.54b

trong { },A B c gi l mc Poisson ca A v . BChuyn sang c hc lng t, cc phng trnh trn vn gi nguyn dng

nu ta thay mc Poisson thnh giao hon t, m ta s gi l mc Poisson lng t:

l l l,i

id x i H xdt

= = l l l,i id p i H pdt

= = 4.55

Thc vy, do mc Poisson tho mn cc tnh cht: { } { }, ,A B B A= { } { } { }, , ,AB C A B C A C B= + { } { } { }, , ,A BC B A C A B C= +

{ } { } { }, , ,A B C A C B C + = + { }{ } { }{ } { }{ }, , , , , ,A B C B C A C A B+ + 0=

4.56

cho nn, nu theo nguyn l tng ng, mc Poisson { },A B tr thnh mt ton t no , gi s k hiu l l l{ },

QA B , th n cng phi tho mn cc iu kin

nh trn. Vi 4 i lng vt l , , ,A B C D tu , ta c:

l l ll{ } l l ll{ } l ll{ } ll l l l{ } l l l{ } l l l l{ } l l l{ } ll

, , ,

, , , ,

Q Q Q

Q QQ Q

AB CD A B CD A CD B

AC B D A B C D C A D B A C DB

= + == + + +

4.57

Tng t, ta c cng c: l l ll{ } l l l l{ } l l l{ } l

ll l l{ } l l l{ } l l l l{ } l l l{ } ll, , ,

, , , ,

QQ Q

Q QQ Q

AB CD C AB D AB C D

C A B D C A D B A B C D A C BD

= + == + + +

4.58

Hai cch khai trin ny phi nh nhau, do :

87


l l l l( ) l l{ } l l{ } ll ll( ), ,QQ

AC C A B D A C BD DB = 4.59 Do l l,A C v l c lp nhau, cho nn h thc trn ch xy ra nu: l l,B D

l l l l( ) l l{ },Q

AC C A A C = l l ll( ) l l{ },

QBD DB B D = 4.60

trong , phi l mt hng s thun o, bi v, ton t l l{ },Q

A B phi

Hermitian. tm hng s , ta ly mt trng hp ring, trong , l l,A B l ton t to v xung lng theo mt trc no . Khi , do { }, 1x p = :

l l ( ) 1x p px i = = 4.61suy ra: 1/ /i i = == = . Nh vy,

Ton t mc Poisson ca hai i lng vt l bng nhn vi giao hon t ca hai ton t vt l tng ng:

/i =

{ } l l{ } l l l l( ) l l, ,Q

i i ,A B A B AB B A A B = = = = 4.62 C th kim tra d dng rng, biu thc ny tho mn tt c cc iu kin ca mc Poisson lng t.

T kt qu trn suy ra rng, cc ton t tho mn h phng trnh Hamilton lng t. Ch rng, biu thc o hm ton phn ca mt i lng vt l:

{ },dA A H Adt t

= + 4.63s phi c dng lng t:

l l l l,d A A i H Adt t

= + = 4.64Cho nn, l /d A dt chnh l ton t tng ng vi . /dA dt

3. Biu din Hamilton Jacobi. C hc c in cng c th din t chuyn ng ca h vt l bng hm

nh ngha trn mi qu o ca h, gi l hm tc dng. Khi , cc phng trnh ng lc s l phng trnh Hamilton- Jacobi:

88


S Ht

= 4.65Cch din t ny c th hin trong c hc lng t bng phng trnh

Schroedinger cho hm sng: li H

t == 4.66

2. 2. Cc bc tranh khc nhau ca c hc lng t

Tuy c hc lng t c nhiu cch din t ng lc khc nhau, tu thuc

vo cch din t c in, nhng v nguyn tc, c th phn ra ba cch sau y: - Hoc l, hm sng ph thuc, cn ton t vt l lA khng ph thuc

vo thi gian, nh vy, ng lc hc s c din t bng phng trnh i vi hm sng m t vect trng thi,

- Hoc l, ton t vt l lA ph thuc, cn hm sng khng ph thuc vo thi gian, nh vy, ng lc hc s c din t bng phng trnh i vi ton t

- Hoc l, c hm sng ln ton t vt l lA u ph thuc vo thi gian, nh vy ng lc hc s c din t bng phng trnh i vi c hm sng ln cc ton t vt l.

Ba cch din t khc nhau ni trn c gi l ba ba bc tranh ca c hc lng t. Cch th nht c gi l bc tranh Schroedinger, cch th hai c gi l bc tranh Heisenberg, cn cch th ba c gi l bc tranh Dirac, hay, bc tranh tng tc.

Cc cch din t khc nhau ca c hc lng t i khi cng c gi l cc biu din. Nh vy, ta c biu din Schroedinger, biu din Heisenberg v biu din Dirac. Biu din trong trng hp ny, chnh l cch din t c th cc bc tranh tin trin theo thi gian ca th gii vt l.

Ni mt cch tng qut, hm sng trng thi ( )t ca mt h lng t l mt vect c di n v trong khng gian Hilbert. S tin trin theo thi gian, ( ) ( )0t t , c th coi nh mt php bin i, chuyn vect n v ny thnh vect n v khc. Nu gi php bin i ny l ton t tin trin theo thi gian v k hiu l l 0( , )U t t , ngha l:

l = 0 0( ) ( , ) ( )t U t t t 4.67th bo ton chun ca vect, n phi l ton t unitary:

89


l l+ = 1U U 4.68Nu php quay trong khng gian ba chiu khng lm thay i di

vect, th ton t tin trin theo thi gian c hiu l php quay trong khng gian Hilbert cc trng thi.

Do c s ca khng gian cc trng thi l h hm ring ca cc ton t vt l, cho nn, nu cc ton t ph thuc thi gian, c s s ph thuc thi gian.

Nh vy, trong bc tranh Schroedinger, h c s c c nh, ta ch quay cc vect trng thi. Trong bc tranh Heisenberg, vect trng thi c gi c nh, ta ch quay cc vect c s. Cn trong bc tranh Dirac, c h c s ln cc vect trng thi u b quay, tuy nhin, mi loi c quay bng cc ma trn khc nhau.

1. Bc tranh Schroedinger Trong cch m t ny, vect trng thi ti thi im bt k c suy ra

t vect trng thi ti thi im : t

0tl = 0 0( ) ( , ) ( )t U t t t 4.69

Khi : ( ) ( ) l ( )0t Ui i t Ht t = = = = t 4.70

trong , lH l mt ton t tuyn tnh no . Do tnh cht bt bin ca xc sut theo thi gian, ta c:

( ) ( ) ( ) ( ) ( ) ( ) = + t td t t dq t dq t dq

dt t t =

( )l ( ) ( )l ( )( )= = * * *i t H t t H t dq = ( ) l l( ) ( )= = ** 0Ti t H H t dq = 4.71

Nh vy, ton t lH l Hermitian: l lH H+ = 4.72

tm ngha ca ton t ny, ta chuyn sang gii hn c in. Khi , phn ng hc ca hm sng c dng:

= =expi S 4.73

vi S l hm tc dng. Ta c:

90


= =Si

t t 4.74

Nh vy, khi chuyn sang gii hn c in, ton t lH tr thnh i lng: St

Mt khc, theo phng trnh Hamilton-Jacobi, y chnh l Hamiltonian c in . Cho nn, ton t H lH ton t ca hm Hamiltonian c in. Kt qu l:

Trong bc tranh Schroedinger, vect trng thi bin thin theo thi gian tun theo phng trnh Schroedinger:

= =i H

t 4.75

Phng trnh Schroedinger l dng lng t ca phng trnh Hamilton Jacobi c in.

Khi bit ton t Hamilton, t phng trnh Schroedinger, ta c th suy ra phng trnh xc nh ton t tin trin theo thi gian. Thc vy, t:

l l ll = = = = =0

0 0( ) ( , ) ( ) ( ) ( , ) ( )t U t ti i t H t HU t tt t

0t 4.76suy ra:

l ll ==0

0( , ) ( , )U t ti HUt

t t 4.77

Nu ton t Hamilton khng ph thuc tng minh vo thi gian, nghim ca phng trnh ny, vi iu kin ban u =0 0( , ) 1U t t , s l:

l l ( ) = =0 0( , ) expiU t t H t t 4.78

Tnh unitary ca ton t tin trin thi gian c bo m t tnh cht Hermitian ca ton t Hamilton.

2. Bc tranh Heisenberg Trong bc tranh Heisenberg, vect trng thi khng thay i theo thi

gian, cho nn, nu mi i lng trong bc tranh Heisenberg ta thm mt ch s H , th s lin h gia vect trng thi trong hai bc tranh s l:

l + = = 0 0( ) ( , ) ( )H t U t t t 4.79

91


S lin h ca ton t lA bt k trong hai bc tranh Schroedinger v Heisenberg s c suy ra t biu thc ca gi tr trung bnh, bi v, gi tr ny khng ph thuc vo bc tranh c th:

l l ll+ = = = * * 0 0( ) ( ) ( , ) ( , )H HA t A t dq U t t AU t t dq

l= * HH HA dq 4.80do :

l l ll+= 0 0( ) ( , ) ( , )HA t U t t AU t t 4.81Nh vy, trong bc tranh Heisenberg, ton t ph thuc vo thi gian.

Tnh o hm ca ton t ny, ta c: l l lll l lll

l lll ll l lll ll l l

+ +

+ + + +

= =

= = = =

= = = ,

H

H H

dA i iU HAU U AHUdt

i i iU HUU AU U AUU HU H A 4.82

Kt qu l: Trong bc tranh Heisenberg, cc ton t vt l bin thin theo thi

gian tun theo phng trnh Heisenberg: l l l = = ,

dA i H Adt

4.83

Phng trnh Heisenberg l dng lng t ca h phng trnh Hamilton. 3. Bc tranh Dirac Trong bc tranh Dirac, ton t Hamilton c chia lm hai phn:

l l m= +0H H H trong l ton t Hamilton t do, cn l 0H mH c gi l ton t Hamilton tng tc. Gi , vect trng thi trong biu din Dirac, thu c t vect

trng thi trong biu din Schroedinger bng php quay

( )I t l +0U . Php quay ny

c xc nh t ton t Hamilton t do : l 0Hl m m ( )+ = = =0 0 0 0 0( ) ( , ) ( ), ( , ) expI

it U t t t U t t H t t0 4.84

Khi , do l = 0 0( ) ( , ) ( )It U t t t , t phng trnh Schroedinger, ta c:

92


l m = + = 0( ) ( ) ( )ti H Ht

t 4.85

l l l m + = + = =0 0

0 00( )( , ) ( ) ( , ) ( ) ( )IItU t ti t i U t t H t H

t tt

l l l l l + = + =0 0 0 00 0( )( , ) ( ) ( , ) ( ) ( )IItH U t t t i U t t H t H t

t

l ml = = 0 00 0( )( , ) ( , ) ( )I Iti U t t H U t t t

t 4.86

Gi l ll+= 0 00( ) ( , ) ( , )I 0A t U t t AU t t , l biu thc ca ton t lA trong bc tranh Dirac, phng trnh trn cho ta:

= =I

I Ii Ht 4.87

Tip theo, ly o hm ca ton t vt l theo thi gian, ta c phng trnh:

l l ll l ll l+ += == =0 0 0 0 0 0( )IdA t i iU H AU U AH U

dt

l l = = 0 ,I Ii H A 4.88

trong l ton t Hamilton t do trong bc tranh Dirac.

l ml+= 0 00 0 0( , ) ( , )IH U t t H U t t0 Kt qu l:

Trong bc tranh Dirac, vect trng thi tho mn phng trnh Schroedinger vi ton t Hamilton tng tc:

= =I

I Ii Ht 4.89

cn cc ton t vt l bin thin theo thi gian tun theo phng trnh Heisenberg, vi Hamilton t do:

l l l = = 0 ,I

I IA i H At

4.90

3. K hiu Dirac

93


3. 1. Vect

phn bit vect trng thi vi hm sng, Dirac a vo cc k hiu sau y.

Mi vect trng thi trong khng gian Hilbert H , s c gi l mt ket, k hiu bng du na ngoc . Ch ci gi tn cc ket s c t vo bn

trong ca du na ngoc . Nh vy, ta s c cc ket , ,... . H { }q s c coi l h ket c s trong biu din to , h { }p s l h ket c s trong biu din xung lng,.

Tp hp cc dng thc tuyn tnh xc nh trn khng gian Hilbert H cng lm thnh mt khng gian vect, c s chiu bng s chiu ca khng gian Hilbert H , v c gi l khng gian i ngu2 ca H v k hiu l . H

Cc vect ca khng gian i ngu c gi l cc bra, k hiu l . Ch ci gi tn cc bra s c t vo bn trong ca du na ngoc . Nh vy, ta s c cc bra , ,...

Cho { }ie l h c s ca . Khi , h c s H { }ke ca c gi l i ngu vi

H{ }ie nu:

( )i ik ke e = 4.91Nu khng ni ngc li, cc c s ca lun lun l i ngu H . HSau y ta s s dng quy c cng theo ch s lp li. Theo quy c ny,

nu ta c tng ly theo mt cp ch s ging nhau, ta s b du tng. Ngc li, khi khng ni khc, trong mt biu thc c hai ch s ging nhau, n s c hiu l ly tng theo tt c cc gi tr kh d ca ch s . Ch s lp li c gi l ch s cm, bi v vai tr ch s ca n khng cn na. V d, trong biu thc:

1 21 2

k kk i k i i i

ke U e U e U e U = + "+ 4.92

k l ch s cm. Tn ca ch s cm c th thay i mt cch tu : , 1 1 2 2k rk i r ie G e G n n m m= = = + +" " 4.93

Khi khng mun dng quy c ny ni r. V d, nu khng mun cng theo trong biu thc k k k , ta s vit:

k k (khng cng theo ) k 4.94

2 Xem: i s, gio trnh dnh cho sinh vin Khoa Vt L, HKH TN, HQG H Ni.

94


Quy c ny khng p dng cho trng hp ch s lin tc. Ngha l, ta s khng b du tch phn trong biu thc sau y:

dq q q 4.95 Xt mt php bin i c s U trong khng gian H :

ki k

ke e = iU 4.96

Php bin i ny s cm sinh mt php i c s i ngu trong : Hi i

ke V e = k 4.97Do:

( ) ( ) ( ) pp p p s r p r s p ri i r i s r s i r i ie e V U e e V U V U VU = = = = = 4.98cho nn, ma trn i c s trong khng gian bra l ma trn nghimch o ca ma trn i c s trong khng gian ket:

1V U = 4.99 Nu ta c ket vi thnh phn l i :

iie = 4.100

quy tc bin i ca thnh phn i i vi php bin i c s s suy ra t h thc:

i k ii k i ke e U e

k = = 4.101aNh vy:

( )1 ik i k iiU U = = 4.101bDo cc thnh phn ca ket bin i bng ma trn nghch o ca ma trn

i c s, cho nn, cc ket c gi l vect phn bin. i vect bra, ta c:

i i ki i k ke V e

ke = = 4.102Ch n (4.99), ta c:

( )1kii k iV U = = 4.103Tc l, thnh phn ca bra bin i bng ma trn i c s . V l , vect bra c gi l vect hip bin.

U

Do H v ng cu mi ket s tng ng vi mt bra qua ng cu i ngu v ngc li:

H

a b a b + + , k i ii k iA e A e 4.104

95


Khng gian i ngu ca khng gian Hilbert, c th c nh ngha thng qua tch v hng. Nh vy, dng na tuyn tnh a s c nh ngha thng qua h thc: ( ) . ,a b a b a b a b a b= = 4.105trong a l nh ca a qua ng cu i ngu.

Trong khng gian Hilbert, ma trn i c s lun c chn l unitary, , cho nn, h c s i ngu v thnh phn ca ket s bin i bng

ma trn U . Trong khi , thnh phn ca bra s bin i bng ma trn U .

1U U+ = +

Ta c th chuyn dch tt c cc cng thc trc y gia cc vect sang ngn ng bra v ket. Nu { }a l h ket ring ca mt ton t vt l, iu kin trc chun ca chng s l: ( )a a a a = 4.106

Khai trin mt ket theo h ket ring { }a s c dng: ( )da a c a = 4.107

vi h s khai trin: ( )c a a = 4.108Nu h { }a l y , mi ket u c khai trin (4.107) vi h s

(4.108): da a a = 4.109

T suy ra, iu kin y ca h { }a s l: 1da a a = 4.110

H s khai trin (1.108) ca mt ket theo c s, c gi l hm sng ca ket trong h c s cho. ( ) ( )c a a a = 4.111

Tng t cho biu din xung lng. Nh vy, thnh phn ca vect trng thi trong Abiu din s l

hm sng: ( )a a = 4.112V d, trong biu din to { }q , ta c hm sng ( )q q = , cn

trong biu din xung lng { }p ta c hm sng ( )p p = . Trong biu 96


din nng lng vi ph lin tc { }E , l ( )E E = . Cn nu ph ca nng lng l gin on, ta thng vit hm sng vi bin nng lng di dng ch s:

nn EE = , hoc nn = .

Nh vy, cng mt vect trng thi , hm sng ca n trong cc biu din khc nhau s khc nhau. Vect trng thi khng ph thuc vo c s, trong khi hm sng li ph thuc vo s la chn c s.

Xc sut, gi tr trung bnh, u c din t thng qua hm sng. Tuy nhin, iu khng c ngha l, chng ph thuc vo biu din. Trong thc t, chng ch ph thuc vo vect trng thi, vic din t chng thng qua hm sng, ch ct thc hin nhng tnh ton c th. V d, nu xt trong Abiu din, mt xc sut i vi i lng A trong trng thi l:

( ) ( ) ( ) __________*a a a a a a a = = = 4.113Nu thay Abiu din bng B biu din, s dng tnh unitary ca php i biu din, ta c: ( ) ( )a a a b UV b b = = = 4.114

3. 2. Ton t trong k hiu Dirac

Trong khng gian vect hu hn chiu, khi la chn mt h c s

H n{ }ie , mi ton t lA s c din t bng mt ma trn ( )kiA : l k

i k iA e e= A 4.115Cc thnh phn ca ma trn, s l tch v hng ca kiA l iA e vi ke :

lki ki k iA A e A e = 4.116

Trong thnh phn ca mt ma trn, ch s trn (hoc ch s th nht) ch hng, trong khi ch s di (hoc ch s th hai) ch ct.

Vi khng gian Hilbert v s chiu, tr trng hp v hn m c, vic sp xp theo hng v ct l v ngha. Khi , cc ch s hng ct ca ma trn s c thay bng cc bin s. V d, yu t ma trn ,q q ca ton t lA trong c s to { }q s l:

l ( ),q A q A q q = 4.117

97


Yu t ma trn kiA thc ra c th hiu l bin xc sut di tc ng ca A h chuyn t trng thi ban u ie sang trng thi cui ke v

2kiA l

mt xc sut h thc hin vic chuyn di .

Mi ton t tuyn tnh lA trong H , s tn ti mt ton t i ngu mA+ trong , sao cho:H , H

m( ) ( )A A + = 4.118Nu ma trn ca lA trong mt h c s ca H l ( )ikA , th ma trn ( )ikB

ca mA+ trong h c s i ngu s l: m( ) ( )l( ) ( ) ( )

* *r ki r ii ik r k r k

i i r i rk r k r k

e A e e B e B e e B

e A e e e A e e A A

+ = = == = = ik

= 4.119

Nh vy, ma trn ca mA+ s l lin hp Hermite ca lA . Nu cho hai ket bt k, , , ta c:

A A += HJJJG 4.120Trong , chiu ca mi tn s ch chiu tc dng ca ton t. Nh vy, khi chuyn ma trn t na bn ny sang na bn kia ca tch v hng, ta phi thay n bng ma trn lin hp Hermite. Nh vy, iu kin (2.31) ca ton t Hermitian A A+= , s c din t bng:

A A A = =JG HJ 4.121i vi ton t Hermitian, trong biu thc A , c th hiu l A tc ng ln vect no ca tch v hng cng c.

Khi xc nh biu din, vic cho ma trn cng tng ng vi vic cho ton t (xem phn cui ca mc 1). Nu iie = l mt trng thi bt k, gi tr trung bnh ca i lng vt l A s hon ton c xc nh bi cc yu t ma trn ikA :

l l* k i ii k kA A e A e A

* ki = = = 4.122

98


Trong trng hp khng gian hu hn chiu, khi bit ma trn ca mt ton t, gi tr ring ca n c suy ra t phng trnh c trng:

det 0i ik k kA = 4.123v cc thnh phn ca vect ring ika ka tng ng vi cc gi tr ring k s l nghim ca phng trnh tuyn tnh thun nht: ( ) 0k k ii k i kA a = (khng cng theo ), k 1,2,...k = 4.124

Trong trng hp khng gian v hn chiu, d l m c, ta cng khng th gii trc tip phng trnh c trng, bi v n c bc v hn. Tuy nhin, nu s dng mt s h thc b tr, ta c th tm c cc yu t ma trn ca mt s ton t khc v t , tm c yu t ma trn hoc gi tr ring ca ton t lA . Cch din t ton t thng qua ma trn v t tm ph cc gi tr ring ca chng c gi l phng php ton t. N hon ton tng ng vi cch gii phng trnh Schroedinger ri tm ph ca cc i lng vt l. Chng ta s ni k vn ny trong chng V, khi ni v moment ng lng.

3.3 Dao ng t iu ho mt chiu

minh ho cho phng php ton t, n gin nht l xt h mt ht khi lng , dao ng mt chiu quanh v tr cn bng. H nh vy, c gi l mt dao ng t iu ho. V bi ton c t ra l: tm cc gi tr kh d ca ton t nng lng.

m

Trc ht, ta tm ton t Hamilton ca h, Khi chn gc to l v tr cn bng, th nng c in s c khai trin di dng:

( ) ( ) ( ) ( ) 20 0 02xU x U U x U = + + +"

Chn ( )0U = 0. Ti v tr cn bng th nng l cc tiu, cho nn, ( )0 0U = , ( )0 0k U = > .

Nh vy, khi ch xt cc dao ng nh, ta c th dng khai trin n s hng lu tha bc hai i vi x . Do F mx kx= = , suy ra:

( ) 2 2 22 2x mU x k x= =

trong , h s /k m = c gi l tn s dao ng. Nh vy, trong c hc lng t, ton t Hamilton cho dao ng t iu

ho, s c dng:

99


l m l2 22 2p m xHm

= +2

4.125

Thng thng, thay cho l ,p x ta dng cc ton t sau y c gi l ton t xung lng v to suy rng:

l l l , pP Q mm

= = x 4.126khi , giao hon t ca chng cng vn l (3.77), (3.78) nh ca lp v x :

l l,Q P i = = 4.127Thng qua cc bin mi, ton t Hamilton s c dng:

l m m( )2 2 212H P Q= + 4.128y chnh l dng tiu chun ca ton t Hamilton cho dao ng t iu ho.

Thay cho vic gii phng trnh Schroedinger, ta s dng h phng trnh ton t (hoc nh l Ehrenfest):

l l l l l,dQ iQ H Qdt

= = = = P l l l l l l2 2,d P i iP H P Qi Q

dt = = = = == =

4.129

Kt hp li, ta c: l l2 0Q Q+ = 4.130

Do Hamiltonian khng ph thuc tng minh vo thi gian, nng lng bo ton, s ph thuc ca vect trng thi vo thi gian c dng:

( ) exp it E t = = 4.131Ni ring, cc trng thi ring ng vi nng lng s l: nE

, exp nin t n E t = = 4.132

trong l s lng t ca nng lng. 0n Do ton t nng lng c xc nh thng qua ton t lQ , cho nn,

tm gi tr ring ca lH , ta hy tnh cc yu t ma trn ca ton t to lQ .

100


Chn h c s l cc vect ring ca ton t nng lng, cc yu t ca ma trn to s c dng:

( ) l ( ) l { }exp expm mn m niQ t m Q n E E t m Q n i t = = = n)

4.133

trong , . Khi : ( /mn m nE E = =( )2, m m m m mn n n n nQ i Q Q Q = = mn 4.134

Chuyn phng trnh (4.130) i vi ton t lQ thnh phng trnh gia cc thnh phn ca ma trn ca n, ta c:

( )22 2 0m m m mn n n nQ Q Q + = = 4.135T suy ra, cc yu t ca ma trn to s bng khng, tr trng hp:

mn = 4.136

Nu xp xp cc mc nng lng, sao cho, 1m mE E + = = 4.137

th mn = khi v ch khi, c s chuyn di gia hai trng thi c s lng t sai khc nhau mt n v:

1mm = 4.138

v nh vy, ch c cc yu t ma trn 1mmQ mi khc khng.

Mt khc, da vo h thc giao hon c bn v (4.1290, ta c: l l l l, ,Q P i Q Q = = = 4.139

suy ra:

ll ll( )ii r i rk k r k r r kk

i QQ QQ i Q Q i Q = = = i i rQ 4.140Ly cc phn t ng cho, i k n= = , ch n (4.138):

r n r n n rn r n r r ni i Q Q i Q Q = = (khng cng theo ) n

1 1 1 1 11 1 1 1 1

n n n n n n n n n n n nn n n n n n n n n n n ni Q Q i Q Q i Q Q i Q Q + + + + + += + 11 =

1

1 1 1

1 1 1 1n n n n n n n nn n n n n n n ni Q Q i Q Q i Q Q i Q Q + + + + + +

Ton t to l Hermitian, v ta s chn biu din sao cho yu t ma trn ca n l thc, khi :

m nn mQ Q=

101


Nh vy, cc yu t khc khng ca ma trn to s tho mn phng trnh:

( ) ( )2 21 1 11 1 1 2n n n n n nn n n n n nQ Q Q Q Q Q + ++ = = = 4.141ngha l, ( lm mt cp s cng vi cng bi l: )21nnQ +

2d =

=

Nu chn yu t ma trn tng ng vi 0n = , 01 0Q = , ta c: ( )2 11 12 2n n nn n nn nQ Q Q = = == = 4.142

Kt qu l, ta c ma trn to Q v : X

l 0 1 0

1 0 22 0 2 0

Q m x

= =

""="

" " " "

4.143

T h thc: l( )mm mn n

n

mnP Q i Q= = 4.144

suy ra, ch c cc yu t 1 1 / 2 0n n

n nP P i n = = = . Ma trn xung lng lP v lp s l:

l l0 1 0

1 0 22 0 2 0

pP im

= =

""="

" " " "

4.145

Ma trn lQ v lP chnh l ton t to v xung lng ca dao ng t iu ho trong biu din nng lng.

T kt qu , ta c th xc nh ph ca ton t nng lng. Thc vy, do h c s l cc vect ring ca ton t Hamilton, cho nn, ma trn ca ton t

102


Hamilton s c dng ng cho. Cc yu t trn ng cho chnh l cc gi tr ring ca ton t Hamilton:

m m( ) ( )2 2 2 21 12 2nn r n n r n rn n r r nnH P Q Q Q Q Q = + = + r n = ( )2 1 2 1 2 1 2 11 1 1 112 n n n n n n n nn n n n n n n nQ Q Q Q Q Q Q Q += + + + + =

( )2 1 1 21 1 2 2 2n n n nn n n n nQ Q Q Q = + = = =+ Ngha l:

12n

E n = + = 4.146Kt qu ny rt ph hp vi thc nghim, v s c tm li c bng

nhiu cch khc nhau trong c cch gii phng trnh Schroedinger.

103

chuong 4 ch.iv- theory of representation

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