i hc hu

Chng 1

C s c hc lng t1.1. Mt s vn m u1.1.2. Ph nguyn t

Mt trong nhng yu cu t ra i vi mi l thuyt v nguyn t l gii thch c s xut hin ph vch ca nguyn t v mt s tnh cht ca chng.

Khi nung nng mt cht (bng ngn la, phng in trong chn khng, h quang...) ti mt nhit ln th n pht sng. V d cho t NaCl vo ngn la n cn th ngn la nhum mu vng thm. nh sng vng y l do nguyn t Na (xut hin trong qu trnh nhit phn NaCl trong ngn la) pht ra. Phn tch nh sng ngn la c cha hi Na bng mt quang ph k ngi ta thy bn cnh ph lin tc ca nh sng ngn la l mt vch m mu vng c bc sng 5892 A0 (vi quang ph c phn gii cao s thy d l mt vch kp). Ph xut hin nh vy gi l ph pht x.

Tri li, nu chiu nh sng trng qua hi Na th trn ph lin tc, v tr tng ng vi vch vng Na l mt vch ti. l ph hp th ca Na. Nguyn t c kh nng hp th nh sng c tn s ng bng tn s nh sng pht x ca n.

Ph nguyn t H vng thy c c cu trc c bit n gin. Balmer (1885) tm thy cc ph vch nguyn t H c bc sng tun theo cng thc n gin:

( =

(1.1)

vi K = 3645,6 . 10-7mm v m = 3,4,5...

Cng thc Balmer c Rydberg (1896) v Ritz (1908) khi qut ho:

= RH ()

(1.2)

n 1 = 1, 2, 3, ...

n 2 = n 1 + 1, n 1 + 2, ...

RH = gi l hng s Rydberg. Thay n 2 = m v n 1 = 2 ta c c cng thc Balmer. Cho n 1 cc gi tr 1,2,3,... v n 2 cc gi tr nguyn ln hn n 1 ta c cng thc biu din ton b ph nguyn t H. Theo Ritz, ngi ta gi cc i lng R/n12 v R/n22 l cc s hng. Nh vy mi mt vch ph ng vi hai s hng. Mi mt gi tr ca n1 c trng cho mt dy ph.

Cc dy ph ca nguyn t H

n 1n 2Dy phVng ph

12,3,...LymanCc tm

23,4,...BalmerNhn thy v gn cc tm

34,5,...PaschenHng ngoi gn

45,6,...BrackettHng ngoi xa

56,7,...PfundHng ngoi xa

1.1.2.Thuyt lng t Planck

a vt l thot ra khi S khng hong t ngoi, nm 1900 nh vt l ngi c l Max Planck a ra thuyt lng t gi l thuyt lng t Planck.

Theo thuyt lng t Planck th: Mt dao ng t dao ng vi tn s ( ch c th pht ra hay hp th nng lng tng n v gin on, tng lng nh mt nguyn vn, gi l lng t nng lng (. Lng t nng lng ny t l vi tn s ( ca dao ng t".

( = h.(

(1.3)

(h = 6,625.10-27erg.sec = 6.625.10-34 J.s)

ngha quan trng ca thuyt lng t Planck l pht hin ra tnh cht gin on hay tnh cht lng t ca nng lng trong cc h vi m. Nng lng ca electron trong nguyn t, nng lng quay, nng lng dao ng ca cc nguyn t hay nhm nguyn t trong phn t ... u nhn nhng gi tr gin on xc nh.

Theo thuyt lng t Planck th nng lng ca dao ng t dao ng vi tn s ( ch c th nhn nhng gi tr gin on:

0, h(, 2h(, 3h(, 4h(, ... nh(ngha l bi s nguyn ln lng t nng lng ( = h(. Do , ta c th biu din E theo cng thc:

E = nh( (n = 0, 1, 2, 3,...)

Mt khc, v nng lng ca dao ng t pht ra hay hp th di dng nng lng bc x nn thuyt lng t Planck cng c ngha l:

nh sng hay bc x ni chung gm nhng lng t nng lng ( = h.( pht i t ngun sng.

V vy, thuyt lng t Planck cn c gi l thuyt lng t nh sng.

1.1.3. Tnh cht sng - ht ca nh sng

H.Hetz (1887) khi lm th nghim chng minh s tn ti ca sng in t trong l thuyt cu MaxWell pht hin ra rng nh sng cc tm c tc dng tr lc cho s phng in trong chn khng. Sau , (1900) Lenard ch ra rng nguyn nhn ca hin tng trn l do nh sng cc tm gii phng electron ra khi b mt catt. Hin tng electron c gii phng ra khi b mt kim loi di tc dng ca nh sng c gi l hiu ng quang in.

einstein (1905) cho rng c th m rng thuyt lng t ca Planck gii thch hiu ng quang in. V vy, Einstein a ra thuyt ht hay thuyt lng t nh sng. Theo thuyt lng t nh sng ca Einstein th nh sng hay bc x ni chung l mt thng lng cc ht vt cht c gi l photon (quang t) hay lng t nh sng vi mt lng t nng lng:

( = h(

(1.3)

Electron trong kim loi hp th hon ton v ngay lp tc ton b nng lng ca photon khi n tng tc vi photon.

Trong nhng iu kin nht nh nh trong cc th nghim giao thoa v nhiu x, bc x in t th hin tnh cht sng ca chng; cn trong iu kin khc, nh trong hiu ng quang in, chng li c bn cht ht. Tnh cht gi l lng tnh sng- ht ca bc x in t.Theo h thc ca einstein, gia khi lng m ca mt vt v nng lng E ca n c h thc:

E = m.c2 (c: vn tc nh sng)

(1.4)

Do , i vi photon ta c: mc2 = h.( = h. hay m =

T suy ra p = m .c =

(1.5)

Nh vy, phng trnh (1.5) cho thy mi quan h ca m (c trng tnh cht ht) v ( (c trng cho tnh cht sng). y l phng trnh quan trng cha ng bn cht nh nguyn ca bc x in t.

1.1.4. Tnh cht sng- ht ca ht vt cht (sng vt cht De Broglie)

Nm 1924, nh vt l Php Louis De Broglie cho rng c th m rng bn cht nh nguyn sng - ht ca bc x in t do Einstein pht hin ra cho mi vt cht. Gi thit ca De Broglie ch yu da trn c s trit hc v s i xng trong t nhin. C th chia th gii vt cht thnh hai phn l bc x v vt cht. Bn cnh thuc tnh sng, bc x cn c thuc tnh ht. Suy ra, ngoi bn cht ht, vt cht cn c tnh cht sng.

S chuyn ng ca mt ht vt cht bt k c th c xem nh mt qu trnh sng c bc sng ( v tn s ( :

( = ; ( = =

(1.6)

m: khi lng ca ht ; p: ng lng ca ht

v: vn tc ht ; h: hng s Plank.

Biu thc (1.6) gi l biu thc De Broglie hay l nhng phng trnh c bn ca sng vt cht De Broglie.

Nu c mt ht vt cht ta c biu thc sng:

((x,t) = a.ei.(Et - px)/ h (1.6) : sng vt cht De Broglie

1.1.5. Nguyn l bt nh Heisenberg

Trong c hc c in khi nghin cu chuyn ng ca cc ht, ngi ta phi ni n qu o ca chng, lc ti mt thi im bt k ta c th xc nh c to v ng lng ca ht.

Trong c hc lng t, khi ni n tnh sng ca ht vt cht th khi nim qu o khng cn ngha na.

Theo h thc De Broglie ta c:

(= ( p =

V ( khng phi l hm ca to , do p khng th l hm ca to . iu ny c Heisenberg pht biu qua h thc bt nh:

To v ng lng ca ht tng ng vi to l khng th ng thi xc nh.

Biu thc bt nh Heisenberg:

(x. (px ( (1.8)

(x: bt nh ca to

(px: bt nh ca ng lng trn phng x.

Bit (px = m. (Vx

Suy ra : ( x. (Vx (

(1.9)

V = const, nn (Vx cng nh (Vx cng chnh xc) th (x cng ln (x cng bt nh) v ngc li. C ngha l ta khng th xc nh c ng thi mt cch chnh xc v tr x v vn tc Vx ca mt electron trong nguyn t. Nu bit Vx th khng th xc nh chnh xc to x ca n, tc l khng tn ti qu o ca electron trong nguyn t.

Nguyn l bt nh Heisenberg cng ng trong trng hp ca h v m, nhng v ht v m th tnh cht sng- ht l rt b nn t c p dng.

T hai tnh cht vt l ca ht vt cht ta c th rt ra tnh cht c trng ca h vi m:

Cc i lng vt l ca ht vi m u gin on.

To x v ng lng ca ht l khng th ng thi xc nh

Chuyn ng ca ht vi m khng c qu o

1.1.6. S khc nhau gia c hc c in v c hc lng t

Da trn cc s liu thc nghim thu c v cc hin tng quan st, ta c th tm tt s khc nhau chnh gia hai loi c hc nh sau:

C hc c in

- Chuyn ng ca ht c qu o

- Cc i lng vt l (nng lng, ng lng, m men ng lng.. .) c th nhn bt c gi tr no.

- Cc i lng c hc u c th xc nh c ng thi. C hc lng t

- Chuyn ng ca ht khng c qu o.

- Cc i lng vt l ch c th nhn nhng gi tr gin on hay c lng t ho.

- To v ng lng tng ng vi to l khng th ng thi xc nh.

1.2. Ton t v hm sng

Do h lng t c cc thuc tnh khc bit vi h v m, nn ngi ta khng th biu din cc i lng vt l ca h ny bng cc biu thc gii tch thng thng nh trong c hc c in m phi dng n mt cng c ton hc mi c kh nng m t bn cht ca h lng t. Mt trong nhng cng c y l ton t tc dng ln hm sng.

1.2.1. Ton t

a- nh ngha: Ton t l mt php ton khi ta tc dng ln mt hm th cho ra mt hm mi.

Thc hin cc php ton c qui c trong ton t A i vi hm s (x ng sau n ta nhn c hm mi (x. Hay ni cch khc (x l kt qu ca s tc ng ton t A ln hm s (x.

K hiu: (x = (x (1.10)

V d: Ton t A hm s hm mi

nhn vi a x

ax

d/ dx

x4 + 5 4x3

Ton t A = nhn vi a c ngha l thc hin php nhn a vo hm s ng sau n.

= d/ dx ngha l ly o hm theo x hm s ng sau n. Ngi ta thng k hiu cc ton t: , , .. .

b. Cc php ton v ton t

1. Php cng ca hai ton t A v B:

Tng cc ton t A v B l ton t C ( = +) sao cho khi tc dng ln hm u (tu ) th bng + tc dng ln hm u .

+= nu u = u + u

V d: = x; = d/ dx ; u = U (x)

= x + d /dx u = xu + du / dx = ( x+ d /dx)u

2. Tch cc ton t: Tch hai ton t A v B l ton t C hay C' sao cho:

= . ( u = [u]

= . ( u = [u]

V d: = x , = d /dx

u = [u] = x.du /dx

u = [u] = d/dx (x.u) = x. du/dx + u ( u

Nu . ( . th ta ni hai ton t , khng giao hon vi nhau,

ta gi [,] = . - . l giao hon t ca hai ton t v .

Nu . = . th ta ni hai ton t v giao hon.

[,] = . - . = 0

3.Lu tha ca ton t: Lu tha ca ton t c nh ngha:

2u = (.)u = (u)

Vy 2 = . l tc dng lin tip hai ln.

V d: = , u(x) = x4

2 u = (du/dx) = (4x3) = 12x2

1.2.2. Ton t tuyn tnh

a. .nh ngha: Ton t c gi l ton t tuyn tnh nu n tho mn biu thc sau:

(au + bv) = au + bv

(1.11)

u,v: hm ; a,b: cc hng s bt k

V d: ton t ca hm f(x) theo x l ton t tuyn tnh v:

(af1(x) + bf2(x) ) = a. f1(x) + b. f2(x)

Mt s ton t tuyn tnh nh: ton t nhn (vi mt s, mt hm s)

+Ton t ( , vi phn: , .. .

+Ton t Laplace: ( =

+Ton t Napla:

+Ton t Hamilton H = - ( + U(x,y,z)

Cc ton t khng tuyn tnh: , ( )m (m 1);

b. Tnh cht ca ton t tuyn tnh

Nu hai ton t , l ton t tuyn tnh (t4) th t hp tuyn tnh ca chng l ton t tuyn tnh v tch ca chng nhn vi mt s cng l ton t tuyn tnh.

, : t4 th (a. + b. ) : t4

(c. . , d.

EMBED Equation.3 ) : t4

c. Hm ring v tr ring ca ton t tuyn tnh

1. nh ngha: Nu kt qu tc ng ca ton t tuyn tnh ln mt hm u bng chnh hm u nhn vi tham s L no , th ta gi u l hm ring v L l tr ring ca ton t :

u = Lu (1.12)

u l hm ring ca , cn L l tr ring ca ng vi hm ring u.

Vd: (eax) = a. eax

hm u(x) = eax l hm ring ca ton t , cn a l tr ring ca ton t v ng vi hm ring eax

Phng trnh (1.12) c gi l phng trnh hm ring- tr ring ca ton t .

2. Tr ring khng suy bin v suy bin

Mt ton t tuyn tnh c th tn ti nhiu hm ring v tr ring khc nhau. Tp hp cc tr ring ca gi l ph cc tr ring. Ph cc tr ring c th l lin tc hoc gin on, hoc mt phn gin on mt phn lin tc.

- Nu ng vi mi hm ring u ch c mt tr ring L th ngi ta ni tr ring l khng suy bin.

- Nu ng vi mt tr ring L ta c k hm ring u th ta ni tr ring L suy bin k ln hay suy bin bc k.

V d: u1 = Lu1

u2 = Lu2

uk = Luk

( L l tr ring suy bin bc k

d. Cc nh l v hm ring v tr ring ca ton t tuyn tnh

1. nh l 1: Nu un l hm ring ca ton t tuyn tnh ng vi tr ring Ln v a l mt hng s tu 0 th aun cng l hm ring ca ng vi tr ring Ln.

un = Lnun (1.13)

(a.un) = Ln(a.un) (1.14)

2. nh l 2: Nu Ln l tr ring suy bin bc k ca ton t :

u1 = Lnu1

u2 = Lnu2

uk = Lnukth t hp tuyn tnh ca k hm ring cng l hm ring ca ng vi tr ring Ln.

(c1u1 + c2u2 +.. . + ckuk) = Ln(c1u1 + c2u2 +.. . + ckuk) (1.15)

3. nh l 3: iu kin cn v hai ton t v c chung hm ring l chng phi giao hon vi nhau.

u = Au

v = Bu

( [, ] = 0 ( u =v

1.2.3. Mt s khi nim v cc h hm

a. H hm trc giao: H hm u, v, w .. . c gi l h hm trc giao nu tch phn ca mt hm no vi lin hp phc ca mt hm khc lun bng 0 trong ton phm vi bin i ca hm s.

u.v* dx = 0, u.w* dx = 0 , v.w* dx = 0 ...

b. Hm chun ho: Hm ( c gi l hm chun ho nu ((*dx = 1.

hay ( 2dx = 1

(1.15)

( cha chun ho: ( 2 dx = N ( N 1)

c c hm ( chun ho, ngi ta chia phng trnh ny cho N:

( 2 dx ( ((*dx = 1

( ) ( (* ) dx = 1

Hm ( = ( l hm chun ho; l tha s chun ho.

c. H hm trc chun

(1, (2, .. ., (m,.. ., (n gi l h hm trc chun nu n chun ho v trc giao vi nhau tng i mt.

((m*(ndx = 1 : nu m = n (1.16)

= 0 : nu m n

d. H hm y : H hm (1, (2, .. ., (m,.. ., (n c gi l h hm y , nu hm ( bt k c th khai trin thnh chui tuyn tnh ca h hm y.

( = C1(1 + C2(2 + .. . + Cm(m + .. . + Cn(n = ( Ci(i (1.17)

Ci : h s khai trin chui

Nu h hm y cng l h hm trc giao th ta c th xc nh c h s khai trin chui.

V d: Mun xc nh Cm th ta nhn phng trnh vi (m* v ly ( ((m*(dx = C1 ((m*(1dx + C2 ((m*(2dx + .. . + Cm ((m*(mdx + .. . + Cn ((m*(ndx

Nu h hm y tho mn tnh cht chun ho th: Cm = ( (m*(dx

e. Hm u ho (hm u n)

Hm ( c gi l hm u ho nu n n tr, hu hn v lin tc trong phm vi bin i ca bin s.

1.2.4 Ton t tuyn tnh t lin hp (ton t Hermite)

a. nh ngha: Ton t c gi l ton t Hermit nu n tho mn h thc sau:

v* u dx = u. * v*dx (1.18)

u,v l cc hm bt k, bng 0 + ( v - (

u*, v*, * l lin hp phc ca u,v,

Cc ton t Hermit:

= x; = U(x,y,z); = -i (ton t ng lng px )

= ; = - ( + U(x,y,z)

= - (b. Cc nh l v hm ring v tr ring ca ton t Hermit

1. nh l 1: Tr ring ca ton t Hermit l tr thc: Ln = Ln*

Tht vy, nu l ton t tuyn tnh Hermit v Ln l tr ring ca th ta c:

(n = Ln(n

(1)

v ( (n* (n d( = ( (n * (n d(

(2)

(1) ( (* (n = (n* n (n ( ( (n* (n d( = ( (n* (n d( = Ln ( (n* (n d(

(3)

Ly lin hp phc ca (1): Ln* (n = Ln* (n* (4)

(4) nhn vi (n v ly ( ta c:

( (n Ln* (n* d( = ( (n Ln* (n* d( = Ln* ( (n (*n d(

(5)

t (2) (3) V (5) suy ra: Ln( (n* (n d( = Ln* ( (n (n* d(hay Ln ( ( 2 d( = Ln* ( ( 2 d( Ln = Ln*

Vy tr ring ca ton t Hermit l tr thc

2. nh l 2: Tp hp tt c cc hm ring khc nhau ca mt ton t Hermit c ph tr ring gin on lm thnh mt hm trc giao.

(n = Ln (n (1)

(m = Lm (m (2)

(Ln Lm)

( (n* (n d( = ( (n * (n* d( (3)

T (1) nhn (m* ri ly ( ta c ( (m* (n d( = ( (m* Ln (n d(

( ( (m* (n d( = Ln ( (m* (n d(

(4)

Ly lin hp phc (2) ri nhn vi (n , sau ly tch phn ta c:

( (n * (m* d( = Lm* ( (m* (n d( = Lm ( (n (m* d( (5)

T (3) so snh (4) v (5) ta c:

Ln ( (m* (n d( = Lm ( (n (m* d( (Ln - Lm ) ( (m* (n d( = 0

( ( (m* (n d( = 0 l iu phi chng minh.

c. Tnh cht ca ton t tuyn tnh Hermit

- Nu l ton t tuyn tnh Hermit th .a (a 0) cng l ton t tuyn tnh Hermit.

V d: Ton t i. l ton t tuyn tnh Hermit th -i cng l ton t tuyn tnh Hermit.

- Nu v l ton t tuyn tnh Hermit th giao hon t .= . cng l ton t tuyn tnh Hermit.

- Ton t A v B l Hermit th tng hoc hiu ca chng cng l ton t tuyn tnh Hermit.

- Nu v l cc ton t Hermit th t hp tuyn tnh ca chng cng l ton t tuyn tnh Hermit.

- Nu (n khng phi l hm ring ca ton t Hermite L, ngha l (n Ln(n th ngi ta gi gi tr Ln thu c l gi tr trung bnh hay k vng ton hc ca v c biu din nh sau:

=

thu c cng l tr thc.

Thng qua cc thuc tnh quan trng ca ton t tuyn tnh Hermite ta thy rng ch c loi ton t ny mi kh nng biu din bn cht ca cc i lng vt l ca h lng t. V cng l l do ti sao ton t Hermite l cng c ton hc trong c hc lng t.

1.3. H tin ca c hc lng t

1.3.1. Tin v hm sng (tin 1) - Nguyn l chng cht cc trng thi

a. Hm sng

1. Ni dung: Mi trng thi ca mt h vt l vi m (h lng t) c c trng bng mt hm xc nh, n tr, hu hn, lin tc ph thuc vo thi gian t v to q, k hiu l hm ( (q,t); gi l hm sng hay hm trng thi ca h .

Mi thng tin v h lng t ch c th thu c t hm sng m t trng thi cu h.

2. ngha vt l v tnh cht ca hm sng

- V hm sng ( (q,t) ni chung l hm phc nn n khng c ngha vt l trc tip, m ch c bnh phng modun ( 2 (tr ny l thc) ca hm sng mi c ngha

l mt xc sut tm thy ht ti to tng ng, chnh l ngha vt l ca hm sng.

- Nu gi dw l xc sut tm thy ht trong mt th tch dv xung quanh mt im no trong khng gian th ta s c: dw = dv

Mt xc sut =

(1.20)

Nu ly tch phn ca trong ton khng gian ta s c xc sut tm thy ht trong ton khng gian, theo l thuyt xc sut th xc sut ny bng 1.

(dv = 1

(1.21)

Biu thc (1.21) mun tho mn tch phn (dv phi c gi tr hu hn, ngha l ( ( 0 nhanh v cc.

y l iu kin chun ho ca hm sng, hm ((q,t) gi l hm chun ho.

Ngoi ra, hm ((q,t) phi tho mn tnh cht n tr, hu hn v lin tc tho mn tnh cht ca mt hm mt v:

1- Tnh n tr: V biu th mt xc sut ca ht v xc sut l mt i lng hon ton xc nh nn ( phi l mt hm n tr ca to , n khng ti mt to xc nh ta s thu c nhiu gi tr xc sut v iu ny hon ton khng c ngha vt l.

2- Tnh hu hn: V xc sut l hu hn nn hm sng ( phi hu hn ti mi v tr.

3- Tnh lin tc: V trng thi ca h lng t phi bin i lin tc trong khng gian, nn hm sng ( m t trng thi ca ht phi l mt hm lin tc.b. Nguyn l chng cht trng thi

Trong c hc lng t xut pht t bn cht ca hm sng ngi ta tha nhn mt nguyn l, gi l nguyn l chng cht trng thi. y l mt nguyn l c bn ca c hc lng t.

Nu cc hm (1, (2,.. ., (n l cc hm sng m t trng thi ca mt h lng t, th t hp tuyn tnh ca chng cng m t c trng thi ca h lng t .

( = C1(1 + C2(2 +.. . + Cn(n : hm trng thi (1.22)

C1 , C2, .. . l nhng h s tu .

Nguyn l chng cht phn nh tnh cht c lp ca mt trng thi ny i vi mt trng tha khc.

1.3.2.Tin v ton t (tin 2)

a. Ni dung: Tng ng vi mi i lng vt l L ca h lng t trng thi ( th c mt ton t Hermit L tng ng

Gia cc ton t ny c cc h thc ging nh nhng h thc i lng vt l trong c hc c in.

b. Mt ton t trong c hc lng t tng ng vi mt i lng vt l trong c hc c in

1. Ton t to : = x

Mt cch tng qut (x,y,z) = q( x,y,z)

2. Ton t xung lng (ng lng) thnh phn

px (

py (

pz (

3. Ton t xung lng

=

= -i

4. Ton t bnh phng xung lng

5. Ton t m men ng lng thnh phn

Mx = ypz - zpy (

My = zpx - xpz (

Mz = xpy -ypx (

6. Ton t th nng

U(x,y,z) (

7. Ton t ng nng

T = (

8. Ton t nng lng (ton t Hamilton)

E = T + U (

Thay cc gi tr ta c:

1.3.3. Tin v tr ring v i lng o c

a. Ph tr ring ca ton t Hermite v nhng gi tr kh d ca cc i lng vt l tng ng

i lng vt l L ca mt h lng t mt thi im ch c th nhn nhng gi tr ring ca ton t tng ng tho mn phng trnh tr ring thi im t:

(n = Ln (n

(1.24)

b. Nhng gi tr ( m i lng vt l L c gi tr xc nh

Nu h lng t trng thi ( m hm ( ny ng nht vi mt hm ring (k no ca ton t Hermite , th trng thi ( i lng vt l L c gi tr xc nh v bng tr ring Lk ca ton t tuyn tnh Hermite .

Nhng trng thi (L m mt i lng vt l L c gi tr xc nh l nhng trng thi tho mn phng trnh tr ring ca ton t tng ng .

(L = L(L

c. Xc sut mt i lng L c mt gi tr Li

Nu h lng t vo trng thi (, m ( khng trng vi mt hm ring no ca th i lng vt l L ca trng thi ( khng c gi tr xc nh. i lng L ch c th nhn mt trong nhng gi tr xc nh Li ca ph tr ring ca ton t , nhng khng bit chc l tr no. V th ngi ta phi xc nh L theo nh lut xc sut.

Xut pht t nguyn l chng cht trng thi v tnh y , trc giao ca h hm ring ca ton t tuyn tnh Hermite ngi ta biu din hm ( m t trng thi ca h thnh chui tuyn tnh theo cc hm ring.

( = C1(1 + C2(2 + .. . + Cn(n = Ci(i (1.25)

Nh vy, trng thi ( c xem l s chng cht nhng trng thi ring Ui ca ton t Hermite . Lc ng vi mi trng thi ring trn, i lng vt l L nhn nhng gi tr xc nh Li l tr ring tng ng vi hm ring Ui.

Xc sut L nhn gi tr Li l W (Li) = .

( = 1 : iu kin chun ho.

Vi W (Li) l xc sut i lng L nhn mt trong nhng gi tr c th c ca Ln.

T (n = Ln (n ( (n* (n = (n* L (n = Ln (n* (n Ln = ( (n* (n d( ( (n* (n d(

Thc t trong c hc lng t t khi tm c (n l mt hm ring ng, m ch tm c hm ring gn ng. Do tr ring (n tm thy l tr trung bnh:

=( (n* (n d( (1.26)

( (n* (n d(

Gi tr trung bnh ny cn gi l k vng ca L.

1.3.4. iu kin hai i lng vt l c gi tr xc nh ng thi trong cng mt trng thi

Ta bit, i lng vt l A ca trng thi (1 c gi tr xc nh nu ( 1 l hm ring ca ton t . i lng vt l B ca trng thi ( 2 c gi tr xc nh nu (2 l hm ring ca . Do , hai i lng vt l A, B ca cng trng thi ( s c gi tr xc nh ng thi nu ( l hm ring chung ca hai ton t , ; khi hai ton t v phi giao hon vi nhau. Ngc li, nu hai ton t giao hon th chng s c chung hm ring v hai i lng vt l tng ng s c gi tr ng thi xc nh.

Vy: iu kin cn v hai i lng vt l ca h lng t c tr xc nh ng thi trong cng mt trng thi l cc ton t ca chng giao hon vi nhau.

Mt s th d:

a. Cc ton t giao hon:

Ton t , , giao hon vi nhau tng i mt

[,] = 0; [,] = 0; [,] = 0

Vy cc to x, y, z ca mt ht c th nhn ng thi nhng gi tr trong cng mt trng thi.

- Ton t thnh phn ng lng px, py, pz giao hon vi nhau tng i mt, nn c gi tr ng thi xc nh trong cng mt trng thi.

b- Cc ton t khng giao hon:

- ng lng v to : Cc ton t to v thnh phn ng lng tng ng vi to khng giao hon, nn tng i mt khng th c gi tr xc nh ng thi. Nhng mt ton t to v ton t thnh phn ng lng ng vi to khc li giao hon. Do , chng li c th ng thi xc nh trong cng mt trng thi.

-Ton t thnh phn momen ng lng: Ton t thnh phn momen ng lng khng giao hon vi nhau tng i mt. Do , cc thnh phn Mx, My, Mz ca momen ng lng khng th c nhng gi tr xc nh.

[x, y] = i z ; [y, z] = i x ; [zx] = i y

Tuy nhin, ton t bnh phng mmen ng lng 2 = x2 + y2 + z2 li giao hon vi mi ton t x, y, z.

[ 2, x] = [ 2,, y] = [ 2, z] = 0

Do , 2 v thnh phn mmen ng lng no l c th ng thi xc nh.

Ta c: 2( = M 2(

z( = Mz(

Mt cch hon ton tng t chng ta cng c th chng minh c ba ton t hnh chiu momen ng spin Sx, Sy, Sz cng mt trng thi khng giao hon vi nhau tng i mt. Ngc li, ton t bnh phng momen ng spin giao hon vi mt trong Sx, Sy, Sz

1.3.5. Tin v phng trnh Schrodinger-Trng thi dng

a. Tin 3 - Phng trnh Schodinger tng qut

Hm sng ((q,t) m t trng thi ca h lng t bin thin theo thi gian c xc nh bi phng trnh Schrodinger tng qut:

(1.27)

i = , : ton t Haminton = (q,t)

( : hm sng m t trng thi ca h theo thi gian ((q,t)

Phng trnh (1.27) do Schrodinger a ra nm 1926 nh mt tin , ngha l khng th suy ra t bt k mt nguyn l no khc. S ng n ca phng trnh ch c th c khng nh bng cc kt qu kim chng khi p dng cho cc h lng t c th.

Phng trnh (1.27) l phng trnh vi phn tuyn tnh thun nht; do nu (1 v (2 l hai nghim c lp ca (1.27) th mi t hp tuyn tnh ( = C1(1 +C2(2 ca chng cng l nghim ca phng trnh.

Nu ( l hm chun ho; (1 , (2 ... l trc chun, cn C 1, C2 ... l nhng s ni chung phc v khng ng thi bng khng th:

C 1 2 + C2 2 + ... + Cn 2 = 1

V vy, phng trnh Schodinger tng qut cng th hin nguyn l chng cht trng thi trong c hc lng t. Do nhng iu , phng trnh Schrodinger tng qut l phng trnh gc v ton t Haminton l ton t quan trng nht ca c hc lng t khng tng i tnh.

b. Phng trnh Schodinger ca cc trng thi dng

Gi s h lng t vo mt trng th U khng ph thuc vo thi gian, ch ph thuc vo to = U(q), th khng ph thuc vo thi gian. Lc ch tc ng ln phn ph thuc to ca hm ( (q,t). Do , hm ((q,t) tch thnh hai phn:

((q,t)= ((q).F(t)

Thay vo phng trnh Schodinger tng qut:

(1.28)

(

(1.29)

Hai v ca ng thc (1.29) ph thuc vo hai bin s khc nhau, nn hai v ch c th bng nhau khi hai v phi bng cng mt hng s ( no :

(1.30)

(1.31)

T (1.31) ( ((q) = (((q)

(1.32)

(1.32) l phng trnh hm ring tr ring ca , m tr ring ca l nng lng ton phn E nn ( = E l tr thc.

Cc hm ((q) l hm ring ca ton t , n m t nhng trng thi nng lng khng bin i theo thi gian E = ( = const. Trng thi c E khng bin i theo thi gian gi l trng thi dng.

Phng trnh Schodinger cho trng thi dng:

((q) = E. ((q)

(1.33)

hay

= 0

(1.34)

Phng trnh (1.33) hoc (1.34) l phng trnh quan trng nht ca c hc lng t. V ho hc lng t ch yu nghin cu cc h trng thi dng.

Gii phng trnh (1.30) = E ta c

F(t) = C.e-i Et / h gi l tha s n sc hay tha s pha ca hm sng.

Nh vy: nghim tng qut ca phng trnh Schrodinger s l:

((q,t) = ((q).F(t)

((q,t) = ( (q). e-iEt / h (

(1.35)

Phng trnh (1.35) cho ta thy trng thi dng, mt xc sut khng ph thuc vo thi gian. Do , khi gii phng trnh Schrodinger cho trng thi dng ta ch cn tm n ((q) l , v ha lng t ch yu nghin cu cc trng thi dng ca phn t.

1.4. Mt s bi ton ng dng

1.4.1. Bi ton vi ht trong hp th mt chiu

Gi s c mt tiu phn (ht) khi lng m chuyn ng trong hp th mt chiu theo phng x vi b rng OA = a. Trong khong 0 ( x ( a th nng ca h khng i. nhng v tr bn ngoi hp (x < 0 v x > a) th c nhng trng lc lm cho th nng ca ht tng v hn. Ni cch khc chuyn ng ca ht b gii hn trong hp:

U = Const = 0 vi 0 ( x ( a

U = ( vi x < 0 v x > a

M hnh ny gi l m hnh hp th mt chiu, trng thi ca ht trong hp th mt chiu l trng thi dng.

Ht chuyn ng trong thnh vch dng ng c th dng m t electron t do trong kim loi hoc electron khng nh c trong cc phn t lin hp.

Ta c phng trnh Schrodinger cho trng thi dng:

= 0

V l hp th mt chiu theo phng x nn:

Suy ra : = 0

t k2=( = 0

(1.36)

y l phng trnh vi phn tuyn tnh bc hai c nghim tng qut:

((x) = A coskx + Bsinkx

(1.37)

Trong A, B l cc hng s cha xc nh.

Ta c th xc nh A bng cch ti iu kin b ca bi ton (x = 0 v x = a).

Ti cc gi tr b (x = 0, x = a) hm sng phi trit tiu, ngha l ( = 0:

((0) = 0 , ( (a) = 0

* ((0) = A cos 0 + Bsin0 = 0 ( A = 0

( ((x) = Bsinkx

* ((a) = Bsinka = 0 ( sinka = 0 ( ka = n( ( n: nguyn)

(B khng th bng 0, v nu B = 0 th ((x) bng 0 vi mi x)

( k = (n = 1,2,3, .. .)

(n khng th bng 0, v n = 0 th k = 0 v ((x) cng bng 0 vi mi x. ng thi n cng khng nhn gi tr m, v khi ta c ((x) = - Bsinka v mt xc sut ca hm sng vn khng thay i).

( ((x) = B sin x

Hng s B cn li c xc nh bng iu kin chun ho:

B =

(thng chn B dng)

Vy hm sng chun ho: (n(x) = sinx

T k2 = v k =

En = n2 (4.18) n: s lng t ( n = 1,2,3,.. .)

T (4.18) ta thy, h ch c th nhn cc gi tr nng lng gin on, ta ni nng lng ca ht c lng t ho. Nh vy, s lng t ho ca nng lng c dn ra mt cch t nhin t yu cu hm sng phi tho mn cc iu kin b. y l im khc bit ca h vi m so vi h v m.

n = 1 : E1 = ; (1 = sin x (x =0, x = a)

n = 2 : E2 = 4. = 4E1 ; ( 2 = sin 2x (x = 0, a, a/2)

n = 3 : E3 = 9. = 9E1 ; ( 3 = sin 3 x (0, a, a/3, 2a/3)

im m ti hm sng ( = 0 ngi ta gi l im nt. Tr nhng im thnh hp, ta thy s im nt ca hm sng ph thuc vo n v bng (n-1).

Gin nng lng hm sng v mt xc sut ca ht trong hp th mt chiu c trnh by hnh sau:

C th rt ra mt s c im v hm sng v mc nng lng ca h:

- Mi hm sng ( n(x) c (n-1) im nt. S im nt tng theo chiu tng ca mc nng lng.

- Xc sut tm thy ht ti mt v tr gia x v dx l : dw = ( 2dx. Xc sut ny c cc i ti nhng v tr khc nhau tu theo trng thi ca h. trng thi c bn

n =1, mt xc sut cc i ti x =a/2.

- Mc nng lng thp nht ca h c gi tr hu hn khc khng E1 = . Ngi ta gi nng lng ny l nng lng im khng. S tn ti nng lng im khng l c trng ca cc h lin kt.

1.4.2. Bi ton vi ht trong hp th 3 chiu

M rng trng hp hp th 1 chiu i vi hp th 3 chiu, vi th nng:

U = Const = 0 trong khong 0 ( x ( a, 0 ( y ( b, 0 ( z ( c

v U = ( ngoi khong .

Phng trnh Schrodinger c dng:

-

(1.40)

E = Ex + Ey + Ez

gii phng trnh (1.40) ta phn li bin s: ( (x,y,z) = ( (x) ( (y) ( (z) (1.41)

a (1.41) vo (1.40) ri chia c hai v cho ( (x) ( (y) ( (z) ta c:

(1.42)

hay

(1.43)

Phng trnh (1.43) c th c xem nh l tng ca 3 phng trnh c dng ging nhau:

(a)

(b)

(c)

Cc phng trnh (a), (b), (c) chnh l phng trnh sng ca ht trong hp th mt chiu m nghim ta bit:

((x) = Axsin

; Ex =

(y = Aysin

;

((z) = Azsin

; Ez =

iu kin chun ho thAx = ; Ay =

; Az =

. Do hm sng chun ho v nng lng ca h l:

sin.sin.sin (1.44)

Enx,ny,nz =

(1.45)

T (1.45) suy ra: Nu mt hay hai cnh ca hp th c di bng s nguyn ln mt cnh khc th s c mt s hm ring (trng thi) khc nhau c cng mt gi tr nng lng nh nhau, tc l tr ring Enx,ny,nz c suy bin. S xut hin tr ring suy bin rt thng gp trong c hc lng t, phn nh tnh i xng ca h kho st.

1.43. Dao ng t iu ho

Chng ta bit rng dao ng t ca mt phn t hai nguyn t, chuyn ng ca cc ht trong mng li tinh th, mt cch gn ng, c xem nh cc dao ng iu ho tuyn tnh.

Khi ht chuyn ng trong trng lc dc theo trc x (theo phng xc nh) th n b tc dng mt lc vi th nng:

U =

(1.46)

trong :

k = m(2 l hng s lc hay h s n hi

m : khi lng ht

x : li dao ng

( = 2(( l tn s gc

Theo c hc c in, nng lng ca h l:

E =

(1.47)

v a (bin ) c th nhn cc gi tr bt k nn E thu c l cc gi tr lin tc.

Theo c hc lng t, thay th nng vo phng trnh Schrodinger, ta c:

(1.48)

t:

EMBED Equation.3

(1.49)

(1.50)

Phng trnh (1.48) c vit li:

+ (( - (2x2) ( = 0

(1.51)

a bin s:

(1.52)

Ly o hm ( theo x ta c:

Hay

(1.53)

(1.54)

Thay (1.53), (.54) vo (1.51) ta c:

(1.55)

Hay

(1.56)

Hm ( phi lin tc, n tr, hu hn i vi mi ga tr ca (. Khi ( kh ln th t s (/( c th b qua, lc phng trnh c dng:

Phng trnh vi phn ny c nghim l:

Khi ( ( ( th ( tng v hn, nghim s khng tho mn iu kin ca hm (. Vy hm sng ( ch c th l:

Nghim ng ca hm ( trong phng trnh l :

y hm H(() phi c xc nh. Mun vy ta t Z = (2/2; Z = ( a phng trnh (1.56) v dng Hermit.

Gii phng trnh ny ngi ta c nghim:

Hn(() = (-1)n

vi n = 0, 1, 2, 3, ...

Nng lng ca h l :E = h((n + )

Nh vy ng vi mi gi tr ca n = 0, 1, 2, ... ta c cc gi tr nng lng c php l 1/ 2, 3/2, 5/2 ... ln nng lng h(, ngha l cc gi tr nng lng ca dao ng t iu ho tuyn tnh lp thnh mt ph gin on ph thuc vo n gi l s lng t dao ng. Mt vi mc nng lng u tin v cc hm sng tng ng c biu din trn th sau:

Kt qu quan trng nht thu c l nng lng c php nh nht E = h(/2 vi n = 0. l nng lng im khng v cng l iu khc vi kt qu thu c ca l thuyt c in. iu ny ph hp vi nguyn l bt nh, v nhng bt nh cn thit v v tr v xung lng sinh v nng lng im khng.

120

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