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
_1125033806.unknown
_1125037773.cdx
_1218481087.unknown
_1219065214.unknown
_1219083751.unknown
_1219084098.unknown
_1249545636.unknown
_1249546275.unknown
_1249546298.unknown
_1249545677.unknown
_1219084368.unknown
_1219084439.unknown
_1219084253.unknown
_1219083871.unknown
_1219083973.unknown
_1219083825.unknown
_1219083435.unknown
_1219083523.unknown
_1219083626.unknown
_1219083461.unknown
_1219083000.unknown
_1219083223.unknown
_1219065227.unknown
_1219064699.unknown
_1219064882.unknown
_1219065192.unknown
_1219064865.unknown
_1219064659.unknown
_1219064676.unknown
_1219064684.unknown
_1219064639.unknown
_1125041005.unknown
_1125042037.unknown
_1125042883.unknown
_1126718790.unknown
_1125056662.unknown
_1125056849.unknown
_1125056956.unknown
_1125056760.unknown
_1125042999.unknown
_1125042145.unknown
_1125042504.unknown
_1125042795.unknown
_1125042317.unknown
_1125042244.unknown
_1125042269.unknown
_1125042204.unknown
_1125042107.unknown
_1125041699.unknown
_1125041886.unknown
_1125041996.unknown
_1125041823.unknown
_1125041464.unknown
_1125041601.unknown
_1125041326.unknown
_1125038387.unknown
_1125038895.unknown
_1125040494.cdx
_1125040730.unknown
_1125039503.unknown
_1125038597.unknown
_1125038851.unknown
_1125038770.unknown
_1125038818.unknown
_1125038478.unknown
_1125038504.unknown
_1125038163.unknown
_1125038289.unknown
_1125038101.unknown
_1125034741.unknown
_1125035236.unknown
_1125036727.unknown
_1125036968.unknown
_1125036988.unknown
_1125036439.unknown
_1125036467.unknown
_1125036597.unknown
_1125036058.unknown
_1125034913.unknown
_1125035012.unknown
_1125035026.unknown
_1125034986.unknown
_1125034770.unknown
_1125034781.unknown
_1125034757.unknown
_1125034166.unknown
_1125034408.unknown
_1125034707.unknown
_1125034726.unknown
_1125034458.unknown
_1125034194.unknown
_1125033971.unknown
_1125034107.unknown
_1125033896.unknown
_1124971004.unknown
_1125031987.unknown
_1125032446.unknown
_1125033064.unknown
_1125033702.unknown
_1125032568.unknown
_1125032200.unknown
_1125032394.unknown
_1125032105.unknown
_1124972077.unknown
_1124972701.unknown
_1124975787.unknown
_1124975929.unknown
_1124972723.unknown
_1124972464.unknown
_1124971706.unknown
_1124971726.unknown
_1124972042.unknown
_1124971225.unknown
_1124970074.unknown
_1124970592.unknown
_1124970926.unknown
_1124970971.unknown
_1124970757.unknown
_1124970129.unknown
_1124970566.unknown
_1124970110.unknown
_1124804793.unknown
_1124889420.unknown
_1124969819.unknown
_1124969903.unknown
_1124970056.unknown
_1124969731.unknown
_1124804994.unknown
_1124805849.unknown
_1124889402.unknown
_1124805498.unknown
_1124805814.unknown
_1124805023.unknown
_1124804905.unknown
_1124804956.unknown
_1124804815.unknown
_1124804609.unknown
_1124804700.unknown
_1124804758.unknown
_1124804658.unknown
_1124799529.unknown
_1124799584.unknown
_1124697030.unknown
_1124697128.unknown
Recommended