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1

M U

M U

Hc phn c hc lng t nng cao l mn hc bt buc i vi hcvin cao hc chuyn ngnh Phng php Ging dy Vt l v chuyn ngnhVt l L thuyt-Vt l Ton, n nhm b sung v nng cao mt s kin thcc hc lng t nh cc phng php tnh gn ng trong c hc lng t,l thuyt tn x lng t, c hc lng t tng i tnh,... Cc kin thc

ny l c s hc vin tip thu cc kin thc v Vt l thng k, Vt lcht rn, C s l thuyt trng lng t,...

Vi mc tiu nh trn, ni dung ca mn hc c xy dng trong 4chng. Chng I khi qut li cc c s ca c hc lng t (c s ton hc,cc tin ca c hc lng t, nguyn l bt nh Heisenberg, phng trnhSchrdinger, s bin i theo thi gian ca gi tr trung bnh cc i lngvt l,...). Chng II trnh by cc phng php gn ng gii phngtrnh Schrdinger thng c s dng trong c hc lng t. Chng III

trnh by l thuyt tn x lng t. Chng IV trnh by khi qut c hclng t tng i tnh, bao gm mt s phng trnh c bn (Phng trnhKlein-Gordon, phng trnh Dirac, phng trnh Pauli,...), mt s khi nimc bn (Mt xc sut tng i tnh v mt dng xc sut tng itnh, spin v mmen t ca ht vi m,...). Ngoi ra, cc hc vin cao hcVt l L thuyt -Vt l Ton cn c 15 tit kho st su hn v cu trccc trng thi nguyn t, l thuyt lng t v bc x, hiu ng Zeemann dthng, cc trng thi nng lng m, tnh bt bin ca phng trnh Dirac.

gip hc vin nm chc cc kin thc ca mn hc, s thi giandnh cho hc vin rn luyn cc k nng vn dng v gii cc bi tp, xminechim 1/4 thi lng ca mn hc.

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2

Mc lc

1 C s ca c hc lng t 41.1 C s ton hc ca c hc lng t . . . . . . . . . . . . . . . 4

1.1.1 Ton t: . . . . . . . . . . . . . . . . . . . . . . . . . . 41.1.2 Cc php tnh trn ton t . . . . . . . . . . . . . . . . 51.1.3 Hm ring, tr ring v phng trnh tr ring ca ton

t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.1.4 Ton t t lin hp tuyn tnh (ton t hermitic) . . . 61.1.5 Cc tnh cht ca ton t hermitic . . . . . . . . . . . 8

1.2 Cc tin ca c hc lng t . . . . . . . . . . . . . . . . . 91.2.1 Tin 1: Trng thi v thng tin . . . . . . . . . . . 91.2.2 Tin 2: Cc i lng ng lc . . . . . . . . . . . . 91.2.3 Tin 3: Php o cc i lng ng lc . . . . . . . 10

1.2.4 Gi tr trung bnh ca bin s ng lc . . . . . . . . . 111.2.5 Tnh h s phn tch ci . . . . . . . . . . . . . . . . . . 11

1.3 S o ng thi hai i lng vt l . . . . . . . . . . . . . . . 121.3.1 S o chnh xc ng thi hai i lng vt l . . . . . 121.3.2 Php o hai i lng ng lc khng xc nh ng

thi. Nguyn l bt nh Heisenberg. . . . . . . . . . . 131.4 Phng trnh Schrdinger . . . . . . . . . . . . . . . . . . . . 15

1.4.1 Phng trnh Schrdinger ph thuc thi gian . . . . . 15

1.4.2 Mt dng xc sut. S bo ton s ht . . . . . . . 161.4.3 Phng trnh Schrdinger khng ph thuc thi gian.Trng thi dng. . . . . . . . . . . . . . . . . . . . . . 17

1.5 S bin i theo thi gian ca cc i lng ng lc . . . . . 191.5.1 o hm ca ton t ng lc theo thi gian . . . . . 19

2 Mt s phng php gn ng trong c hc lng t 222.1 Nhiu lon dng trong trng hp khng suy bin . . . . . . . 232.2 L thuyt nhiu lon dng trong trng hp c suy bin . . . 26

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C hc lng t nng cao Ch.1: C s ca c hc lng t 3

2.2.1 L thuyt nhiu lon khi c hai mc gn nhau . . . . . 262.2.2 L thuyt nhiu lon dng khi c suy bin: . . . . . . . 31

2.3 Hiu ng Stark trong nguyn t Hydro . . . . . . . . . . . . . 35

2.4 Nhiu lon ph thuc thi gian . . . . . . . . . . . . . . . . . 392.5 S chuyn di lng t ca h vi m sang cc trng thi midi nh hng ca nhiu lon . . . . . . . . . . . . . . . . . 42

2.6 Nguyn t Hli . . . . . . . . . . . . . . . . . . . . . . . . . . 442.7 Phng php trng t hp Hartree-Fok . . . . . . . . . . . . 48

2.7.1 Nguyn l bin phn . . . . . . . . . . . . . . . . . . . 482.7.2 Phng php trng t hp Hartree-Fok . . . . . . . . 52

3 L thuyt tn x lng t 57

3.1 Bin tn x v tit din tn x . . . . . . . . . . . . . . . . 573.1.1 Tit din tn x . . . . . . . . . . . . . . . . . . . . . . 573.1.2 Bin tn x . . . . . . . . . . . . . . . . . . . . . . 593.1.3 Tn x n hi ca cc ht khng c spin . . . . . . . 60

3.2 Tn x n hi trong php gn ng Born . . . . . . . . . . . 653.3 Phng php sng ring phn . . . . . . . . . . . . . . . . . . 68

4 C hc lng t tng i tnh 74

4.1 Phng trnh Klein-Gordon (K-G) . . . . . . . . . . . . . . . 754.2 Phng trnh Dirac . . . . . . . . . . . . . . . . . . . . . . . . 764.3 Mt xc sut v mt dng xc sut trong l thuyt Dirac 814.4 Nghim ca phng trnh Dirac i vi ht chuyn ng t do 834.5 Spin ca ht c m t bng phng trnh Dirac . . . . . . . 854.6 Chuyn t phng trnh Dirac sang phng trnh Pauli. M-

men t ca ht. . . . . . . . . . . . . . . . . . . . . . . . . . . 87

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4

Chng 1

C s ca c hc lng t

1.1 C s ton hc ca c hc lng t

1.1.1 Ton t:

a) nh ngha: Ton t l mt php ton tc dng vo mt hm ny thbin i thnh mt hm khc.

Ta gi A l mt ton t nu

A(x) = (x). (1.1)

V d: Cc ton t :

+ Php nhn vi x2

A(x) = x2(x),

trong trng hp ny A ph thuc bin s x.+ Php ly o hm vi bin s x:

A(x) =d(x)

dx

+ Php nhn vi mt s phc C:

A(x) = C(x),

y, A khng ph thuc vo bin x v php ly o hm theo x. c bitnu:

C = 0 : A(x) = 0, A l ton t khng,

C = 1 : A(x) = (x), A l ton t n v.

+ Php ly lin hip phc:

A(x) = (x).

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b) Ton t tuyn tnh: Ton t A c gi l ton t tuyn tnh nu ntho mn tnh cht sau:

A(c11 + c22) = c1A1 + c2A2. (1.2)

Trong h thc trn, 1 v 2 l hai hm bt k, c1 v c2 l hai hng sbt k.

V d: A = (d/dx) l ton t tuyn tnh v

d

dx(c11 + c22) = c1

d1dx

+ c2d2dx

.

Cn ton t ly lin hip phc khng phi l ton t tuyn tnh v

A(c11 + c22) = (c11 + c22) = c11 + c22 = c1A1 + c2A2

= c1A1 + c2A2.

1.1.2 Cc php tnh trn ton t

Cho ba ton t A, B, C. ta nh ngha cc php tnh ton t sau:a) Tng hai ton t: S c gi l tng ca hai ton t A, B, k hiu

l

S A + B nu (x), S(x) = A(x) + B(x). (1.3)b) Hiu hai ton t: D c gi l hiu hai ton t A, B, k hiu

D A B nu (x), D(x) = A(x) B(x). (1.4)c) Tch hai ton t: P AB l tch ca hai ton t A v B nu

P (x) = (AB)(x) = A

B(x)

. (1.5)

Tch ca hai ton t ni chung l khng giao hon, ngha l AB = BA.Chng hn, choA =

d

dx, B = x

th ta cAB(x) =

d

dx(x(x)) = (x) + x

d(x)

dx,

cnBA(x) = x

d(x)

dx = AB(x) = (x) + x

d(x)

dx,

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r rng BA = AB, nn A, B khng giao hon nhau.Nu A = x2, B = x th

AB(x) = x3(x) = BA(x)

hai ton t A, B giao hon nhau.d) Giao hon t ca hai ton t A v B c nh ngha l [A, B]

AB BA. Nu A v B giao hon th AB = BA, do giao hon t cachng bng khng, ngha l [A, B] = 0. Nu hai ton t khng giao hon th[A, B] = AB BA = 0 hay [A, B] = 0.

1.1.3 Hm ring, tr ring v phng trnh tr ring ca ton t

Xt mt ton t A, khi cho A tc dng ln mt hm (x) no , ta cth thu c chnh hm nhn vi mt hng s:

A(x) = a(x). (1.6)

(1.6) l mt phng trnh, dng ca (x) c th thu c t vic gii phngtrnh trn.

Ta bo (x) l hm ring vi tr ring a ca ton t A. V vic giiphng trnh (1.6) c th cho ta bit cc hm ring v tr ring ca ton tA. Nu c s hm ring c cng mt tr ring a, th ta bo ton t A c trring suy bin bc s. Cc tr ring c th bin thin gin on hoc lin tc.

Trong c hc lng t, hm ring phi tho mn cc iu kin chunsau:

- Hm (x) phi tn ti, xc nh trn ton min bin thin ca ccbin c lp.

- Trong min tn ti, hm (x) v o hm bc nht ca n d(x)/dxphi hu hn, lin tc (tr mt s im c bit).

- Hm (x) phi xc nh n tr

1.1.4 Ton t t lin hp tuyn tnh (ton t hermitic)

Ton t tuyn tnh A+ c gi l ton t lin hp tuyn tnh vi tont tuyn tnh A nu:

1(x), 2(x), V 1(x)A2(x)dx = V A+1(x)

2(x)dx. (1.7)

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Nu A+ = A th ta bo A l ton t t lin hp tuyn tnh, hay tont hermitic, ngha l:

V 1(x)A2(x)dx = V A1(x) 2(x)dx. (1.8)Nu ta a ra k hiu mi v tch v hng hai hm sng

1(x)|2(x) =V

1(x)2(x)dx, (1.9)

theo (1.8) c vit li nh sau:

1(x)

|A

2(x)

=

A

1(x)

|2(x)

.

V d 1: A = (d/dx) c phi l ton t hermitic khng?Mun bit, ta tnh+

Adx =

+

d

dxdx.

t u = , dv = (d/dx).dx, th

+ Adx = |x=+x= +

d

dx dx,

v cc hm (x), (x) 0 khi x nn |x=+x= = 0,+

Adx = +

d

dxdx =

+

d

dx

dx =

+

A

dx.

Vy A = (d/dx) khng phi l ton t hermitic.V d 2: A = i(d/dx) c phi l ton t hermitic khng?

Ta c:+

Adx = i+

d

dxdx =

+

id

dx

dx =

+

i

d

dx

dx,

+

Adx =+

A

dx.

Vy A = i(d/dx) l ton t hermitic.

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1.1.5 Cc tnh cht ca ton t hermitic

a) Tr ring ca ton t hermitic l s thc.Gi thit ton t hermitic A c tr ring gin on vi phng trnh

tr ringAn = ann.

Ta c: n|An = An|n v A hermitic, ngha l:

ann|n = an|n = (an an)n|n = 0.V n|n = 0 nn an = an: an l s thc.b) Hm ring tng ng vi hai tr ring phn bit th trc giao vi

nhau. Thc vy, theo nh ngha ca ton t hermitic th:

1|A2 = A1|2 = a21|2 = a11|2, = (a2 a1)1|2 = 0,v a2 = a1 nn (a2 a1) = 0. Vy:

1|2 = 0 : 1, 2 trc giao vi nhau.Tm li, nu cc hm ring ca ton t hermitic A c chun ho th

ta c:

Ph tr ring gin on : m|n = mn, (1.10)Ph tr ring lin tc : a|a = (a a). (1.11)

Trong , mn, (a a) l cc hm Dirac.c) Cc hm ring ca ton t hermitic lp thnh mt h hm c s

trc giao v trong khng gian Hilbert cc hm sng, ngha l vi mt hmsng bt k (x) trong khng gian Hilbert, ta c:

i vi ph tr ring gin on : (x) =n

cnn(x). (1.12)

i vi ph tr ring lin tc : (x) =a

caa(x)da. (1.13)

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1.2 Cc tin ca c hc lng t

Trong c hc lng t, ht khng c hnh dung nh l mt cht im

chuyn ng theo mt qu o xc nh m n c hnh dung nh l mtb sng nh x trong mt min ca khng gian ti mt thi im v b sngthay i theo thi gian. Ti mt thi im ta ch c th ni v xc sut tm thy ht trong mt phn t th tch ca khng gian, hay ni khc i lxc xut to ca ht c gi tr nm trong khong no . Ni chungv cc bin s ng lc khc cng vy, ta ch c th ni v xc sut mtbin s ng lc c gi tr nm trong khong no ch khng th ni vgi tr xc nh ca bin s ng lc ti mt thi im nh trong c hc cin.

V c s khc bit ni trn nn trong c hc lng t bin s ng lckhng phi c m t bng mt s nh trong c hc c in. Chng ta phitm mt cch m t khc th hin c nhng c tnh ca cc quy lutlng t. Nhng nghin cu v ton t cho thy c th dng cng c tonhc ny m t bin s ng lc trong c hc lng t. Chng ta tha nhnmt s gi thit v ni dung cch m t nh nhng tin . Nhng tin y khng c mu thun nhau v cho cc kt qu ph hp vi thc nghim.

1.2.1 Tin 1: Trng thi v thng tin

" Trng thi vt l ca mt h lng t th tng ng vi mt hm sngchun ho."

Ta k hiu (x, t) l hm sng ca h lng t thi im t v ti vtr to x ( hay ng vi bin ng lc x).

Hm sng c chun ho khi

(x, t)

|(x, t)

= V (x, t)(x, t)dx = 1. (1.14)

Nh vy, (x, t) v c(x, t) cng chung mt trng thi nu cc = |c|2 =1.

1.2.2 Tin 2: Cc i lng ng lc

" Tng ng vi mt i lng ng lc A trong c hc lng t l mtton t hermitic A."

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V gi tr bng s ca bin ng lc l thc nn tr ring ca ton ttng ng vi bin ng lc phi thc, do ton t tng ng vi binng lc phi hermitic. Ton t A hermitic nn c mt h cc vect ring

trc giao chun ho {i(x, t)} tng ng vi ph cc tr ring thc {ai},i = 1, 2,...,n. Theo , mt trng thi bt k ca h lng t s c khaitrin theo cc hm ring nh sau:

(x, t) =ni=1

cii(x, t). (1.15)

1.2.3 Tin 3: Php o cc i lng ng lc

Nu h lng t trng thi biu din bi hm sng (x) th xcsut khi o bin ng lc A thu c gi tr ai s l |ci|2 = pi. R rng

ni=1

pi =ni=1

|ci|2 = 1 (1.16)

c suy t tnh cht trc giao, chun ho ca cc hm ring.Nh vy php o lm nhiu lon trng thi. Nu (x) = i(x), ta c

A(x) = Ai(x) = aii(x) vi xc sut |ci|2 = pi = 1.Ch rng theo tin 3 th

(i) Khng th tin on chnh xc kt qu php o mt ilng ng lc ca h vi m c trng thi (x) hon ton xc nh.

(ii) Nu tin hnh hai php o ring bit nhng ging nhau trncng mt h c trng thi ban u trc mi ln o l (x) hon ton gingnhau th kt qu hai ln o ny khng nht thit phi trng nhau.

Ta chp nhn tnh khng tin on c v tnh khng ng nhtca qu trnh o nh l mt thuc tnh vn c ca t nhin.

Trong trng hp ph tr ring lin tc th

(x) =

a

c(a)a(x)da (1.17)

v xc sut dW(a) i lng A c gi tr trong khong t a n a + da l

dW(a) =|c(a)

|2da. (1.18)

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1.2.4 Gi tr trung bnh ca bin s ng lc

Xt bin s ng lc A c ton t hermitic tng ng A, tr trung bnhA ca n trng thi (x) ng vi trng hp ph tr ring gin on

{ai

}A =

ni=1

piai =ni=1

ai|ci|2 =V

(x)A(x)dx (1.19)

v V

(x)A(x)dx =V

i

j

cii (x)Acjj(x)dx

= i j ci cj V i (x)Aj(x)dx=i

j

ci cjaj

V

i (x)j(x)dx

=i

j

ci cjajij

=

i|ci|2ai.

Trng hp ph tr ring lin tc, ta c

A =

a

adW(a) =

a

|c(a)|2ada

1.2.5 Tnh h s phn tch ci

Theo tin 3, mun tnh xc sut o A c gi tr ai th ta phixc nh cho c h s phn tch ci. Mun vy, ta nhn lng lin hip phcca hm ring i(x) l i (x) vi hm sng (x) ri ly tch phn theo bins x, ta c

V

i (x)(x)dx =k

V

i (x)ckk(x)dx =k

ckik = ci, (1.20)

gi tr ny ca ci hon ton xc nh vi sai km hng s nhn.

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1.3 S o ng thi hai i lng vt l

1.3.1 S o chnh xc ng thi hai i lng vt l

Xt hai bin s ng lc L v M c biu din bi hai ton t L vM. H trng thi c biu din bi hm sng m y cho rmr ta hiu ngm l hm theo bin s x. Chng ta s xt trong iu kin nohai bin ng lc c th o c chnh xc ng thi. Theo tin 3, muncho bin ng lc L c gi tr xc nh th = L,k l hm ring ca L ngvi tr ring Lk. Ngha l

L = LL,k = LkL,k.

Ta o ng thi i lng M vi L, tc l lc h trng thi = L,k.Mun cho M cng c gi tr xc nh Mk th phi l hm ring ca M,ngha l = M,k . Theo

M = MM,k = MkM,k .

Nh vy, hai ton t L v M phi c chung hm ring:

= L,k = M,k .

y chnh l iu kin ng thi o c chnh xc hai i lng nglc L v M. V ta c th rt ra nh l sau:

iu kin t c v hai i lng ng lc o c ng thi lton t tng ng ca chng giao hon vi nhau.

Chng ta s chng minh nh l ny sau y.a) iu kin t c: Nu L, M c chung hm ring k th hai ton t

L, M giao hon c vi nhau.Ta c

LMk = LM k = MkLk = MkLkk,MLk = M

Lk

= LkMk = LkMkk.

Suy raLM k = MLk,

hay

LM MLk = 0 = LM ML = 0 = LM = ML.

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R rng L v M giao hon vi nhau.a) iu kin : Nu hai ton t giao hon th chng c chung hm

ring.

Gi l hm ring caL, ngha l

L = L,ML

= M

L

= M(L) = L

M

.

V M v L giao hon nnML

=

LM

= L

M

.

R rng M l mt hm ring ca ton t L vi tr ring L. Nhvy, v u l hm ring ca L vi cng tr ring L. Khi khng c suybin th chng trng nhau, nhng v hm ring ca cc ton t hermitic cxc ng sai km nhau mt hng s nhn nn

= hng s.,

hay M = hng s. = M., ngha l cng l hm ring ca ton t M.

1.3.2 Php o hai i lng ng lc khng xc nh ng thi.Nguyn l bt nh Heisenberg.

Trong trng hp tng qut nu hai ton t L, M theo th t biu dinhai i lng ng lc L, M khng giao hon c vi nhau th khng tho c chnh xc ng thi L v M. By gi ta xt xem nu o ng thihai bin ng lc y th chnh xc t n mc no.

Do L v M l nhng ton t hermitic khng giao hon c vi nhaunn L, M = iP , (1.21)trong P l mt ton t hermitic, P = 0.

Gi L v M l tr trung bnh ca L v M trng thi (x). Xt lch

L = L L; M = M M (1.22)Nhng i lng ny theo th t c biu din bi cc ton t hermitic

L = L L; M = M M (1.23)

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Ta c giao hon t

L,

M

=

L L, M M

=

L, M

= iP . (1.24)

Xt tch phn:I() =

V

|

L iM|2dx 0 (1.25)trong l mt thng s thc, tch phn ly trong ton b min bin thinV ca x.

I() =

V

(

L i

M)

(

L i

M)dx

= V

(L iM)+(L iM)dxv tnh cht hermitic, L = L+, M = M+, do (L iM)+ =L + iM, nn

I() =

V

L + iM)(L iMdxI() =

V 2L2 iLM ML+ M2dx

I() =

V

2L2 i L, M+ M2dxtheo (1.24), th

I() =

V

2

L

2+ P +

M

2

dx, suy ra

I() = 2

L2 + P + M2 0.Mun cho I() 0 th tam thc bc hai theo trn phi c bit thc

= P2 4

L2

M2

0, ngha l

L2

M2

P

2

4hay

L2

M2

L, M

2

4. (1.26)

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y l cng thc cho bt nh khi o ng thi hai bin ng lcL v M, n c gi l h thc bt nh Heisenberg. t

L = L2, M = M2, (1.27)h thc bt nh c th vit di dng khc

L.M P

2hay , L.M

L, M2

. (1.28)

V d: Nu chn L = x = x : ton t to ,

M = px = i x

: ton t xung lng theo phng x.

th[x, px] = i,

suy ra h thc bt nh Heisenberg cho to v xung lng

x.px 2

. (1.29)

Nh vy ta khng th ng thi o chnh xc to v xung lng camt ht vi m. Sai s mc phi khi o tun theo h thc bt nh Heisenberg(1.29).

ngha vt l: Vic khng o c chnh xc ng thi to v xunglng ca ht vi m chng t rng n lng tnh sng ht. Ht vi m khngc qu o xc nh. l mt thc t khch quan do bn cht ca s vtch khng phi v kh nng hiu bit s vt ca ta b hn ch hoc my okm chnh xc. V h thc bt nh l biu thc ton hc ca lng tnh

sng ht ca ht vi m.

1.4 Phng trnh Schrdinger

1.4.1 Phng trnh Schrdinger ph thuc thi gian

Trong c hc lng t, do lng tnh sng ht ca cc i tng vim nn trng thi ca ht c c trng bi hm sng (r, t).V vy, cn

c phng trnh m t din bin ca hm trng thi theo thi gian. Phng

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trnh ny c Schrdinger a ra nm 1926 v c gi l phng trnhSchrdinger ph thuc thi gian

i(r, t)

t =H

(r, t

),

(1.30)trong H l Hamiltonian ca h

H = T + U = 2

2m2 + U(r, t) (1.31)

y l phng trnh vi phn hng hai theo khng gian v hng nhttheo thi gian. V nguyn tc tm nghim ca phng trnh, ta phi bitc hm sng ti thi im t0 (iu kin u) v bit c hai iu kin

bin lin quan n to (x0, t0) = 0, v d(x,t)dx x=x0 = 0.1.4.2 Mt dng xc sut. S bo ton s ht

n gin, ta s vit tt , theo th t thay cho (r, t), (r, t).T phng trnh (1.30), ta suy ra phng trnh lin hip phc ca n

i

t= H

H = H+

. (1.32)

Nhn cho hai v ca (1.30) v pha tri v nhn cho hai v ca (1.32)cng v pha tri ri tr cho nhau v theo v, ta c

i

t+

t

= H H . (1.33)

Thay H = 2/2m2+U v lu (/t)() = (/t)+(/t),ta c

i

t() =

2

2m 2 2 , (1.34)m

( ) = + 2 2,nn ta c th vit li (1.34) nh sau

t() +

i

2m ( ) = 0. (1.35)

t

= ||2

(1.36)

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l mt xc sut tm thy ht to r ti thi im t. V

j(r, t) =i

2m( ) (1.37)

l vect mt dng xc sut. ln ca j(r, t) c ngha nh l dng httrung bnh qua mt n v din tch t vung gc vi phng chuyn ngtrong mt n v thi gian.

Theo phng trnh (1.35) c dng ca phng trnh lin tc m tnh lut bo ton s ht vi m:

j + t

= 0. (1.38)

1.4.3 Phng trnh Schrdinger khng ph thuc thi gian. Trngthi dng.

Ta xt mt ht vi m chuyn ng trong trng th U(r) khng binthin theo thi gian v do c nng lng khng thay i theo thi gian.Gi E l gi tr nng lng ca ht v ta k hiu E(r) l hm sng ng vitrng thi c nng lng E. Ta c th vit phng trnh tr ring ca nnglng nh sau

HE(r) = EE(r) (1.39)vi H = (2/2m)2 + U(r) nn ta c th vit (1.39) di dng khc:

2

2m2 + U(r)

E(r) = EE(r) (1.40)

Trong trng hp ny hm sng E(r, t) = E(r).f(t) c vit didng phn ly bin s. Theo , phng trnh Schrdinger (1.30), vi lu Hkhng ph thuc tng minh vo thi gian t, c vit li

E(r)if

t= f(t)HE(r)

iftf(t)

=HE(r)

E(r)= E,

Nh vy, ta c hai phng trnh c lp

if

t= E.f(t), (1.41)

HE(r) = E.E(r). (1.42)

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Phng trnh (1.41) cho ta nghim

f(t) = CeiEt. (1.43)

Cn (1.42) chnh l phng trnh cho ta hm ring v tr ring ca ton tnng lng. Gi s nng lng ca h c gi tr gin on En, n = 0, 1, 2,...,lc ta vit li (1.42) nh sau

Hn(r) = En.n(r). (1.44)

trong n(r) l vit tt ca En(r). Nh vy, nghim ring y ca htvi m ng vi trng thi dng c nng lng hon ton xc nh En l

n(r, t) = n(r)eiEnt

. (1.45)Nghim tng qut ca phng trnh Schrdinger trng thi dng

trong trng hp ph gin on

(r, t) =n

cne i

Entn(r) =

n

Cn(t)n(r), vi Cn(t) cne iEnt.(1.46)

Trng hp ph tr ring lin tc, hm sng c dng

(r, t) = cEe iEtE(r)dE = CE(t)E(r)dE, vi CE(t) cEe iEt.(1.47)

Cc h s cn, cE c th c xc nh t iu kin u.Ni tm li, mt h lng t trng thi dng c cc tnh cht sau:a) Hm sng ph thuc thi gian ca trng thi dng xc nh n tr

bi gi tr nng lng ca trng thi .b) trng thi dng, mt xc sut v mt dng xc sut khng

ph thuc vo thi gian.c) trng thi dng, tr trung bnh ca mt i lng ng lc cton t tng ng khng ph thuc r rt vo thi gian th khng i theothi gian.

d) Xc sut o gi tr ca mt i lng ng lc trng thi dngkhng ph thuc thi gian.

Nghim ca phng trnh Schrdinger khng ph thuc thi gian ccc tnh cht c bn sau:

a) Hm (r, t) phi n tr.

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b) Hm (r, t) phi lin tc. Trong trng hp th nng U(r) ginon th hm sng (r, t) v o hm ca n vn lin tc ti nhng imgin on . Tuy nhin, nhng min m th nng U th hm sng

v o hm ca n gin on.c) Nu th nng U khng tin n v cng th hm sng (r) phi huhn trong ton b khng gian. iu ny cng c tho mn trong trnghp U ti mt im no nhng khng qu nhanh (U 1rs , s 2).

1.5 S bin i theo thi gian ca cc i lng nglc

1.5.1 o hm ca ton t ng lc theo thi gianTa c tr trung bnh ca mt i lng ng lc L trng thi (x)

L =

(x)L(x)dx, (1.48)

trong x bao gm tt c cc bin s kh d v (x) c chun ho.Ton t L c th ph thuc thi gian nn L cng c th ph thuc thi gian.Ta tnh o hm ca tr trung bnh L theo thi gian

dL

dt=

(x)L

t(x)dx +

(x)

tL(x)dx +

(x)L

(x)

tdx. (1.49)

Lu rng, theo phng trnh Schrdinger (1.30), ta c

(x)

t= i

H(x) v

(x)t

=i

H(x), (1.50)

do phng trnh (1.49) c th vit li

dL

dt=

(x)

L

t(x)dx+

i

H(x)

L(x)dx+

(x)L

i

H(x)

dx,

dL

dt=

(x)

L

t(x)dx +

i

H(x)

L(x)dx

(x)LH (x)dx

,

dL

dt=

(x)

L

t(x)dx +

i

(x)

HL LH

(x)dx

,

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dL

dt=

(x)

L

t+

i

H, L

(x)dx. (1.51)

Ta nh ngha o hm ton t L theo thi gian dL/dt l ton t cxc nh sao cho

dL

dt=

dL

dt

=

(x)

dL

dt

(x)dx. (1.52)

i chiu (1.52) vi (1.51), ta thu c cng thc ca o hm tont theo thi gian, c gi l phng trnh Heisenberg:

dLdt

= Lt

+ iH, L . (1.53)

Trong c hc c in, tch phn chuyn ng l mt i lng khngthay i trong qu trnh chuyn ng. Trong c hc lng t cng c tchphn chuyn ng, l khi (dL/dt) = 0, i lng L khng thay i theothi gian v l tch phn chuyn ng. Da vo phng trnh Heisenberg(1.53), nu L l tch phn chuyn ng th

Lt + iH, L = 0. (1.54)

Trng hp c bit ng ch : khi L khng ph thuc tng minhvo thi gian, ta c (L/t) = 0, phng trnh (1.54) tr thnh

H, L

= 0, (1.55)

ngha l khi ton t L khng ph thuc r rt vo thi gian v giao honvi ton t nng lng H th i lng ng lc L tng ng l tch phnchuyn ng.

Theo (1.52), nu L l tch phn chuyn ng th (dL/dt) = 0 hayL = const.: tr trung bnh ca tch phn chuyn ng khng ph thuc thigian.

Ta c th chng minh xc sut p(Ln, t) tch phn chuyn ng L cgi tr bng Ln khng ph thuc vo thi gian. Thc vy, L, H giao hon vinhau nn chng c hm ring chung n(x)

Ln(x) = Lnn(x) v

Hn(x) = Enn(x),

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(x, t) =n

Cn(t)n(x), trong Cn(t) = cneiEnt = Cn(0)e

iEnt.

Theo tin 3 ca c hc lng t

p(Ln, t) = |Cn(t)|2 = |Cn(0)|2 = const.y l iu phi chng minh.

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22

Chng 2

Mt s phng php gn ng trongc hc lng t

Bi ton trong c hc lng t l gii phng trnh Schrdinger

H = E

2

2m2 + U(r, t)

= E

tm nghim E v . Nghim chnh xc ca phng trnh ch c th tmc trong mt s tng i nh cc trng hp n gin nht. S phctp ca vic gii phng trnh ph thuc vo dng ca th nng v s chiukhng gian trong bi ton cn gii. Phn ln cc bi ton ca c hc lng

t dn ti nhng phng trnh rt phc tp v dng ton hc, v khng thgii c chnh xc. Do phi ng dng nhng phng php gn ng gii bi ton, ngha l phi tm mt cch gii gn ng cc hm ring vtr ring ca n. Gn y, do c xut hin my tnh in t nn cc phngphp gii gn ng bng s cc bi ton c hc lng t c tm rt quantrng.

Trong chng ny, chng ta s kho st mt phng php gn ngthng c dng trong c hc lng t, l l thuyt nhiu lon. Thut

ng nhiu lon c vay mn trong thin vn hc ch nh hng camt hnh tinh ny ln qu o ca mt hnh tinh khc. Ni dung ca phngphp nhiu lon ln lt c kho st nh sau.

Gi s Hamiltonian ca h vi m ang xt c dng

H = H0 + V , (2.1)

trong V l ton t hiu chnh nh (ton t nhiu lon) cho ton t khngnhiu lon H0. iu kin coi V l nh so vi H0 s ni sau. xc

nh, ta xt trng hp ph gin on. Gi thit bi ton tm hm ring (0)

n ,

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C hc lng t nng cao Ch.2: Mt s phng php gn ng trong c hc lng t 23

tr ring E(0)n ca ton t khng nhiu lon H0 t phng trnh

H0(0)n = E

(0)n

(0)n (2.2)

c gii chnh xc. By gi cn phi tm nghim gn ng ca phngtrnh

H =

H0 + V

= E, (2.3)

ngha l phi tm cc biu thc gn ng cho cc hm ring n v cc trring En ca ton t H.

2.1 Nhiu lon dng trong trng hp khng suy bin

Trong tit ny, gi thit tt c cc tr ring ca H l khng suy bin,chng ta tm nghim ca (2.3) di dng khai trin theo cc hm ring trcgiao chun ho ca ton t H0

=k

Ck(0)k . (2.4)

Thay (2.4) vo (2.3), c xt n (2.2), ta thu c

k

Ck E(0)k + V(0)k = k

CkE(0)k .

Nhn hai v ca ng thc mi tm c vi (0)m v ly tch phn theo tonmin ca cc bin c lp, ng thi xt n tnh trc giao chun ho cacc hm (0)k , th ta c

Cm

E E(0)m

=

kVmkCk, m = 1, 2, 3,... (2.5)

trong

Vmk =

V

(0)m (x)V (0)k (x)dx (2.6)

l phn t ma trn ca ton t nhiu lon c tnh theo cc hm sng cabi ton khng nhiu lon. H phng trnh (2.5) hon ton tng ng viphng trnh (2.3). N chnh l phng trnh Schrdinger trong biu dinnng lng. By gi, ta s dng gi thit coi ton t V l nh theo ngha l

cc mc nng lng v hm sng trong bi ton nhiu lon s gn vi cc gi

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tr tng ng ca bi ton khng nhiu lon. V th, ta s tm chng didng chui vi tham s b 1.

En = E(0)n + E

(1)n +

2E(2)n + ... (2.7)Cm = C

(0)m + C

(1)m +

2C(2)m + ... (2.8)

v Vmn = vmn, V = v (2.9)

Chng ta tm hiu chnh cho mc nng lng th n v hm sng tng ngca bi ton nhiu lon. Ta xt gn ng cp khng, tc l khng nhiu lon(V = 0), ta c H = H0 v = 0, hm sng n =

(0)n , ngha l

n = (0)n = k C

(0)k

(0)k =

(0)n =

C

(0)k = kn (2.10)

v E = En = E(0)n , do En = E(0)n + E

(1)n +

2E(2)n + .... (2.11)

Thay (2.10) v (2.11) vo (2.5) vi lu E = En, ta cmn + C

(1)m +

2C(2)m + ...

E(0)n E(0)m + E(1)n + 2E(2)n + ...

=

kvmk

kn + C

(1)k +

2C(2)k + ...

,

(2.12)Hay

mn

E(0)n E(0)m

+

mnE(1)n + C

(1)m

E(0)n E(0)m

vmn

+2

mnE

(2)n + C

(1)m E

(1)n + C

(2)m

E(0)n E(0)m

k

vmkC(1)k

+ ... = 0.

Suy ra, ta c cc phng trnhmnE

(1)n + C

(1)m

E(0)n E(0)m

vmn = 0, (2.13)

mnE(2)n + C

(1)m E

(1)n + C

(2)m

E(0)n E(0)m

k

vmkC(1)k = 0. (2.14)

Phng trnh (2.13) cho

Khi m = n, ta thu c E(1)n = vnn, (2.15)

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Khi m = n, ta thu c C(1)m =vmn

E(0)n E(0)m

. (2.16)

Phng trnh (2.14) cho ta khi m = n

E(2)n = C(1)n

vnn E(1)n

+k=n

vnkC(1)k =

k=n

vnkC(1)k v theo (2.15) vnn = E

(1)n .

Vn dng (2.15) v (2.16), ta suy ra

E(2)n =k=n

vnkvkn

E(0)n E(0)k

=k=n

|vnk|2E

(0)n E(0)k

. (2.17)

Cn khi m = n, (2.14) choC(2)m

E(0)n E(0)m

= C(1)m E(1)n +

k

vmkC(1)k .

Lu (2.15) v (2.16), ta thu c

C(2)m = vnnvmn

E

(0)n E(0)m

2 +

C(1)n vmn

E(0)n E(0)m

+k=n

vmkvkn

E

(0)n E(0)m

E

(0)n E(0)k

.

By gi ta tm gi tr ca C(1)n v C(2)n . Chng c th thu c t iu kinchun ho c xt n (2.4)

V

(x)(x)dx = 1 k

|Ck|2 = 1. (2.18)

Thay khai trin (2.7) v (2.8) vo (2.18), ta thu c

k|kn + C(1)k + 2C(2)k |2 = 1.

k

kn + C

(1)k +

2C(2)k

kn + C

(1)k +

2C(2)k

= 1.

k

kn + kn

C

(1)k + C

(1)k

+ 2

kn

C

(2)k + C

(2)k

+ |C(1)k |2

= 1.

Cn bng cc i lng cng cp b v tri v v phi s rt ra c

C(1)n + C(1)n = 0 v C

(2)n + C

(2)n +k |C

(1)k |2 = 0. (2.19)

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T cc h thc (2.19), suy ra cc phn t o ca cc h s khai trinC

(1)n , C

(2)n l cc i lng tu . Do , khng hn ch tnh tng qut, ta c

th chn chng l thc v c gi tr

C(1)n = 0 v C(2)n = 12

k

|C(1)k |2 = 12k=n

|vkn|2E

(0)n E(0)k

2 . (2.20)Theo gi tr ca C(2)m tr thnh

C(2)m = vnnvmn

E(0)n E(0)m

2 +k=n

vmkvknE

(0)n E(0)m

E

(0)n E(0)k

. (2.21)Nh vy nng lng ca h nhiu lon c vit n mc chnh xc

cp hai l

En = E(0)n + vnn +

2k=n

|vnk|2E

(0)n E(0)k

, (2.22)

da vo (2.9), ta c

En = E(0)n + Vnn +

k=n|Vnk|2

E(0)n

E

(0)k

. (2.23)

Cn hm sng nu vit n mc chnh xc cp mt s l

n = (0)n +

k=n

vkn

E(0)n E(0)k

(0)k =

(0)n +

k=n

Vkn

E(0)n E(0)k

(0)k . (2.24)

Biu thc (2.24) cho thy rng s hng hiu chnh cp mt tht s b nu(|Vkn|/|E(0)n E(0)k |) 1, ngha l iu kin v ton t V nh l

|Vkn| |E(0)n E

(0)k |, n = k. (2.25)

2.2 L thuyt nhiu lon dng trong trng hp c suybin

2.2.1 L thuyt nhiu lon khi c hai mc gn nhau

T cc cng thc (2.23) v (2.24), ta thy rng nu trong s cc tr

ring E(0)

n ca

H0 c hai mc nng lng gn bng nhau th cc hiu chnh

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cho hm sng v cc mc nng lng E(0)n s ln v ta khng dng c cccng thc . Tuy nhin, nu s cc tr ring gn nhau ln cn mc n ca H0khng nhiu th c th thay i phng php tnh sao cho c trong trng

hp ny vn c th kh c s xut hin cc s hiu chnh ln. Chng tach xt trong trng hp n gin l c hai mc nng lng gn nhau.Gi s H0 c hai tr ring E

(0)1 v E

(0)2 gn nhau, tng ng vi cc

hm ring (0)1 v (0)2 , cn tt c cc tr ring khc xa chng. Trong php

tnh gn ng cp khng, ta tm nghim di dng

(0) = a(0)1 + b

(0)2 (2.26)

Thay gi tr ny ca (0) vo trong phng trnh

H(0) = E(0), H = H0 + V ,

chng ta thu c

aH(0)1 + bH

(0)2 = E

a

(0)1 + b

(0)2

. (2.27)

Nhn (2.27) vi (0)1 v ly tch phn, ta c

aH11 + bH12 = aE; H11 = V (0)1 H(0)1 dx,H12 = V (0)1 H(0)2 dx (2.28)Tng t vi (0)2 , ta c

aH21 + bH22 = bE; H21 =

V

(0)2 H

(0)1 dx,H22 =

V

(0)2 H

(0)2 dx. (2.29)

Ta c:Hmn = V

(0)m H(0)n dx = E

(0)n mn + Vmn. (2.30)

Hai phng trnh (2.28) v (2.29) c bin i thnh(H11 E)a + H12b = 0H21a + (H22 E)b = 0

(2.31)

cho h phng trnh c nghim khng tm thng (a = 0, b = 0),th nh thc ca n phi bng khng, ngha l

E

2

(H11 + H22) E+ H11H22 H12H21 = 0. (2.32)

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Gii phng trnh ta thu c cc nghim

E1 =

12

H11 + H22 +

(H11 H22)2 + 4|H12|2

E2 =

12

H11 + H22

(H11 H22)2 + 4|H12|2

,

(2.33)

trong ta lu H12 = H21 do H l ton t hermitic.Ta xt hai biu thc ca (2.33) trong hai trng hp gii hn1. Nu H11 H22 |H12|, th theo (2.30) c ngha l

E(0)1 + V11 E(0)2 + V22 E(0)1 E(0)2 |V12|.Nh vy, iu kin (2.25) c tho mn v l thuyt nhiu lon trong tittrc c th ng dng c. Nu trong php gn ng ta c th b qua4|H12|2 trong s hng di cn s bc hai (2.33), th ta s c gi tr gnng cp mt trong php nhiu lon thng thng:

E1 = H11 = E(0)1 + V11; E2 = H22 = E

(0)2 + V22.

Trong php gn ng chnh xc hn, ngha l 1 + 1 + /2, ta thu cE1 =

1

2

H11 + H22 + H11 H22 + 2|H12|

2

H11 H22

,

E1 = H11 +|H12|2

H11 H22 = E(0)1 + V11 +

|V12|2E

(1)1 E(1)2

. (2.34)

Tng t, ta c

E2 = E

(0)

2 + V22 |V21

|2

E(1)1 E(1)2 , (2.35)trong E(1)i = E

(0)i + Vii, i = 1, 2.

2. Nu H11 H22 |H12|, trong trng hp ny, vi chnh xc ncc s hng c b cp mt

E1,2 =H11 + H22

2

|H12| + (H11 H22)2

8|H12|

. (2.36)

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Chng ta nghin cu xem hiu cc gi tr nng lng xc nh bi cccng thc (2.33) v hiu H11 H22 c quan h vi nhau nh th no. Munvy, t:

H11 = H0 + x; H22 = H0 x (2.37)Trong l mt h s khng i, x l bin c lp. Theo :H11 H22 = 2x v H11 + H22 = 2H0.

Tin hnh nhng php thay th tng ng trong (2.33), kt qu thu cnh sau:

E1 = H0 +

2x2 + |H12|2E2 = H0

2x2 + |H12|2.

(2.38)

Trn hnh v 2.1 c biu din cc th ca cc hm trong (2.38)(ng lin nt) v cc hm (2.37) (ng chm chm) ng vi mt gi trc nh no ca |H12|. Hiu cc tung ca cc ng lin nt v ccng chm chm gn nht cho ta hiu chnh cp hai ca cc gi tr nnglng. rng, hiu chnh cp hai bao gi cng lm tng khong cch giacc mc. V th i khi ngi ta gi l s y ca cc mc, c hiu llm tng khong cch gia cc mc gn nhau, xut hin do c xt n ccs hng b b qua trong Hamiltonian bi ton n gin ho hn. Tronghnh 2.1, ta nhn thy rng ngay c khi hiu H11 H22 = 0 th

E1 E2 = 2|H12| = 2|V12|.By gi ta tm hm sng tng ng vi cc nng lng E1 v E2. Munvy, cn xc nh cc h s a v b trong cng thc (2.26). T (2.31), ta c

a

b

=H12

E H11

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Th cc gi tr ca E bng E1 v E2 c xc nh cc biu thc (2.33)

a

b1,2=

2H12

(H11 H22)1 1 + 2H12H11H222, (2.39)

cc ch s 1 v 2 theo th t ng vi du + v ng trc du cn. t

tg2 =2H12

H11 H22 , (2.40)

cng thc (2.39) c dng mi

ab1,2 =tg2

1 1 + tg22.T rt ra a

b

1

= cotg,a

b

2

= tg. (2.41)

H thc chun ho cho hm sng (2.26) yu cu

a2 + b2 = 1, (2.42)

hai phng trnh (2.41) v (2.42) cho ta rt raa1 = cos , b1 = sin ; a2 = sin , b2 = cos . (2.43)

Thay cc kt qu ny vo cng thc (2.26), ta thu c cc hm sng chunho tng ng vi cc gi tr nng lng E1 v E2:

1 = (0)1 cos +

(0)2 sin

2 = (0)1 sin + (0)2 cos .(2.44)

Theo (2.40) khi bt ng thc H11 H22 |H12|, c nghim ngth tg2 0, do 1 =

(0)1 , cn 2 =

(0)2 ,

ngha l cc hm mi trng vi cc hm ban u. Khi bt ng thc H11 H22 |H12|, c tho mn th tg2 , ngha l = /4, cng thc(2.44) tr thnh

1 =

12

(0)1 +

(0)2

2 =

1

2 (0)1 (0)2 .

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T iu ni trn, suy ra rng trong s cc gi tr nng lngE1, E2, E(0)3 , E

(0)4 ,...

s khng c cc gi tr gn nhau. Do c th dng cc gi tr ny cng cchm tng ng ca chng 1, 2,

(0)3 ,

(0)4 ,... lm cc i lng gn ng cp

khng khi cn tnh cc hm sng theo cng thc (2.24) trong php tnhgn ng cp mt v cc hiu chnh cho nng lng trong php gn ngcp hai theo cng thc (2.23).

Phng php ny cng c th dng c khi E1 = E2, ngha l khi cmc suy bin bc hai vi hai hm (0)11 v

(0)12 . Tt c cc cng thc ny vn

cn ng, nu hiu (0)1 l (0)11 v

(0)2 l

(0)12 .

2.2.2 L thuyt nhiu lon dng khi c suy bin:

n gin, ta xt trc tip trng hp suy bin bi hai. C th lmt mc nng lng En ca h tng ng vi hai hm sng n1 v n2 clp tuyn tnh vi nhau. Ta c th chn sao cho n1, n2 trc chun. Vi nxc nh, ta gi thit

V

nndx = . (2.45)

Gi H l Hamiltonian ca h,

H =

H0 +

V . (2.46)

Ta cn tm tr ring E v hm ring ca H, ngha l phi tm nghim caphng trnh

H = E. (2.47)

Do c suy bin bi hai nn phng trnh (2.47) c th vitH0

(0)n1 = E

(0)n

(0)n1

H0(0)n2 = E

(0)n

(0)n2 .

(2.48)

Ta tm E v vi iu kin trc chun (2.45). Biu din hm didng t hp tuyn tnh

= C1(0)n1 + C2

(0)n2

E = E(0)n + E(1).

(2.49)

Thay (2.49) vo (2.47) v vn dng (2.46), (2.48), ta c

H0 + VC1(0)n1 + C2(0)n2 = E(0)n + E(1)C1(0)n1 + C2(0)n2 ,

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C1V (0)n1 + C2V

(0)n2 = C1E

(1)(0)n1 + C2E

(1)(0)n2 .

Nhn hai v ng thc trn vi (0)n , vi = 1, 2, ri ly tch phn trn tonmin gi tr ca x, ta c

C1V

(0)n V (0)n1 dx+C2

V

(0)n V (0)n2 dx = C1E

(1)V

(0)n (0)n1 dx+C2E

(1)V

(0)n (0)n2 dx

hayV1C1 + V2C2 = C1E

(1)1 + C2E(1)2, = 1, 2.

lu rng V =V

(0)n V

(0)ndx, ta suy ra

V11 E(1)C1 + V12C2 = 0V21C1 + V22 E(1)C2 = 0. (2.50)H phng trnh (2.50) c nghim khc khng khi nh thc lp bi cc hs ca cc n C1, C2 bng khng, ngha l

V11 E(1) V12

V21 V22 E(1)

= 0

Khai trin nh thc s thu c mt phng trnh bc hai theo E(1).Gii phng trnh, ta c hai nghim

E(1)1,2 =

1

2

V11 + V22

(V11 + V22)

2 4 (V11V22 V12V21)

E(1)1,2 =

1

2

V11 + V22

(V11 V22)2 + 4|V12|2

. (2.51)

Tm li, i vi h khng nhiu lon

H =

H0, ch c mt mc nng lngE(0)n cho hai hm sng (0)n1 v (0)n2 . Khi h c nhiu lon H = H0 + V, mc

nng lng ca h tch thnh hai mcE1n = E

(0)n + E

(1)1

E2n = E(0)n + E

(1)2

(2.52)

Xt trng hp c bit khi V11 = V22, V12 = V21, th

E

(1)

1 = V11 + |V12|; E(1)

2 = V11 |V12|. (2.53)

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ng vi hai gi tr E(1)1 v E(1)2 s c hai cp gi tr cho C1 v C2.

a)Vi E(1)1 = V11 + V12, h phng trnh (2.50) tr thnh

V12C1 + V12C2 = 0V12C1 V12C2 = 0.

Kt hp vi iu kin chun ho hm sng, ta suy ra

C1 = C2 =1

2,

nh vy hm sng ng vi mc nng lng E(1)1 l

n1 =1

2(0)n1 + (0)n2 .

b)Vi E(1)2 = V11 V12, h phng trnh (2.50) tr thnhV12C1 + V12C2 = 0

V12C1 + V12C2 = 0.

Trong trng hp ny ta tm c

C1 = C2 = 12

,

hm sng tng ng vi mc nng lng E(1)2 l

n2 =1

2

(0)n1 (0)n2

.

Mc nng lng khng cn suy bin na. Nh vy nhiu lon lm mtsuy bin.By gi ta xt l thuyt nhiu lon khi c suy bin bi n 2. C th

l t vn nh sau: Cn tm nghim ca phng trnh

H = E (2.54)

trong

H = H0 + V; H0(0)p = E

(0)(0)p , p = 1, 2,...,n. (2.55)

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Mt mc E(0) ng vi n hm (0)p . Gii hn tm cc hiu chnh nng lngtrong php gn ng cp mt v cc hm sng trong php gn ng cpkhng, ngha l

E = E(0)

+ E(1)

, =

pCp

(0)p .

(2.56)Thay (2.56) vo (2.54) v vn dng (2.55), (2.56), ta vit c :

H0 + Vp

Cp(0)p =

E(0) + E(1)

p

Cp(0)p , hay

pCpV

(0)p = E

(1)

pCp

(0)p , (2.57)

nhn hai v ca (2.57) vi (0)m , ri ly tch phn trn ton min gi tr cabin x, ta c

np=1

Cp

Vmp E(1)mp

= 0, (2.58)

cho m = 1, 2,...,n, ta thu c h n phng trnh dng (2.58) vi n n sC1, C2,...,Cn. Mun cho cc nghim khng tm thng th nh thc lp bicc h s ca cc n phi bng khng

V11 E(1) V12 V13 ... V 1n

V21 V22 E(1) V23 ... V 2n... ... ... ... ...

Vn1 Vn2 Vn3 ... Vnn E(1)

= 0 (2.59)

Khai trin nh thc (2.59), ta c mt phng trnh bc n ca E(1). Phngtrnh trn gi l phng trnh th k (thut ng mn trong thin vn hc).

N c n nghim thc. Nu tt c cc nghim ca phng trnh th k ukhc nhau th mc nng lng E(0) bi n ca bi ton suy bin s tch thnhn mc nng lng khc nhau E(0)p , p = 1, 2,...,n

Ep = E(0) + E(1)p , (2.60)

mi mc Ep ng vi mt hm sng

p = k Ck(0)pk . (2.61)

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Trong trng hp ny suy bin bi n b mt hon ton.Nu mt hay mt s nghim ca phng trnh th k (2.59) l nghim

bi s th suy bin bi n b mt mt phn. Cc hm sng pk vi cc nghim

bi Epk, k = 1, 2,...,s, ca phng trnh (2.59) khng c xc nh n trbi cc phng trnh. Tuy nhin, bao gi cng c th chn chng sao chochng c trc giao vi nhau. Cc hm sng ng vi cc nghim khc nhauca phng trnh (2.59) cng trc giao vi nhau. Trong phn ny, ta nhnthy rng nhiu lon lm mt suy bin. Thng thng khi c nhiu lon,cc tr ring ca ton t H0 s khng suy bin hoc bi suy bin gimi. iu ny c lin quan mt thit n tnh i xng ca Hamiltonian ivi mt lp xc nh cc php bin i to ca h. Thng thng, nhiulon V khng c cng tnh i xng vi H0, do Hamiltonian tng hpH = H0 + V s khng c tnh i xng nh trc v mc nng lng ca ns khng suy bin. Nh vy, nhiu lon lm mt s suy bin.

2.3 Hiu ng Stark trong nguyn t Hydro

Khi nguyn t c t trong mt in trng th cc vch quang phca n s b tch ra. Hin tng ny c Stark pht hin vo nm 1913.Hiu ng Stark ch c th gii thch bng c hc lng t. Trong phn ny, tagii hn kho st hiu ng Stark bc nht, c trng cho nguyn t ngdng Hydro. i vi cc nguyn t ny, cc mc nng lng khng nhngsuy bin theo m m cn suy bin theo na. Chnh s suy bin theo gy ra hiu ng Stark bc nht. Cn i vi cc nguyn t khng phi ngdng Hydro, s suy bin theo ni chung khng c, do khng quan stc hiu ng Stark bc nht. Vi mc nng lng th nht (n = 1, = 0)khng c suy bin nn khng c s tch mc, do ta s xt s tch mcnng lng th hai ca nguyn t Hydro (n = 2).

Do in trng ngoi Etrong cc th nghim vo khong 104106V/cm,nh hn rt nhiu so vi in trng gy bi ht nhn Enh = e/a2 5.109V/cm, trong a l bn knh qu o Bohr th nht, nn ta c thdng l thuyt nhiu lon kho st hiu ng Stark. y, ton t nhiulon l ton t th nng ca in t trong in trng ngoi V

V = eEz. (2.62)

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trng thi khng nhiu lon, in t c mc nng lng

E(0)2 =

R

4, (2.63)

trong R l hng s Rydberg. Mc nng lng ny (n = 2) tng ng vin2 = 4, hm sng

(0)1 = 200 = R20(r)Y00 =

14

R20(r), (2.64)

(0)2 = 210 = R21(r)Y10 =

3

4R21(r)cos , (2.65)

(0)

3 = 211 = R21(r)Y11 = 38 R21(r)sin ei, (2.66)(0)4 = 211 = R21(r)Y11 =

3

8R21(r)sin ei. (2.67)

Thay cc bin bng to Descartes, cc hm sng c dng

(0)1 = f1(r) =

14

R20(r), (2.68)

(0)2 = zf2(r); f2(r) = 34 R21(r)r , (2.69)Do

rsinexp(i) =

x2 + y2exp(i) = x + iy;

rsinexp(i) =

x2 + y2exp(i) = x iynn

(0)3 = f2(r)

x + iy2

, (2.70)

(0)4 = f2(r)

x iy2

. (2.71)

Hm sng tng qut nht ng vi mc nng lng E(0)2 l

=4k=1

Ck(0)k . (2.72)

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Trong trng hp ny, bi suy bin l 4 nn xc nh cc h s Ck vcc hiu chnh bc nht E(1)2 cho mc nng lng E2 ca trng thi nhiulon, ta p dng (2.58) tm c h 4 phng trnh sau:

V11 E(1)2 C1 +V12C2 +V13C3 +V14C4 = 0V21C1+

V22 E(1)2

C2 +V23C3 +V24C4 = 0

V31C1 +V32C2+

V33 E(1)2

C3 +V34C4 = 0

V41

C4 +

V42

C2 +

V43

C3+ V44 E(1)2 C4 = 0

(2.73)

trong

Vij =

V

(0)i V

(0)j dV = eE

V

(0)i z

(0)j dV. (2.74)

Ch cc phn t ma trn V12 = V21 l khc khng v chng l cc hmchn ca c ba to x,y,z

V12

= V21

= eEV f1(r)f2(r)z2dV, (2.75)

cn cc phn t ma trn V khc u trit tiu v biu thc di du tchphn ca chng u l hm l i vi ba to x, y v z.

Thay vo (2.75) cc hm f1(r), f2(r) ly t (2.68) v (2.69) vi lu rng

R20(r) =12a3

1 r2a

exp

r2a

,

R21(r) =1

6a3 r2a expr2a .

(2.76)

Ta tnh tch phn (2.75) trong to cu vi dV = r2 sin drdd,

V12 =eE

8a3

0

0

20

exp

r2a

1 r

2a

expr2a r

r

2az2r2 sin ddrd,

m

0 2

0

z2 sin dd = r2

0 2

0

cos2 sin dd =4

3r2.

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a vo bin mi = r/a, ta thu c

V21 = V12 =eEa12

0

exp()

1 2

4d = 3eEa. (2.77)

cho cc nghim C1, C2, C3, C4 khng tm thng th nh thc

E(1)2 3aeE 0 0

3aeE E(1)2 0 0

0 0 E(1)2 0

0 0 0 E(1)2

= 0 (2.78)

Khai trin nh thc, ta thu c phng trnh bc bn ca E(1)2

E(1)2

2

E(1)2

2 9a2e2E2

= 0. (2.79)

V tm c 4 nghim ca phng trnh (2.79)

E(1)

21

=

3aeE

; E(1)

22

= 3aeE

; E(1)

23

= E(1)

24

= 0. (2.80)

Mi nghim tng ng vi mt b hon ton xc nh ca cc h s

E(1)21 C11 = C21, C31 = C41 = 0,

E(1)22 C12 = C22, C32 = C42 = 0,

E(1)23 C13 = C23 = 0, C33 = 0; C43 = 0,

E(1)24 C14 = C24 = 0, C34 = 0; C44 = 0.

(2.81)Nh vy ng vi mc nng lng

E21 = E(0)2 + E

(1)21 = E

(0)2 3aeE (2.82)

ta c hm sng trong php gn ng cp khng

1 = C11(0)1 + C21

(0)2 = C11 (200 + 210) . (2.83)

iu kin chun ho hm sng cho

V 11dV = 1,

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ta tm c C11 = 1/

2, vy

1 =1

2(200 + 210) . (2.84)

Tng t mc nng lng

E22 = E(0)2 + E

(1)22 = E

(0)2 + 3aeE (2.85)

tng ng vi hm sng trong php gn ng cp khng

2 =1

2(200 210) . (2.86)

Cc mc nng lng E23 = E24 = E(0)2 tng ng vi trng thi 3 =211 (m = 1), hay 4 = 211 (m = 1), hay t hp tuyn tnh ca chngv C13 = C23 = C14 = C24 = 0. Cn C33, C34, C43 v C44 vn cha xc nh.

Nh vy s suy bin b kh mt phn, v mc nng lng ban u E(0)2ch tch thnh ba mc khc nhau. S minh ho c trnh by hnh 2.2.

2.4 Nhiu lon ph thuc thi gianXt mt h c nng lng ph thuc thi gian. Ta k hiu ton t nhiu

lon l mt hm ca thi gian V(t). Hamiltonian ca h trong trng hpny c dng

H = H0 + V(t). (2.87)

Trong trng hp ny, nng lng ca h khng bo ton, do khngc cc trng thi dng.

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Phng trnh Schrdinger ca h c dng

i(x, t)

t= H0(x, t) + V(t)(x, t). (2.88)

Ta s gii phng trnh bng phng php bin thin hng s do Diraca ra nm 1926. Gi

(0)n (x, t) = (0)n (x)e

iE(0)n t (2.89)

l cc hm sng trng thi dng bit ca h khng nhiu lon. Cc hmny tho mn phng trnh khng nhiu lon

i

(0)n (x, t)

t = H0(0)n (x, t) = E(0)n (0)n (x, t). (2.90)

Gi s c mt nhiu lon nh V(t) tc dng ln h. Hm sng cn tm(x, t) ca h nhiu lon tho mn phng trnh (2.88). Dng tng qut cahm sng

(x, t) =k

Ck(t)(0)k (x, t). (2.91)

V cc hm sng (0)k (x, t) to thnh mt h cc hm ring ca ton t

hermitic H0, nn mt khai trin nh trn bao gi cng thc hin c. Cch s khai trin Ck(t) ch ph thuc thi gian v khng ph thuc to .

Thay (2.91) vo (2.88) v ch n (2.90), ta c

ik

(0)k (x, t)

dCk(t)

dt=k

Ck(t)V(t)(0)k (x, t).

Nhn bn tri hai v vi (0)m (x, t) ri ly tch phn theo to , ta c

idCm(t)dt

= k

Vmk(t)Ck(t), (2.92)

trong

Vmk(t) = ei(E

(0)m E(0)k )t

V

(0)m (x)V(t)(0)k (x)dx = e

imktvmk(t),

mk =1

E(0)m

E

(0)k ; vmk(t) = V

(0)m (x)V(t)

(0)k (x)dx, (2.93)

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vi Vmk(t) l cc phn t nhiu lon bao gm c tha s thi gian.H phng trnh (2.92) l h phng trnh chnh xc. N tng ng

vi phng trnh (2.88), v tp hp cc h s Ck(t) xc nh hon ton hm

sng (x, t). Tuy nhin, gii phng trnh (2.92) khng n gin hn giiphng trnh xut pht (2.88). n gin ho phng trnh (2.92), ta cndng tnh cht nhiu lon V(t) l nh. Gi thit rng ban u khi t 0, h trng thi ring (0)n , theo

Ck(0) = kn. (2.94)

Bt u t t = 0, h chu tc dng ca mt nhiu lon nh, do hm sng(0)n ca trng thi ban u ph thuc t vo thi gian. V th, cc h s Ck(t)

ti thi im t > 0 c tm di dngCk(t) = kn + C

(1)k (t) + C

(2)k (t) + .... (2.95)

Hiu chnh C(1)k (t) c cng cp b vi phn t nhiu lon Vmk(t) (hayvmk(t)), C

(2)k (t) l bc hai i vi phn t nhiu lon,.... Thay khai trin

(2.95) vo (2.92), ta tm c cc phng trnh cng bc nhiu lon:- Bc nht

idC(1)

m (t)dt

= k

vmk(t)eimktnk = vmn(t)eimnt, (2.96)

khi ta b qua tt c cc s hng c cp b cp hai v cao hn canhiu lon. Ly tch phn (2.96), ta c

C(1)m (t) =1

i

t0

vmn(t)eimntdt. (2.97)

- Bc hai:i

dC(2)m (t)

dt=k

vmk(t)eimktC(1)k . (2.98)

Gii phng trnh ny bng cch th kt qu (2.97) vo v phi caphng trnh (2.98), ta thu c C(2)m (t). Tip tc lp li cho phng trnhnhiu lon bc 3, bc 4,...

Khi nhiu lon V(t) nh th ta c th gii hn php tnh gn ngbc nht.

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2.5 S chuyn di lng t ca h vi m sang cc trngthi mi di nh hng ca nhiu lon

Mt trong nhng bi ton quan trng nht ca c hc lng t l victnh xc sut chuyn di ca h t trng thi lng t ny sang trng thilng t khc. Gi s c mt h trng thi nng lng xc nh E(0)n vc m t bi mt hm sng xc nh (0)n . Nu khi t 0, h chu tc dngca mt nhiu lon V(t), th ti thi im t > 0, h s nm trong mt trngthi mi m t bi hm sng

(x, t) =

kCk(t)

(0)k .

iu ny c ngha l ti thi im t > 0 h c th mt trng thi bt kno trong s cc trng thi dng kh d ca n. Theo cc quy lut tngqut ca c hc lng t, xc sut tm thy h trong trng thi lng t mc xc nh bng |Cm|2. V ti t = 0 h trng thi dng n nn |Cm(t)|2xc nh xc sut chuyn di ca h t trng thi n sang trng thi m trongkhong thi gian t, wmn(t) = |Cm(t)|2 |Cmn(t)|2. y ch s th hai khiu trng thi u.

Nh vy nhiu lon l nguyn nhn gy ra s di chuyn ca h t mttrng thi lng t ny sang mt trng thi khc. Nt c trng ca qutrnh ny khng c s tng t trong vt l c in. l nhiu lon gyra s di chuyn t mt trng thi dng vi nng lng xc nh sang mttrng thi mi, trong nng lng khng c gi tr xc nh. Trng thicui ca h c m t bi hm sng (x, t) v l mt trng thi xc nh(theo nh ngha ca c hc lng t).

S di chuyn ny khng c thc hin bng bc nhy m din ratheo thi gian. T (2.97), ta d dng tnh c xc sut di chuyn ca h

t trng thi dng (0)n (x) sang trng thi dng (0)m (x), (m = n) trongkhong thi gian t 0 t c nhiu lon tc ng trong gn ng bc nht

wmn(t) = |C(1)mn(t)|2 =1

2

t0

vmn(t)eimntdt2 (2.99)

Lu rng (2.99) ch ng khi vmn(t), t nh hiu chnh C(1)mn nh so vi

n v.

a) Ta xt trng hp c bit khi nhiu lon khng ph thuc thi

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gian, ngha l V(t) = V(0), theo vmn(t) = vmn(0), nn

C(1)mn(t) =1

ivmn(0)

t

0

eimntdt = vmn(0)ei(E

(0)m E(0)n )t 1

E(0)m

E

(0)n

(2.100)

wmn(t) = |C(1)mn(t)|2 = 4|vmn(0)|2sin2

t2

E

(0)m E(0)n

E(0)m E(0)n

2 (2.101)Nu tm xc sut chuyn t trng thi ban u n n tt c cc trng thikhc th ta ly tng mi gi tr ca m

wn(t) = m wmn(t) = m |C(1)mn(t)

|2 = 4m |vmn(0)|

2sin2

t2

E

(0)m E(0)n

E(0)m E(0)n 2 .(2.102)

Nu cc trng thi cui c ph lin tc th du tng thay bng du tch phntheo bin vi phn

E

(0)m

dE

(0)m

wn(t) = 4|vmn(0)|2

sin2t2

E

(0)m E(0)n

E

(0)m E(0)n

2

E(0)m

dE(0)m , (2.103)

trong E(0)m l hm mt trng thi trong khong nng lng E(0)m E

(0)m + dE

(0)m . Khi tnh tch phn, ta xem |vmn(0)|2 v

E

(0)m

khng i v

dng cng thc

sin2(x)

x2dx =

th s thu c xc sut chuyn di t trng thi n n tt c cc trng thikhc

wn(t) = 2t

|vmn(0)|2E(0)m . (2.104)b) Mt trng hp quan trng khc l nhiu lon tun hon n sc:

V(t) = V0 exp(it). (2.105)Xc sut chuyn di lng t c dng

wmn(t) = |C(1)mn(t)|2 = 4|vmn(0)|2sin2

t2

E

(0)m E(0)n

E(0)m E(0)n

2 , (2.106)

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khi = E(0)m E(0)n (2.107)

th xc sut c gi tr cc i. R rng nu nng lng ca nng lng

kch thch bng hiu hai mc nng lng E(0)m E

(0)n th xc sut chuyn dit trng thi n n trng thi m ca h lng t c gi tr cc i. l

tnh cht cng hng ca s kch thch bng bc x. Vic kho st hm

fm(E(0)m , t) =

sin2[ t2

(E(0)m E(0)n )]

12(E

(0)m E(0)n )

2t

cho ta iu kin

E(0)m

t

Lu rng t yu cu bt nh nng lng ca trng thi cuiE

(0)m phi nh so vi nng lng ca trng kch thch, ta rt ra bt

ng thc

t 1

,

E(0)m .t

2

(2.108)

ngha l E(0)m nu thi gian tc ng ca nhiu lon ln so vi chuk ca nhiu lon.Khi t th da vo cng thc

limtsin2 t

2t= () : hm delta Dirac,

vi = (E(0)m E(0)n )/(2) v lu tnh cht (ax) = (x)/a, ta suy ra

wmn(t) =2

|vmn(0)|2 t

E(0)m E(0)n . (2.109)

Nh vy khi xt trong mt thi gian di th ch c chuyn di lng tnu i s ca hm delta Dirac bng khng, tc l nng lng bc x kchthch ng bng hiu hai mc nng lng.

2.6 Nguyn t Hli

Nguyn t Hli gm ht nhn dng mang in tch +2e v hai in t

chuyn ng xung quanh ht nhn. Chn h quy chiu c gc to ht

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nhn Hli, do ht nhn ng yn trong h quy chiu ny. Ta vit cHamiltonian ca h hai in t di dng

22m21 22m22 2e2r1 2e2

r2 +

e2

r12 = E, (2.110)trong r1, r2 theo th t l khong cch gia ht nhn Hli vi hai in tv r12 l khong cch gia hai in t. Hai s hng th ba, th t m t thnng tng tc gia

hai in t vi ht nhn, s hng cui cng m t nng lng tng tcCoulomb gia hai in t. Trong biu thc ca Hamiltonian nu trn, ta b qua mt s hiu ng gy bi mmen t spin v mmen t qu o ca

in t,.... Do nghim ca phng trnh (2.110) cn c tm di dngtch ca cc hm to (r1, r2) v hm spin (1, 2)

= (r1, r2) (1, 2) . (2.111)

thu c cc hm sng v nng lng ca trng thi c bn ca Hlitrong php gn ng tm gi l tho ng, ta dng phng php nhiu lon.Khi tng tc gia hai in t e2/r12 trong (2.110) c xem nh mtnhiu lon. Theo , trong php gn ng cp khng, phng trnh (2.110)

c dng

2

2m21

2

2m22

2e2

r1 2e

2

r2

(0) = E(0)(0), (2.112)

Bng phng php phn ly bin s vi s b qua tng tc gia hai in tnh trn, ta vit li Hamiltonian ca h

H = H1 + H2; H1 = 2

2m21

2e2

r1; H2 =

2

2m22

2e2

r2. (2.113)

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Gi 1 (r1) , 2 (r2) theo th t l hm sng ca ch in t 1, 2. Ta vit c

1 (r1) = (n,,m,sz)1; 2 (r2) = (n,,m,sz)2 (2.114)

v H11 (r1) = E11 (r1) ; H22 (r2) = E22 (r2) . (2.115)

Hm v nng lng E ca h gm c hai ht trong trng ht nhn s l

= 1 (r1) 2 (r2) = (n,,m,sz)1(n,,m,sz)2; E = E1 + E2.

Thc vy

H =

H1 + H2

1 (r1) 2 (r2)

= H11 (r1) 2 (r2) + H21 (r1) 2 (r2)= E11 (r1) 2 (r2) + E21 (r1) 2 (r2)= (E1 + E2) 1 (r1) 2 (r2) , r rng

H = E.

Ta c th k hiu tp hp cc s lng t (n,,m,sz)i ni, do i(ri) = ni(ri),... Tng qut, trng thi ca c h c to khng gian lr1, r2,... s l chng cht cc trng thi trn

(r1, r2) = n1,n2

C(n1, n2)n1(r1)n2(r2). (2.116)

l dng tng qut ca hm sng ca h ti mt to khng gian xc nh.By gi nu coi tng tc V = e2/r12 gia hai in t vi nhau l nhiu

lon th

H = H1 + H2 +e2

r12= H0 + V . (2.117)

Khi V = 0, nng lng ca h l E0 v 0 l hm sng tng ng khi khng

c nhiu lon.Trong php gn ng cp mt ca nhiu lon, mc nng lng trng

thi c bn c cho bi cng thc

E = E0 + E(1) = En + Em + E

(1). (2.118)

En, Em l cc mc nng lng ca in t 1 v in t 2 ln lt trngthi n v m c xc nh bi

(0)

=

(0)

1 = n(r1)m(r2), (2.119)

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(0) = (0)2 = n(r2)m(r1). (2.120)

Nh vy theo nguyn l khng phn bit cc ht ng nht , (0)1 v (0)2 u

tho mn cc phng trnh

H0(0)1 = E0

(0)1 ; H0

(0)2 = E0

(0)2 .

y l trng hp suy bin cp hai. Khi c ton t nhiu lon V = 0, ta c:H = E, H = H0 + V . (2.121)

Ta s tm hm sng trong php gn ng cp khng v mc nng lngtrong php gn ng cp mt. Ta vit c

= k

Ck(0)k = C1(0)1 + C2(0)2 , E = E0 + E(1). (2.122)

Vi lu

Vij = Vji =

(0)i V

(0)j dVidVj =

(0)i

e2

r12(0)j dVidVj ,

theo :V11 = V22; V12 = V21 = V

12 = V

21. (2.123)

ta tm E(1) bng cch cho nh thc sau bng khngV11 E(1) V12

V21 V22 E(1)

= 0. (2.124)Vn dng kt qu (2.123), ta suy ra hai nghim

E(1)

1

= V11 + V12; E(1)

2

= V11

V12.

Cui cng, suy ra nng lng h

E1 = E0 + V11 + V12

E2 = E0 + V11 V12

(2.125)vi E0 = En + Em. t

V11 = V22 K; V12 = V21 A, l cc i lng thc. (2.126)

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1,2 = E1,2 E0 = K A, (2.127)vi E = E0 + E(1), theo (2.50) ta c h phng trnh

(E0 + V11 E) C1 + V12C2 = 0V12C1 + (E0 + V22 E) C2 = 0.

(2.128)T suy ra

(K ) C1 + AC2 = 0

(K ) C2 + AC1 = 0.

(2.129)

Thay = K+ A, ta thu c C1 = C2 = 1/

2.Thay = K A, ta thu c C1 = C2 = 1/2.Suy ra hm sng i xng

s (r1, r2) =1

2

(0)1 +

(0)2

, Es = En + Em + K+ A, (2.130)

hm sng phn xng

a (r1, r2) =1

2 (0)1

(0)2 , Ea = En + Em + K A. (2.131)

Phng php ny khng cho chnh xc cao, so vi thc nghim thvo khong 20%. Ta s xt bi ton ny theo phng php khc c chnhxc cao hn, l phng php trng t hp Hartree-Fok.

2.7 Phng php trng t hp Hartree-Fok

2.7.1 Nguyn l bin phn

Phng trnh Schrdinger di dng tng qut H = E c th thuc t nguyn l bin phn

H E

dq = 0. (2.132)

Thc vy, tr trung bnh ca nng lng trng thi dng l

E = Hdq (2.133)

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Vi iu kin chun ho cho hm l

dq = 1, ta cn tm cchm sao cho E t cc tr. E l mt i lng ph thuc vo v cgi l phim hm. Ta tm cc tr ca phim hm ny. y l mt bi ton

bin phn c iu kin. Mun vy, ta dng phng php tha s bt nhLagrange. Ni dung phng php ny sau: Gi s cn tm cc tr ca phimhm f = f(x1, x2,...,xn) vi iu kin i(x1, x2,...,xn) = 0, i = 1, 2,...,n.Lagrange a bi ton tm cc tr f c iu kin v bi ton tm cc trkhng c iu kin ca hm F = f +

i ii, trong i c gi l cc

tha s bt nh Lagrange. Ta tm cc tr ca cc hm F vi bin x. Khi cc im m F t cc tr cng l nhng im m f t cc tr vi iu kinno .

Hm F t cc tr khi bin phn ca n F = 0.Chuyn bi ton (2.132) v bi ton cc tr ca F

F =

Hdq E

dq 1

. (2.134)

Trong s hng th nht v th hai ng vai tr ca f vi ii,

cn E l tha s bt nh Lagrange. Php tnh bin phn cho ta

F =

Hdq E

dq =

H E

dq = 0. (2.135)

Ly bin phn v c lp vi nhau, ta c

H E

dq = 0, (2.136)

H E

dq= 0 =

H E

dq, (2.137)

(2.137) vi ly tu l khc khng, suy ra

H E = 0; hay H = E, (2.138)(2.136) vi ly tu , nn suy ra

H E

= 0; hay H = E, (2.139)

Tm li, bi ton bin phn (2.135) khng c iu kin tng ng vibi ton bin phn c iu kin

Hdq= 0, (2.140)

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dq = 1. (2.141)

Gi tr cc tiu ca (2.140) vi iu kin (2.141) l gi tr u tin ca nnglng, tc l mc nng lng c bn E0. Ta s chng minh phim hm

I(0) =

0H0dq = E0

l tr ring cc tiu ca H. Thc vy, khai trin theo h hm ring vtrc chun {n} ca H:

=

n=0ann

v dng iu kin (2.141), ta thu cHdq=

n=0

|an|2En E0n=0

|an|2 = E0 = min{En} (2.142)

Nh vy, vic tnh nng lng E0 trong trng thi c bn quy v victnh cc tiu ca phim hm. Hm sng thc hin vic tnh cc tiu lhm sng 0 trng thi c bn.. T iu kin cc tiu (2.140), ta xt tip

cc i lng 1, E1, 2, E2,.... Cc hm sng ca cc trng thi dng kchthch n tip sau khng nhng ch tho mn iu kin chun ho m cnphi tho mn iu kin trc giao

ndq = 0. (2.143)

Cc hm n ny thc hin cc cc tr, ch khng phi cc tiu. Ta c thho iu kin (2.143):

Hm 1 buc phi trc giao vi 0,Hm 2 buc phi trc giao vi 0, 1 ,Hm 3 buc phi trc giao vi 0, 1, 2 ,...............................................................................,Hm n buc phi trc giao vi 0, 1,...,n1.Nh vy bit c cc hm 0, 1,...,n1 ta tm tip cc trng thi

tip theo sau v buc chng phi tho mn

||2dq = 1, mdq = 0, m = 0, 1, 2,...,n 1.

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Thc t vic tnh E0 quy v vic tm cc hm , c gi l cc hm th.Cc phng n c th trong phng php bin phn ny khc nhau cchchn cc hm th. Thng thng ngi ta chn hm th ph thuc vo cc

thng s ,,.... Tm cc tiu ca phim hm J(,,...)

J(,,...) =

(q,,,...)H (q,,,...)dq.

tm , , ta tnhJ

=

J

= ... = 0

v thu c 0, 0,.... Ta tm c hm th E = J(0, 0,...) rt gn E0 v

hm sng0(q, 0, 0,...) rt gn vi 0. Ta tip tc tm E1, 1, E2, 2 khi bit E0, 0E1 = min

1H1dq,

vi iu kin|1|2dq = 1, 10dq = 0.

E2 = min

2H2dq,

vi iu kin |2|2dq = 1, 21dq = 20dq = 0. c th, ta xt th d sau:Dng phng php bin phn tm tr ring v hm ring ca ton

t H ca dao ng iu ho mt chiu

H = 2

2m

d2

dx2+

m2

2x2,

ta tm hm th. Trc ht c nhn xt l hm th phi tho mn cc iukin chun, trong c iu kin

0 khi x

.

Hm sng ca trng thi c bn khng c mt (xem phn dao ngiu ho) c dng

(x, ) = A exp

1

2x2

iu kin

||2dq = 1 cho ta A = (/)1/4, vyJ() =

Hdx =

1

4

2

m+

m2

.

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J()

= 0 = 0 = m

Trong trng thi c bn E0 = J(0) = ()/2 v hm sng

0 = (x, 0) =m

1/4 expmx2

.Tm tip E1, 1. Hm 1 phi trc giao vi 0. Chn

1(x, ) = Bx exp

1

2x2

,

trong B2 = (2/

)3/2, do iu kin chun ho

|1|2dx = 1. Cn

J() = 1H1dx,vi iu kin cc tiu

J()

= 0 = 0 = m

.

Cui cng ta thu c

E1 = J(0) =3

2,

1 =

2

1/2 m

3/4x exp

mx2

2

.

2.7.2 Phng php trng t hp Hartree-Fok

a) Phng php trng t hp Hartree:

nghin cu h nhiu in t, ngi ta dng rng ri phng php

trng t hp. Ni dung phng php ny nh sau: Trong php gn ng cpkhng, tt c cc in t c coi nh chuyn ng c lp vi nhau trongtrng ht nhn. Da vo cc hm sng trong php gn ng cp khng, tatm c mt in tch v trng tnh in trung bnh gy ra bi tt ccc in t.

Trong php gn ng tip theo, mi in t c coi nh chuyn ngtrong trng ht nhn v trng gy bi cc in t cn li. Nghim caphng trnh Schrdinger trong trng ny cho ta hm sng trong php gn

ng cp mt.

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thu c phng trnh Schrdinger trong trng t hp, ngi tadng phng php bin phn. c th, ta xt nguyn t Hli v gi thitrng mi in t u trng thi s. Ta cng khng yu cu phi i xng

ho h hm sng ca h cc in t. Trong php gn ng cp khng, c haiin t c m t bng cc hm sng thc 1(r1) v 2(r2), cn hm sngca nguyn t c dng

= 1(r1)2(r2). (2.144)

Trong php gn ng (2.144), ta ly bin phn cc hm 1 v 2 c lp vinhau. Php tnh cho

1

2

H E

12dV2

dV1 = 0,

2 1 H E12dV1 dV2 = 0. (2.145)Do tnh tu ca cc bin phn, ta suy ra

2

H E

12dV2 = 0,

1

H E

12dV1 = 0.

(2.146)

Thay H t (2.110), ta i ti cc phng trnh sau, gi l phng trnht hp Hartree

22m21 2e2

r1+

22e2

r12dV2

1 = E11,

22m

22 2e2

r2+

21e2

r12dV1

2 = E22,

(2.147) y ta k hiu E1 = E H22, E2 = E H11 v

Hii =i 2

2m2i 2e

2

riidVi, i = 1, 2,

trong E l nng lng ton phn ca h hai in t trong trng htnhn, E1, E2 l nng lng ca cc in t ring l.

Cc phng trnh (2.147) chng t trong th nng ca mi in t cxut hin cc s hng b sung

g1(r1) = 22e2

r12dV2 = e

2(r2)

r12dV2, (2.148)

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g2(r2) =

21

e2

r12dV1 = e

1(r1)

r12dV1, (2.149)

trong i = e|i|2, i = 1, 2 l mt in tch gy ra bi mt in t tiim c to ri.Nng lng ton phn ca h bng

E =

HdV,

E =

12

2

2m21

2

2m22

2e2

r1 2e

2

r2+

e2

r12

12dV1dV2

E = E1 + E2 G. (2.150)

Trong G = e2

21

22

r12dV1dV2 =

12r12

dV1dV2 (2.151)

l nng lng tng tc tnh in gia cc in t. T (2.147), ta nhn thytrong biu thc ca E1, E2 u c mt nng lng tng tc gia cc int, do trong tng E1 + E2, nng lng ny c tnh n hai ln. Nhvy nng lng E ca h phi l E1 + E2 G. Nu h gm N in t, bnglp lun tng t, ta thu c phng trnh t hp Hartree cho in t th

i trong trng thi lng t ni:

2

2m2i + U(ri) +

k

ei

ek

|nk|2|ri rk|dVk

ni = Enini. (2.152)

a) Phng php trng t hp Hartree-Fok

Phng php trng t hp c xt n s i xng hay phn xng cahm sng c gi l phng php trng t hp Hartree-Fok. Trong trng

hp n gin nht ca h hai in t, tt c php tnh trn u c chuynd dng cho hm sng i xng ho

s(1, 2) =1

2[1(1)2(2) + 2(1)1(2)] , (2.153)

a(1, 2) =1

2[1(1)2(2) 2(1)1(2)] . (2.154)

Trong php gn ng cp khng, trng thi c bn ca nguyn t

Hli, hai in t trng thi ng dng Hydro 1s. Trng thi ny c k

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hiu ngn gn di dng (1s)2. Trong (... ) c nu trng thi in t, cn sm nu s cc in t trong trng thi . Mt s biu din nh th cgi l cu hnh in t. Trng thi kch thch th nht ca nguyn t Hli s

tng ng vi cu hnh (1s)1

(2s)1

. Cc hm sng s(1, 2) v a(1, 2) thucv s Young [2] v [1,1]. Hm sng ton phn (tch ca hm to vhm spin) phi l phn xng, do hm to i xng s phi ng vitrng thi c cc spin i song (spin ton phn bng khng), l cc paratrng thi; cn hm sng to phn xng a ng vi trng thi spin c ccspin song song (spin ton phn bng 1), do l cc ortho trng thi.

Trong php gn ng cp khng, cc para v ortho trng thi vi cuhnh (1s)1(2s)1 c cng nng lng. Tuy nhin, nu xt n tng tc giacc in t, th nng lng ca cc trng thi ny s khc nhau: nng lngca para trng thi para hi ln hn nng lng ca ortho trng thi orth.C th thy c iu t nhng nhn nh nh tnh n gin: T dngpara(1, 2) v orth(1, 2), suy ra rng khi hai in t c to trng nhauth hm orth(1, 2) = 0 cn hm para(1, 2) cc i. Nh vy trong trng thiorth(1, 2), cc in t thng xa nhau hn so vi khi chng trong trngthi para(1, 2). Do nng lng y Coulomb trung bnh ca cc in ttrong trng thi orth(1, 2) b hn nng lng trong trng thi para(1, 2).Th th s khc nhau v nng lng ca cc trng thi para v ortho ca

cu hnh (1s)1(2s)1 l h qu ca s tng giao trong chuyn ng ca ccin t, xut hin t cc iu kin v tnh i xng ca cc hm sng ivi s hon v cc to khng gian. Nu khng xt n tnh i xng cacc hm sng, th khng c s khc bit nng lng nh trn.

Chn hm a (c spin ton phn S=1) lm hm th sao cho hm nyl gn ng tt nht vi hm thc. Dng nguyn l bin phn, ta xt

min

HdV vi

dV = 1.

Bi ton rt v

H E

dV1dV2 = 0,

H E

dV1dV2 = 0,

Thay = a v H bng biu thc trong (2.110) ri ly bin phn c lp

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1, 2, ta thu c hai phng trnh

2

2m2 E 2e

2

r+ H22 + G22

1(r) [H21 + G12] 2(r) = 0, (2.155)

2

2m2 E 2e

2

r+ H11 + G11

2(r) [H12 + G12] 1(r) = 0, (2.156)

vi

Gik(r1) =

i(r2)k(r2)

e2

r12dV2,

Hik = i 22m2 2e2r kdV.cc phng trnh (2.155), (2.156) thu c bng phng php trng t hpHartree-Fok. So snh chng vi cc phng trnh t hp Hartree (2.147), tanhn thy trong cc phng trnh Hartree-Fok c thm cc tch phn traoi, l cc tch phn dng Gik. Phng php trng t hp c ng dngrng ri tnh hm ring v nng lng ca cc nguyn t phc tp vcho kt qu rt ph hp vi thc nghim. Tuy nhin, vic gii phng trnhHartree-Fok phi dng phng php tnh s v s dng my tnh.

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57

Chng 3

L thuyt tn x lng t

S tn x l s va chm ca cc ht. y l mt qu trnh rt c bntrong vt l vi m. y va chm c hiu l tng tc trong qu trnhdch chuyn i vi nhau. trng thi ban u hai ht t khong cch rtxa tin li gn nhau, trong qu trnh y tng tc lm thay i trng thichuyn ng ca chng. Sau qu trnh, hai ht li chuyn ng ri xa nhaura cho ti lc tng tc gia chng tr thnh khng ng k. Ta gi trngthi ny l trng thi cui cng ca qu trnh tn x. Nu cc ht trngthi cui cng ch khc vi trng thi u v xung lng m khng c thayi v loi ht cng nh trng thi bn trong th tn x gi l tn x nhi. Nu c s thay i v loi ht hoc v trng thi bn trong th tn xc gi l tn x khng n hi.

Thng thng, thun tin, thay cho din bin ca qu trnh tn xtheo thi gian, ngi ta xt bi ton dng tng ng vi gi thit cho rngc mt dng lin tc cc ht bay t v cc n tng tc vi tm tn x,sau bin thnh mt dng cc ht tn x t tm bay ra mi pha. Mt trong dng phi nh c th b qua tng tc gia cc ht ti. Trongbi ton tn x dng, nu bit c trng lc tn x, ta s tnh c dngcc ht tn x (ti khong cch v cng so vi tm tn x) nh l hm ca

dng cc ht ti.

3.1 Bin tn x v tit din tn x

3.1.1 Tit din tn x

S tn x c c trng bng tit din tn x vi phn d(, )

d(, ) =dNtx(, )

jt, (3.1)

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C hc lng t nng cao Ch.3: L thuyt tn x lng t 58

trong dNtx(, ) l s cc ht tn x trong mt n v thi gian trong gckhi d(, ) ly theo phng (, ); jt l mt dng cc ht ti. Ta chntrc z theo phng chuyn ng ca cc ht ti (hnh 3.1).

Gi jtx(r,,) l mt dng cc ht tn x ti cc khong cch r lnso vi tm tn x, ta c:

dNtx(, ) = jtx(r,,)dS,

trong dS l din tch vi cp vung gc vi bn knh vect vch t tm tnx di cc gc (, ). ln ca dS v phn t gc khi tng ng d cmi lin h

dS = r2d. (3.2)

Do , tit din tn x vi phn c xc nh bng cng thc

d =jtxjt

dS =jtxjt

r2d. (3.3)

Trong c hc lng t, jtx v jt theo th t c gi l mt dngxc sut tn x v mt dng xc sut ti.

T (3.3), ta thu c i lng

=1

jt

Sr

jtx(r,,)dSr =txjt

. (3.4)

c gi l tit din tn x hiu dng ton phn. Trong , dsr = r2d l

ln ca phn t din tch vi cp ti khong cch r so vi tm tn x ng vi

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gc khi d; tx =

jtxdSr l dng ht tn x qua mt mt kn bao quanhtm tn x, mt ly tch phn c gi thit cch tm tn x, do c thcoi ti mi im ca mt ny, cc ht tn x bay theo phng xuyn tm.

Theo (3.4), tit din tn x hiu dng ton phn l t s gia xc suttn x ton phn ca ht (trong mt n v thi gian) v mt dng xcsut trong chm ht ti.

Khi tng tc gia cc ht ch ph thuc vo khong cch gia chng,th bi ton v chuyn ng ca hai ht c th quy v hai bi ton chuynng mt ht: Mt bi ton xt chuyn ng ca mt ht vi khi lng rtgn m = m1m2/(m1 + m2) i vi tm qun tnh. Cn bi ton th hai xtchuyn ng t do ca tm qun tnh. Nghim ca bi ton th nht cho tagc tn x trong h tm qun tnh. Vic chuyn t h tm qun tnh sangh phng th nghim c thc hin bng cc cng thc

tg1 =m2 sin

m1 + m2 cos , 2 =

2

, (3.5)

trong 1 l gc tn x ca ht th nht, 2 l gc git li ca ht th haic xc nh trong h phng th nghim, cn l gc lch ca ht th nhttrong h tm qun tnh. Trong chng ny, ta ch kho st s tn x trongh tm qun tnh ca cc ht va chm.

3.1.2 Bin tn x

Chng ta xt bi ton tn x dng ca ht tn x ti tm ca trnglc. Chn tm tn x c nh ti gc to . Trc z hng theo phng cacc dng ht ti.

xa tm lc, ht ti chuyn ng t do, do nng lng ca htbao gi cng dng, c ph lin tc do khng b lng t ho, hm sng ca

ht c dng sng phng t = eikz. (3.6)

Tng tc ca ht vi tm lc c th c m t bng hm th U(r),gi thit rng hm ny ch khc khng trong mt min khng gian hu hnc r a , m ta gi l min tc dng lc. Trong min ny, ht b tn x, do hm sng ca ht thay i.

Tuy nhin, sau ht tn x li bay ra xa khi tm lc, n li chuynng t do. V dng ht tn x bao gi cng c phng i qua tm tn x,

nn chuyn ng ca ht tn x phi c m t bng mt sng cu phn

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k

tx = A(, )eikr

r, (3.7)

trong r,, l cc to cu; A(, ) c gi l bin tn x, nichung ph thuc vo gc , . Trong tn x n hi, k c gi tr nh nhautrong cc biu thc (3.6) v (3.7).

3.1.3 Tn x n hi ca cc ht khng c spin

trong min tc dng ca lc chuyn ng (r a), ht tn x tuntheo phng trnh

2

2m02(r) + U(r)(r) = E(r), (3.8)

trong m0 l khi lng ca ht tn x. Chia hai v (3.8) cho 2/(2m0) vchuyn v s hng th nht bn phi ca phng trnh, t

k2 =2m0E

2=

p2

2, (3.9)

ta suy ra dng mi ca (3.8)

2 + k2(r) = 2m02

U(r)(r). (3.10)

Nghim ca phng trnh (3.10) ti cc khong cch xa so vi tm tn x(r a U(r) = 0), bng tng cc hm t v tx

= eikz+ A(, )eikr

r. (3.11)

Trong biu thc (3.11), s hng th nht bn phi c vit trong to

Descartes, m t chuyn ng ca ht ti; cn s hng th hai c vittrong h to cu m t chuyn ng ca cc ht tn x.Mt dng ca cc ht ti

jt =

2m0i(tt tt ) . (3.12)

Do ch ph thuc vo z nn nu gi ez l vect n v theo phng z, th

jt =ez

2m0i tt

z t

t

z =k

m0= v0, (3.13)

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trong v0 l vn tc ht ti. R rng hm sng t c chun ho sao chomt dng cc ht ti v tr s bng vn tc ca ht ti v cc.

Gradient trong h to cu c xc nh bng cng thc

= r

er + 1r

e + 1r sin

e. (3.14)

Ta ch xt thnh phn xuyn tm jr ca dng ht tn x, nn

jr =

2m0i(txtx txtx) =

er2m0i

tx

txr

txtx

r

. (3.15)

Thay tx = A(, )eikr/r, ta thu c

jr = km0r2

|A(, )|2 . (3.16)

Th kt qu tnh c (3.13) v (3.16) vo (3.3), ta thu c biu thc chotit din tn x vi phn

d(, ) = |A(, )|2 dSr2

= |A(, )|2 d. (3.17)

Nh vy vic xc nh tit din tn x vi phn quy v vic tm bin tn

x. Vic tnh bin tn x thng c tin hnh nh sau: Tm nghimca phng trnh Schrdinger cho chuyn ng ca ht trong trng ca tmtn x. Ti cc khong cch xa tm, nghim c dng (3.11). Khi h sca nhn t exp

ikr

/r cho ta bin tn x cn tm.

Da vo phng php hm Green, c th vit nghim ca phng trnh(3.10) di dng

= 0 +

G(r, r)

2m02

U(r)(r)d3r, (3.18)

trong G(r, r) l hm Green m ta s xc nh sau, cn 0 l nghim caphng trnh khng c v sau

2 + k2

0 = 0,

nghim ca phng trnh ny c dng sng phng exp(ikr) = exp(ikz). tF(r) = (2m0U/

2)(r), phng trnh (3.10) tr thnh

2 + k2(r) = F(r), (3.19)

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t

(r) =

Vq

Aqqd3q, (3.20)

trong q(r) =

eiqr

(2)3/2. (3.21)

Coi q l mt vect no vi cc thnh phn qx, qy, qz cn d3q = dqxdqydqz.H hm q(r) l mt h hm trc chun, ngha l

q(r)q(r)d3r = (q q). (3.22)

tm Aq, ta thay (3.21) vo v tri ca (3.19)2 + k2

Aqqd

3q =

2 (Aqq(r)) + k2Aqq(r) d3q. (3.23)Dng (3.19) v ch rng dng tn x xuyn tm nn

2 eiqr = 2r2

eiqr = q2eiqr,

ton t 2

= (2

)/(r2

) khng tc dng ln Aq(, ), ta vit li (3.23)2 + k2

Aqqd

3q =1

(2)3/2

2 + k2

Aqe

iqrd3q. (3.24)

Theo phng trnh (3.19) tr thnh

1

(2)3/2

Aq

k2 q2 eiqrd3q = F(r). (3.25)Nhn hai v (3.25) vi q

(r) =

1

(2)3/2 exp(iqr) ri ly tch phn theo d3

rtrn ton vng gi tr ca r, ta c

1

(2)3

Aq

k2 q2 ei(qq)rd3qd3r = 1(2)3/2

F(r)eiq

rd3r,

1

(2)3

Aq

k2 q2 ei(qq)rd3qd3r = Aq k2 q2 1(2)3

ei(qq

)rd3r

d3q,

= Aq k2 q2 (q q) d3q,

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= Aq

k2 q2 ,Theo (3.25) tr thnh

Aq k2 q2 = 1(2)3/2 F(r)eiqrd3r,hay

Aq =1

(2)3/21

k2 q2

F(r)eiqrd3r,

chuyn q thnh q ta c biu thc cho bin tn x

Aq =1

(2)3/21

k2

q2

F(r)eiqrd3r (3.26)

Thay kt qu tnh c ny vo (3.20) v lu (3.21), ta c

(r) =

1

(2)3/21

k2 q2F(r)eiqr

eiqr

(2)3/2d3qd3r, (3.27)

Lu rng r trong

...d3r l bin tch phn, cn r trong (r) l to cahm . T (3.27), ta vit c

(r) = F(r) 1(2)3 eiq(rr)

k2 q2d3q d3r. (3.28)t

G(r, r) =1

(2)3

eiq(rr

)

k2 q2d3q, (3.29)

th (3.28) c th vit li thnh

(r) = F(r)G(r, r)d3r. (3.30)

By gi ta xt ngha hm Green.Dng ca (r) tu thuc vo dng ca s hng F(r), cn hm Green

G(r, r) l mt hm c th tnh c. C th nu cho

F(r) = (r0 r),th

(r) = (r0 r)G(r, r)d3r = G(r, r0).

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Nh vy, hm Green G(r, r0) l nghim ca phng trnh

2 + k2

(r) = (r r0)

Biu thc ca hm Green theo (3.29) c dngG(r, r) =

1

(2)3

eiq(rr

)

k2 q2d3q,

vect q th hin bin tch phn, r, r l hai bn knh vect tu . Ta chnthnh phn qz trng phng vi r r, cn d3q = q2dqsin dd

q(r r) = q|r r| cos = q cos ,trong ta t =

|r

r

|. Do

G(r, r) =1

(2)3 eiq cos

k2 q2q2 sin dqdd = e

ik

4,

G(r, r) = eik|rr|

4|r r|. (3.31)

Vy (3.30) by gi c th vit

r(r) = 14 F(r

)eik|rr

|

|r

r

|d3r. (3.32)

y l nghim ring ca phng trnh Schrdinger c v sau. Nghim tngqut ca phng trnh l tng ca hai nghim

(r) = eikz+ r(r) = eikz 1

4

F(r)

eik|rr|

|r r|d3r. (3.33)

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Xt trng hp min tc dng ca lc l mt h c kch thc nh|r| |r|, ti min U(r) = 0, ngoi min U(r) = 0. T hnh v 3.2,ta c |r r| = r r cos v

r cos = r.r

rnn |r r| r r.r

r

Do

(r) = eikz+ r(r) = eikz 1

4

F(r)

eikrr.rr

r r.r

r

d3r. (3.34)Nu ta xem |r r| r v gi n = r/r, k.(r/r) = kn = kr l vect

sng hng theo phng bn knh vect, n c trng cho phng truynsng ca sng cu phn k. V tn x n hi nn |kn| = k. Thay F(r) =(2m0U/

2)(r) v cc khai trin trn vo (3.33), ta c

(r) = eikz eikr

4r

2m0U(r

)2

(r)eikr

d3r, (3.35)

hay

(r) = eikz+ A(, )eikr

r

, (3.36)

trong

A(, ) = m022

U(r)(r)ei

krd3r. (3.37)

V mt l thuyt, nu bit c bin tn x A(, ), ta s tnh c titdin tn x hiu dng. Mt khc, trong A(, ) c th nng tng tc U(r),do nu bit c U(r) ta s tnh c A(, ) v tnh c tit din viphn d. V mt thc nghim, ta o c d, nn tnh c bin tn x

A(, ), t tnh c th nng U(r). l phng php thc nghim kho st nhng th nng tng tc U(r) cha bit.

3.2 Tn x n hi trong php gn ng Born

Tuy tm c biu thc tim cn ca hm sng, nhng vn chathu c dng c th ca bin tn x. Thc vy, theo cng thc (3.37),bin tn x li c biu din qua hm sng (r). Vic gii chnh xc

phng trnh Schrdinger v vic tm A(, ) trong phn ln cc bi ton

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u gp kh khn v mt ton hc. Do , trong l thuyt nhiu lon ngita dng rng ri cc phng php gn ng. Mt trong nhng php gnng quan trng nht l phng php gn ng Born. Gi thit c bn ca

phng php ny l ch coi th nng tng tc U ca ht tn x vi tmtrng lc l nh. Do c th coi th nng tng tc U nh mt nhiu lonnh, theo chuyn ng ban u ca ht t thay i, phng trnh tchphn (3.35) c th gii c d dng bng phng php gn ng lin tip.

Trong php gn ng cp khng, ta b qua s hng ca hm sng ccha th nng:

0(r) = eikz = ei

k0r, (3.38)

trong k0 = kn0 = kez. Trong php gn ng cp mt, thay cho hm sng

v phi ca (3.35), ta a vo hm sng cp khng 0(r). Ngha l:

(r) = eikz m0eikr

22r

U(r)ei(

k0k)rd3r, (3.39)

Trong php gn ng cp mt ny, bin tn x bng

A(, ) = m022

U(r)eir

d3r, (3.40)

trong ta k hiu = k0 k. (3.41)

T hnh v 3.3, mun ca vect va chm c xc nh bng hthc

= k

|n

n0

|= 2k sin

2

=2m0v

sin

2

. (3.42)

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Mt cch tng ng, vect P = c gi l vect truyn xung lng. Nuth nng khng ph thuc vo cc gc, ngha l U = U(r) th trong (3.40)c th thc hin php ly tch phn theo cc gc

A(, ) = m022

0

U(r)r2dr0

eir cos sin d

20

d,

A(, ) = 2m02

0

U(r)sin(r)

rr2dr. (3.43)

Trong php gn ng cp mt, bin tn x c xc nh bng thnng lu tha mt. Trng hp ang xt c th nng i xng cu, bin tn x khng ph thuc vo gc . Thay biu thc (3.40) vo (3.17), ta c

biu thc ca tit din tn x vi phn c gi l cng thc Born:

d = |A(, )|2d = m20

424

0

U(r)eir

d3r2 d

d =4m204

0

U(r)sin(r)

rr2dr

2 d. (3.44)Cng thc ny c ng dng nhiu trong vt l ht nhn.

Trng hp gi tr nh ca gc tn x th

d =4m204

0

U(r)r2dr2 d (3.45)

khng ph thuc vo vn tc ht.Tip tc cc php tnh gn ng k tip, ngha l thay (3.39) vo v

phi ca (3.35), ta c th tm c hm sng v bin tn x trong phpgn ng cp hai, c xc nh bi tch phn ca bnh phng th nng

tng tc. Mt cch tng t, ta c th tm c cc hiu chnh vi cc cptip theo.By gi, ta xt iu kin c th ng dng c cng thc Born. Th

nng tng tc ca ht tn x vi trng tn x trong php gn ng Bornc gi thit l nh v c th xem nh l mt nhiu lon. T (3.10), ta cth chng minh c rng th nng U(r) c xem l nhiu lon nu mttrong hai iu kin sau c thc hin:

|U

| 2

m0a2(vi ka

1) (3.46)

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hay

|U| va

=2

m0a2ka (vi ka 1), (3.47)

trong a l bn knh tng tc ca trng U, cn U c cp ln catrng trong min tn ti c bn ca n.Biu thc 2/(m0a2) v cp ln bng su cc tiu ca ging th

bn knh a, ti c xut hin mc nng lng. T rt ra ngha ngin ca iu kin c th ng dng c php gn ng Born cho cc httn x chm. C th t iu kin (3.46), suy ra rng nng lng tng tctrung bnh phi nh so vi th nng cc tiu ca ht trong ging th, ti c hnh thnh trng thi lin kt. Khi iu kin (3.46) c thc hin, phpgn ng Born c ng dng cho tt c cc vn tc.

3.3 Phng php sng ring phn

Ngoi l thuyt gn ng kho st, ngi ta cn pht trin mt lthuyt tn x chnh xc gi l l thuyt tn x pha hay l thuyt tn x ccsng ring phn. Trong l thuyt ny, ngi ta khng c mt gi thit nocho th nng tng tc U. V vy, n c ng dng vi mi gi tr nnglng ca cc ht tn x. S chung ca l thuyt tn x cc sng ringphn khng khc vi s m t trong cc tit trc. Hm sng ca httn x xa tm c dng

(r) = eikz+ A()eikr

r, (3.48)

do trng tn x i xng xuyn tm theo phng z nn bin tn xkhng th ph thuc vo gc . C hm sng cng khng ph thuc vo. Tuy U(r) = U(r), nhng nghim ca phng trnh Schrdinger ca hkhc vi nghim ca hm sng trong trng xuyn tm ch l y chngta ch xt chuyn ng v hn v cc nghim phi tho mn cc iu kinbin sao cho dng iu ca nghim (vi r ) phi c xc nh bi cngthc (3.48).

Ta bit trong trng i xng xuyn tm, nghim tng qut caphng trnh Schrdinger c dng

(r,,) =

,mbmR(r)Ym(, ), (3.49)

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trong bm l cc h s khng i xc nh bng cc iu kin bin v iukin chun ho. Do khng ph thuc vo gc nn hm sng ch cn li ccs hng trong tng c m = 0, ngha l ch c cc hm cu

Y0() =2 + 1

4P(cos ), (3.50)

trong P(x) l a thc Legendre xc nh bi cng thc

P(x) =1

!2d

dx

(x2 1) . (3.51)Do cng thc (3.49

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