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PHNG TRNH SCHROEDINGER

Nam 1926, nha vat ly ngiAo Schroedinger a a ramot phng trnh cho phep xac

nh c ham song mo tatrang thai cua mot he lngt. Phng trnh nay ong motvai tro can ban trong c hoclng t

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HM SNGBiu thc sng phng n sc ti im M cchngun O mt on :

Vctsng xc nh theo vct n v ca phngtruyn sng:

Hm sng dng phc:v

= OMr

)r.ktsin(A)v.T

r.2tsin(A)t,r(

=

=

)]rkt(iexp[A)t,r(

=

k

n

2

k

=

)}rktsin(i)rkt{cos(A)t,r(

+=

}sini{cosAAe i +=

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4. Quan h gia sng Broglie v vi ht chuyn

ng tdocnng lngv xung lngTnh tn s gc:

Cn vctsng:

Hm sng vit di dng:

mvP = == chhE

.Ehc

.

h

2c22

=

=

==

Pn

h

h

2n

2k =

=

=

)]r.kt(iexp[A)t,r(

=

]rPEt)[i

exp(A

=

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1. ngha thn g k ca hm sng

Theo thuyt sng nh sng:

Thuyt ht nh sng: ht photon to ra I t l s photon qua 1m2

trong 1 s gi l mt ht:

V Hm sng phc m t trng thi vi m ca ht chuyn ngnhanh c bnh phng ca bin :

2. iu kin chun ha: Xc sut tm thy ht trong th tch V btk m ht ctr l 1.0.

3. iu kin ca hm sng:

1- Gii ni.2- n tr.3- Lin tc.4- oh mbc nht ca hm sng phi lin tc.

2i.i2 *Aee.AAI ===

2i.i2 AAee.A*.)t,r(p ====

2A*)t,r( ==

1dV)t,r(*).t,r(V

=

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TON T (OPERATOR)

1. Ton t: nh x tc dng ln mt hm bin hm thnh mt hm khc:

V d :

)t,z,y,x(g)t,z,y,x(fA =

xt4)zyx2(A 2 =+

2. Mt s ton tthng dngA-Ton t o hm:V d: dx

dA = 2)zyx2(

dx

d)zyx2(A 22 =+=+

321 ez

ey

ex

dGra

+

+

==

3

2

21

22 eyeyz2e2)zyx2()zyx2(dgra

++=+=+

2

2

2

2

2

2

zyxA

++==

2

22

2

22

2

222

z

)zyx2(

y

)zyx2(

x

)zyx2()zyx2(A

+

+

++

+

=+

z2)zyx2(A2

=+

zyx2)z,y,x(f 2+=

C-Ton tLaplace:V d :

B-Ton tgrad:

V d :

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A. PHP TON CHO TON T

1. PHP CNG:V d :

CBA =+

zxyx22)z,y,x(f)x

dx

d()zyx2(C 222 ++=+=+

2. PHP TRV d:

DBA =

zxyx22)zyx2(D 222 =+

)fB(Af)B.A( =

zyx4)}zyx2(x{dx

df)B.A( 22 +=+=

)fA(Bf)A.B( =

xB;dxdA ==

3. PHP NHN

V d :

zyx2)z,y,x(f 2+=

DEAB ==

x2)}zyx2(

dx

d{xf)AB( 2 =+= f)B.A(f)A.B(

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B. GIAO HON T

1. nh ngha:V d :

A.BB.A =

0)yz2(dx

d

)}zyx2(dy

d

{dx

d

)zyx2(B

.A 22

==+=+

z,y,x

dydB;

dxdA ==

2. Cc ton tgiao hon c

zyx2)z,y,x(f 2+=

0)2(dy

d)}zyx2(

dx

d{

dy

d)zyx2(A.B 22 ==+=+

dz

d;

dy

d;

dx

d2

2

2

2

2

2

dz

d;

dy

d;

dx

d

xy

;

yx

22

3. Cc ton tkhng giao hon c

dz

d;z

dy

d;y

dx

d;x ...

zy;

yx

22

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Bi tp : Xem cc TT sau c th giao hon c vi nhau ?

2. T hp ton tgiao hon cKhi m

)D

C

)(B

A

( ++ADDAACCA ==

BDDBBCCB ==

321 e

z

e

y

e

x

dGra

+

+

==

2

2

2

2

2

2

zyxA

+

+

==

321 ezeyexr

++=rdGra

++

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D. HM RING V TR RING CA TON T

1. nh ngha:V d : Ta c tm hm ring tr ring

)x(f)x(fA =

2. Dng nh ngha

dx

da =

3. Chuyn v:

)x(fdx

)x(df)x(fa ==

dx)x(f

)x(df=

4. Ly tch phn x.)c(lnc)x(flndx)x(f

)x(df1 =+=

5. Bin i x11 e)x(fcx.)x(fcln ==x

2ec)x(f=

6. Kt lun: C nhiu tr ring khc nhau c nhiu hmring khc nhau

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E. TON TTLIN HP TUYN TNH HERMITTE

1. nh ngha:Ta c l cc hm bt k

l TT Hermitte:

2. V d Xt ton t

Xt v tri : Dng tch phn tng phn:

V phi:

So snh: l Hermitte th ta c:

Kt lun: cc hm fi(x) khi nhn ln nhau bng khng

Gi l t rc giao

)x(f),x(f 21

dx)x(f].A)[x(fdx)x(fA)x(f 1221 =

dxdiA =

])dxfdx

d

f(ff[idx)x(fdx

d

)x(fi 122121

=

dx)x(fdx

d)x(fidx)x(f].

dx

di)[x(f 1212

=

0)x(f).x(f 21 =

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Tnh cht TT hermitte

1. N c tr ring l cc gi tr thc.

2. Cc hm ring l trc giao:

3. Cc hm ring to thnh mt h : mthmbtk c khai trin thnh t hp TT cc hm trcgiao

===

KLkhi0

KLkhi1)KL()x(f).x(f KL

)x(fC)x(Un

1k

kk=

=

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PHNG TRNH SCHRODINGERCc tin trong Clng t

1. Mi i lng a trong CH c in tng ng mt TTHermitte trong CH Lng tsao cho tr ring ca l sthc bng chnh gi tr ca i lng a.

V d l ton tnng lng c tr ring l E

2. H thc ca cc TT c hnh thc ging ht nhcc

i lng c in tng ng

H

r,z,y,x

]P.xr[L

=

V d: TT ta l php nhn

TT mmen xung lng

Hai TT giao hon th chng c cng hm ring v khngtun theo nguyn l bt nh.

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Cc ton tthng dng trong Chc lng t

1. TT ta = Tng ng php nhn r,z,y,x

2. Cc ton txung lng

4. Ton tnng lng:

Ton tth nng

xiPx = y

iPy =

ziPz =

]z.ey.ex.e[iiP321

+

+

==

3. Ton txung lng ton phn

)z,y,x(Um2

PE

2

+=

)zyx

(m2m2

P

2

2

2

2

2

222

++=

)x,y,x(U)z,y,x(U =

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PHNG TRNH SCHRODINGER ngha

1. Hm ring v tr ring ca ton tnng lng.

Nu nng lng l khng i

2. PT Schodinger khng ph thuc t

Gii c:- Tr ring l mc nng lng

- Hm ring m t trng thi

H (x, y, z, t) E (x, y, z, t)

iEt iEt(r, t) A exp( ) (r) A exp( ) (x, y, z) =

H (r) H (x, y, z) E (x, y, z) =2

[ U(x, y, z)] (x, y, z) E. (x, y, z)2m

+ =

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GII PHNG TRNH SCHRODINGER

MC CH KHI GII1. TM TR RING: Tc l xc nh cc mc nng lngv xem n c b gin an khng (lng tha)2. TM HM RING: Dng tnh xc sut nhng ni tmthy ht (m my in t). Xc nh hm mt xc sut

CC LU KHI GII1. BIT DNG TON T TH NNG: Thng l

mt php nhn. Nu n gin th U=02. CHIU CA KHNG GIAN: 1D/ 2D/ 3D. n gin lmt chiu khi

3. C khi phi tch khng gian lm nhiu vng khcnhau tm hm sng cho tng vng.

2 2 2

2

d

2m 2m dx =

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V HT TRONG H TH VUNG

U(x) = 0, 0

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iu kin ca hm sng1. Sin(x) va Cos(x) hu han

2. Lien tuc: (0) = (a) = 0

(0) =A sin(k0) +B cos(k0 ) = 0 B = 0 (x) =A sin(kx)

(a) =A sin(ka) = 0 sin(ka) = 0 ka =n, n = 1, 2, 3,

( ) 22 2nn n 22 2 2

2 2

n 2 2

2mEk n k n

a a

hE n n

2ma 8ma

= = = =

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Kt qu:

Kt lun v cc hm sng bc n:

1- Ta chng minh cc hm sng l trc giao.

2- Xc sut tm thy ht t l vi mc nng lng thn

(x) A sin kx A ? = =

a

2A1a

2

1Adx)kx(sinA

22

a

0

2===

Theo schun ha hm sng :

n n

2 2 n x(x) sin(k x) sin( )

a a a

= =

a

m n m n

0

2/ sin(k x)sin(k x)dx 0(m n)

a < >= =

x

U(x)

a

x

U(x)

a

x

U(x)

a

n=1 n=2 n=3

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Kt lun v mc nng lng:1- Nng lng b lng tha2- Nng lng t l vi bnh phng cc s

nguyn3- E1 l mc thp nht4- TE2 ln trn l mc kch thch

5- Khong cch cc mc khng u

)1n2(ma2]n)1n[(ma2EEE 2

2222

2

22

n1n +

=+

== +

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Kt qu: nghim tng qut l t hp tuyn tnh cc nghim

n n n

2 n xf (x) c c sin( )

a a

=

Kt qu: nghim c yu t thi gian

n n

2 n x(x, t) (x)exp( iEt) c sin( ) exp( iE t)

a a

=

)tma2

niexp()

a

xnsin(c

a

2

2

22

n

Kt qu: cho trng hp 3D ht trong hp vung

2

222

2

222

2

222

321 mc2

nx

mb2

ny

ma2

nx

EEEE

+

+

=++=

nx,ny,nz

2 nx x 2 ny y 2 nz z(x, y, x) sin( ) sin( ) sin( )

a a b b c c

=

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3.3 Hiu ng ng hmGii bi ton ht chuyn ng vt qua ro th c U

cao hn nng lng ca n.

O X

U0

Min 3Min 2Min 1

Khi 0 x U = 0: min 1Khi a x 0 U = U0: min 2Khi x a U = 0: min 3

TheoTheo cc hohocc llngng tt tata thathayy hahattvanvan cocokhakha nangnang xuyenxuyen quaqua raraoo thethetheotheo momott hiehieuu ngng gogoii lala::HieHieuu ngngngngngangamm..

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PhPhngngtrtrnhnh SchrodingerSchrodingerooii vviicacaccmiemiennnanayycocodadangng::

miemienn I:I:

vvii ::

miemienn II:II:

vvii::

miemienn III:III:

2 2I

1 I2d + K = 0dx

21 22mEK =

h

22II

2 II 2

d

+ K = 0dx2

2 02

2m K = (U - E)

h

1

2

2III

III2

d+ K = 0

dx

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NghieNghiemm totongng quaquatt cucuaa cacaccphphngngtrtrnhnhnanayycocodadangng::

TrongTrong

oo

::

vava

lala

aa

cc

trtrngng chocho sosongng ttii vava sosongng phaphannxaxa trentren bb x = 0.x = 0.

1 1iK x -iK x

I 1 1(x) = A e + B e

2-K x K x II 2 2

(x) = A e + B e

1 1iK (x-a) -iK (x-a) III 3 3

(x) = A e + B e

1iK x

1

A e 1-iK x

1

B e

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aacc trtrngng chocho sosongngtruyetruyenn quaqua raraoo thethe

momo tata sosongng phaphann xaxatt vovo cccc trtr veve..

CaCacc hahangng soso : A: A11, B, B11 , A, A33, B, B33cc gogoii lala cacacc bienbien oo sosongng .. VV

rarangng vovo cccc khongkhong coco ss phaphannxaxa sosongng, do, do oo tata aatt BB33 = 0= 0

1iK (x-a)

3A e

1-iK (x-a)

3B e

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TTnhnhhehesosoxuyenxuyenraraoo::

TaTa

nhnh

nghngh

aa

hehe

soso

truyetruye

nn

quaqua

raraoo DD lala tt soso gigiaa bbnhnh phphngngbienbien oo sosongng truyetruyenn quaqua hahangngraraoo vava bbnhnh phphngng bienbien oo sosongngttii.. 2

1

2

A

D =A

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CanCan cc vavaoo ieieuu kiekienn bienbien

sausau::

I I

' '

I(0) = (0)

II II I

' '(a) = (a)

I I I (0) = (0)

II III (a) = (a)

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TaTa coco thethe rurutt rara cacacc hehe ththcc sausau::

AA11 + B+ B11 = A= A22 + B+ B22

iKiK11( A( A11BB11) =) = --KK22( A( A22BB22))

AA22ee--KK22aa + B+ B22 ee

KK22aa = A= A33

--KK22( A( A22ee--KK22aa

BB22eeKK22aa

= iK= iK11AA33

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tm nghim, dng iu kin bin, v c 6 bin ng cc

min nhng ch c 4 DK bin phi b mt h s B3 vi githuyt sng khng phn x v cng.Vn ta quan tm l sng c qua ro khng?

H s truyn qua D: l t s gia bnh phng bin sngtruyn qua hng ro th v bnh phng bin sng ti tihng ro th.

???0AAD 2

1

2

3 =

Kt qa thu c0)ak2exp(

)n1(

n16D 222

2

+

=

n kk

EU E

= =

1

2 0

k m U E22 0

2

2= ( )

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T cac ieu kien lien tuc, co the suy ra rang

he so truyen qua co dang gan ung.

0D exp[ (2a / ) 2m(U E)] 0

2

2.m(U E)D exp( 2.T.a), T = =

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Trong trng hp mot hang rao co dang bat ky, ta co thephan tch no thanh nhng hnh bac thang, khi o bieu thccua D co dang

D exp[ (2a / ) 2m(U(x) E)dx

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V d: Nu hiu nng lng cho l E-U0=1,28.10-31 J, khi

ta c th dng l thuyt tnh sph thuc ca h struyn qua D vo rng h th a.

a(m) 10-10 1,5.10-

10

2.10-10 5.10-

10D 0,1 0,03 0,008 5.10-7

H s truyn qua D ch ng k khi rng h th a l

rt nh, khi ht th hin tnh cht sng ca vi ht viu khng th c vi c c ht vm.

ng dng:

1- Gii thch pht x lnh electron trong kim loi2-Phn r ht anpha tnhn c 2 prtn v 2 Ntrn.

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3.3 DAO NG TIU HATrong 1D : H chu tc ng lc tunhon f=-kx, nn ng nng U=kx2/2

Phng trnh Schrodinger mt chiu:

Xt hai ton ttng v gim:

Ly php nhn 2 ton t vit lI PT Schrodinger

2

xm

2

xm

2

xk)x(U

22222 =

==

2 2 22

n n n2

d m( i x ) (x) E (x)

2m dx 2

+ =

+ a,a

]xim

dx

d

i

[

m2

1a =

n n n n

1H (x) {(a a ) } (x) E (x)

2

+ + =

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NGHIM CA DAO NG TIU HANghim trng thi cbn khi Nu t c dng h bc s khng cn sngPhng trnh xc nh:

Gii c nghim:

Dng iu kin chun ha Bin sng l V vit lih m cbn:

Hm trng thi m

0a (x) 0 =0 0

1 da (x) [ im x] (x) 0

i dx2m

= =2

0 0

m(x) A exp( x )

2

=

4/1

0

mA

=

1/4

2

0

m m(x) exp( x )

2

=

1/4

m m 2

m 0

m m(x) (a ) (x) (a ) exp( x )

2

+

= =

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3.3 DAO NG TIU HATa chng minh c lun im sau:

Nu U(x) l nghim ring tha PT Schrodingervi tr ring E th hm +(x) cng l nghim ring

ca PT Schrodinger vi nng lng ring l +EHm -(x) cng l nghim ring ca PTSchrodinger vi nng lng ring l E

Kt qa v mc nng lng

1- Cc nng lng cch u nhau mt on

2- Mc nng lng thp nht c gi tr dng

v l nng lng nhit 0K. ??

3- Mc thJ bt k c gi tr

= E

= 2

1E0

+= )5,0j(Ej

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Gii bi ton cho dao ng t iu ha mt chiukhng ph thuc thi gian

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Kt qu: nghim c yu t thi gian m m m(x,t) (x)exp( iE t) Kt qu: cho trng hp 3D ht trong hp vung

222222321 zm

2

1ym

2

1xm

2

1)z(U)y(U)x(U)z,y,x(U ++=++=

Kt qu: V nng lng

)nznynxN()

2

3nznynx(EN ++=+++=

nx,ny,nz nx ny nz(x, y, x) X (x).Y (y).Z (z) =Kt qu: V hm sng

Lc ny c ssuy bin: Cng mt mc nng lng s cnhiu trng thi khc nhau do cc gi tr nx, ny v nz to ra.

73

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nx ny nz

Trng thi 1 2 0 0

Trng thi 2 0 2 0

Trng thi 3 0 0 2

Trng thi 4 1 1 0

Trng thi 5 0 1 1

Trng thi 6 1 0 1

V d vi mc =+= 2

7)

2

32(EN