Mô hình Ising trong mô phỏng vật lý

  • View
    11

  • Download
    3

Embed Size (px)

Transcript

  • 1M hnh Ising

  • 2Mt s kin thc v thng k

  • 3Mt s kin thc v thng k

    Khng gian pha

    Xt mt h c in N ht

    Trng thi ca h c xc nh bi ta r v xung lng p ca tt c cc ht

    Khng gian pha: 6N bin, = (r,p) hoc (q,p)

    S thay i trng thi theo thi gian tun theo cc phng trnh c hc c in

    qk=H pk

    , pk=Hqk

    H=KV p

  • 4

    Chuyn ng ca h theo thi gian m t bi mt qu o trong khng gian pha (t)

    Do tnh tt nh ca cc phng trnh Newton, qu o ny khng bao gi ct chnh n!

    Poincare: nu i lu th h c th quay tr v trng thi ban u!

    Poincare recurrence time > tui v tr i vi h v m

  • 5Mt s kin thc v thng k

    i lng o c A()

    Gi tr o c bng thc nghim l gi tr trung bnh theo thi gian

    Gibbs: ly trung bnh theo tp hp vi phn b cn thit!

    (): mt xc sut trng thi iu kin v m nht nh: NVE, NVT, NPT...

    Aobs=Atime=A t time=1

    tobs0

    t obs

    A t dt

    Aobs=Aens=

    A

    Tp hp thng k

  • 6Mt s kin thc v thng k

    Tp hp: bao gm cc bn sao ca h nhiu trng thi khc nhau

    (,t) mt xc sut

    nh l Louville:

    s h trong tp hp khng thay i theo thi gian

    tp hp chuyn ng theo thi gian trong khng gian pha nh mt cht lng c nn bng 0!

    d dt=0

    t

    =i=1

    N

    ri ri pi pi

  • 7Mt s kin thc v thng k

    Khi t v cng ln, ta c tp hp cn bng:

    khi , khng ph thuc thi gian!

    v ta c

    H ergodic: any point in phase space is accessible from any other point

    H non-ergodic: some region of phase space is not accessible from outside

    t

    =0

    Atime=Aens

  • 8Mt s kin thc v thng k

    Trng s & hm phn hoch:

    ty thuc vo cch ly trng s ta c cc tp hp khc nhau

    M phng Monte Carlo: cho php to ra mt tp hp cc trng thi theo mt xc xut cho trc, khi

    =Q1w

    Q=

    w

    A=Q1

    A w

    A=1

    Kk=1

    K

    Ak

  • 9Mt s kin thc v thng k

    Tp hp vi chnh tc N,V,E = constants

    Phng php ng lc hc phn t (MD): to ra tp vi chnh tc (E=constant), ng thi bo ton xung lng tng cng

    QNVE=

    H E

    QNVE=1

    N !

    1

    h3N dr dpH r , pE

    S=k B lnQNVE entropy

  • 10

    Mt s kin thc v thng k

    Tp hp chnh tc N,V,T = constants

    w ()=eH ()/k BT

    QNVT=eH ()/k BT

    F=k BT lnQNVT

    QNVT=1

    N !

    1

    h3N dpe

    K /k BTdr eV pr /k BT

    Nng lng t do Helmholtz

    Z NVT=

    QNVT=1

    N ! h2/2mk BT 3N /2 Z NVE

    dr eV pr /k BTTch phn cu hnh

  • 11

    Mt s kin thc v thng k

    Tp hp ng nhit ng p

    N,P,T=constants

    w =eH PV /k BT

    QNPT=V

    eHPV /k BT=

    V

    eP /k BT QNVT

    G=k BT lnQNPT

    Z NPT=dV ePV /k BTdr eV pr /k BTNng lng t do Gibbs

  • 12

    Mt s kin thc v thng k

    Tp hp chnh tc ln

    ,V,T=constants

    w =eH N /k BT

    QVT=N

    eH N /k BT=

    N

    e N /k BT QNVT

    PV=k BT lnQVT phng trnh trng thi

  • 13

    Mt s kin thc v thng k

    nh lut ng phn

    Mi bc t do ng vi kch thch nng lng kT

    S bc t do =

    Nc l s rng buc (constraint)

    pk H pk =k BT qkHqk =k BT

    3NN c

  • 14

    Mt s kin thc v thng k

    Nhit tc th Nhit o c bng thc nghim l nghit

    trung bnh theo thi gian

    Trong m phng c th tnh nhit t mt trng thi vi m ca h

    T nh lut ng phn ta c:

    Nhit tc th:

    K =i=1

    N pi2

    2mi =3N2 k BT

    T=2K

    3NkB

    =1

    3NkBi=1

    N pi2

    mi

  • 15

    Mt s kin thc v thng k

    Trong trng hp c Nc rng buc:

    Nhit trung bnh:

    T=2K

    3NN ck B=

    1

    3NN ck Bi=1

    N pi2

    mi

    T= T

  • 16

    Mt s kin thc v thng k

    p sut tc th T trng thi vi m ca h c th tnh c p sut

    tc th

    T nh lut ng phn ta c:

    suy ra:

    Lc tng cng bng ngoi lc + ni lc:

    qk pk =k BT pk= f ktot=

    qk

    V p

    1

    3 i=1N

    rif itot =N kBT

    f itot=f i

    extf iinternal

  • 17

    Mt s kin thc v thng k

    Ngoi lc cn bng vi p sut ln cc bc tng:

    Hm virial

    p sut tc th:

    1

    3 i=1N

    rif iext =PV

    W1

    3i=1

    N

    rif iinternal=

    1

    3i=1

    N

    riri V p

    PV=N kBTW

    P= k B TW

    V= Pideal gas Pex

    P= k BTW

    V= Pideal gas Pexhoc

  • 18

    Mt s kin thc v thng k

    Tng tc cp

    W=1

    3i

    i j

    rif ij=1

    3i

    i j

    ri rij v rij

    V p=i j

    v rij

    W=1

    3i

    i j

    w rij

    w r =rdv r dr

    hm virial cho tng tc cp

  • 19

    Mt s kin thc v thng k

    Nhit dung ring

    N,V,T=constants

    N,P,T=constants

    E=H

    E 2=H 2H 2

    Cv=H 2H 2

    k BT2

    C p=H 2H 2

    k BT2

  • M hnh Ising

    M hnh Ising l g? V sao n quan trng?

    M hnh Ising l m hnh ton hc c t theo tn ca nh Vt l Ernst Ising (Ngi c).

    M hnh Ising l m hnh dng m t hin tng chuyn pha st t m ch s dng cc spin-up v down

  • M hnh Ising Ising gii bi ton 1D nm 1924 trong lun

    vn Tin s ca mnh (thun tu Ton). Trng hp mng vung 2D c th gii chnh xc c bng gii tch (Onsager, 1944)

    n nay, bi ton v m hnh Ising c p dng trong rt nhiu lnh vc: vt l, sinh hc (lin quan n t) n cc vn x hi (m hnh n gin 2 la chn)

    M hnh Ising l m hnh chun th xem mt thut ton trong khun kh p dng ca m hnh c hiu qu khng

  • St tCc domain t sp xp thng hng theo mt hngThng thng, cc domain khngsp xp thng hng theo mt hng

    Tuy nhin, cc domain c th c p

    Ti nhit thp th cu hnh n nh l cu hnh vi tt c spinu hng ln hoc hng xung (2 cu hnh)

    Nhit Curie (nhit ti ton b tnh st t bin mt). Vi st l 1043 K

    im ti hn: l im xy ra s chuyn pha (loi II)

  • Gin pha

    Nhit thp Nhit cao

  • M hnh

    Universality Class l mt lp ca cc h Vt l c chung mt tnh cht ng m khng ph thuc vo cc tnh cht ng lc ca h. V d: h hp kim 2 cht, h 2 cht lng trn ln, hay h siu chy ca Helium trong 3 chiu u thuc vo mt lp

    M hnh Ising ch s dng cc vector UP v DOWN nhng li m t c rt nhiu pha khc nhau ca vt cht

    - Hp kim 2 cht - Trn 2 cht lng - Cht lng v kh trn ln

    - Siu chy ca Helium- Hin tng siu dn trong kim loi

  • M hnh IsingGii tch

    Ising 1924

    Onsager 1944

    Gii s, v d pp Monte Carlo

    Nhit cao

    2-D

    3-D

    Nhit thp

    1-D

  • 26

    M hnh Ising

    E=J ij

    si s jHi=1

    N

    si

    cc cp ln cn gn nht

    J - nng lng tng tc trao iJ > 0 - st t (ferromagnet)J < 0 - phn st t (anti-ferromagnet)

    s=1

    M=i=1

    N

    si

    cm t (susceptibility)

    m=1

    Ni=1

    N

    si

    C H=E2E 2

    k BT2

    T=M 2M 2

    k BT

    H spin trn mng Ernst Ising (1924)

    t ha (magnetization)

    i khi s dng B thay v H

  • Gii tch m hnh Ising 1 chiu

    The partition function is given by

    Z =+1

    s1=1

    +1s2=1

    ...+1

    sN=1eEI{Si} (3)

    One Dimensional Ising Model and Transfer MatricesLet us consider the one-dimensional Ising model where N spins are on a chain. We

    will impose periodic boundary conditions so the spins are on a ring. Each spin onlyinteracts with its neighbors on either side and with the external magnetic field B. Thenwe can write

    EI{Si} = JNi=1

    SiSi+1 BNi=1

    Si (4)

    The periodic boundary condition means that

    SN+1 = S1 (5)

    The partition function is

    Z =+1

    s1=1

    +1s2=1

    ...+1

    sN=1exp

    [

    Ni=1

    (JSiSi+1 +BSi)

    ](6)

    Kramers and Wannier (Phys. Rev. 60, 252 (1941)) showed that the partition functioncan be expressed in terms of matrices:

    Z =+1

    s1=1

    +1s2=1

    ...+1

    sN=1exp

    [

    Ni=1

    (JSiSi+1 +

    1

    2B (Si + Si+1)

    )](7)

    This is a product of 2 2 matrices. To see this, let the matrix P be defined such thatits matrix elements are given by

    S|P |S = exp{[JSS +

    1

    2B(S + S )

    ]}(8)

    where S and S may independently take on the values 1. Here is a list of all the matrixelements:

    +1|P |+ 1 = exp [(J +B)]1|P | 1 = exp [(J B)]+1|P | 1 = +1|P | 1 = exp[J ] (9)

    Thus an explicit representation for P is

    P =

    (e(J+B) eJ

    eJ e(JB)

    )(10)

    2

    The partition function is given by

    Z =+1

    s1=1

    +1s2=1

    ...+1

    sN=1eEI{Si} (3)

    One Dimensional Ising Model and Transfer MatricesLet us consider the one-dimensional Ising model where N spins are on a chain. We

    will impose periodic boundary conditions so the spins are on a ring. Each spin onlyinteracts with its neighbors on either side and with the external magnetic field B. Thenwe can write

    EI{Si} = JNi=1

    SiSi+1 BNi=1

    Si (4)

    The periodic boundary condition means that

    SN+1 = S1 (5)

    The partition function is

    Z =+1

    s1=1

    +1s2=1

    ...+1

    sN=1exp

    [

    Ni=1

    (JSiSi+1 +BSi)

    ](6)

    Kramers and Wannier (Phys. Rev. 60, 252 (1941)) showed that the partition functioncan be expressed in terms of matrices:

    Z =+1

    s1=1

    +1s2=1

    ...+1

    sN=1exp

    [

    Ni=1

    (JSiSi+1 +

    1

    2B (Si + Si+1)

    )](7)

    This is a product of 2 2 matrices. To see this, let the matrix P be defined such thatits matrix elements are given by

    S|P |S = exp{[JSS +

    1

    2B(S + S )

    ]}(8)

    where S and S may independently take on the values 1. Here is a list of all the matrixelements:

    +1|P |+ 1 = exp [(J +B)]1|P | 1 = exp [(J B)]+1|P | 1 = +1|P | 1 = exp[J ] (9)

    Thus an explicit representation for P is

    P =

    (e(J+