Prof. Dr. I. Nasser Phys 630, T Aug-15 One_dimensional_Ising_Model
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1 EXACT OE-DIMESIOAL ISIG MODEL The one-dmensonal Isng model conssts of a chan of spns, each spn nteractng only wth ts two nearest neghbors. The smple Isng problem n one dmenson can be solved drectly n several ways. Frst the chan s consdered as open ended and the Hamltonan n the form: The partton functon s gven by H J J 0 K e (7.5.) K where K J. The exponental can be factored as a product of terms of the form e, each of whch can be wrtten as (ote that: can only be + or -, cosh( x ) x and snh( x ) x ) K e cosh K snh K cosh K snh K * (7.5.) ( y)cosh( K) where y tanh( K). Here we have used: c c ' e ' * e cosh c 'snh c c e ' whch holds because can only + or -. The partton functon (7.5.) then becomes and the ndcated summaton yelds (...). ext, summaton over gves another factor and a product of 3 terms. Contnued summatons fnally produce the result cosh K (7.5.4) Average energy and the specfc heat: In the thermodynamc lmt, the free energy per spn s gven by:
2 f k BT lm ln k BT ln cosh K, ln U E lm J tanh( K) (7.5.5) and the heat capacty C s E cosh C kbk K (7.5.6) T The energy and heat capacty are smoothly varyng, always fnte functons of temperature, exhbtng no phase transton. Thus the molecular-mean-feld- approxmaton s ncorrect, no matter how plausble, for a one-dmensonal system, and ts valdty n n-dmensons then s mmedately to be doubted. Another drect technque for the open, one-dmensonal chan makes use of a change n varables f,, where t may be seem from Eq. (7.5.) f or (7.5.3) that s ndependent of. For the open chan, the 's are ndependent, and can be wrtten as K K K e e e cosh K where each sum on a cosh K, and the fnal result s the product of such ndependent factors, n agreement wth Eq. (7.5.4). For closed chans n one dmenson and for Isng lattces of hgher dmenson, no such smple technque wll work because the 's are no longer ndependent. yelds a factor Second When the chan s closed, wth S S,: For the open ended chan, the Hamltonan has the form: The partton functon s gven by H J S S J 0 K SS e S When the chan s closed, wth S S, drect evaluaton of becomes slghtly more dffcult, but other procedures are often smpler for the closed chan than for the open one. For the closed chan, becomes:
3 cosh K ys S S where K J and y tanh( K).We work out the product and sort terms n powers of y : cosh K y SS S S3 S S S 3 y S S S S y S S S S S S The terms, lnear n y contan products of two dfferent (neghborng) spns, lke SS. The SS 0, because there are sum over all spn confguratons of ths product vansh, S or S two confguratons wth parallel spns ( SS ) and two wth antparallel spns ( SS ). Thus, the term lnear n y vanshes after summaton over all spn confguratons. For the same reason also the sum over all spn confguratons, whch appear at the term proportonal to y, vansh. In order for a term to be dfferent from zero, all the spns n the product must appear twce (then, S ). Ths condton s fulflled only n the last term, S whch after summaton over all spn confguratons gves y. Therefore the partton functon of the Isng model of a lnear chan of spns s: cosh K y cosh K for open chan. In the lmt of very a result that dffers from that of large, however, the fnte J and. ote that: -, S - ss () s 0, s SS y contrbuton becomes vanshngly small, snce y tanh K 0 - v- S or S S 4640 for all SS SjSj v- s s = even, 0 = odd s 3
4 Isng Model and Transfer Matrx Exact solutons of the Isng model are possble n and dmensons and can be used to calculate the exact crtcal exponents for the two correspondng unversalty classes. In one dmenson, the Isng Hamltonan becomes: H J s s h s, J, h whch corresponds to spns on a lne. We wll mpose perodc boundary condtons on the spns so that s s 0, s or. Thus, the topology of the spn space s that of a crcle, see the fgure. Wth the defntonsk Where s s s e J and H h o o o, the partton functon s then: K ( s s s s s s ) H ( s s s s ) s s s s s s H ( ) K ( s s ) h( ) H ( ) K ( s s ) h( ) H ( ) K ( s s ) H ( ) P P e e e P 0,,,0 P, e s s H ( ) K ( s s ) H ( ) In order to carry out the spn sum, let us defne a matrx P wth matrx elements: sps' e s s' H ( ) K ( ss') H ( ) K H K H P e, P e, P P e The matrx P s called the transfer matrx. Thus, the matrx P s a x matrx gven by K, 4
5 s s K H K s e e P K K H s e e From the matrx rules, the larger and smaller egenvalues are calculated as: K H K H K Tr( P) e e e cosh( H) : K H K H K K K K det( P) e e e e e e snh( K) Solvng for, one fnds: K 4K e cosh( H ) snh ( H ) e The trace of P s gven by: =Tr( P ) / Free energy: In the thermodynamc lmt, the free energy per spn s gven by: f k BT lm ln k BT ln As K f k T lm ln k T ln e cosh( H) snh( H) J h B whch s the energy per spn as expected. B The magnetzaton: becomes snh( H)cosh( H) snh( H ) 4K f ln snh ( H) e m h ( H) 4K cosh( H) snh ( H) e Whch s regular as H 0, snce cosh( H ) and snh( H ) 0, tself vanshes. Thus, there s no magnetzaton at any fnte temperature n one dmenson, hence no nontrval crtcal pont. Example: It was that the exact egenvalues of the perodc Isng model s gven by: K 4K e cosh( H ) snh ( H ) e, / kt, K J and H h For H 0,smplfy the expresson: K 4K K 4K K K e cosh( H ) snh ( H ) e e e e e, / kt, cosh( K), snh( K), Consequently 5
6 cosh( K) snh( K) / coshk y coshk Where y tanh( K ) Then fnd the followng: ; - F ln( ) n the extreme lmts,.e. T 0 and T. Ans: F ln( ) kt ln cosh K - E ln( ) Ans: E ln( ) J tanh K E - C n the extreme lmts,.e. T 0 and T, and T fnd T at maxmum C. Ans: C E K K T cosh( K) ote: J / T J / T e e cosh K ; F ln ln cosh K ln e e ln ln J / T J / T As T 0 J J / T F e T The frst term, even though t looks lke t blows up at T 0, s actually regular. It smply says that the ground state energy s J per spn. It could be removed by a constant shft of energy, for example. The second term s sngular. So the sngular part of the free energy behaves as: J / T Fsngular e The correlaton length s 6
7 J / T e ln ln tanh K The last approxmate equalty works at T 0. Ther product, n the lmt T 0 s thus Fsngular Whch s a unversal number (does not depend on parameters) H.W. Wth the egenvector of the Isng matrx n -dmensons, calculate the magnetzaton per spn, the correlaton functon, and the correlaton length, and check f they behave n a sensble way.(go to the dscusson n sectons 3.3 and 3.4 of Goldenfeld) H.W. Wrte down the transfer matrx for the one-dmensonal spn- Isng model n zero feld whch s descrbed by the Hamltonan H J s s, J 0, s,0 0 Hence calculate the free energy per spn of ths model and show that t has the expected behavor n the lmts 0 T and T. [Answer:.] Whle the one-dmensonal Isng model s a relatvely smple problem to solve, the twodmensonal Isng model s hghly nontrval. It was only the pure mathematcal genus of Lars Onsager that was able to fnd an analytcal soluton to the two-dmensonal Isng model. Ths, then, gves an exact set of crtcal exponents for the d and n unversalty class. To date, the three-dmensonal Isng model remans unsolved. Here, the Onsager results wll be stated as: In the thermodynamc lmt, the fnal result at zero feld s: where The energy per spn s where The magnetzaton, then, becomes 7
8 for T Tc and 0 fort Tc, ndcatng the presence of an order-dsorder phase transton at zero feld. The condton for determnng the crtcal temperature at whch ths phase transton occurs turns out to be eart T c, the heat capacty per spn s gven by Thus, the heat capacty can be seen to dverge logarthmcally ast The crtcal exponents computed from the Onsager soluton are T. c whch are a set of exact exponents for the d and n unversalty class. 8
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