Assignment 6. Tyler Shendruk December 6, 2010

Transcription

1 Assignmen 6 Tyler Shendruk December 6, 1 1 Harden Problem 1 Le K be he coupling and h he exernal field in a 1D Ising model. From he lecures hese can be ransformed ino effecive coupling and fields K and h respecively. Unforunaely, he ransformaion is ieraive and akes he form x x (1 + y) = (x + y) (1 + xy) y y (x + y) = 1 + xy (1) () where x = e 4K and y = e h. We can show ha his ieraive scheme has wo fixed poins by asking ourselves wha are fixed poins? They are poins when he ieraive scheme has arrived a some value and won change no maer how many more imes we ierae i.e. when x = x and y = y we call hese value x, y. Le s sar wih y. A he fixed poin y y (x + y) = 1 + xy y = y (x + y ) 1 + xy y (1 + xy ) = y (x + y ) = y (x + y ) y (1 + xy ) = y (x + y 1 xy ) = y (1 y ) (1 x) y =, 1 (3) For x we are only asked abou one fixed poin x =. Bu ha s obvious when 1

2 we se x = x = x : x = x = x (1 + y) (x + y) (1 + xy) x (1 + y) (x + y) (1 + x y) (1 + y) = x (1 (x + y) (1 + x y) [ = x (x + y) (1 + x ) (1 + y) ] ) = x [ x 1 + y x y + x y y ] = x (x 1) [ 1 + y (x + 1) + y ] x =, 1 (4) We can show ha x = 1 is a fixed line by subsiuing i ino he ieraive scheme and seeing ha he lef over y erms cancel ou. Therefore, he lef over combinaions give he fixed poins (x, y ) = (, ), (, 1). (5) To see hese, we swich o a graphical represenaion. Fig. 1 is a plo of he difference beween x and x, and y and y. The oal daa is in he flow diagram of he vecors bu since I someimes find i hard o judge he magniude of hese arrows when here are so many of hem, I ploed he magniude as a color plo in he background (he color scale is normalized in each one, see he color legends o he righ of each figure). From Fig. 1a he x = 1 fixed line is clear as a dark blue sripe. (a) Renormalizaion flow diagram. (b) Renormalizaion flow diagram zoomed ino he fixed poins. Figure 1: Normalized difference beween x, x and y, y. Colour gives he magniude while arrows give he enire vecor. The fixed poins aren as obvious from Fig. 1a so we zoom ino small x o ge Fig. 1b which shows he fixed poins very clearly. Noice one hing abou

3 hese figures is ha y only echniqually exiss beween [, 1] by y = e h. The range in Fig. 1 was jus o help you visualize. The ruely accepable range is shown in Fig.. Figure : Normalized difference beween x, x and y, y over physically accepable range. Shendruk Problem.1 1-D.1.1 Problem.1.a The Ising model is concepually simple bu ha doesn mean ha i is always easy o ge good daa ou of i. For insance, consider a 1-D ferromagne of N = 5 spins wih no exernal field presen. From Fig. 3a, very rapidly he ho and cold sars agree when = 1. Bu when = 1 far longer is required. An equilibraion ime of, seems sufficien since he ho sar and he cold sar agree, even for such cold emperaures (he colder i is he slower i equilibraes). Bu here is clearly a problem: Neiher he energy flucuaions in Fig. 3a nor he energy iself in Fig. 3b have plaeaued. Obviously, his is no equilibraed properly. The rae of change wih ime of energy and energy flucuaions is no due o oo shor of an equilibrium ime. The sysem is jus oo small or he emperaure is oo large. Le s keep all oher parameers fixed and increase he number of spins o N = 1. This is shown in Fig. 4. Afer long imes (abou = 1 ime seps for = 1 and say 3 for = 1), he energy and 3

4 1 8 Cold sar: Ho sar: Cold sar: Ho sar: E 6 4 E (a) Energy flucuaions (b) Insananeous energy. Figure 3: Equilibraion for 1-D chain of N = 5 spins forming a ferromagne (J = 1) in no exernal field (B = ). i s flucuaions have reached a poin we migh reasonably say heir values are no rapidly evolving. The larger he sysem he more sable i is. Anoher (ofen easier) way o do his is o add a field. A small magneic field will help sabilize he sysem and in fac we will find ha i is essenial o geing good daa in -D bu i s a rade off since i will change some of he analyic resuls. 4

5 .1. Problem.1.b Bu we have a new problem! The values of E and E for he ho and cold sar prey much agree in Fig. 4 bu he sysems for which = 1 do no have agreeing ho and cold sars. Therefore, we mus conclude ha one of (or boh) haven reached heir equilibrium. = 5 ime seps is no huge for one or wo simulaions bu once we sar ploing hermodynamic variables as a funcion of emperaure we are going o need many, many simulaions o ge accepable resoluion. We mus comprimise. 1. 1e8.8.6 Cold sar: Cold sar: E Ho sar: E 4 Ho sar: (a) Energy flucuaions (b) Insananeous energy. Figure 4: Equilibraion for 1-D chain of N = 1 spins forming a ferromagne (J = 1) in no exernal field (B = ). Fig. 5 shows he ime evoluion of a 1-D ferromagne Ising chain of N = 5 spins wih no exernal field applied. The energy and i s flucuaions do no descend or climb as hey did for he smaller sysem shown in Fig. 3 and he differences beween ho and cold sars are no as pronounced as in Fig. 4. The E agreemen beween ho and cold sars for low emperaures is sill no grea bu his ells us o expec greaer uncerainy of our hermodynamic funcions a lower emperaures..5 1e7 Cold sar: =1 =1 Ho sar: =1 =1 1 Cold sar: =1 E 1. E 3 =1 Ho sar: =1 = (a) Energy flucuaions (b) Insananeous energy. Figure 5: Equilibraion for 1-D chain of N = 5 spins forming a ferromagne (J = 1) in no exernal field (B = ). From our discussion and Fig. 5, we asser ha N = 5 wih an equilibraion ime of eq = 3 is reasonable. 5

6 .1.3 Problem.1.c I look a N = 5 a =.75 wih a iny B =.1 afer equilibraion in Fig. 6. The ferromagne in Fig. 6a has clear domains of correlaed spin up or down regions while Fig. 6b is almos grey since he black and he whie are so ani-correlaed. (a) Ferromagneic sysem of J = 1 (b) Ferromagneic sysem of J = 1 Figure 6: Visual comparision beween ferromagneic (lef) and aniferromagneic (righ) sysems. Time is represened by he y-axis. The emperaure is low enough ha few changes happen over he viewing ime. 6

7 .1.4 Problem.1.d We plo he energy, he magneizaion, he hea capaciy, and he suscepibiliy (which is found very similarily o he hea capaciy and I should have asked you o plo). We do his for a ferromagne wih an exchange energy J, wih periodic boundary condiions, in a weak exernal field B =.5. We iniialize wih a ho sar and equilibrae for eq = 3 ime seps. We do a hundred simulaions from = o 1. The expecaion values I ll compare agains are E J = N anh ( J k B T C = (J/k BT ) cosh (J/k B T ) ) M = Ne J/kBT sinh (B/k B T ). e J/k BT sinh (B/k B T ) + e J/k BT To sar wih we use N = 5 spins as suggesed in one of he previous secions bu o improve he agreemen we redo all simulaions for wice as many spins (N = 1). To compare beween he wo we can jus scale by N E/N.6.8 E/N (a) Energy as a funcion of k B T.. (b) Energy flucuaions as a funcion of k B T. I is fairly eviden ha increasing N beyond N = 5 does no significanly increase he accuaracy. In Fig. 7a we sor of capure some of he levelling off a small emperaures bu oherwise he daa don differ. Furhermore, he poor agreemen of specific hea C a low emperaures indicaes ha we do need some improvemen. Le s increase he field o B =.5 which is prey large by no huge. We will ge worse agreemen wih heory; however, he daa shouldn be as noisy. According o Fig. 9 he noise is much beer (Look a he C/N scale) bu sill no grea. 3 Shendruk Problem D For -D we use a sysem of N = 5 spins in a small magneic field of B =.5 compared o he exchange energy J = 1. Speaking of which we know ha all 7

8 M/N.6.4 χ/n (a) Magneizaion.. (b) Suscepibiliy is he variance of he average of he magneizaion. Figure 7: Magneizaion and suscepibiliy as a funcion of k B T C/N C/N 1 1 Figure 8: Specific hea as a funcion of k B T. he ineresing suff happens below = 5 so we simulae over a smaller range of han we did in 1-D. We expec here o be a phase ransiion a a criical emperaure of k B T c J = ln ( 1 + ).7. (6) The criical emperaure T c comes ino he heoreical expecaion value of he magneizaion and he specific hea M = ( 1 C = 8 ( ln πt c [ ( ( sinh ln 1 + ) Tc T T c e (1+π/4) 1 T/T c )] 4 ) 1/8 I ve included Fig. 1, a picure of he domains near he criical emperaure T c. Near he criical emperaure here is no characerisic size o he domains i.e. iny domains and huge domains are equally likely. In a manner of speaking i s fracal in naure. ). 8

9 B =.5.4 B = C/N C/N kb T/J kb T/J Figure 9: Specific hea as a funcion of kb T bu now wih an applied field of B =.5. Figure 1: Domains for a -D Ising model near he ransiion emperaure. Noice he fracal naure. 9