8.334: Statistical Mechanics II Spring 2014 Test 2 Review Problems & Solutions

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1 8334: Statstcal Mechancs II Sprng 014 Test Revew Problems & Solutons The test s closed book, but f you wsh you may brng a one-sded sheet of formulas The ntent of ths sheet s as a remnder of mportant formulas and defntons, and not as a compact transcrpton of the answers provded here If ths prvlege s abused, t wll be revoked for future tests The test wll be composed entrely from a subset of the followng problems, as well as those n problem sets 3 and 4 Thus f you are famlar and comfortable wth these problems, there wll be no surprses! ******** 1 Scalng n fluds: Near the lqud gas crtcal pont, the free energy s assumed to take the scalng form F/N = t α g(δρ/t β ), where t = T T c /T c s the reduced temperature, and δρ = ρ ρ c measures devatons from the crtcal pont densty The leadng sngular behavor of any thermodynamc parameter Q(t, δρ) s of the form t x on approachng the crtcal pont along the sochore ρ = ρ c ; or δρ y for a path along the sotherm T = T c Fnd the exponents x and y for the followng quanttes: Any homogeneous thermodynamc quantty Q(t, δρ) can be wrtten n the scalng form δρ t x Q Q(t, δρ) = g Q Thus, the leadng sngular behavor of Q s of the form t x Q f δρ = 0, e along the crtcal sochore In order for any Q to be ndependent of t along the crtcal sotherm as t 0, the scalng functon for a large enough argument should be of the form t β so that x Q /β lm g Q (x) = x, x x Q Q(0, δρ) (δρ) y Q, wth y Q = β (a) The nternal energy per partcle (H)/N, and the entropy per partcle s = S/N Let us assume that the free energy per partcle s f = F N = t α g δρ t β, T 1 T c t and that T < T c, so that = The entropy s then gven by f 1 f δρ s = = = g S, T V T c t ρ T t β c 1 t 1 α

2 so that x S = 1 α, and y S = (1 α)/β For the nternal energy, we have ( ) (H) (H) 1 α δρ f = T s, or T c s(1+t) t g H, N N t β therefore, x H = 1 α and y H = (1 α)/β (b) The heat capactes C V = T s/ T V, and C P = T s/ T P The heat capacty at constant volume S s t α ( ) δρ C V = T = = g CV, T t t β V ρ T c so that x CV = α and y CV = α/β To calculate the heat capacty at constant pressure, we need to determne frst the relaton δρ(t) at constant P For that purpose we wll use the thermodynamc dentty P δρ t ρ = t P P The pressure P s determned as ( ) F ρ f α β δρ P = = ρ c t g P, V δρ t β δρ t whch for δρ t β goes lke ( P t α β 1+A δρ ) t β, and consequently P t 1 α β t ρ P t α β δρ t In the other extreme of δρ t β, ( ) P δρ ( α β)/β t 1+B, and δρ 1/β P δρ (1 α β)/β t ρ P δρ ( α β)/β δρ t, where we have agan requred that P does not depend on δρ when δρ 0, and on t f t 0 From the prevous results, we can now determne β 1 β δρ t = δρ t t P δρ (β 1)/β = t δρ 1/β

3 From any of these relatonshps follows that δρ t β, and consequently the entropy s s t 1 α The heat capacty at constant pressure s then gven by α α C P t, wth x CP = α and y CP = β (c) The sothermal compressblty κ T = ρ/ P T /ρ, and the thermal expanson coeffcent α = V/ T P /V Check that your results for parts (b) and (c) are consstent wth the thermodynamc dentty C P C V = TVα /κ T The sothermal compressblty and the thermal expanson coeffcent can be computed usng some of the relatons obtaned prevously 1 ( ) 1 ρ 1 P 1 α+β δρ κ T = = = t g κ, ρ P ρ ρ ρ 3 t β T c T c wth x κ = α +β, and y κ = (α +β )/β And 1 V 1 ρ α = = t β 1, V T P ρt c t P wth x α = β 1, and y α = (β 1)/β So clearly, these results are consstent wth the thermodynamc dentty, (C P C V )(t, 0) t α, or (C P C V )(0, δρ) δρ α/β, and α κ T (t, 0) t α, or α κ T (0, δρ) δρ α/β (d) Sketch the behavor of the latent heat per partcle L, on the coexstence curve for T < T c, and fnd ts sngularty as a functon of t The latent heat L = T (s + s ) s defned at the coexstence lne, and as we have seen before ( ) δρ ± Ts ± = t 1 α g s t β The densty dfference between the two coexstng phases s the order parameter, and vanshes as t β, as do each of the two devatons δρ + = ρ c 1/v + and δρ = ρ c 1/v 3

4 L t=0 t of the gas and lqud denstes from the crtcal crtcal value (More precsely, as seen n (b), δρ P=constant t β ) The argument of g n the above expresson s thus evaluated at a fnte value, and snce the latent heat goes to zero on approachng the crtcal pont, we get L t 1 α, wth x L = 1 α ******** The Isng model: The dfferental recurson relatons for temperature T, and magnetc feld h, of the Isng model n d = 1+ǫ dmensons are dt T = ǫ T + dl dh =dh dl, (a) Sketch the renormalzaton group flows n the (T,h) plane (for ǫ > 0), markng the fxed ponts along the h = 0 axs The fxed ponts of the flow occur along the h = 0 axs, whch s mapped to tself under RG On ths axs, there are three fxed ponts: () T = 0, s the stable snk for the low temperature phase () T, s the stable snk for the hgh temperature phase () There s a crtcal fxed pont at (T = ǫ, h = 0), whch s unstable All fxed ponts are unstable n the feld drecton (For h 0, the lnearzed recurson relatons lead to vertcal flows for T = T Hgher order terms wll cause the flows to bend as ndcated) (b) Calculate the egenvalues y t and y h, at the crtcal fxed pont, to order of ǫ Lnearzng T = T + δt, around the crtcal fxed pont yelds dδt = ǫ δt + T { δt = ǫδt dl y t =+ ǫ, = dh =(1+ ǫ)h y h =1+ ǫ dl 4

5 (c) Startng from the relaton governng the change of the correlaton length ξ under ( ) renormalzaton, show that ξ(t, h) = t ν g ξ h/ t Δ (where t = T/T c 1), and fnd the exponents ν and Δ Under rescalng by a factor of b, the correlaton length s reduced by b, resultng n the homogenety relaton ξ(t, h) = bξ(b y t t, b y h h) Upon selectng a rescalng factor such that b y t t 1, we obtan ( ν ξ(t, h) = t g ξ h/ t Δ), wth 1 1 y h 1 ν = =, and Δ = = +1 y t ǫ y t ǫ (d) Use a hyperscalng relaton to fnd the sngular part of the free energy f sng (t, h), and hence the heat capacty exponent α Accordng to hyperscalng ( d/y f t sng (t, h) ξ(t, h) d = t g f h/ t Δ) Takng two dervatves wth respect to t leads to the heat capacty, whose sngularty for h = 0 s descrbed by the exponent 1+ǫ 1 α = dν = = +1 ǫ ǫ 5

6 (e) Fnd the exponents β and γ for the sngular behavors of the magnetzaton and susceptblty, respectvely The magnetzaton s obtaned from the free energy by f d y h m = t β, wth β = = 0 h h=0 (There wll be correctons to β at hgher orders n ǫ) The susceptblty s obtaned from a dervatve of the magnetzaton, or f y h d 1+ǫ 1 χ = t γ, wth γ = = = +1 h y t ǫ ǫ h=0 y t (f) Startng the relaton between susceptblty and correlatons of local magnetzatons, calculate the exponent η for the crtcal correlatons ((m(0)m(x)) x (d +η) ) The magnetc susceptblty s related to the connected correlaton functon va χ = d d x (m(0)m(x)) c Close to crtcalty, the correlatonsdecay as a power law (m(0)m(x)) x (d +η), whch s cut off at the correlaton length ξ, resultng n χ ξ ( η) t ( η)ν From the correspondng exponent dentty, we fnd γ = ( η)ν, = η = y t γ = y h + d = d = 1 ǫ (g) How does the correlaton length dverge as T 0 (along h = 0) for d = 1? For d = 1, the recurson relaton for temperature can be rearranged and ntegrated, e ( ) 1 dt 1 =, = d = dl T dl T We can ntegrate the above expresson from a low temperature wth correlaton length ξ(t ) to a hgh temperature where 1/T 0, and at whch the correlaton length s of the order of the lattce spacng, to get ( ) ( ) ξ = ln = ξ(t ) = a exp T a T 6

7 ******** 3 Longtudnal susceptblty: Whle there s no reason for the longtudnal susceptblty to dverge at the mean-feld level, t n fact does so due to fluctuatons n dmensons d < 4 Ths problem s ntended to show you the orgn of ths dvergence n perturbaton theory There are actually a number of subtletes n ths calculaton whch you are nstructed to gnore at varous steps You may want to thnk about why they are justfed Consder the Landau Gnzburg Hamltonan: K t βh = d d x ( m ) + m + u(m ), descrbng an n component magnetzaton vector m (x), n the ordered phase for t < 0 ( (a) Let m (x) = m+φl (x) ) ê l +φm t (x)ê t, and expand βh keepng all terms n the expanson Wth m (x) = (m + φ l (x))ê l + φm t (x)ê t, and m the mnmum of βh, ( ) { t β + um + d d K ( ) ( ) 4 t H =V m x ( φ ) + φm + +6um l t φ l ( ) } t ( ) ( ) + +um φm m m m t +4um φ 3 l + φ l φt + u φ 4 l +φ l φ t + φt Snce m = t/4u n the ordered phase (t < 0), ths expresson can be smplfed, upon droppng the constant term, as { K ( 3 βh = d d x ( φ m l ) + φ m ) ) t tφ 4 l + um (φ l + φ l φ t ( +u φ 4 m l +φ m t + l φ ) } φt (b) Regard the quadratc terms n φ l and φm t as an unperturbed Hamltonan βh 0, and the lowest order term couplng φ l and φm t as a perturbaton U; e U = 4um d d xφ l (x)φm t (x) Wrte U n Fourer space n terms of φ l (q) and φm t (q) We shall focus on the cubc term as a perturbaton = 4um d d m U xφ l (x)φ t (x), 7

8 whch can be wrtten n Fourer space as d d q d d q U = 4um φ l ( q q )φ m t (q) φ m t (q ) d d (π) (π) (c) Calculate the Gaussan (bare) expectaton values (φ l (q)φ l (q )) 0 and (φ t,α (q)φ t,β (q )) 0, and the correspondng momentum dependent susceptbltes χ l (q) 0 and χ t (q) 0 From the quadratc part of the Hamltonan, { ( ) } βh 0 = d d x 1 [ ] K ( φ l ) + φ m t tφ l, we read off the expectaton values d (π) δ d (q + q ) (φ l (q)φ l (q )) 0 = Kq t d δ d (π) (q + q )δ αβ (φ t,α (q)φ t,β (q )) = 0 Kq, and the correspondng susceptbltes 1 χ l (q) 0 = Kq t 1 χ t (q) 0 = Kq m m (d) Calculate (φ m t(q 1 ) φ t (q ) φ m t(q 1 ) φ t (q )) 0 usng Wck s theorem (Don t forget that mφ t s an (n 1) component vector) Usng Wck s theorem, ( ) m m m m φ t (q 1 ) φ t (q )φ t (q 1 ) φ t (q ) (φ t,α (q 1 )φ t,α (q )φ t,β (q 1 )φ t,β (q )) = (φ t,α (q 1 )φ t,α (q )) (φ t,β (q 1 )φ t,β (q )) + (φ t,α (q 1 )φ t,β (q 1 )) (φ t,α (q )φ t,β (q )) + (φ t,α (q 1 )φ t,β (q )) (φ t,α (q )φ t,β (q 1 )) 0 0 Then, from part (c), { ( ) d m m m m (π) δ d (q 1 + q )δ d (q 1 + q ) φ t (q 1 ) φ t (q )φ t (q 1 ) φ t (q ) = (n 1) 0 K q 1 q 1 } δ d (q 1 + q 1 )δ d (q + q ) δ d (q 1 + q )δ d (q 1 + q ) +(n 1) +(n 1), q1 q q1 q 8

9 snce δ αα δ ββ = (n 1), and δ αβ δ αβ = (n 1) (e) Wrte down the expresson for (φ l (q)φ l (q )) to second-order n the perturbaton U Note that snce U s odd n φ l, only two terms at the second order are non zero Includng the perturbaton U n the calculaton of the correlaton functon, we have ( ) U φ l (q)φ l (q )e φ l (q)φ l (q ) 1 U + U /+ 0 0 (φ l (q)φ l (q )) = = (e U ) 0 ((1 U + U /+ )) 0 Snce U s odd n φ l, (U) 0 = (φ l (q)φ l (q )U) 0 = 0 Thus, after expandng the denomnator to second order, we obtan 1 U ( U 3 = 1 + O ), 1+(U /) (φ l (q)φ l (q )) = (φ l (q)φ l (q )) ( φl (q)φ l (q )U (φ l (q)φ l (q )) 0 U ) 0 0 (f) Usng the form of U n Fourer space, wrte the correcton term as a product of two 4 pont expectaton values smlar to those of part (d) Note that only connected terms for the longtudnal 4 pont functon should be ncluded Substtutng for U ts expresson n terms of Fourer transforms from part (b), the fluctuaton correcton to the correlaton functon reads G F (q, q ) (φ l (q)φ l (q )) (φ l (q)φ l (q )) 0 1 d d d d d d d d ( q 1 q q 1 q = (4um ) φ l (q)φ l (q )φ l ( q m m 1 q )φ t (q 1 ) φ t (q ) d d d d (π) (π) (π) (π) ) m m 1 U φ l ( q 1 q )φ t (q 1 ) φ t (q ) (φ l (q)φ l (q )) 0, 0 e G F (q, q ) s calculated as the connected part of 1 d d q 1 d d q d d q d d q (4um ) 1 (φ l (q)φ l (q d d d d )φ l ( q 1 q )φ l ( q 1 q )) 0 (π) (π) (π) (π) ( ) m m m m φ t (q 1 ) φ t (q )φ t (q 1 ) φ t (q ), where we have used the fact that the unperturbed averages of products of longtudnal and 9 0 0

10 transverse felds factorze Hence 1 d d q 1 d d q d d q ) 1 d d q ( ) m m m m G F (q, q = (4um ) φ t (q 1 ) φ t (q )φ t (q 1 ) φ t (q ) d d d d (π) (π) (π) (π) 0 { (φ l (q)φ l ( q 1 q )) 0 (φ l (q )φ l ( q 1 q )) 0 } + (φ l (q)φ l ( q 1 q )) 0 (φ l (q )φ l ( q 1 q )) 0 1 d d d d d d d d ( ) q 1 q q 1 q = (4um ) φm m m m t (q 1 ) φ t (q )φ t (q 1 ) φ t (q ) d d d d (π) (π) (π) (π) 0 (φ l (q)φ l ( q 1 q )) 0 (φ l (q )φ l ( q 1 q )) 0 Usng the results of parts (c) and (d) for the two and four ponts correlaton functons, and snce u m = ut/4, we obtan { d d d d d d d d d q 1 q q1 q (π) δ d (q 1 + q )δ d (q 1 + q ) G F (q, q ) = 4u ( t) (n 1) d d d d (π) (π) (π) (π) K q1 q 1 } δ d (q 1 + q 1 )δ d (q + q )+δ d (q 1 + q )δ d (q 1 + q ) +(n 1) q 1 q d d (π) δ d (q q 1 q )(π) δ d (q q 1 q ), Kq t Kq t whch, after dong some of the ntegrals, reduces to { 4u ( t) δ d (q)δ d (q ( ) d d ) q 1 G F (q, q ) = (n 1) K 4t q 1 δ d (q + q } ) d d q 1 +(n 1) (Kq t) q (q + q 1 ) 1 (g) Ignore the dsconnected term obtaned n (d) (e the part proportonal to (n 1) ), and wrte down the expresson for χ l (q) n second order perturbaton theory From the dependence of the frst term (proportonal to δ d (q)δ d (q )), we deduce that ths term s actually a correcton to the unperturbed value of the magnetzaton, e [ ( (n 1)u d d )] q 1 m m 1, Kt q 1 and does not contrbute to the correlaton functon at non-zero separaton The spatally varyng part of the connected correlaton functon s thus d (π) δ d (q + q ) 8u ( t) δ d (q + q ) d d q 1 (φ l (q)φ l (q )) = + (n 1) Kq, t K (Kq t) q (q + q 1 ) 10 1

11 leadng to 1 8u ( t) (n 1) d d q 1 1 χ l (q) = + Kq t K d (Kq t) (π) q (q + q 1 ) 1 (h) Show that for d < 4, the correcton term dverges as q d 4 for q 0, mplyng an nfnte longtudnal susceptblty In d > 4, the above ntegral converges and s domnated by the large q cutoff Λ In d < 4, on the other hand, the ntegral clearly dverges as q 0, and s thus domnated by small q 1 values Changng the varable of ntegraton to q 1 = q 1 /q, the fluctuaton correcton to the susceptblty reads Λ/q d d q 1 d d q 1 ( 0 ) d 4 1 d 4 1 χ l (q) F q = q + O q, d d 0 (π) q (qˆ+ q ) 0 (π) q (qˆ+ q ) whch dverges as q d 4 for q 0 NOTE: For a translatonally nvarant system, (φ (x)φ (x )) = ϕ (x x ), whch mples q x+q x (φ(q)φ(q )) = d d xd d x e (φ(x)φ(x )) q (x x )+(q+q ) x = d d (x x )d d x e ϕ (x x ) d = (π) δ d (q + q )ψ (q) Consder the Hamltonan d d q βh = βh+ d d xh (x)φ(x) = βh+ h (q)φ( q), d (π) where βh s a translatonally nvarant functonal of φ (a one-component feld for smplcty), ndependent of h (x) We have d d q m (x = 0) = (φ(0)) = (φ(q)), d (π) and, takng a dervatve, m d d q = (φ(q )φ(q)) h (q) (π) d 11

12 At h = 0, the system s translatonally nvarant, and m h (q) = ψ (q) h=0 Also, for a unform external magnetc feld, the system s translatonally nvarant, and βh = βh+ h d d xφ (x) = βh+ hφ (q = 0), yeldng m d d q χ = = φ (q )φ(q = 0) = ψ (0) h d (π) ( ) ******** 4 Crystal ansotropy: Consder a ferromagnet wth a tetragonal crystal structure Couplng of the spns to the underlyng lattce may destroy ther full rotatonal symmetry The resultng ansotropes can be descrbed by modfyng the Landau Gnzburg Hamltonan to [ ] K βh = d d x ( m ) t ( ) + m r + u m + m1 + vm1 m, n w here m (m 1,,m n ), and m L = =1 m (d = n = 3for magnetsnthreedmensons) Here u > 0, and to smplfy calculatons we shall set v = 0 throughout (a) For a fxed magntude m ; what drectons n the n component magnetzaton space are selected for r > 0, and for r < 0? r > 0 dscourages orderng along drecton 1, and leads to order along the remanng (n 1) drectons r < 0 encourages orderng along drecton 1 (b) Usng the saddle pont approxmaton, calculate the free energes (lnz) for phases unformly magnetzed parallel and perpendcular to drecton 1 In the saddle pont approxmaton for m (x) = mê 1, we have [ ] t + r 4 lnzsp = V mn m + um, m J where V = d d x, s the system volume The mnmum s obtaned for { 3 0 for t + r > 0 (t + r)m +4um = 0, = m = J (t + r)/4u for t + r < 0 For t + r < 0, the free energy s gven by lnzs p ( t + r) f sp = = V 16u 1

13 When the magnetzaton s perpendcular to drecton 1, e for m(x) = mˆ e for = 1, the correspondng expressons are [ ] { t J for t > 0 lnz sp = V mn m + um, tm +4um = 0, m =, t/4u for t < 0 and the free energy for t < 0 s m t f sp = 16u (c) Sketch the phase dagram n the (t, r) plane, and ndcate the phases (type of order), and the nature of the phase transtons (contnuous or dscontnuous) The saddle pont phase dagram s sketched n the fgure (d) Are there Goldstone modes n the ordered phases? There are no Goldstone modes n the phase wth magnetzaton algned along drecton 1, as the broken symmetry n ths case s dscrete However, there are (n ) Goldstone modes n the phase where magnetzaton s perpendcular to drecton 1 (e) For u = 0, and postve t and r, calculate the unperturbed averages (m 1 (q)m 1 (q )) 0 and (m (q)m (q )) 0, where m (q) ndcates the Fourer transform of m (x) The Gaussan part of the Hamltonan can be decomposed nto Fourer modes as [ ] n d d q K t + r t βh 0 = q m(q) + m 1 (q) + m (q) (π) d = 13

14 From ths form we can easly read off the covarances (π) d δ d (q + q (m 1 (q)m 1 (q ) )) 0 = t + r + Kq (π) d δ d (q + q ) (m (q)m (q )) 0 = t + Kq J (f) Wrte the fourth order term U u d d x(m ), n terms of the Fourer modes m (q) J d d q Substtutng m (x) = exp(q x)m (q) n the quartc term, and ntegratng over (π) d x yelds n ( ) d d q 1 d d q d d q 3 U = u d d x m = u m (q 1 )m (q )m j (q 3 )m j ( q 1 q q 3 ) (π) 3d,j=1 (g) Treatng U as a perturbaton, calculate the frst order correcton to (m 1 (q)m 1 (q )) (You can leave your answers n the form of some ntegrals) In frst order perturbaton theory (O) = (O) 0 ((OU) 0 (O) 0 (U) 0 ), and hence n d d q 1 d d q d d q 3 (m 1 (q)m 1 (q )) = (m 1 (q)m 1 (q )) 0 u (π) 3d,j=1 ( c ) (m 1 (q)m 1 (q )m (q 1 )m (q )m j (q 3 )m j ( q 1 q q 3 )) 0 } (π) d δ d (q + q { ) u d d [ ]} k 4(n 1) 4 8 = t + r + Kq t + r + Kq (π) d t + Kk t + r + Kk t + r + Kk The last result s obtaned by lstng all possble contractons, and keepng track of how many nvolve m 1 versus m 1 The fnal result can be smplfed to } (π) d δ d (q + q { ) 4u d d [ ]} k n 1 3 (m 1 (q)m 1 (q )) = 1 + t + r + Kq t + r + Kq (π) d t + Kk t + r + Kk (h) Treatng U as a perturbaton, calculate the frst order correcton to (m (q)m (q )) Smlar analyss yelds n d d q 1 d d q d d q 3 (m (q)m (q )) = (m (q)m (q )) 0 u (π) 3d,j=1 ( c ) (m (q)m (q )m (q 1 )m (q )m j (q 3 )m j ( q 1 q q 3 )) 0 (π) d δ d (q + q { ) u d d [ ]} k 4(n 1) 4 8 = t + Kq t + Kq (π) d t + Kk t + r + Kk t + Kk (π) d δ d (q + q { ) 4u d d [ ]} k n +1 1 = 1 + t + Kq t + Kq (π) d t + Kk t + r + Kk 14

15 () Usng the above answer, dentfy the nverse susceptblty χ 1, and then fnd the transton pont, t c, from ts vanshng to frst order n u Usng the fluctuaton response relaton, the susceptblty s gven by χ = d d d d q x (m (x)m (0)) = (m (q)m (q = 0)) (π) d { } 1 4u d d [ ]} k n +1 1 = 1 + t t (π) d t + Kk t + r + Kk Invertng the correcton term gves The susceptblty dverges at d d [ ] k n +1 1 χ 1 = t +4u + + O(u ) (π) d t + Kk t + r + Kk d d [ ] k n +1 1 t c = 4u + + O(u ) (π) d Kk r + Kk (j) Is the crtcal behavor dfferent from the sotropc O(n) model n d < 4? In RG language, s the parameter r relevant at the O(n) fxed pont? In ether case ndcate the unversalty classes expected for the transtons The parameter r changes the symmetry of the ordered state, and hence the unversalty class of the dsorderng transton As ndcated n the fgure, the transton belongs to the O(n 1) unversalty class for r > 0, and to the Isng class for r < 0 Any RG transformaton must thus fnd r to be a relevant perturbaton to the O(n) fxed pont ******** 5 Cubc ansotropy Mean-feld treatment: Consder the modfed Landau Gnzburg Hamltonan [ n 4 βh = d d K x ( m) t ] + m + u(m ) + v m, for an n component vector m(x) = (m 1,m,,m n ) The cubc ansotropy term L, breaks the full rotatonal symmetry and selects specfc drectons n 4 =1 m (a) For a fxed magntude m ; what drectons n the n component magnetzaton space are selected for v > 0 and for v < 0? What s the degeneracy of easy magnetzaton axes n each case? 15 =1

16 v 0 > e ^e ^ e ^1 e ^1 m m m m m m m m order < v dagonal 0 cubc axs order In the fgures below, we ndcate the possble drectons of the magnetzaton whch are selected dependng upon the sgn of the coeffcent v, for the smple case of n = : Ths qualtatve behavor can be generalzed for an n-component vector: For v > 0, m les along the dagonals of a n-dmensonal hypercube, whch can be labelled as m m = (±1, ±1,, ±1), n and are consequently n -fold degenerate Conversely, for v < 0, m can pont along any of the cubc axes ê, yeldng m = ±mê, whch s n-fold degenerate (b) What are the restrctons on u and v for βh to have fnte mnma? Sketch these regons of stablty n the (u, v) plane The Landau-Gnzburg Hamltonan for each of these cases evaluates to t 4 v 4 βh = m + um + m, f v > 0 n, t 4 4 βh = m + um + vm, f v < 0 mplyng that there are fnte mnma provded that v u + > 0, for v > 0, n u + v > 0, for v < 0 Above, we represent schematcally the dstnct regons n the (u, v) plane (c) In general, hgher order terms (eg u 6 (m ) 3 wth u 6 > 0) are present and ensure stablty n the regons not allowed n part (b); (as n case of the trcrtcal pont dscussed n earler problems) Wth such terms n mnd, sketch the saddle pont phase dagram n the (t, v) plane for u > 0; clearly dentfyng the phases, and order of the transton lnes 16

17 u + v = 0 u allowed v unphyscal u + v = 0 n We need to take nto account hgher order terms to ensure stablty n the regons not allowed n part b) There s a trcrtcal pont whch can be obtaned after smultaneously solvng the equatons t+4(u+v)m +6u 6 m 4 = 0 }, = t (u+v) (u+v) =, m = t+(u+v)m +u 6 m 4 = 0 u 6 u6 The saddle pont phase dagram n the (t,v) plane s then as follows: (d) Are there any Goldstone modes n the ordered phases? There are no Goldstone modes n the ordered phases because the broken symmetry s dscrete rather than contnuous We can easly calculate the estmated value of the 17

18 transverse fluctuatons n Fourer space as (π) d φ t (q)φ t ( q) = vt, Kq + u+v from whch we can see that ndeed these modes become massless only when v = 0, e, when we retreve the O(n) symmetry ******** 6 Cubc ansotropy ε expanson: (a) By lookng at dagrams n a second order perturbaton expanson n both u and v show that the recurson relatons for these couplngs are du =εu 4C [ (n+8)u ] +6uv dl, dv [ ] =εv 4C 1uv+9v dl where C = K d Λ d /(t+kλ ) K 4 /K, s approxmately a constant Let us wrte the Hamltonan n terms of Fourer modes d d q t+kq βh = m (q) m ( q) (π) d d d q1d d q d d q 3 +u m (q 1 )m (q ) m j (q 3 )m j ( q 1 q q (π) 3d 3 ) d d q 1 d d q d d q 3 +v m (q 1 )m (q ) m (q 3 )m ( q (π) 3d 1 q q 3 ), where, as usual, we assume summaton over repeated ndces In analogy to problem set 6, after the three steps of the RG transformaton, we obtan the renormalzed parameters: t =b d z t K =b d z K, u = b 3d z 4 ũ v =b 3d z 4 ṽ wth t, K, ũ, and ṽ, are the parameters n the coarse graned Hamltonan The dependence of ũ and ṽ, on the orgnal parameters can be obtaned by lookng at dagrams n a second order perturbaton expanson n both u and v Let us ntroduce dagrammatc representatons of u and v, as j j u v 18

19 Contrbutons to u Contrbutons to v q 1 q q j j q 1 + q q q 3 q 4 n u q q 1 j q 3 K d d (t + K ) l 6 6 v K d d q j (t + K ) l q4 q 1 + q q q 1 q q j q 3 q 1 q q 1 + q l q q 4 4 u d Kd (t + K ) l j q 3 K d d l 6 4 uv q l (t + K ) q 1 + q q q 4 q 1 q q j q3 m q 1 + q l q q u d Kd (t + K ) l q 1 q q + q q 1 q j q 3 6 uv j q 4 d Kd (t + K ) l where, agan we have set b = e δl The new coarse graned parameters are { ũ = u 4C[(n+8)u +6uv]δl, ṽ = v 4C[9v +1uv]δl whch after ntroducng the parameter ǫ = 4 d, rescalng, and renormalzng, yeld the recurson relatons du =ǫu 4C[(n+8)u +6uv] dl dv =ǫv 4C[9v +1uv] dl (b) Fnd all fxed ponts n the (u,v) plane, and draw the flow patterns for n < 4 and n > 4 Dscuss the relevance of the cubc ansotropy term near the stable fxed pont n each case From the recurson relatons, we can obtan the fxed ponts (u,v ) For the sake of smplcty, from now on, we wll refer to the rescaled quanttes u = 4Cu, and v = 4Cv, n terms of whch there are four fxed ponts located at u = v = 0 Gaussan fxed pont ǫ u = 0 v = Isng fxed pont 9 u ǫ = v = 0 O(n) fxed pont (n+8) ǫ ǫ(n 4) u = v = Cubc fxed pont 3n 9n 19

20 Lnearzng the recurson relatons n the vcnty of the fxed pont gves ( ) ( )( ) d δu ǫ (n+8)u 6v 6u δu A = = d l δv 1v ǫ 1u 18v δv u,v As usual, a postve egenvalue corresponds to an unstable drecton, whereas negatve ones correspond to stable drectons For each of the four fxed ponts, we obtan: 1 Gaussan fxed pont: λ 1 = λ = ǫ, e, ths fxed pont s doubly unstable for ǫ > 0, as ( ǫ ) 0 A = 0 ǫ Isng fxed pont: Ths fxed pont has one stable and one unstable drecton, as ǫ A = 3 ǫ 4 3 0, ǫ correspondng to λ 1 = ǫ/3 and λ = ǫ Note that for u = 0, the system decouples nto n nonnteractng 1-component Isng spns 3 O(n) fxed pont: The matrx ǫ ǫ 6 (n+8) A =, (n 4) 0 ǫ (n+8) has egenvalues λ 1 = ǫ, and λ = ǫ(n 4)/(n+8) Hence for n > 4 ths fxed pont has one stable an one unstable drecton, whle for n < 4 both egendrectons are stable Ths fxed pont thus controls the crtcal behavor of the system for n < 4 4 Cubc fxed pont: The egenvalues of (n+8) A = 3 ǫ, (n 4) n 4 4 n 3 are λ 1 = ǫ(4 n)/3n, λ = ǫ Thus for n < 4, ths fxed pont has one stable and one unstable drecton, and for n > 4 both egendrectons are stable Ths fxed pont controls crtcal behavor of the system for n > 4 0

21 v n<4 v n> 4 u u q q 1 nu K d d (t + K ) l q1 q j q q 4u K d d (t + K ) l K d d ~t = t+4 (t + K ) [(n+)u+3v] d dt dl = t+4 K d (t + K ) [(n+)u+3v] q q 1 q K d d 6v (t + K ) l In the (u,v) plane, v = 0 for n < 4, and the cubc term s rrelevant, e, fluctuatons restore full rotatonal symmetry For n > 4, v s relevant, resultng n the followng flows: (c) Fnd the recurson relaton for the reduced temperature, t, and calculate the exponent ν at the stable fxed ponts for n < 4 and n > 4 Up to lnear order n ǫ, the followng dagrams contrbute to the determnaton of t: After lnearzng n the vcnty of the stable fxed ponts, the exponent y t s gven by y t = 4C[(n+)u +3v ],= ν = 1 = y t 1 + (n+) 4(n+8) ǫ+o(ǫ ) for n < (n 1) 6n ǫ+o(ǫ ) for n > 4 (d) Is the regon of stablty n the (u, v) plane calculated n part (b) of the prevous problem enhanced or dmnshed by ncluson of fluctuatons? Snce n realty hgher order terms wll be present, what does ths mply about the nature of the phase transton for a small negatve v and n > 4? All fxed ponts are located wthn the allowed regon calculated n 1b) However, not all flows startng n classcally stable regons are attracted to stable fxed pont If the RG 1

22 flows take a pont outsde the regon of stablty, then fluctuatons decrease the regon of stablty The domans of attracton of the fxed ponts for n < 4 and n > 4 are ndcated n the followng fgures: u + v = 0 u doman of attracton v unphyscal u + v n = 0 n > 4 Flows whch are not orgnally located wthn these domans of attracton flow nto the unphyscal regons The couplng constants u and v become more negatve Ths s the sgnal of a fluctuaton nduced frst order phase transton, by what s known as the Coleman Wenberg mechansm Fluctuatons are responsble for the change of order of the transton n the regons of the (u,v) plane outsde the doman of attracton of the stable fxed ponts ******** 7 Cumulant method: Apply the Nemejer van Leeuwen frst order cumulant expanson to the Isng model on a square lattce wth βh = K <j> σ σ j, by followng these steps: (a) For an RG wth b =, dvde the bonds nto ntra cell components βh 0 ; and nter cell components U The N stes of the square lattce are parttoned nto N/4 cells as ndcated n the fgure below (the ntra-cell and nter-cell bonds are represented by sold and dashed lnes respectvely) The renormalzed Hamltonan βh [σ α] s calculated from βh [σ 1 ( U α] = lnz 0 [σ α]+ U 0 ) 0 U 0 +O ( U 3), where 0 ndcates expectaton values calculated wth the weght exp( βh 0 ) at fxed [σ α ] (b) For each cell α, defne a renormalzed spn σ α = sgn(σ1 α +σ α +σ α 3 +σ α 4 ) Ths choce 4 becomes ambguous for confguratons such that =1 σ α = 0 Dstrbute the weght of these confguratons equally between σ α = +1 and 1 (e put a factor of 1/ n addton

23 ntra-cell bonds nter-cell bonds boundary of new unt blocks to the Boltzmann weght) Make a table for all possble confguratons of a cell, the nternal probablty exp( βh 0 ), and the weghts contrbutng to σ α = ±1 The possble ntracell confguratons compatble wth a renormalzed spn σ α = ±1, and ther correspondng contrbutons to the ntra-cell probablty exp( βh ), are gven below, 0 0 = 1 Weght Weght e 4 K e 4K 1 resultng n Z [σ ] = ( N/4 0 α e 4K +6+e 4K) = ( ) e 4K +6+e 4K α (c) Express U 0 n terms of the cell spns σ α ; and hence obtan the recurson relaton K (K) The frst cumulant of the nteracton term s U 0 = K σ α σ β1 +σ α3 σ β4 = K 0 α,β α,β σα 0 σ β1 0, 3

24 0 = 1 Weght Weght e 4K e 4K α β where, for σ α = 1, e 4K +(3 1)+0+0 e 4K + σ α 0 = = (e 4K +6+e 4K ) (e 4K +6+e 4K ) Clearly, for σ α = 1 we obtan the same result wth a global negatve sgn, and thus As a result, e 4K + σ α 0 = σ α (e 4K +6+e 4K ) βh [σ α] = N 4 ( ) ( 4K ln 4 e e K +6+e 4K) + +K e 4K +6+e 4K α,β σ σ, α β correspondng to the recurson relaton K (K), ( ) 4K e + K = K e 4K +6+e 4K (d) Fnd the fxed pont K, and the thermal egenvalue y t To fnd the fxed pont wth K = K = K, we ntroduce the varable x = e 4K Hence, we have to solve the equaton x+ 1 =, or x+6+x 1 ( ) 1 x 4 ( 6 ) x 1 = 0,

25 whose only meanngful soluton s x 796, resultng n K 05 To obtan the thermal egenvalue, let us lnearze the recurson relaton around ths non-trval fxed pont, K [ ] 4K 4K 4K = b y e t y e e, t = 1+8K, y 100 K t 6 K e 4K + e 4K +6+e 4K (e) In the presence of a small magnetc feld h σ, fnd the recurson relaton for h; and calculate the magnetc egenvalue y h at the fxed pont In the presence of a small magnetc feld, we wll have an extra contrbuton to the Hamltonan h e 4K + σα, 0 = 4h σ (e 4K +6+e 4K α ) Therefore, α, e 4K + h = 4h (e 4K +6+e 4K ) (f) Compare K, y t, and y h to ther exact values The cumulant method gves a value of K = 05, whle the crtcal pont of the Isng model on a square lattce s located at K c 044 The exact values of y t and y h for the two dmensonal Isng model are respectvely 1 and 1875, whle the cumulant method yelds y t 1006 and y h 15 As n the case of a trangular lattce, y h s lower than the exact result Nevertheless, the thermal exponent y t s fortutously close to ts exact value ******** 8 Mgdal Kadanoff method: Consder Potts spns s = (1,,,q), on stes of a hypercubc lattce, nteractng wth ther nearest neghbors va a Hamltonan βh = K δs,s j < j> α (a) In d = 1 fnd the exact recurson relatons by a b = renormalzaton/decmaton process Indentfy all fxed ponts and note ther stablty In d = 1, f we average over the q possble values of s 1, we obtan { q K q 1+e f σ 1 = σ e K(δ σ s 1 + δ 1 s1σ ) = = e g +K δ σ1σ, q +e K f σ 1 = σ s 1 =1 from whch we arrve at the exact recurson relatons: q 1+e K e K =, e g = q q +e K +e K 5

26 To fnd the fxed ponts we set K = K = K As n the prevous problem, let us ntroduce the varable x = e K Hence, we have to solve the equaton q x = 1+x, or x +(q )x (q 1) = 0, q +x whose only meanngful soluton s x = 1, resultng n K = 0 To check ts stablty, we consder K 1, so that ) q +K +K K K ln( K, q +K +K q whch ndcates that ths fxed pont s stable In addton, K s also a fxed pont If we consder K 1, e K 1 e K, = K = K ln < K, whch mples that ths fxed pont s unstable * * (b) Wrte down the recurson relaton K (K) n d dmensons for b =, usng the Mgdal Kadanoff bond movng scheme In the Mgdal-Kadanoff approxmaton, movng bonds strengthens the remanng bonds by a factor d 1 Therefore, n the decmated lattce we have e K q 1+e d 1 K = q +e d 1 K (c) By consderng the stablty of the fxed ponts at zero and nfnte couplng, prove the exstence of a non trval fxed pont at fnte K for d > 1 In the vcnty of the fxed pont K = 0, e for K 1, d K K q K, and consequently, ths pont s agan stable However, for K, we have e 1 [( ) ] K exp d d 1 K, = K = d 1 K ln K, whch mples that ths fxed pont s now stable provded that d > 1 As a result, there must be a fnte K fxed pont, whch separates the flows to the other fxed ponts * * * 6

27 (d) For d =, obtan K and y t, for q = 3, 1, and 0 Let us now dscuss a few partcular cases n d = For nstance, f we consder q = 3, the non-trval fxed pont s a soluton of the equaton +x 4 x =, or x 4 x 3 x+ = (x )(x 3 1) = 0, 1+x whch clearly yelds a non-trval fxed pont at K = ln 069 The thermal exponent for ths pont K [ ] y t e 4K e K 16 = = =, = y t 083, K K e 4K + 1+e K 9 whch can be compared to the exact values, K = 1005, and y t = 1 By analytc contnuaton for q 1, we obtan e K = e4k 1+e K The non-trval fxed pont s a soluton of the equaton x 4 x =, or (x 3 1+x x +1) = (x 1)(x x 1) = 0, whose only non-trval soluton s x = (1 + 5)/ = 16, resultng n K = 048 The thermal exponent for ths pont K [ ] = y t = 4 1 e K, = y t 061 K K As dscussed n the next problem set, the Potts model for q 1 can be mapped onto the problem of bond percolaton, whch despte beng a purely geometrcal phenomenon, shows many features completely analogous to those of a contnuous thermal phase transton And fnally for q 0, relevant to lattce anmals (see PS#9), we obtan e K = 1+e4K, +e K for whch we have to solve the equaton x = 1+x4, +x or x 4 x 3 +x 1 = (x 1) 3 (x+1) = 0, K K K + > K, whose only fnte soluton s the trval one, x = 1 For q 0, f K 1, we obtan ndcatng that ths fxed pont s now unstable Note that the frst correcton only ndcates margnal stablty (y t = 0) Nevertheless, for K, we have e 1 K exp[k], = K = K ln K, whch mples that ths fxed pont s stable * * 7

28 ******** 9 The Potts model: The transfer matrx procedure can be extended to Potts model, where the spn s on each ste takes q values s = (1,,,q); and the Hamltonan s βh = K N =1 δ s,s +1 +Kδ sn,s 1 (a) Wrte down the transfer matrx and dagonalze t Note that you do not have to solve a q th order secular equaton as t s easy to guess the egenvectors from the symmetry of the matrx The partton functon s Z = < s 1 T s >< s T s 3 > < s N 1 T s N >< s N T s 1 >= tr(t ), {s } where < s T s j >= exp ( Kδ s,s j ) s a q q transfer matrx The dagonal elements of the matrx are e K, whle the off-dagonal elements are unty The egenvectors of the matrx are easly found by nspecton There s one egenvectors wth all elements equal; the correspondng egenvalue s λ 1 = e K +q 1 There are also (q 1) egenvectors orthogonal to the frst, e the sum of whose elements s zero Ths correspondng egenvalues are degenerate and equal to e K 1 Thus Z = λ N α = α ( e K +q N 1 ) +(q 1) ( e K 1 ) N N (b) Calculate the free energy per ste Snce the largest egenvalue domnates for N 1, lnz ( = ln e K +q 1 ) N (c) Gve the expresson for the correlaton length ξ (you don t need to provde a detaled dervaton), and dscuss ts behavor as T = 1/K 0 Correlatons decay as the rato of the egenvalues to the power of the separaton Hence the correlaton length s [ ( )] 1 [ ( 1 λ e K )] 1 +q 1 ξ = ln = ln e K 1 In the lmt of K, expandng the above result gves e K ( ) 1 1 ξ = exp q q T λ ******** 8

29 MIT OpenCourseWare Statstcal Mechancs II: Statstcal Physcs of Felds Sprng 014 For nformaton about ctng these materals or our Terms of Use, vst: