Übungen zur Theoretischen Physik Fa WS 17/18
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1 Krlsruher Institut für ehnologie Institut für heorie der Kondensierten Mterie Übungen zur heoretishen Physik F WS 17/18 Prof Dr A Shnirmn Bltt 4 PD Dr B Nrozhny Lösungsvorshlg 1 Nihtwehselwirkende Spins: () Sttionry sttes re hrterized by the sets of quntum numbers σ } with the orresponding energies Eσ } = µ σ z = M ere M is the mgnetiztion in the diretion of the field, whih we hve hosen s the z xis σ = ±1 re the z-omponents of the individul spins We hoose units with k B = 1 (b) he sttistil sum is Z = σ } e Eσ}/ = σ } N e µσ/ = =1 N e µσ/ = =1 σ } 2 osh µ ] N he free energy is the logrithm of the sttistil sum F = ln Z = N ln 2 osh µ ] () he entropie: ( ) F S = = N ln 2 osh µ ] µn tnh µ he mgnetiztion: he speifi het: M = = ( ) S = = µn tnh µ µ osh 2 µ he speifi het for onstnt mgnetiztion n be lulted inverting the funtion M(, ): ( ) S(, (M, )) M = M ( ) ( ) ( ) S(, (M, )) S(, (M, )) (M, ) = + = M
2 (d) In the limit of lrge fields, M(µ ) µn In the limit of smll fields, M(µ ) µ 2 N herefore the suseptibility is given by the Curie lw ( ) M χ( ) = lim = µ2 N 2 eisenberg Modell für 2 Gitterplätze: Aording to the generl rules of ddtion of ngulr momentum, the system of two moment n be desribed by set of four quntum numbers he two ommon possibilities re either (i) s 2 1, s 2 2, s z 1, s z 2, or (ii) s 2 1, s 2 2, S 2, S z, where S = s 1 + s 2 Suppose we hoose the first possibility hen for fixed s 2 1 = s 1 (s 1 +1), s 2 2 = s 2 (s 2 +1), the quntities s z i tke 2s i + 1 vlues eh, so tht the totl number of sttes being (2s 1 + 1)(2s 2 + 1) For two spins 1, s i = 1, we hve 9 possible sttes In this problem, it is more onvenient to hoose the seond representtion, sine s 1 s 2 = 1 2 S(S + 1) s 1(s 1 + 1) s 2 (s 2 + 1)] he 9 possible sttes orrespond to the following vlues (fixing s 1 = s 2 = 1): S =, S z =, s 1 s 2 = 2, S = 1, S z =, ±1, s 1 s 2 = 1, S = 2, S z =, ±1, ±2, s 1 s 2 = 1 () In the bsene of the field, the sttistil sum is given by Z = e Js 1 s 2 / = 5e J/ + 3e J/ + e 2J/ = 6 osh J/ + 2e J/ + e 2J/ S,S z } he free energy is he speifi het is C V = 2 F 2 = J 2 3 e 2J/ F = ln Z ( ) ( ) ( ) ] J J 1 e J/ 5 J + 1 e 3J/
3 (b) In the presene of the field, the sttistil sum is Z = e (Js 1 s 2 +µs z )/ = S,S z } = e J/ ( osh µ ) ( 2µ + 2 osh + e J/ osh µ ) + e 2J/ he free energy is he mgnetiztion is M = F = 4µ Z F = ln Z osh J sinh µ + ej/ sinh 2µ ] he suseptibility is χ( ) = 4µ2 Z osh J + 2eJ/ ], where Z is the bove sttistil sum in the bsene of the field his suseptibility beomes Curie-like only t high tempertures J 3 Msselose reltivistishe eilhen: () he sttistil sum of system of non-interting prtiles ftorizes: Z = ZN 1 N! he single-prtile sttistil sum is given by (here = = 1) Z 1 = V () 2 e p / = V pe p/ dp = V 2 2 ze z dz = V 2 2 ene, the sttistil sum is he free energy is F = ln Z = Z = V N N!() N 2N ln } V ln N! ( ) 2N = 2N ln ( ) ] } 1/2 V N ln N + N, or F = 2N ln ( ) ] 1/2 V N N
4 (b) he entropy is given by the derivtive S = V,N = F + 2N Similrly, the pressure is,n = N V () he eqution of stte reltes the internl energy of the system to temperture he internl energy is U = V () 2 pθ(p F p) = V p F p 2 dp = V p3 F 6π his should be relted to the prtile density N V = () θ(p 2 F p) = 1 p F pdp = p2 F 4π ene p F = 4πN/V nd the internl energy n be expressed s U = 4N 4πN 3 V he pressure n be evluted lso s ( ) U,N = U 2V Compring with the previous result, we find the eqution of stte 4 Ideles Gs: P V = N U = 2N () he sttistil sum of system of non-interting prtiles ftorizes: Z = ZN 1 N! he single-prtile sttistil sum is given by (here = = 1) Z 1 = V d 3 p / = V () 3 e p 4 2 p 2 e p4 / dp = V 3/4 2 3/4 z 2 e z4 dz = V ( ) 3/4 8π Γ(3/4) 2
5 ene, the sttistil sum is he free energy is Z = V N N! F = ln Z = Γ(3/4) N ln 3N = 4 ln = 3 4 N ln (b) he entropy is given by the derivtive S = Similrly, the pressure is ] N ( ) 3N/4 V Γ(3/4) ( V Γ(3/4) ( ) ] } 3/4 ln N! ) ] 4/3 ( ) ] 4/3 V Γ(3/4) N N V,N,N = F N = N V N ln N + N () he eqution of stte reltes the internl energy of the system to temperture he internl energy is U = V d 3 p () 3 p4 θ(p F p) = V 2 his should be relted to the prtile density ene N V = d 3 p () θ(p 3 F p) = 1 2 p F = 3 6π 2 N/V nd the internl energy n be expressed s U = 3N ( 6π 2 N 7 V he pressure n be evluted lso s ( ) U,N p F p F ) 4/3 = 4U 3V p 6 dp = V p7 F 14π 2 p 2 dp = p3 F 6π 2 Compring with the previous result, we find the eqution of stte P V = N U = 3 4 N },
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