1.4 Equipartition Theorem

1.4 Equipartition Theorem Equipartition Theorem: If the dynamics of the system is described by the Hamiltonian: 2 H = Aζ + H0 where ζ is one of the general coordinates q1,p1, ,q3N ,p3N , and H0 and A are independent of ζ, then ··· 1 Aζ2 = k T 2 B βH where is the thermal average, i.e., the average® over the Gibbs measure e− . Thishi result can be easily seen by the following calculation: Aζ2e βHd3N qd3N p Aζ = − h i e βHd3N qd3N p R − 2 βAζ2 βH RAζ e− dζe− 0 [dpdq] = 2 e βAζ dζe βH0 [dpdq] R − − Aζ2e βAζ2 dζ = R − e βAζ2 dζ R − ∂ βAζ2 = R ln e− dζ −∂β Z ∂ 1 Ax2 x = ζ β = ln β− 2 e− dx −∂β ∙ Z ¸ ³ p ´ 1 = 2β 1 = k T 2 B where dζ [dpdq]=d3N qd3N p, i.e., [dpdq] stands for the phase-space volumn element without dζ. Note that if 2 2 H = Aipi + Biqi wheretherearetotalM terms in theseX sums andXAi and Bi are constant, independent of pi,qi , then { } 1 H = kBT M h i 2 µ ¶ 1 i.e., each quadratic component of the Hamiltonian shares 2 kBT of the total energy. For example, the Hamiltonian for interacting gas particles is p2 H = i + U (r r ) 2m i j i ij − X X 7 where U (ri rj) is the potential energy between any two particles. The equipartition theo- − 3 rem tells us that the average kinetic energy is 2 kBT for every particle (since there are three translational degrees of freedom for each particle). Furthermore, for an ideal gas, i.e. gas particlesdonotinteract,wehavetheaveragetotalenergy 3 U = k TN 2 B where N is the total number of particles. Then, the specific heat for this system is ∂U CV ≡ ∂T µ ¶V ∂ 3 = k TN ∂T 2 B µ ¶ 3 = k N 2 B 8 1TheIsingModel For example, Fe or Ni exhibit macroscopic magnetic field via spontaneous polarization in the same direction for temperatures below the so-called critical temperature Tc, where Tc is referred to as the Curie temperature. For T>Tc, the spins are randomly orientated, there is no macroscopic magnetic field. 1.1 Model – Ising spins The Ising model is a caricature of spin dynamics. It teaches us many interesting aspects of spin dynamics and can be used to contrast real spin systems. Especially, two dimension Ising model is a nontrivial example of phase transitions that can be worked-out exactly. Themodel:weconsiderann-dimensional lattice (usually with periodic boundary condi- tions). For two dimensions, for example, the lattice structure can be cubic or hexagonal and it has N sitesintotal.Ateachsitethereisaspindescribedbythevariablesi ,i=1, 2, ,N with its value: ··· +1 spin up s = i 1 spin down ½ − Then, a set of si,i=1, ,N constitutes a configuration of the system. The energy in a configuration is{ ··· } N E si = Jijsisj H si { } − − i,j i=1 Xh i X where J is the exchange energy between two spins and H is a constant external magnetic field. i, i denotes the nearest next neighbor interactions. If the exchange interaction is isotropic,h theni Jij = J and N E si = J sisj H si { } − − i,j i=1 Xh i X For J>0, it describes ferromagnetism, in which two neighboring spins like to line up in the same direction. For J<0, it is antiferromagnetism, in which two neighboring spins tend to be in the opposite direction in order to lower the energy of the system. We consider only the case J>0 below. The partition function for the Ising model is βE si Q (H, T)= e− { } s s ··· s X1 X2 XN N there are total 2 termsinthesummationandeachsi, ranging over 1 is independent from another. ± 1 Thermodynamic functions can be obtained via the Helmholtz free energy: A (H, T)= kBT log Q (H, T) − For example, the internal energy is 2 ∂ A u (H, T)= kBT − ∂T k T µ B ¶ the heat capacity is ∂ c (H, T)= u (H, T) ∂T and the magnetization is ∂A M (H, T)= −∂H Itcanbeeasilyseenthat N M (H, T)= si * i=1 + X where istheensembleaverage. h· · · i The so-called spontaneous magnetization is M (0,T) =0 6 for the case of Jij > 0, i.e., there is a finite magnetization in the absence of the external field. 1.1.1 Transfer-Matrix Method Here we discuss only the 1D Ising model. The method can be generalized to 2D with considerably difficulty. Consider a chain of N spins, i.e., s1 sN+1 ≡ For the configuration s1, sN ,sk = 1, the energy of the system is { ··· } ± N N E = J sksk+1 H sk − − k=1 k=1 X X 2 and the partition function is β N (Js s +Hs ) Q (H, T)= e k=1 k k+1 k s s ··· s S X1 X2 XN Now we can evaluate this explicitly using a matrix formulation. First, using the periodic boundary condition, we rewrite Q in a symmetric form: β N Js s + 1 H(s +s ) Q (H, T)= e k=1[ k k+1 2 k k+1 ] s s ··· s S X1 X2 XN It turns out that there is a natural matrix formulation underlying this expression. Define a 2 2 matrix P with its elements × 1 β[Jss0+ 2 H(s+s0)] Pss0 = e where s, s0 takesthevalueof 1 independently. More explicitly, we have ± β(J+H) P++ = e β(J H) P+ = e − − βJ P+ = P + = e− − − i.e., β(J+H) βJ e e− P = βJ β(J H) e− e − µ ¶ Therefore, Q (H, T)= Ps1s2 Ps2s3 PsN s1 s s ··· s ··· X1 X2 XN = P N s1s1 s1 X ¡ ¢ = TrPN (a consequence of periodic BCs) N N = λ+ + λ − where λ+,λ are the eigenvalue of P. − βJ 2 4βJ 1/2 λ = e cosh (βH) sinh (βH)+e− ± ± h i λ+ λ for all values of H. P is referred to¡ as the transfer matrix.¢ ≥ − As N , only λ+ is relevant since →∞ 1 λ N log Q (H, T)=logλ +log 1+ + N + λ Ã µ − ¶ ! log λ+ as N → →∞ Physical results: 3 1. The Helmholtz free energy per spin is 1 2 4βJ 1/2 A (H, T)= J kBT log cosh (βH)+ sinh (βH)+e− N − − h ¡ ¢ i 2. The magnetization per spin is 1 sinh (βH) M (H, T)= N 2 4βJ 1/2 sinh (βH)+e− £ ¤ For T>0, ∀ 1 M (H =0,T)=0 N therefore is no spontaneous magnetization for 1-D Ising model. 1.1.2 Symmetry Breaking A more general model for spins is, for example, that the energy of the system is E = J Si Sj · i,j Xh i 2 where Si is a vector with S S =S , where S is a fixed number. Then the total magnetic moment is determined by · βE Tr e− M M = Tr(e βE) h i ¡ − ¢ In general, E is invariant under rotation in the absence of external field. Therefore, M =0 h i since M and M occur with equal probability. A deep question then is how come we have ferromagnetism?− This answer lies the fundamental concept of spontaneous symmetry breaking. 4 Spontaneous Symmetry Breaking Despite the fact that a Hamiltonian is invariant under certain symmetry groups, the ground state of the system does not have to possess the same symmetry of the Hamiltonian. That is, a ground state can be degenerate. For example, a ferromagnet ground state is not rotationally invariant – any one of the degenerate ground state is a physical solution. Once a system magnetizes along a certain direction, it cannot make a transition to another direction (Although, for this transition, there is no energy required, it has almost zero probability since all spins have to rotate simultaneously exactly thesameamount). How can we mathematically capture the spontaneous symmetry breaking? In the case of spontaneous magnetizatoin, we can add an exteranl field H,thenletH 0 in the end: → β(E MH) M 1 Tr Me− − h i lim lim β(E MH) V H 0 V V ³ − ´ ≡ → →∞ Tr e− where it is important to have the correct order of¡ limiting¢ processes, which reflects the underlying thermodynamic limit. If the limit H 0 is taken before the thermodynamic limit, then we have → β(E MH) 1 Tr Me− − lim lim β(E MH) =0 V H 0 V ³ − ´ →∞ → Tr e− i.e., there would be no spontaneous magnetization.¡ ¢ For the Ising model, the Hamiltonian is invariant under the discrete, reflectional symmetry: Si Si →− For a finite external field H, we have the following phase diagram: 1.2 Curie-Weiss Model (Mean-field model) The energy of the Curie-Weiss model is given by J N N E S = S S H S N i N i j j { } − i,j=1 − j=1 X X 5 which can also be expressed as 2 1 N N E S = JN S H S N i N i j { } − Ã i ! − j=1 X X or 1 N N E S = J S S H S (1) N i i N j j { } − i Ã j ! − j=1 X X X where H is the external magnetic field. Note that 1. due to the long-rang nature of the interaction (i.e., everying spin interacts with every other spin with an equal strength of coupling in this model), we can equivallently view the model as a spin interacting with its N neighbors in an N-dimensional lattice. 2. The bracked term in Eq.

Load more