Statistical Thermodynamics Solution Exercise 8 HS Solution Exercise 8

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1 Statistical Thermodynamics Solution Exercise 8 HS 05 Solution Exercise 8 Problem : Paramagnetism - Brillouin function a According to the equation for the energy of a magnetic dipole in an external magnetic field, we get the following energy spectrum for a single paramagnetic center: Ĥ µ ˆ µ B 0 g J µ B ˆ J B0 g J µ B Ĵ Z B 0, (. ɛ g J µ B m J B 0. (. Please note that the J ˆ Z -operator is given unitless } in eq.., so that its eigenvalue is directly m J. The Boltzmann operator Ĥ k B describes the density matrix of such a center in T thermal equilibrium. In the eigenframe of} the ĴZ operator ( B 0 is along Z the density matrix is diagonal with elements. The partition function for a single paramagnetic g J µ B m J B 0 center is thus (see eq. 6.9 in the lecture notes Z i T r ( Ĥi gj µ B m J B 0 (.3 For a system of non-interacting, distinguishable paramagnetic centers the overall Boltzmann operator can be represented as a direct product of one-center Boltzmann operators. Using the property of the trace of a direct product (see eq.. in the problem set, we obtain the -center system partition function: z T r ( Ĥ } Ĥ } ] gj µ B m J B 0 Zi Ĥ } (.4 The system partition function could also be deduced from the total Hamiltonian. For noninteracting particles the total Hamiltonian can be constructed from the Hamiltonians of the individual subsystems (see the 6th postulate of quantum mechanics in PC III, Ĥ Ĥ Ĥ Ĥ ˆ + ˆ Ĥ. (.5 To calculate the partition function from that total Hamiltonian, for non-interacting particles (Ĥ, Ĥ] 0 the following relation can be used ( Ĥ Ĥ ( Ĥ ( Ĥ. (.6 page of 8

2 Statistical Thermodynamics Solution Exercise 8 HS 05 g b Substituting J µ B B 0 k B in equation.3 by q, the one-center partition function Z T i can be rewritten as Z i ( mj gj µ B B 0 changing the index of summation m J to l m J + J + we obtain ( J+ Z i q J l q l q m J, (.7 q J qj+, (.8 q using relation.3 of the problem set. Reformulation of eq.8 and using the definitions of the hyperbolic functions leads to Z i q / q J / q J+/ q J / q J+/ q q / q / qj+/ q J / sinh (J + / x], q / q / sinh x/] (.9 with x g J µ B B 0 k B. Accordingly, the partition function for paramagnetic centers is T ] sinh (J + / x] z Zi (.0 sinh x/] c Let s show that equations.4 and.5 of the problem set are equal. Using the relations for z (equation.4 and Z i (equation.3 evaluation of problem set equation.5 leads to: βv ln z ln z x d ln z x B 0 βv x B 0 βv dx B 0 d ln Z i x dz i x, βv dx B 0 βv Z i dx B 0 (. note that ln(z is only dependent on x, so that the partial differential is changed to a total differential. Further and βv x B 0 βv B 0 gj µ B B 0 k BT V gjµ B g Jµ B, (. V page of 8

3 (J + / coth (J + / x] coth x/]. (.5 Statistical Thermodynamics Solution Exercise 8 HS 05 Z i dz i dx J (m J x m J (m J x J (m J x m J P (m J x. m J (m J x (.3 Accordingly, βv ln z B 0 g Jµ B V m J P (m J x g Jµ B V m J V µ z M Z, (.4 and the equality of.4 and.5 is obtained. To calculate the equilibrium magnetization we will use above equation.9 for the partition dz function and relation.5 of the problem set. i Z i is then given as dx Z i dz i dx sinh x/] sinh (J + / x] d dx sinh (J + / x] sinh x/] sinh x/] (J + / sinh x/] cosh (J + / x] / cosh x/] sinh (J + / x] sinh (J + / x] sinh x/]] (J + / cosh (J + / x] sinh (J + / x] cosh x/] sinh x/] In literature often the variable ξ Jx is used. Using this notation and inserting equation.5 into., we obtain: M Z ( ] J + J + V g Jµ B J coth ξ ]] ξ J J J coth J V g Jµ B JB J (ξ. (.6 B J (ξ is called Brillouin function. It describes the dependency of the magnetization on the external magnetic field B 0 and the effective spin J at any arbitrary temperature. d In the limit of weak magnetic field, when the energy levels of the magnetic dipole are much smaller than the thermal energy, one can use the approximation coth (x /x + x/3. This allows us to simplify equation.6 to the form page 3 of 8

4 Statistical Thermodynamics Solution Exercise 8 HS 05 M Z V g J + Jµ B J ( J J+ + J ξ V g Jµ B J ξ ] + ξ V g J (J + Jµ B Jx J J ( J+ J ξ (J + J ξ J } V g (J + Jµ B Jx J 4J V g + 4J Jµ B Jx J ]} ] ξ 3 J ]} J ξ J + ]} ξ J 3 (.7 V J(J + g J µ B 3 B 0. Thus, for the magnetic susceptibility at weak magnetic fields we get M Z χ lim B 0 0 B 0 V J(J + g J µ B. (.8 3 Problem : Ising model in one dimension - Transfer matrix method - Magnetism a Due to the interactions between the quantum magnetic dipoles in such a circular chain, we cannot directly analyse the problem as a combination of individual dipoles. Thus, when computing the partition function, we must compute the total energy of all dipoles for each microscopic state. One microscopic state of such a system would be characterized by a set of states σ, σ,..., σ } of all spins, with all possible combinations. Following, the summation in the partition function can be done as a nested series of sums with all possible states for the first spin counted for every state of the second spin and all those combinations counted for each state of the third spin and so on. The energy of each microscopic state would consist of the Zeeman energies h B σ i of the spins as well as all pairwise exchange interactions Jσ i σ i+ between neighboring spins. For the exchange interactions it is clear that if for each spin we count the interaction with the next one (for the -th spin - with the first one, then summing over all spins would result into counting each exchange term for each pair of spins exactly once. In the problem set equation. the Zeeman term is written in such a way that for it is counted twice for the i-th spin, at the steps i and i +, so that a factor / has to be added to the Zemann term. These considerations lain the validity of the equation. in the problem set. b In equation.6 of the problem set the terms σ i σ i make sure that for each microscopic state of the spin system the same quantum state of the spin i is counted in both steps of the summation, where this spin is involved. Since each of the matrix elements of the transfer matrix ˆP takes the form e x i, the product of all such elements would result in a term e x i with x i values that sum to E (σ, σ,..., σ /(. page 4 of 8

5 Statistical Thermodynamics Solution Exercise 8 HS 05 c Summation over both states of i-th spin results in an identity matrix: ˆ σ i σ i. Thus σ i, z (, h B, T σ σ ˆP σ σ ˆP σ 3 σ ˆP σ σ ˆP σ σ σ σ ˆP σ Tr ˆP } σ (. d The eigenvalues of the transfer matrix ˆP can be found in any representation from the equation det ˆP λˆ e β(j+hb λ e βj e βj e β(j hb λ e βj λ(e β(j+h B + e β(j h B + λ e βj e βj e βj λe βj cosh(βh B + λ 0. (. By solving this quadratic equation we obtain λ ± ( e βj cosh (βh B ± 4e βj cosh (βh B 4(e βj e βj e (cosh βj (βh B ± cosh (βh B + e 4βJ e (cosh βj (βh B ± sinh (βh B + e 4βJ. (.3 In its eigenframe the transfer matrix can thus be written as ( λ+ 0 ˆP, (.4 0 λ ( λ ˆP λ, (.5 and the trace is then equal to Tr ˆP λ + +λ. If h B is not infinitely small, then λ + > λ, and therefore, for very large numbers of magnetic dipoles (statistical limit λ + >> λ. Thus, the partition function reduces to a form z (, h B, T λ +. page 5 of 8

6 Statistical Thermodynamics Solution Exercise 8 HS 05 e Calculating the Helmholtz free energy gives F (h B, T ln Z (h B, T ]]} ln e βj cosh (βh B + sinh (βh B + e 4βJ ]} J + ln cosh (βh B + sinh (βh B + e 4βJ ] J ln cosh (βh B + sinh (βh B + e 4βJ (.6 and thus the average magnetic moment is given by m µ F h B (βh B ln λ +} sinh (βh B cosh (βh B/ sinh (βh B + e 4βJ. cosh (βh B + sinh (βh B + e 4βJ (.7 f For h B 0 cosh (βh B and sinh (βh B 0, so that lim m 0 (.8 h B 0 ote that including the λ term in the calculation would result into zero magnetization as well. Thus, the exact calculation predicts that no magnetic ordering is possible in such an interacting circular chain at finite temperature without external magnetic field. However, the system of magnetic dipoles is paramagnetic in the presence of an external magnetic field. Problem 3: Mean field theory calculation for the Ising model - Ferromagnetism a If we want to sum over all i, j without restricting j to be larger than i, we will count each pair i, j} twice. Thus the sum counting i, j should be divided by two compared to problem set equation.. This gives for the system partition function Z (, h, T σ β σ σ J 0 σ i σ j + h B i,j σ i ]}. (3. ote that the Zeeman term is only counted once in the definition of the Hamiltonian in equation 3. compared to equation. of the problem set. b By using the relation σ i σ j (σ i m (σ j m σ i σ j σ i σ j + m (σ i + σ j m m (σ i + σ j m, (3. page 6 of 8

7 Statistical Thermodynamics Solution Exercise 8 HS 05 and inserting it into equation 3., we obtain Z (, h B, T σ β σ σ J 0 ( m (σi + σ j m + h B σ i ]}. (3.3 i,j Let s consider the double sum over i and j. The term m will be counted n 0 times, while n 0 is the number of nearest neighbours for each spin. Thus, J 0 ( m n 0 J 0 ( m Jm. (3.4 i,j The term m (σ i + σ j will be counted for each site n 0 times when the index of the site matches with i (sum over j and another n 0 times, when the index of the site matches with j (sum over i. Thus, J 0 i,j m (σ i + σ j ] n 0J 0 mσ i + n 0J 0 mσ j n 0 J 0 j mσ i Jm σ i (3.5 and following for the partition function Z (, h B, T σ β Jm + Jm σ σ σ i + h B σ i ]}. (3.6 We can now rewrite the Hamiltonian in the form Ĥ Jm + (Jm + h B σ i, (3.7 which can be interpreted as a system of non-interacting spins in an external magnetic field B ext Jm + h B. c The factor e Jm is identical for every term in the sum determining the -particle partition function, and can thus be taken out. The remaining sum is identical to the sum calculated in the Problem. Accordingly, the partition function can be factorized in analogy to the calculations for the non-interacting paramagnetic centers in Problem. z(, h B, T Z i (h B, T ] e βjm e β(jm+h B + e β(jm+h B ] e βjm cosh (β (Jm + h B ] (3.8 Calculation of the Helmholtz free energy results to F (h B, T ln Z i (h B, T Jm ln cosh (β (Jm + h B ]. (3.9 page 7 of 8

8 Statistical Thermodynamics Solution Exercise 8 HS 05 d As described in Problem, the average magnetic moment can be computed as m F h B. We can thus write m F h B sinh (β (Jm + h B] cosh (β (Jm + h B ] tanh (β (Jm + h B]. (3.0 And in absence of an external magnetic field (h B 0 we obtain m tanh (βjm. (3. e The trivial solution of the equation 3. is m 0. The non-trivial solutions exist in case the slope of tanh (βjm at m 0 is larger than one. In this case the tanh function first rises above the line f(m m and symmetrically crosses f(m m at ±m 0. Figure 3- shows the tanh function for the three regimes: βj >, βj, and βj <. tanh(βjm m 0 0 m 0 βj > βj βj < m f(m m Figure 3-: tanh function fot the three regimes: βj >, βj and βj < This means, that for βj > additional non-zero solutions of m ±m 0 exist in the absence of an external magnetic field (h B 0. This phenomenon is called ferromagnetism. However, note that the mean-field approach is an approximation. For instance, ferromagnetic ordering is wrongly predicted for the linear chain of interacting spins (compare to the exact solution of Problem. evertheless, this approach is capable of correctly describing the physics of a number of magnetic ordering phenomena that would be too complicated to be solved exactly. Further, the dependence on the parameters h B, β, and J is correctly predicted by this approximation. page 8 of 8