Distinguishable non-interacting systems : Exercises

Introduction

The following questions all involve systems of non-interacting particles that sit on lattice sites. An exact expression for th canonical partition function for these systems of so-called independent-disinguishable particles can be found and as such it is possible to calculate the equations of state and average energies of these systems exactly.

Example problems

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Problem 1

Consider a system of $N$ distinguishable, non-interacting spins in a magnetic field $H$. Each spin will have an associated magnetic moment $\mu$, which can point either parallel or anti-parallel to the field. The state of the system will be characterised by the set of spin variables $\{n\}$, where $n_i = \pm 1$ represents the fact that spin $i$ can be in one of two possible configurations (spin-up or spin-down). The total energy, $E$, of all the spins will be given by: \[ E = -\sum_{i=1}^N n_i \mu H \] as $n_i\mu$ will be the value of the magnetic moment that is pointing in the direction of the field.

Consider an statistical mechanical ensemble in which the number of spins, $N$, the magnetic field, $H$, and the temperature, $\beta = \frac{1}{k_BT}$, are kept fixed. Write an expression for the partition function for this ensemble. You will need the definition of the hyperbolic trig function $\cosh(\theta) = \frac{ e^\theta + e^{-\theta}}{2}$.

Lets start by considering the partition function for a single spin. We are in the canonical ensemble. We have constant $N$, constant $H$ and constant $T$. In the canonical ensemble for a (quantum) system that occupies a discrete set of levels we have: \[ Z_c(N,H,T) = \sum_i e^{-\beta E_i} \] For a single spin this is simply: \[ Z_c(1,H,T) = \sum_{s_1=0}^1 e^{-\beta z(s_1) \mu H} = e^{\beta \mu H} + e^{-\beta \mu H} = 2 \cosh(\beta \mu H) \] where here $z(s_1)$ is a function for which $z(0)=-1$ and $z(1)=1$. Let's now consider a system with $N$ spins. We now will have multiple summation signs in our expression for the partition function as shown below: $$ Z_c(N,N,T) = \sum_{s_1=0}^1 \sum_{s_2=0}^1 \sum_{s_3=0}^1 \dots \sum_{s_N=0}^1 \exp\left(\beta \sum_{j=1}^N z(s_j)\mu H\right) $$ The first sum here run over the two spin states the first spin can take, the second sum runs over the two spins the second spin can take, the third runs over the three spin states the third spin can take and so on. Once again $z(s)$ is a function for which $z(0)=-1$ and $z(1)=$. We progress with this problem by recognising that the exponential of a sum in the final term can be rewritten as a product of exponentials as shown below: $$ Z_c(N,N,T) = \sum_{s_1=0}^1 \sum_{s_2=0}^1 \sum_{s_3=0}^1 \dots \sum_{s_N=0}^1 \prod_{j=1}^N e^{z(s_j) \mu H} $$ Each of the terms in this product depends on only one of the spin variables. We can thus rewrite the following sum over products as shown below: $$ Z_c(N,N,T) = \left[ \sum_{s_1=0}^1 e^{z(s_1) \mu H} \right] \left[ \sum_{s_2=0}^1 e^{z(s_2) \mu H} \right] \left[ \sum_{s_3=0}^1 e^{z(s_3) \mu H} \right] \dots \left[ \sum_{s_N=0}^1 e^{z(s_N) \mu H} \right] $$ Notice that each of the sums in the above is the result is the same as the single spin parition function , $Z_c(1,H,T)$, which we derived above. We can thus write: $$ Z_c(N,H,T) = 2^N \cosh^N(\beta \mu H) $$

Consider the system of $N$ distinguishable, non-interacting spins in a magnetic field $H$ that were introduced in the previous exercise and use the expression that you obtained for the partition function to calculate the average internal energy of the system of $N$ spins and the average magnetisation, $\langle M \rangle = \langle \sum_{i=1}^N \mu n_i \rangle$, of the spins. Use the expressions you derive to discuss the behavior of the spins in the limit as limit as $T\rightarrow \infty$ and in the limit as $T\rightarrow 0$.

In the canonical ensemble the ensemble average for the energy is given by:: \[ \langle E \rangle = -\left( \frac{\partial \ln(Z_c) }{\partial \beta} \right) \] So given the expression we have just derived for the canonical partition function of this system, $Z_c(N,H,T)=2^N \cosh^N(\beta \mu H)$, in this case we have: \[ \begin{aligned} \langle E \rangle & = -\frac{\partial }{\partial \beta} \left\{ N\ln(2) + N\ln\left[\cosh(\beta \mu H)\right] \right\} = -\frac{\partial }{\partial \beta} N\ln\left[ \frac{ e^{\beta \mu H} + e^{-\beta \mu H}}{2} \right] \\ & =- \frac{2N}{e^{\beta \mu H} + e^{-\beta \mu H} }\left[ \frac{\mu H e^{\beta \mu H} - \mu H e^{-\beta \mu H}}{2}\right] \\ & = -N \mu H \frac{ e^{\beta \mu H} - e^{-\beta \mu H}}{e^{\beta \mu H} + e^{-\beta \mu H} } = -N\mu H \frac{\sinh(\beta \mu H)}{\cosh(\beta\mu H)} = -N\mu H \tanh(\beta \mu H) \end{aligned} \] To get the average magnetisation we note based on the formula in the queation that: $$ \langle M \rangle = - \frac{ \langle E \rangle }{ H } $$ The magnetisation is thus: $$ \langle M \rangle = \mu H \tanh(\beta \mu H) $$

In the limit as $T \rightarrow \infty$: $\beta \rightarrow 0$. Hence $\tanh(\beta \mu H) \rightarrow 0$, $\langle E \rangle \rightarrow 0$ and $\langle M \rangle \rightarrow 0$. In the infinite temperature limit we can thus conclude that the number of spin up particles equals the number of spin down particles.

In the limit as $T\rightarrow 0$: $\beta \rightarrow \infty$. Hence $\tanh(\beta \mu H) \rightarrow 1$, $\langle E \rangle \rightarrow -N\mu H$ and $\langle M \rangle \rightarrow N\mu$ . In the zero temperature limit we can thus conclude that all the spins align with the field because they all want to be in the state with the lowest energy.

Derive an expression for the entropy of a system of $N$ distinguishable, non-interacting spins in a magnetic field $H$ as a function of $\beta$, $H$ and $N$.

In the canonical ensemble the entropy is given by: \[ \frac{S}{k_B} = \ln Z_c + \frac{ \langle E \rangle }{k_B T} \] So given the expression we have just derived for the canonical partition function of this system, $Z_c(N,H,T)=2^N \cosh^N(\beta \mu H)$, in this case we have: \[ S = k_B N\ln\left[2\cosh(\beta \mu H)\right] - \frac{N\mu H}{T} \tanh(\beta \mu H) \]

Problem 2

A lattice gas consists of a lattice of $N$ sties. Each of these sites can be either empty in which case the energy due that site is zero or occupied in which case the energy dues to that site is $\epsilon$. Use this information to calculate the canonical partition function for a system of $N$ lattice gas sites.

The canonical partition function is given by $Z_c = \sum_j = e^{-\beta E_j}$ so for a single lattice gas site this is a sum over the two possible states: \[ Z_c(1) = \sum_{s_1=1}^2 e^{-\beta E(s_1)} = e^{-\beta 0} + e^{-\beta \epsilon} = 1 + e^{-\beta \epsilon} \] where $E(1)=0$ for $E(2)=\epsilon$. For N particles the partition function is given by the following sum \[ \begin{aligned} Z_c(N) & = \sum_{s_1=1}^2 \sum_{s_2=1}^2 \dots \sum_{s_N=1}^2 e^{-\beta \sum_{j=1}^N E(s_j} \\ & = \sum_{s_1=1}^2 \sum_{s_2=1}^2 \dots \sum_{s_N=1}^2 \prod_{j=1}^N e^{-\beta E(s_j)} \\ & = \left[ \sum_{s_1=1}^2 e^{-\beta E(s_1)} \right]\left[ \sum_{s_2=1}^2 e^{-\beta E(s_2)} \right] \dots \left[ \sum_{s_N=1}^2 e^{-\beta E(s_N)} \right] & = (1 + e^{-\beta \epsilon})^N \end{aligned} \]

Evaluate the average energy $\langle E \rangle$ and the average number of lattice sites that are occupied when the the lattice gas described in the previous parts is at a temperature of $T$. N.B. Notice from the first part of the question that the energy of a microstate $E$ is related to the number of occupied sites, $n$, via $\mathbf{E} = n\epsilon$.

The average energy can be calculated using $\langle E \rangle = - \frac{\partial \ln Z_c }{\partial \beta}$. For the lattice gas we thus have: \[ \langle E \rangle = - \frac{\partial }{\partial \beta} N\ln(1+e^{-\beta \epsilon}) = \frac{N\epsilon e^{-\beta \epsilon} }{1+e^{-\beta \epsilon}} \] From the information in the question the total energy can be written in terms of the occupation as follows $E = N_{occ} \epsilon$. Hence: \[ \langle N_{occ} \rangle = \frac{\langle E \rangle }{\epsilon} = \frac{Ne^{-\beta \epsilon} }{1+e^{-\beta \epsilon}} \]

Suppose now that each occupied lattice site has a magnetic moment of magnitude $\mu$, which in the presence of an applied magnetic field, $H$, can adopt one of two orientations. These two orientations have energies $\epsilon - \mu H$ and $\epsilon + \mu H$. Calculate the canonical partition function for this new system.

The essential difference between this new system and the previous one is that each site can now be in one of three states. It can be unoccupied, occupied with a spin up moment or occupied with a spin down moment. We thus write the single particle canonical partition function as a sum over these three possible states: \[ \begin{aligned} Z_c(1) = \sum_j e^{-\beta E_j} & = e^{-\beta 0} + e^{-\beta(\epsilon + \mu H)} + e^{-\beta(\epsilon - \mu H} \\ & = 1 + e^{-\beta\epsilon} \left( e^{-\beta \mu H} + e^{\beta \mu H} \right) \\ & = 1 + 2e^{-\beta\epsilon}\cosh(\beta \mu H) \end{aligned} \] Then since the sites are independent and distinguishable we have: \[ Z_c(N) = [ 1 + 2e^{-\beta\epsilon}\cosh(\beta \mu H) ]^N \] for reasons that have been explained in the answers to the previous parts above.

Evaluate the average energy and the average magnetisation of the system that was introduced in the previous question by taking derivatives of the logarithm of the partition function

The average energy can be calculated using: \[ \begin{aligned} \langle E \rangle = - \frac{\partial \ln Z_c }{\partial \beta} & = - \frac{\partial }{\partial \beta} N \ln[ 1 + 2e^{-\beta\epsilon}\cosh(\beta \mu H) ] \\ & = -N \frac{1}{1 + 2e^{-\beta\epsilon}\cosh(\beta \mu H)} \frac{\partial }{\partial \beta} \left[ 1 + 2e^{-\beta\epsilon}\cosh(\beta \mu H) \right] \\ & = -N\frac{1}{1 + 2e^{-\beta\epsilon}\cosh(\beta \mu H)} \left[ -2\epsilon e^{-\beta\epsilon}\cosh(\beta \mu H) + 2\mu H e^{-\beta\epsilon} \sinh(\beta \mu H) \right] \\ & = \frac{2N\left[\epsilon e^{-\beta\epsilon}\cosh(\beta \mu H) - \mu H e^{-\beta\epsilon} \sinh(\beta \mu H)\right]}{1 + 2e^{-\beta\epsilon}\cosh(\beta \mu H)} \\ & = \frac{2N\left[\epsilon \cosh(\beta \mu H) - \mu H \sinh(\beta \mu H)\right]}{e^{\beta\epsilon} + 2\cosh(\beta \mu H)} \end{aligned} \] We get to last line by multiplying top and bottom by $e^{\beta \epsilon}$. From this final result the average magnetisation can be calculated using: \[ \begin{aligned} \langle M \rangle = \frac{\partial \ln Z_c }{\partial (\beta H) } & = \frac{\partial }{\partial (\beta H) } N \ln[ 1 + 2e^{-\beta\epsilon}\cosh(\beta \mu H) ] \\ & = N \frac{1}{1 + 2e^{-\beta\epsilon}\cosh(\beta \mu H)} \frac{\partial }{\partial (\beta H) } \left[ 1 + 2e^{-\beta\epsilon}\cosh(\beta \mu H) \right] \\ & = N \frac{1}{1 + 2e^{-\beta\epsilon}\cosh(\beta \mu H)} \left[ 0\cosh(\beta \mu H) + 2e^{-\beta\epsilon} \mu \sinh(\beta \mu H) \right] \\ & = \frac{2N\mu \sinh(\beta \mu H)}{e^{\beta\epsilon} + 2\cosh(\beta \mu H)} \end{aligned} \]

Problem 3

A simple model for a polymer can be constructed by considering a linear chain composed of $N$ non-interacting, molecular units. These units can exist in one of two states: state 1 has a length $l_1$ and energy $\epsilon_1$, while state 2 has length $l_2$ and energy $\epsilon_2$. The work done when the polymers length is changed by an infinitesimal amount $\textrm{d}L$ is given by $\textrm{d}w=f\textrm{d}L$, where $f$ is the tensile force acting on the polymer. Given this use the laws of classical thermodynamics to derive an expression for the infinitesimal change in entropy, $\textrm{d}S$, for infinitesimal changes in the internal energy, $\textrm{d}E$, and polymer length, $\textrm{d}L$.

\[ \textrm{d}S = \frac{1}{T} \textrm{d}E - \frac{f}{T}\textrm{d}L \]

Use statistical mechanics to derive an expression for the infinitesimal change in entropy $\textrm{d}S$ that would accompany an infinitessimal change in the number of polymber units, the average energy or the average length of the chain. Use your result and the result from classical thermodynamics that you found in the previous part to derive an expression for the partition function for an $N$-monomer chain in an ensemble in which the temperature, $T$, the number of monomer units, $N$, and the tensile force, $f$ on the polymer are fixed. then explain why the partition function for an $N$-monomer chain is given by:

The final partition function is: \[ Z(N,T,f) = \left[ e^{-\frac{\epsilon_1-f l_1}{k_B T}} + e^{-\frac{\epsilon_2-f l_2}{k_B T}}\right]^N \]

Derive expressions for the average length of a polymer consisting of $N$ monomers and the polymers average energy. These expressions should be a function of the temperature, $T$, and tensile force, $f$.

Average energy: \[ \langle E \rangle = \frac{N\left( \epsilon_1 e^{-\beta(\epsilon_1-f l_1)} + \epsilon_2 e^{-\beta(\epsilon_2-f l_2)} \right)}{e^{-\beta(\epsilon_1-f l_1)} + e^{-\beta(\epsilon_2-f l_2)}} \] Average length: \[ \langle L \rangle = = \frac{N\left( l_1 e^{-\beta(\epsilon_1-f l_1)} + l_2 e^{-\beta(\epsilon_2-f l_2)} \right)}{e^{-\beta(\epsilon_1-f l_1)} + e^{-\beta(\epsilon_2-f l_2)}} \]

Problem 4

The energy of an atom with angular momentum $J\hbar$ in an applied magnetic field $H$ is given by $U=g\mu H m$, where $g$ is the so called electron g-factor, $\mu$ is the electron's magnetic moment or Bohr magneton and $m$ is the atom's magnetic quantum number, which can take values $m=-J, -J+1,\dots,-1,0,1 \dots, J-1, J$. Write an expression the canonical partition function for a system composed of $N$ of such atom on a lattice. In doing this question you should assume that the atoms do not interact

\[ Z_c = \left[ \sum_{m=-J}^J e^{\beta g \mu H m} \right]^N \]

Write an expression for the average energy of the lattice gas introduced in the previous part in terms of the Brillouin function: $$ B_J(x) = \frac{\partial }{\partial x} \ln\left( \sum_{m=-J}^J e^{xm} \right) = \frac{2J +1}{2J} \coth\left( \frac{2J+1}{2J} x \right) - \frac{1}{2J} \coth\left( \frac{1}{2J}x\right) $$

\[ \langle E \rangle = -N g \mu H B_J(\beta g \mu H) \]