The Grand Canonical Ensemble : Exercises

Introduction

The grand canonical ensemble is used to model system with a fixed number temperature, a fixed pressure/volume and a variable number of atoms. In this exercise you are going to derive an expression for the probability of being in a particular microstate within this ensemble, an expression for the grand canonical partition function and the ensemble average for the number of atoms in the system. In doing this you will start from the expressions for the probability of being in a particular micostate and the partition function in the generalised partition function. You will then follow arguments similar to those presented in the videos on the canonical partition function and the isothermal isobaric ensemble. You should thus make sure you are familiar with the derivations presented in those videos before attempting the exercises below.

Example problems

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Problem 1

Use the fact that only the volume is held fixed in the isothermal, isobaric ensemble to write an expression for the probability of being in a microstate for the grand canonical ensemble. N.B. Use the symbol $\beta \xi$ for the Lagrange multiplier that appears in front of the number of atoms $N(\mathbf{x}_i,\mathbf{p}_i)$.

$$ p_i = \frac{ e^{-\beta E(\mathbf{x}_i,\mathbf{p}_i)} e^{-\beta \xi N(\mathbf{x}_i,\mathbf{p}_i)} }{ Z_{GC} } $$

Now write an expression for the grand canonical partition function.

$$ Z_{GC} = \sum_i e^{-\beta E(\mathbf{x}_i,\mathbf{p}_i)} e^{-\beta \xi N(\mathbf{x}_i,\mathbf{p}_i)} $$

Now write an expression for $\frac{\textrm{d}S}{k_B}$ in terms of $\textrm{d}V$, $\textrm{d}\langle E \rangle$ and $\textrm{d}\langle N \rangle$ that holds when you are considering an ensemble in which the volume is the only extensive quantity that is being held fixed.

$$ \frac{\textrm{d}S}{k_B} = - \left[ \beta \left\langle \frac{\partial E}{\partial V} \right\rangle + \beta \xi \left\langle \frac{\partial E}{\partial N} \right\rangle \right] \textrm{d}V + \beta \textrm{d}\langle E \rangle + \beta \xi \textrm{d}\langle N \rangle $$

Use the result from part the previous part and the result from classical thermodynamics $\textrm{d}E = T\textrm{d}S - P\textrm{d}V + \mu \textrm{d}N$ to show that $\xi = - \mu$

Rearranging the equation in the question we have: $$ \textrm{d}S = \frac{1}{T} \textrm{d}E + \frac{P}{T}\textrm{d}V - \frac{\mu}{T} \textrm{d}N $$ Comparing coefficients of $\textrm{d}N$ in this equation and $\frac{\textrm{d}S}{k_B} = - \left[ \beta \left\langle \frac{\partial E}{\partial V} \right\rangle + \beta \xi \left\langle \frac{\partial E}{\partial N} \right\rangle \right] \textrm{d}V + \beta \textrm{d}\langle E \rangle + \beta \xi \textrm{d}\langle N \rangle$: $$ -\frac{\mu}{T} = k_B \beta \xi \qquad \rightarrow \qquad -\beta \mu = \beta xi \qquad \rightarrow \qquad \mu = -\xi $$ In doing this remember that $\beta = \frac{1}{k_B T}$

What thermodynamic potential can be calculated from the grand canonical partition function and how is this done?

The grand potential can be calculated from the grand canonical partition function as shown below: $$ \frac{S}{k_B} = \psi + \sum_k \lambda_k \langle B^{(k)} \rangle \qquad \rightarrow \qquad \frac{S}{k_B} = \psi + \frac{E}{k_B T} - \frac{N\mu}{k_B T} \quad \textrm{for Grand Canonical ensemble} $$ Rearranging this expression gives: $$ k_B T \psi = ST - E + N\mu = -\Omega \qquad \rightarrow \qquad \Omega = -k_B T \ln Z_{GC} $$ Here we must remember that the grand canonical potential, $\Omega$, is given by $\Omega = E - TS + \mu N$

Show that: $$ \langle N \rangle = \left( \frac{\partial \ln Z}{\partial (\beta \mu)} \right) $$ where $Z$ is the grand canonical partition function.

We start here from: $$ 1 = e^{\psi} e^{-\psi} = e^{-\psi} Z = e^{-\psi} \sum_i e^{-\beta E(\mathbf{x}_i,\mathbf{p}_i)} e^{\beta \mu N(\mathbf{x}_i,\mathbf{p}_i)} = \sum_i e^{-\beta E(\mathbf{x}_i,\mathbf{p}_i) + \beta \mu N(\mathbf{x}_i,\mathbf{p}_i) - \psi } $$ Differentiating both sides of this expression with respect to $(\beta \mu)$ gives: $$ \begin{aligned} 0 & = \sum_i \left( N(\mathbf{x}_i,\mathbf{p}_i) - \frac{\partial \psi}{\partial (\beta \mu)} \right) e^{-\beta E(\mathbf{x}_i,\mathbf{p}_i) + \beta \mu N(\mathbf{x}_i,\mathbf{p}_i) - \psi } \\ & = \sum_i N(\mathbf{x}_i,\mathbf{p}_i) \frac{e^{-\beta E(\mathbf{x}_i,\mathbf{p}_i)} e^{\beta \mu N(\mathbf{x}_i,\mathbf{p}_i)}}{e^\psi} - \sum_i \left( \frac{\partial \psi}{\partial (\beta \mu)} \right) \frac{e^{-\beta E(\mathbf{x}_i,\mathbf{p}_i)} e^{\beta \mu N(\mathbf{x}_i,\mathbf{p}_i)}}{e^\psi} \\ & = \sum_i N(\mathbf{x}_i,\mathbf{p}_i) P_i - \left( \frac{\partial \psi}{\partial (\beta \mu)} \right) \sum_i P_i \\ 0 & = \langle N \rangle - \left( \frac{\partial \psi}{\partial (\beta \mu)} \right) \end{aligned} $$ Note the that energy of a microstate {\bf does not} depend on $\mu$ the derivatives of these quantities are zero.

Show that: $$ \left\langle (N - \langle N \rangle)^2 \right\rangle = \left( \frac{\partial^2 \ln Z}{\partial (\beta \mu)^2} \right) $$ where $Z$ is the grand canonical partition function.

We differentiate the first result we arrived at when we differentiated $Z$ with respect to $(\beta \mu)$ in the last question with respect to $(\beta \mu)$ once more. This brings us to: $$ \begin{aligned} 0 & = \sum_i \left( N(\mathbf{x}_i,\mathbf{p}_i) - \frac{\partial \psi}{\partial (\beta \mu)} \right)^2 e^{-\beta E(\mathbf{x}_i,\mathbf{p}_i) + \beta \mu N(\mathbf{x}_i,\mathbf{p}_i) - \psi } - \sum_i \frac{\partial^2 \psi}{\partial (\beta \mu)^2} e^{-\beta E(\mathbf{x}_i,\mathbf{p}_i) + \beta \mu N(\mathbf{x}_i,\mathbf{p}_i) - \psi } \\ & = \sum_i \left( N(\mathbf{x}_i,\mathbf{p}_i) - \langle N \rangle \right)^2 \frac{e^{-\beta E(\mathbf{x}_i,\mathbf{p}_i)} e^{\beta \mu N(\mathbf{x}_i,\mathbf{p}_i)}}{e^\psi} - \frac{\partial^2 \psi}{\partial (\beta \mu)^2} \sum_i \frac{e^{-\beta E(\mathbf{x}_i,\mathbf{p}_i)} e^{\beta \mu N(\mathbf{x}_i,\mathbf{p}_i)}}{e^\psi} \\ & = \sum_i \left( N(\mathbf{x}_i,\mathbf{p}_i) - \langle N \rangle \right)^2 P_i - \frac{\partial^2 \psi}{\partial (\beta \mu)^2} \sum_i P_i \\ 0 & = \left\langle (N - \langle N \rangle)^2 \right\rangle - \frac{\partial^2 \psi}{\partial (\beta \mu)^2} \end{aligned} $$

Problem 2

Suppose you are given the partition function, $Z(M)$, for a system with a fixed magnetization, $M$. Explain why the partition function for a system in which the magnetic field, $H$, rather than the magnetism, is fixed is given by: $$ Z(H) = \sum_i Z(M_i) e^{\beta H M_i} $$ Hence demonstrate that $\langle ( \delta M)^2 \rangle = \left( \frac{\partial M}{\partial (\beta H)} \right)_{N}$ where $\langle ( \delta M)^2 \rangle = \langle M^2 \rangle - \langle M \rangle^2$

To prove that the partition function is the one given in the question we first begin by noting that classical thermodynamics tells us that: $$ \textrm{d}E = T\textrm{d}S - P\textrm{d}V + H \textrm{M} $$ we can rearrange this expression to give: $$ \textrm{d}S = \frac{1}{T} \textrm{d}E + \frac{P}{T} \textrm{d}V - \frac{H}{T} \textrm{d}M $$ Now suppose that we are in an ensemble in which only the value of the volume is held fixed. In other words, the energy and the magnetisation of the system is allowed to fluctutuate, but the average values of these quantities is not allowed to be infinite. Statistical mechanics tells us that: $$ \frac{\textrm{d}S}{k_B} = -\beta \langle \frac{\partial H}{\partial V} \rangle \textrm{d}V + \beta \textrm{d}\langle E \rangle + \beta \xi \textrm{d}\langle M \rangle $$ If we compare the coefficients of $\textrm{d}E$ and $\textrm{d}M$ in this expression and the previous one we find that: $$ \begin{aligned} k_B \beta = \frac{1}{T} \qquad \rightarrow \qquad \beta = \frac{1}{k_B T} \\ k_B \beta \xi = - \frac{H}{T} \qquad \rightarrow \qquad \xi = - H \end{aligned} $$ Given the definition of the generalised partition function: $$ Z = \sum_j e^{\sum_k \lambda_k B_j^{(k)} } $$ we can thus write that the partition function $Z(H)$ is given by: $$ Z(H) = \sum_j e^{-\beta E_j} e^{\beta H M_j} $$ Now suppose that we had fixed the magnetization, $M$, of the system. The partition function in this case would have been: $$ Z(M) = \sum_n e^{-\beta E_n} = \sum_j e^{-\beta E_j} \delta( M_j - M ) $$ Critically the first sum in the above (the one over $n$) consider only those microstates that have a magnetism of $M$. We can calculate $Z(M)$ by summing over all microstates including those that have a magnetisation, $M_j$, that is not equal to $M$, as we did when calculating $Z(H)$, by including a function: $$ \delta( M_j - M ) = \begin{cases} 1 & \textrm{if} \quad M_j = M \\ 0 & \textrm{otherwise} \end{cases} $$ in the sum as we did in the second summation above. We recover the expression in the question for $Z(H)$ by recognising that: $$ Z(H) = \sum_j e^{-\beta E_j} e^{\beta H M_j} = \sum_i e^{\beta H M_i} \sum_j e^{-\beta E_j} \delta( M_i - M_j ) = \sum_i e^{\beta H M_i} Z(M_i) $$ We now define $\psi$ as $Z(H)=e^\psi$ and note that. \[ 1 = e^{-\psi}e^{\psi} = e^{-\psi} \sum_j e^{\beta H M_j} = \sum_j e^{\beta H M_j - \psi} \] Notice that Z(M_i) does not depend on $\beta H$. We can thus differentiate both sides of this expression above with respect to $\beta H$ to give: \[ \begin{aligned} 0 &= \frac{\partial }{\partial (\beta H)} \sum_j Z(M_j) e^{\beta H M_j - \psi} = \sum_j Z(M_j) \frac{\partial }{\partial (\beta H)} e^{\beta H M_j - \psi} \\ & = \sum_j Z(M_j) e^{\beta H M_j - \psi}\left( M_j - \frac{\partial \psi}{\partial (\beta H)} \right) \\ & = \frac{1}{Z(H)} \sum_j M_j Z(M_j) e^{\beta H M_j } - \frac{\partial \psi}{\partial (\beta H)} \frac{1}{Z(H)} \sum_j Z(M_j) e^{\beta H M_j - \psi} \\ & = \langle M \rangle - \frac{\partial \psi}{\partial (\beta H)} \end{aligned} \] so obviously $\frac{\partial \psi}{\partial (\beta H)} = \langle M \rangle$. If we now take second derivatives of $1 = e^{-\psi}e^{\psi}$ we find \[ \begin{aligned} 0 & = \frac{\partial }{\partial (\beta H)} \sum_j Z(M_j) e^{\beta H M_j - \psi}\left( M_j - \frac{\partial \psi}{\partial (\beta H)} \right) = \sum_j Z(M_j) \left[ e^{\beta H M_j - \psi}\left( M_j - \frac{\partial \psi}{\partial (\beta H)} \right)^2 - e^{\beta H M_j - \psi} \frac{\partial^2 \psi}{\partial (\beta H)^2} \right] \\ & = \sum_j Z(M_j) e^{\beta H M_j - \psi} \left\{ \left[ M_j - \langle M_j \rangle \right]^2 - \frac{\partial^2 \psi}{\partial (\beta H)^2} \right\} = \langle (\delta M)^2 \rangle - \frac{\partial^2 \psi}{\partial (\beta H)^2} \end{aligned} \] We thus have: $ \langle (\delta M)^2 \rangle = \frac{\partial^2 \psi}{\partial (\beta H)^2}$ where $\psi=\ln Z_c$. But remember $\frac{\partial \psi}{\partial (\beta H)} = \langle M \rangle$ so we can rewrite this second derivative as a first derivative i.e. as $\langle (\delta M)^2 \rangle = \left( \frac{\partial \langle M \rangle }{\partial (\beta H)^2 } \right)$. If we do this at constant $\beta$ we get the result required by the question.

The Grand Canonical Ensemble : Exercises

Introduction

Example problems

Problem 1

Problem 2

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