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In Bose-Einstein statistics, the grand canonical partition function is $$\mathcal{Z}=1+e^{-\beta(\epsilon-\mu)}+e^{-2\beta(\epsilon-\mu)}+e^{-3\beta(\epsilon-\mu)}+\cdots$$

In Maxwell-Boltzmann statistics, the grand canonical partition function is $$\mathcal{Z}=1+e^{-\beta(\epsilon-\mu)}+\frac{e^{-2\beta(\epsilon-\mu)}}{2!}+\frac{e^{-3\beta(\epsilon-\mu)}}{3!}+\cdots$$

Why is there this difference and how does it relate to the respective assumptions of $N\simeq Z_1$ and $N \ll Z_1$ (where $Z_1$ is the canonical partition function for a single particle)? Are there additional assumptions needed to justify this difference?

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The difference lies in the way we count the number of states of the system in quantum and classical cases.

The formulas you wrote are actually for the grand canonical partition functions for a single energy state, not for the whole system including all the energy states. The total grand canonical partition function is $$\mathcal{Z} = \sum_{all\ states}{e^{-\beta(E-N\mu)}} = \sum_{N=0}^\infty\sum_{\{E\}}{e^{-\beta(E-N\mu)}}$$

Now, if the particles are bosons, then the energy eigenstates are countable as $\{\epsilon_i\}$ and $\mathcal{Z}$ would be $$\mathcal{Z} = \sum_{\{n_i\}}e^{-\beta\sum_{i}{n_i(\epsilon_i-\mu)}} = \sum_{\{n_i\}}\prod_ie^{-\beta n_i(\epsilon_i-\mu)}=\prod_i \mathcal{Z_i}^{B-E}$$ where $n_i$ is the number of particles in $i$-th energy state, thus $\sum_i{n_i}=N$, and $$\mathcal{Z_i}^{B-E} = \sum_{n=0}^\infty e^{-n\beta(\epsilon_i-\mu)}$$ Here, $\mathcal{Z_i}^{B-E}$ is the grand canonical partition function for one energy eigenstate with energy $\epsilon_i$ in Bose-Einstein statistics.

On the other hand, in classical regime the energy of a particle can take any energy. In this case, one point in the $6N$-dimensional phase space denotes one state of system. Therefore, the energy states are not countable as there is an infinite number of points in the phase space within any phase space volume. To count the states we take the "semi-classical" approach by taking the phase space volume of one state of the system to be $(2\pi\hbar)^{3N}$. We can then integrate over the whole phase space and divide the integral by this unit volume to get the number of states. However, as the particles are assumed to be indistinguishable, any permutation of the system configuration (the set of $\{\vec{x}_n,\ \vec{p}_n\}$) would actually be the same state of the system. Therefore, when we integrate over the whole phase space volume we overcount the total number of states by $N!$. That's why we need to divide the integral by Gibbs factor $N!$. For a system of non-interacting particles, the N-particle canonical partition function then can be written as $Z_N = \frac{Z_1^N}{N!}$ where $Z_1$ is the canonical partition function for one particle.

Now, the grand canonical partition function for a classical system would be $$\begin{align} \mathcal{Z}&=\sum_{N=0}^\infty{\int_0^\infty dE\ \Omega(E,N)e^{-\beta(E-\mu N)}} =\sum_{N=0}^\infty e^{\beta\mu N} Z_N = \sum_{N=0}^\infty e^{\beta\mu N} \frac{Z_1^N}{N!}\\ \end{align}$$

If we want to derive the partition function $\mathcal{Z_i}$ for a single energy state similar to the Bose-Einstein statistics, we can assume the energy of a single particle to be discrete and countable as $\{\epsilon_i\}$. This can be achieved by dividing the one particle phase space into s of unit volume and assigning one representative energy to every unit volume section. Then, the grand canonical partition function is, $$ \begin{align} \mathcal{Z} &= \sum_{N=0}^\infty \frac{e^{\beta\mu N}}{N!} (\sum_i e^{-\beta\epsilon_i})^N\\ &=\sum_{N=0}^\infty \frac{e^{\beta\mu N}}{N!} \sum_{\{n_i\},\sum n_i = N} \frac{N!}{\prod_i n_i!} e^{-\beta\sum_i n_i\epsilon_i}\\ &=\sum_{\{n_i\}}\prod_i \frac{1}{n_i!} e^{-\beta n_i(\epsilon_i-\mu)}\\ &=\prod_i \sum_{n=0}^\infty \frac{1}{n!} e^{-\beta n(\epsilon_i-\mu)} = \prod_i \mathcal{Z_i}^{M-B} \end{align}$$

This $\mathcal{Z_i}^{M-B}$ is the single energy state grand canonical partition function in Maxwell-Boltzmann statistics.

Maxwell-Boltzmann statistics is the classical limit for Bose-Einstein statistics. The condition for the system to be classical is the single state occupation number $\bar{n}$ to satisfy $\bar{n} \ll 1$, in other words, the total number of single particle states $M$ should satisfy $N \ll M$. As, the single particle partition function $Z_1$ is actually a weighted sum over all the states, $N\ll Z_1$ will satisfy $N \ll M$ for the system to be classical.

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General quantum remarks - identical v. non-identical particles

For a given quantum statistical mechanical system with Hilbert space $\mathcal H$, hamiltonian $\hat H$, and number operator $\hat N$, the grand canonical partition function is defined as follows: \begin{align} Z(\beta, \mu) = \mathrm{tr}(e^{-\beta(\hat H - \mu \hat N)}) \end{align} where the trace is being taken over $\mathcal H$.

The difference between the statistics lies in the fact that for systems of identical particles, such as systems of bosons, the Hilbert space is smaller than if the particles are non-identical, so the trace is taken over a different vector space.

Suppose, for example, that we have a system of non-identical, non-interacting bosons. Suppose, for the sake of simplicity, that the single particle Hilbert space is two-dimensional and that the single-particle Hamiltonian has two distinct eigenvalues $E_1$ and $E_2$ with corresponding orthonormal eigenstates $|1\rangle$ and $|2\rangle$. Let $|0\rangle$ denote the state with no particles in it, then the Hilbert space of the system is spanned by the following energy eigenstates: \begin{align} \text{0-particles}:&\qquad|0\rangle\\ \text{1-particle}: &\qquad|1\rangle \\ &\qquad|2\rangle\\ \text{2-particles}:&\qquad |1\rangle|1\rangle\\ &\qquad|1\rangle|2\rangle\\ &\qquad|2\rangle|1\rangle\\ &\qquad|2\rangle|2\rangle\\ \text{3-particles}: &\qquad |1\rangle|1\rangle|1\rangle\\ &\qquad |1\rangle|1\rangle|2\rangle\\ &\qquad |1\rangle|2\rangle|1\rangle\\ &\qquad |1\rangle|2\rangle|2\rangle\\ &\qquad |2\rangle|1\rangle|1\rangle\\ &\qquad |2\rangle|1\rangle|2\rangle\\ &\qquad |2\rangle|2\rangle|1\rangle\\ &\qquad |2\rangle|2\rangle|2\rangle\\ &\qquad\vdots \end{align} In fact, the $N$-particle subspace is spanned by $2^N$ distinct states. However, if the particles are identical bosons, then the Hilbert space is spanned only by the symmetric combinations of states in each $N$-particle subspace, so listing the states gives: \begin{align} \text{0-particles}:&\qquad|0\rangle\\ \text{1-particle}:&\qquad |1\rangle \\ &\qquad |2\rangle\\ \text{2-particles}:&\qquad |1\rangle|1\rangle\\ &\qquad \tfrac{1}{\sqrt{2}}\big(|1\rangle|2\rangle+ |2\rangle|1\rangle\big)\\ &\qquad|2\rangle|2\rangle\\ \text{3-particles}: &\qquad |1\rangle|1\rangle|1\rangle\\ &\qquad \tfrac{1}{\sqrt{3}}\big(|1\rangle|1\rangle|2\rangle + |1\rangle|2\rangle|1\rangle + |2\rangle|1\rangle|1\rangle\big)\\ &\qquad \tfrac{1}{\sqrt{3}}\big(|1\rangle|2\rangle|2\rangle + |2\rangle|1\rangle|2\rangle + |2\rangle|2\rangle|1\rangle\big)\\ &\qquad |2\rangle|2\rangle|2\rangle\\ &\qquad\vdots \end{align} and it is easy to show that in general, the subspace with $N$ identical bosons is spanned by $N+1$ distinct states.

In general, when the single-particle Hilbert space is greater than two dimensional, the counting may get more difficult, but the moral is the same. You sum over less states when the particles are identical because the Hilbert space only includes appropriately symmetric or anti-symmetric states.

Bose-Einstein statistics - partition function

It can be shown (see https://physics.stackexchange.com/a/101456/19976) that the quantum grand canonical partition function for a system of non-interacting, identical bosons is \begin{align} Z = \prod_{i\in I}\sum_{n=0}^\infty x_i^n , \qquad x_i := e^{-\beta(\epsilon_i-\mu)} \end{align} where $I$ is some index set whose elements label single particle energy states (elements of an orthonormal energy eigenbasis of the Hilbert space), $i$ is the index that labels these states, and $\epsilon_i$ is the energy of state $i$.

Now, suppose that for each state $i$, we define a quantity $Z_i$ as follows: \begin{align} Z_i = \sum_{n=0}^\infty x_i^n, \tag{$\star$} \end{align} which we interpret as the "partition function for the $i$th energy eigenstate," then the full partition function for the system can be written as a product of these $Z_i$'s; \begin{align} Z = \prod_{i\in I} Z_i. \end{align} Now, it can also be shown (again see https://physics.stackexchange.com/a/101456/19976) that the ensemble average occupancy $\langle n_i\rangle$ of state $i$ is given by \begin{align} \langle n_i \rangle = x_i \frac{\partial}{\partial x_i} \ln Z. \end{align} Now, notice what happens when we insert the expression for the full partition function $Z$ as a product of the $Z_i$'s into this expression; \begin{align} \langle n_i\rangle &= x_i \frac{\partial}{\partial x_i} \ln \left(\prod_{j\in I} Z_j\right) \\ &= x_i \frac{\partial}{\partial x_i} \sum_{j\in I} \ln Z_j \\ &= x_i \sum_{j\in I} \frac{\partial}{\partial x_i} \ln Z_j \\ &= x_i \frac{\partial}{\partial x_i} \ln Z_i \end{align} where in the last step, we used the fact that each $Z_j$ only depends on the corresponding $x_j$, its partial with respect to $x_i$ vanishes unless $i=j$. In other words, what we have shown is that

The ensemble average occupancy of state $i$ can be computed entirely from the partition function $Z_i$ associated with that state.

Now, what does each single state partition function $Z_i$ look like? Well, going back to the definition, we notice that it's simply a geometric series; \begin{align} Z_i = \sum_{n=0}^\infty x_i^n = (e^{-\beta(\epsilon_i-\mu)})^0 + (e^{-\beta(\epsilon_i-\mu)})^1 + (e^{-\beta(\epsilon_i-\mu)})^2 + \cdots \end{align} which is exactly what you wrote. The physical interpretation of this is that the index $n$ represents the number of particles occupying the given energy state $i$ with energy $\epsilon_i$, so if we think of this state as its own little system, then its energy levels are $n\epsilon_i$ with no degeneracy, where $n=0,1,2,3, \cdots$.

Maxwell-Boltzmann statistics - the classical limit

When we take the classical limit of the quantum ensemble average occupancies, we obtain the Maxwell-Boltzmann ensemble average occupancy; \begin{align} n_i = e^{\beta(\epsilon_i-\mu)}. \end{align} Aside from the fact that the single-state partition function you wrote down for the Maxwell-Boltzmann distribution yields this limit of the quantum ensemble average occupancies, I'm not sure at present how one can argue that it's "correct." In fact, I'm not entirely sure there is another valid criterion by which to judge whether it's correct.

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  • $\begingroup$ So then why is the grand canonical partition function larger when the bosons ARE identical (Bose-Einstein) than when they are not (Maxwell-Boltzmann)? Wouldn't it be smaller, since you are summing over less states? $\endgroup$ – Joshua Meyers May 13 '14 at 18:01
  • $\begingroup$ Or is my expression for Maxwell-Boltzmann wrong? (I am starting to doubt it) $\endgroup$ – Joshua Meyers May 13 '14 at 18:02
  • $\begingroup$ @JoshuaMeyers Is the system you're considering a system of non-interacting particles? If so, what are the single-particle energy levels and their degeneracies? $\endgroup$ – joshphysics May 13 '14 at 18:27
  • $\begingroup$ Yes, they are non-interacting. Do the single-particle energy levels and their degeneracies need to be specified? The derivations of Bose-Einstein and Maxwell-Boltzmann statistics that I have seen don't make explicit reference to them. $\endgroup$ – Joshua Meyers May 13 '14 at 18:57
  • $\begingroup$ The formulas I have yield the correct results for $\bar{n}=-k_{B}T\frac{\partial}{\partial\epsilon}\ln\mathcal{Z}$. That is why I think they are correct. $\endgroup$ – Joshua Meyers May 13 '14 at 19:02

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