11
$\begingroup$

My textbook says that microcanonical ensemble, canonical ensemble and grand canonical ensemble are essentially equivalent under thermodynamic limit. It also derives Fermi-Dirac and Bose-Einstein distribution from grand canonical ensemble.

My question is then: How to derive Fermi-Dirac and Bose-Einstein distribution using canonical ensemble. The expression of canonical ensemble $$\rho\propto e^{-\beta E}$$

seems to imply Boltzmann distribution only.

$\endgroup$
20
$\begingroup$

I don't really see the answer in the other answer so let me do the calculation here. Your general Boltzmann Ansatz says that the probability of a state $n$ depends on its energy as $$ p_n = C \exp(-\beta E_n) $$ where $\beta = 1/kT$. Fermions are identical particles that, for each "box" or one-particle state they can occupy (given e.g. by $nlms$ in the case of the Hydrogen atom-like states), admit either $N=0$ or $N=1$ particles in it. Higher numbers are forbidden by the Pauli exclusion principle. The energies of the multi-particle state with $N=1$ and $N=0$ in a particular one-particle state $nlms$ differ by $\epsilon$. Consequently, $$ \frac{p_1}{p_0} = \frac{C\exp(-\beta (E+\epsilon))}{\exp(-\beta E)} = \exp(-\beta \epsilon) $$ where I used the Boltzmann distribution. However, the probabilities that the number of particles in the given one-particle state is equal to $N=0$ or $N=1$ must add to one, $$ p_0 + p_1 = 1.$$ These conditions are obviously solved by $$ p_0 = \frac{1}{1+\exp(-\beta\epsilon)}, \qquad p_1 = \frac{\exp(-\beta\epsilon)}{1+\exp(-\beta\epsilon)}, $$ which implies that the expectation value of $n$ is equal to the right formula for the Fermi-Dirac distribution: $$\langle N \rangle = p_0\times 0 + p_1 \times 1 = p_1= \frac{1}{\exp(\beta\epsilon)+1} $$ The calculation for bosons is analogous except that the Pauli exclusion principle doesn't restrict $N$. So the number of particles (indistinguishable bosons) in the given one-particle state may be $N=0,1,2,\dots $. For each such number $N$, we have exactly one distinct state (because we can't distinguish the particles). The probability of each such state is called $p_n$ where $n=0,1,2,\dots$.

We still have $$\frac{p_{n+1}}{p_n} = \exp(-\beta\epsilon) $$ and $$ p_0 + p_1 + p_2 + \dots = 1 $$ These conditions are solved by $$ p_n = \frac{\exp(-n\beta\epsilon)}{1+\exp(-\beta\epsilon)+\exp(-2\beta\epsilon)+\dots } $$ Note that the ratio of the adjacent $p_n$ is what it should be and the denominator was chosen so that all the $p_n$ from $n=0,1,2\dots$ sum up to one.

The expectation value of the number of particles is $$ \langle N \rangle = p_0 \times 0 + p_1 \times 1 + p_2\times 2 + \dots $$ because the number of particles, an integer, must be weighted by the probability of each such possibility. The denominator is still inherited from the denominator of $p_n$ above; it is equal to a geometric series that sums up to $$ \frac{1}{1-q} = \frac{1}{1-\exp(-\beta\epsilon)} $$ Don't forget that $1-\exp(-\beta\epsilon)$ is in the denominator of the denominator, so it is effectively in the numerator.

However, the numerator of $\langle N \rangle$ is tougher and contains the extra factor of $n$ in each term. Nevertheless, the sum is analytically calculable: $$ \sum_{n=0}^\infty n \exp(-n \beta\epsilon) = - \frac{\partial}{\partial (\beta\epsilon)} \sum_{n=0}^\infty \exp(-n \beta\epsilon) =\dots$$ $$\dots = - \frac{\partial}{\partial (\beta\epsilon)} \frac{1}{1-\exp(-\beta\epsilon)} = \frac{\exp(-\beta\epsilon)}{(1-\exp(-\beta\epsilon))^2} $$ This result's denominator has a second power. One of the copies gets cancelled with the denominator before and the result is therefore $$ \langle N \rangle = \frac{\exp(-\beta\epsilon)}{1-\exp(-\beta\epsilon)} = \frac{1}{\exp(\beta\epsilon)-1} $$ which is the Bose-Einstein distribution.

You could also obtain another version of the Boltzmann distribution for distinguishable particles by a similar calculation. For such particles, $N$ could take the same values as it did for bosons. However, the multiparticle state with $N$ particles in the one-particle state would be degenerate because the particles are distinguishable. There would actually be $N!$ multiparticle states with $N$ particles in them. The sum would yield a Taylor expansion for the same exponential.

Note added later: the derivation above was for $\mu=0$. When the chemical potential is nonzero, all appearances of $\epsilon$ have to be replaced by $(\epsilon-\mu)$. Of course that one may only talk about a well-defined value of $\mu$ when we deal with a grand canonical potential; it is impossible to derive a formula depending on $\mu$ from one that contains no $\mu$ and assumes it's ill-defined. The derivation above was meant to show that the difficult $1/(\exp\pm 1)$ structures do appear from a simpler $\exp(-\beta E)$ Ansatz because I think it's the only nontrivial thing to be shown while discussing the relations between the Boltzmann and BE/FD distributions. If that derivation proves the same link as the textbook does, then I apologize but I think there is "nothing else" of a similar kind to be proven.

$\endgroup$
  • 2
    $\begingroup$ Lubos: you are doing the derivation of the expected value, which is nice, but the person's confusion was "why go grand canonical?" This is the entire content of my answer. Also, your derivation is also implicitly doing it grand canonical, because the factor of chemical potential is missing, so it must have been absorbed into the multiplicative factor for extra particles. All in all, you gave an inferior answer, and downvoted my answer, so -1. $\endgroup$ – Ron Maimon Dec 21 '11 at 19:29
  • 1
    $\begingroup$ Neat derivation, I haven't seen that one before. $\endgroup$ – David Z Dec 21 '11 at 21:56
  • $\begingroup$ This derivation is in the grand canonical ensemble!! The canonical ensemble is defined as one in which the number of particles is fixed. This therefore does not answer the question at all. In fact, it completely misses the point, that BE statistics is a property of the GC ensemble. The main use of the GC ensemble is to calculate averages like this. $\endgroup$ – Mark Mitchison Aug 22 '16 at 20:01
  • $\begingroup$ Dear Mark, the "derivation" of the microcanonical distribution is simple - it's just a bracket in energy. The hard thing is the canonical one, which is what I did. When the chemical potential is nonzero, one may derive the grand canonical potential in this way, by using $\epsilon-\mu$ instead of $\mu$, as I said. If one wants a nonzero $\mu$ but a "non grand" canonical potential, then it's simply the derivation above combined with a bracket (allowed interval) on $N$. There is nothing nontrivial missing in my answer. Also, you're wrong that BE statistics is a property of an ensemble. $\endgroup$ – Luboš Motl Aug 25 '16 at 9:05
  • $\begingroup$ The BE statistics is a term describing the behavior of bosons (or any particle species in the boson category) in any physical conditions that include a possibly higher number of these particles. The BE statistics has implications for all the three distributions mentioned here. It is misleading to identify BE statistics just with the canonical or just with the grand canonical distribution. $\endgroup$ – Luboš Motl Aug 25 '16 at 9:06
3
$\begingroup$

The "Boltzmann distribution" is a classical function $e^{-\beta E}$ on the classical phase space. The classical phase space is not a quantum notion, but a classical one. The quantum state space for a collection of non-interacting Bosonic particles is given by first writing down a basis for one particle states, in terms of the collection $|k\rangle$ of all possible momenta (you can work in a periodic box where this collection is discrete, to avoid the complications of continuously labelled states). Then the most general basis state of the bosonic field is

$$|\psi\rangle = |n(k)\rangle$$

where $n(k)$ is an integer for each allowed $k$, telling you how many particles have this momentum. The canonical ensemble is the statement that each such state occurs with probability proportional to

$$ e^{-\beta E(n(k))} = e^{-\beta \sum_k E(k) n(k)} $$

subject to the constraint of fixed particle number

$$ \sum_k n(k) = N $$

In order to mathematically analyze the constraint of fixed particle number, it is most convenient to introduce a chemical potential, which acts as a Lagrange multiplier in the large $N$ limit, to fix the number of particles. But it is not strictly necessary, because you can consider all the n's restricted by the sum condition.

For fermions, $n(k)$ is either 0 or 1, but all the remaining relations are unchanged. The physics is entirely different--- for large $\beta$, the $N$ particles are forced to occupy the states where $E(k)<E_f$ where $E_f$ is adjusted to make the particle count come out right.

The description of multiparticle states in this way gives the Fock space. In Fock space for free particles, the classical expression $e^{-\beta E}$ gives the probability of being in any one of the orthogonal particle occupation number basis states.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.