# 11. The Microcanonical Ensemble

- Page ID
- 9160

The goal of equilibrium statistical mechanics is to calculate the diagonal elements of \(\hat{\rho}_{en}\) so we can evaluate average observables \(<A> =Tr\{\hat{A}\hat{\rho}_{en}\}=A\) that give us fundamental relations or equations of state. Just as thermodynamics has its potentials \(U\), \(A\), \(H\), \(G\) etc., so statistical mechanics has its ensembles, which are useful depending on what macroscopic variables are specified. We first consider the microcanonical ensemble because it is the one directly defined in postulate II of statistical mechanics.

In the *microcanonical *ensemble \(U\) is fixed (Postulate I), and other constraints that are fixed are the volume \(V\) and mole number \(n\) (for a simple system), or other extensive parameters (for more complicated systems).

## 1. Definition of the partition function

The ‘partition function’ of an ensemble describes how probability is partitioned among all the available microstates compatible with the constraints imposed on the ensemble.

In the case of the microcanonical ensemble, every microstate has the same energy and the same probability. According to postulate II, this probability is given by \(p_i=\rho_{ii}^{(eq)}=q/W(U)\) for each microstate “\(i\)” at energy \(U\). The microcanonical partition function is \(W(U)\). Using just this, we can evaluate equations of state and fundamental relations.

## 2. Calculation of thermodynamic quantities from W(U)

Example 11.1 |
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Consider again the model system of a box with \(M=\dfrac{V}{V_0}\) volume elements \(V_0\) and \(N\) particles of volume \(V_0\), so each particle can fill one volume elements. The particles can randomly hop among unoccupied volume elements to randomly sample the full volume of the box. This is a simple model of an ideal gas. As shown in the last chapter, \[W=\dfrac{M!}{(M-N)!N!}\] for identical particles, and we can approximate this, if \(M<<N\) by \[ W \approx \dfrac{1}{N!} \left(\dfrac{V}{V_0}\right)^N\] since \(M!/(M-N!) \approx M^ \[ S =k_B\ln W \approx S_0 + NK_B\ln \left(\dfrac{V}{V_0}\right)\] where \(S_0\) is independent of volume. This gives the volume dependence of the entropy of an ideal gas. Note that by taking the derivative \[ \dfrac{\partial S}{\partial V} = k_B\dfrac{N}{V} = \dfrac{P}{T}\] we can immediately derive the ideal gas law \(PV = Nk_BT = nRT\). |

Example 11.2 |
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The above model does not give us the energy dependence, since we did not explicitly consider the energy of the particles, other than to assume there was enough energy for them to randomly hop around. We now remedy this by considering the energy levels of particles in a box. The result will also demonstrate once more that increases so ferociously fast that it is equal to W with incredibly high accuracy for more than a handful of particles. Let the total energy \(U\) be randomly distributed among particles in a box of volume \(L^ , where i=1,2,3 are the x,y,z coordinates of particle #1, and so forth to i=3
The wavelengths l=L/2, L, 3L/2 , the energy for a bunch of particles in a box. W(U) is the number of states at energy U. Looking at the figure again, all the energy levels are “dots” in a 3N-dimensional cartesian space, called the “state space”, or “action space” or sometimes “quantum number space.” The surface of constant energy U is the surface of a hypersphere of dimension \(3N-1\) in state space. The reason is that the above equation is of the form constant = \(x^2+ y^2+ ...\) where the variables are the quantum numbers. The number of states within a thin shell of energy U at the surface of the sphere is . is the total number of states inside the sphere, which at a first glance would seem to be much larger than W(U), the states in the shell. In fact, for a very high dimensional hypervolume, a thin shell at the surface contains all the volume, so in fact, is essentially equal to \(W(U)\) and we can just calculate the former to a good approximation when N is large. If this is hard to believe, consider an analogous example of a hypercube instead of a hypersphere. Its volume is L Now back to our hyper . The (1/2) , where R~√U, so U is raised to the 3N/2 power, where N is the number of particles, 3 is because there are three modes per particle, and the 1/2 is because of the energy of a free particle depends on the square of the quantum number. Thus, where the constant . This equation relates the energy of an ideal gas to its temperature. 3 |

Example 11.3 |
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The Hamiltonian for this system in a magnetic field is given by , where the extra term at the end is added so the energy equals zero when all the spins are pointing down. At energy . This is our usual formula for permutations; the right side is in terms of Gamma functions, which are defined even when This formula has a potential problem built-in: clearly, when For large N, and temperature neither so low that , nor so high that , we can use the Stirling expansion ln
after canceling terms as much as possible. We can now calculate the temperature and obtain the equation of state In this equations, at . So even at infinite temperature the energy can only go up to half the maximum value, where This observation does not mean that it is impossible to get all the spins to point up. It is just not an equilibrium state at any temperature between T=0 and T=∞. Such non-equilibrium states with more spins up (or atoms excited) than down are called “inverted” populations. In lasers, such states are created by putting the system (like a laser crystal) far out of equilibrium. Such a state will then relax back to an equilibrium state, releasing a pulse of energy as the spins (or atoms) drop from the excited to the ground state. The heat capacity of the above example system is peaked at 4 so we can calculate thermodynamic quantities as input for thermodynamic manipulations. As we shall see in detail later (actually, we saw it in the previous example!), in any real system the heat capacity must eventually approach , where |

Example 11.4 |
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indeed agrees with the intuitive concept of temperature. Consider two baths within a closed system, so . If we know
for each bath, then at equilibrium because the maximum number of states is already occupied. For this to be true for any infinitesimal energy flow
At equilibrium, the temperatures are equal, fitting our thermodynamic definition that “temperature is equalized when heat flow is allowed.” |