Microcanonical ensemble for system of harmonic oscillators

[everyday847]

Homework Statement

A system consists of 3N (N >> 1) independent, identical, but distinguishable one-dimensional oscillators. This is relevant in that the atoms in a solid are sitting around their equilibrium positions. Assume that every atom constitutes an independent oscillator and all oscillators are characterized by the angular frequency ω. From quantum mechanics, the allowed energies of a 1D oscillator with angular frequency ω is given by

[tex]\epsilon = \hbar \omega/2 + 3\hbar \omega/2 + ... + (n +1/2) \hbar \omega + ...[/tex]

where \hbar \omega / 2 is the ground state energy of the oscillator. For the 3N-oscillator system, given that the total energy is given as follows

[tex]E = M\hbar \omega + (3/2) N\hbar \omega[/tex]

where M is an integer that describes the extent of excitations in the system and
can be taken to be M >> 1 .

a) Find density of states, Ω(M, N) corresponding to the total energy
in Eq. (2).
b) Using microcanonical ensemble, show that the internal energy
can be written as follows.

[tex]\bar{E} = 3N\left(\frac{\hbar \omega}{2} + \frac{\hbar \omega}{e^{\hbar \omega / kT}-1} \right)[/tex]

Homework Equations

Clearly the above definitions are relevant. Also relevant ought to be the fundamental equation of the microcanonical ensemble: that the energy derivative of \ln \Omega is equal to beta.

The Attempt at a Solution

Well, there are M excitations distributed among 3N degrees of freedom. But I can't really discuss the number of ways they can be distributed (thus enumerating the number of states) within the microcanonical ensemble: one needs the grand canonical ensemble because the number of excitations is fixed.

I can certainly handle the second part using a canonical ensemble because it's straightforward to take the CPF of a harmonic oscillator, take the 3N power, and pump. Microcanonical, not so much. Actually enumerating the states? Certainly not sure I can do that.

I'm actually fairly muddled here. Can I just assume that \Omega is a Gaussian? If so, can I simply then approximate that, since beta is approximately 3N/\bar{E}, \bar{E} is approximately 3N divided by the energy derivative of \ln \Omega? That all seems wrong.

 

[mark726]

Hello! First of all, it is important for a scientist to always approach a problem with a clear understanding of the relevant equations and concepts. In this case, the equations that are relevant are the ones you have listed: the fundamental equation of the microcanonical ensemble and the definition of the density of states.

To start, let's look at part (a) of the problem. The density of states, Ω(M, N), is defined as the number of states that are available for a given energy level. In this case, we are looking at a system with 3N independent, identical oscillators. Each oscillator can have a range of energies given by the equation you have provided. What we need to do is to find the number of ways that M excitations can be distributed among the 3N oscillators.

To do this, we can use the fundamental equation of the microcanonical ensemble, which states that the energy derivative of lnΩ is equal to beta. In this case, beta is given by 3N/ E, where E is the total energy of the system. So, we have:

d(lnΩ)/dE = 3N/E

Integrating both sides, we get:

lnΩ = 3N lnE + constant

Now, we need to find the constant. This can be done by using the fact that the total number of states in the system is given by the sum of all the states for each individual oscillator. So, we have:

Ω = Ω1 * Ω2 * ... * Ω3N

where Ω1 is the number of states for the first oscillator, Ω2 is the number of states for the second oscillator, and so on. Since all the oscillators are identical, we can say that Ω1 = Ω2 = ... = Ω3N. So, we have:

Ω = (Ω1)^3N

Substituting this into our equation for lnΩ, we get:

lnΩ = 3N lnE + 3N lnΩ1 + constant

Now, we can use the definition of the density of states to find the value of lnΩ1. This is given by:

lnΩ1 = ln(1 + e^(-beta * epsilon))

Substituting this into our equation, we get:

lnΩ = 3N lnE + 3N