# Introductory Statistical Mechanics - counting number of microstates

1. Aug 3, 2011

### ausdreamer

1. The problem statement, all variables and given/known data

Consider a system composed of 2 harmonic oscillators with frequencies w and 2w respectively (w = omega). The total energy of the system is U=q * h_bar * w, where q is a positive negative integer, ie. q = {1, 3, 5, ...}.

Write down the number of microstates of the system for each value of q.

2. Relevant equations

-

3. The attempt at a solution

The energy of the first harmonic oscillator with frequency w is: E_1 = j1 * h_bar * w.
The energy of the second harmonic oscillator with frequency 2w is: E_2 = 2 * j2 * h_bar * w.

So now the total energy of the system is given by U = (j1 + 2j2) * h_bar * w = q * h_bar * w.

Say q = 1. So there's only1 microstate for this energy level because 1) my lecturer said that j1, j2 are integers and can represent the number of particles the harmonic oscillator, and by writing out a table for values of j1 and 2j2 we just get:

| j1 | 2j2|
----------
| 1 | 0 |
----------

Now say q = 3. By writing out the table of possible microstates we get the table:

| j1 | 2j2|
----------
| 3 | 0 |
| 1 | 2 |
----------

So for q=3, there are 2 possible microstates of the system. Repeating this a few more times, I get a table which looks like this:

(let g = number of microstates for energy q)

| q | g |
--------
| 1 | 1 |
| 3 | 2 |
| 5 | 3 |
| 6 | 4 |
| 7 | 5 |
. .
. .
--------

And so writing g(q) (number of microstates as a function of energy q) I get:

g(q) = CIELING(q/2)

However, this is apparently wrong according to my lecturer. Can someone see where I went wrong in my reasoning? Thanks

2. Aug 3, 2011

### vela

Staff Emeritus
I don't see why q would be confined to odd values. Also, your expression for the energy of the oscillators is incorrect. They should be
\begin{align*}
E_1 &= \hbar\omega(n_1 + 1/2) \\
E_2 &= 2\hbar\omega(n_2 + 1/2)
\end{align*}
where ni=0, 1, ....

3. Aug 3, 2011

### ausdreamer

Well the problem stated q as positive odd integers, and also my lecturer said we can ignore the 1/2h_bar * w term since all we're interested in is the difference in energy not E itself.

4. Aug 3, 2011

### vela

Staff Emeritus
Clearly if j1=0 and j2=1, you'd have q=2, which isn't odd. Either the problem is wrong or you're not accurately conveying the original problem statement.