Introductory Statistical Mechanics - counting number of microstates

[ausdreamer]

Homework Statement

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.

Homework Equations

-

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

 

[vela]

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 n_i=0, 1, ...

 

[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.

 

[vela]

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.

 

[paranormal]

Dear student,

Your attempt at finding the number of microstates for each value of q is a good start. However, there are a few things that need to be clarified.

Firstly, the values of j1 and j2 represent the number of energy quanta in each oscillator, not the number of particles. So for q = 1, there is only one possible microstate because both oscillators can only have one energy quantum each (j1 = j2 = 1). Similarly, for q = 3, there are only two possible microstates because the first oscillator can have 3 energy quanta (j1 = 3) and the second oscillator can have 0 or 2 energy quanta (j2 = 0 or 1).

Secondly, the table you have created does not account for the fact that the oscillators are distinguishable. This means that the order in which the energy quanta are distributed between the oscillators matters. For example, for q = 6, the possible microstates are (j1 = 6, j2 = 0) and (j1 = 0, j2 = 3), but your table only shows (j1 = 3, j2 = 0) and (j1 = 1, j2 = 2).

To account for these factors, we can use the formula for the number of microstates in a system with distinguishable particles and distinguishable energy levels:

g(q) = (q + N - 1)! / (q! * (N - 1)!)

where N is the number of energy levels (in this case, 2) and q is the total energy of the system.

Using this formula, we can find the correct values for g(q):

| q | g |
--------
| 1 | 1 |
| 3 | 3 |
| 5 | 6 |
| 6 | 10 |
| 7 | 15 |
. .
. .
--------

I hope this helps clarify your understanding of the problem. Keep up the good work in your studies of statistical mechanics!

Best regards,