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Introductory Statistical Mechanics - counting number of microstates

  1. Aug 3, 2011 #1
    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. jcsd
  3. Aug 3, 2011 #2

    vela

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    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, ....
     
  4. Aug 3, 2011 #3
    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.
     
  5. Aug 3, 2011 #4

    vela

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    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.
     
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