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Introductory Statistical Mechanics  counting number of microstates 
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#1
Aug311, 03:37 AM

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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
Aug311, 11:52 AM

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PF Gold
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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 n_{i}=0, 1, .... 


#3
Aug311, 02:42 PM

P: 23

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
Aug311, 03:24 PM

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PF Gold
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Introductory Statistical Mechanics  counting number of microstates
Clearly if j_{1}=0 and j_{2}=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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