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Statistical Physics - counting states

  • Thread starter LiorE
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  • #1
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1. Homework Statement [/b]
There are N 3-dimensional quantum harmonic oscillators, so the energy for each one is:
[tex]E_i = \hbar \omega (\frac{1}{2} + n_x^i + n_y^i + n_z^i)[/tex]. What is the total number of states from energy E_0 to E, and what is the density of states for E?

The Attempt at a Solution



In class they've shown that for a 1-D harmonic oscillator:

[tex] \Omega(E) = \frac{(\frac{E-E_0}{\hbar \omega} + N-1)!}{(\frac{E-E_0}{\hbar \omega})!(N-1)!} \delta E[/tex].

How did they get that? I Don't understand how it becomes (roughly) [tex] (\sum_i n_i + N) choose N [/tex]. And so I really have no idea how to generalize it to 3D (even though I guess it should be just the same).
 

Answers and Replies

  • #2
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In the one dimensional case:

E_0 = N hbar omega/2

Then

n_1 + n_2 + ...n_N = R (1)

where R = (E - E_0 )/(h omega)

So, you need to count the number of solutions of eq. (1). Then because we always specify a small energy interval delta E and count the number of states in that inteval, you need to multiply the answer by delta E/(h omega), because this is the number of possible values of R when E is specified with an uncertainty delta E.

Suppose I want to paint R objects with N colors. If n_j is the number of objects with color j, then all the n_j sum up to R, so any coloring is a solution of (1). Also, any solution of (1) defines a coloring. We can count the number of colorings by imagening a bog box with N compartments filled with the N types of paint. There are then N - 1 separation walls. We can then schematically denote any coloroing as a string:

oo|oooo|o|o...

where the "o" are the objects and the "|" are the separations between the compartments. The number of possible strings is thus given by:

Binomial[R + N-1,R] = (R+N-1)!/[R! (N-1)!]
 

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