Equation (9-27) gives the density of states for a system of oscillators but ignores spin. The result, simply one state per energy change ofbetween levels, is incorrect if particles are allowed different spin states at each level, but modification to include spin is easy. From Chapter 8, we know that a particle of spinis allowedspin orientations, so the number of states at each level is simply multiplied by this factor. Thus,

$D\left(E\right)=(2s+1)/h{\omega }_0$.

(a) Using this density of states, the definition$\text{Nh}{\omega }_0/(2s+1)={\epsilon }_1$, and

$N={\int }_0^{\infty }\mathcal{N}\left(E\right)D\left(E\right)dE$

calculate the parameterin the Boltzmann distribution (9-31) and show that the distribution can thus be rewritten as

$N(E{)}_{B\text{oltz}}=\frac{\epsilon }{k_{\text{B}}T}\frac{1}{e^{E/k_{\text{B}}T}}$

(b) Argue that if$k_{B}T>>\epsilon$,the occupation number is much less than 1 for all E.

Maxwell-Boltzmann statistics:

$\mathcal{N}\left(E\right)=\frac{B}{e^{E/k_{B}T}},$

(a)

To find B, we will make use of the normalization condition defined as:

$N={\int }_0^{\infty }\mathcal{N}\left(E\right)D\left(E\right)dE$ ……. (1)

Where , is the occupation number following the Maxwell-Boltzmann statistics:

$\mathcal{N}\left(E\right)=\frac{B}{e^{E/k_{B}T}},$ …….. (2)

The density of states for this system of harmonic oscillators with intrinsic spin:

$D\left(E\right)=\frac{2s+1}{\hslash {\omega }_0}$ ……. (3)

Put the equation (2) and(3) in equation (1), we get

$\begin{array}{rcl}N& =& {\int }_0^{\infty }\frac{B}{e^{E/k_{B}T}}·\frac{2s+1}{\hslash {\omega }_0}dE\\ & =& B\frac{(2s+1)}{\hslash {\omega }_0}{\int }_0^{\infty }{e}^{-E/k_{B}T}dE\\ & =& B\frac{(2s+1)}{\hslash {\omega }_0}\left({\left.e^{-E/k_{B}T}\left(-k_{B}T\right)\right|}_0^{\infty }\right)\\ & =& B\frac{(2s+1)}{\hslash {\omega }_0}\left(k_{B}T\right)\\ & & \end{array}$

Solving for B we obtain:

$B=\frac{N\hslash {\omega }_0}{2s+1}·\frac{1}{k_{B}T}$

According to the problem:

$\mathcal{E}=\frac{N\hslash {\omega }_0}{2s+1}$

Thus, the expression for B is:

$B=\frac{\mathcal{E}}{k_{B}T}$

Put this expression for B into eq. (2), the occupation number is thus:

$\mathcal{N}(E{)}_{\text{boltz}}=\frac{\mathcal{E}}{k_{B}T}\frac{1}{e^{E/k_{B}T}}$

(b)

Assuming that,$k_{B}T>>\epsilon$ the denominator of N becomes extremely large and the magnitude of$\mathcal{E}$ pales in comparison with the size of $k_{B}T$. As a result, the occupation number is much less than 1 for all possible values of E, and that $\mathcal{N}\to 0$holds true if$k_{B}T\to \infty$

In high temperature systems, the harmonic oscillators are uniformly distributed across individual energy levels. In such cases, the occupation number is effectively less than 1.

$B=\frac{\mathcal{E}}{k_{B}T}$