Show that in the limit$\mathit{h}{\mathit{\omega }}_{\mathbf{0}}\mathbf{\ll }{\mathit{k}}_{\mathbf{\text{B}}}\mathit{T}$. Equation (9.15) becomes (9.28).

Theexpression for the average energyE of a system of harmonic oscillators is given as follows,

$\overline{E}=\frac{\hslash {\omega }_0}{e^{\hslash {\omega }_0/k_{B}T}-1}$

The expression for the firstorder Taylor expansion of the function$e^x$is given by:

$e^x\approx 1+x$

To carry out the approximation of E in the limit $k_{B}T>>\hslash {\omega }_0$, recall that the first-order Taylor expansion of the function$e^x$ that is a reasonable approximation for$x<<1$

Now, consider the exponential in the expression for E which is$e^{\hslash {\omega }_0/k_{B}T}$ . Consider$x=\hslash {\omega }_0/k_{B}T$ , and $x<<1$then the expression becomes as follows,

$\hslash {\omega }_0<<k_{B}T$

which is the assumption needed. So, the Taylor’s expansion becomes as follows,

$e^{\hslash {\omega }_0/k_{B}T}\approx 1+\frac{\hslash {\omega }_0}{k_{B}T}$

Substitute the above value in the expression for the average energy E,.

$\begin{array}{rcl}\overline{E}& \approx & \frac{\hslash {\omega }_0}{\left(1+\frac{\hslash {\omega }_0}{k_{B}T}\right)-1}\\ & \approx & \frac{\hslash {\omega }_0{k}_{B}T}{\hslash {\omega }_0}\\ & \approx & k_{B}T\\ & & \end{array}$

Thus, the given expression is verified.