(a) Find the chemical potential and the total energy for distinguishable particles in the three dimensional harmonic oscillator potential (Problem 4.38). Hint: The sums in Equations5.785.78and5.795.79can be evaluated exactly, in this case−−no need to use an integral approximation, as we did for the infinite square well. Note that by differentiating the geometric series,

$\frac{\mathbf{1}}{\mathbf{1}\mathbf{-}\mathbf{x}}\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{x}}^{\mathbf{n}}$

You can get

$\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left(\frac{x}{1-x}\right)\mathbf{=}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\left(n+1\right){\mathit{x}}^{\mathbf{n}}$

and similar results for higher derivatives.

(b)Discuss the limiting caserole="math" localid="1658400905376" ${\mathit{k}}_{\mathbf{B}}\mathit{T}\mathbf{}\mathbf{\ll }\sout{\mathbf{h}}\mathit{\omega }$.
(c) Discuss the classical limit,role="math" localid="1658400915894" ${\mathit{k}}_{\mathbf{B}}\mathit{T}\mathbf{}\mathbf{\gg }\sout{\mathbf{h}}\mathit{\omega }$, in the light of the equipartition theorem. How many degrees of freedom does a particle in the three dimensional harmonic oscillator possess?

The chemical energy per mole of a substance is its "chemical potential." Gibbs free energy is defined here as chemical energy, and the substance can either be a single, pure substance or a system of several substances.

(a)

From problem 4.38:

$$\begin{gathered} E_n=\left(n+\frac{3}{2}\right)\sout{h}\omega ,n=0,1,2,...;d_n \\ =\frac{1}{2}\left(n+1\right)\left(n+2\right) \end{gathered}$$

From Eq. 5.103

$$\begin{gathered} n\left(\in \right)=e^{-\left(e-\mu \right)/k_{B}T} \\ {N}_n=\sum _{n=0}^{\infty }{N}_n=\frac{1}{2}{e}^{\left(\mu -\frac{3}{2}\sout{h}\omega \right)/kT}\sum _{n=0}^{\infty }\left(n+1\right)\left(n+2\right)x^n,\mathrm{Where},x\mathit{=}{e}^{\mathit{-}\sout{\mathit{h}}\mathit{\omega }\mathit{/}{\mathit{k}}_{\mathit{B}}\mathit{T}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\frac{\mathit{1}}{\mathit{1}\mathit{-}\mathit{x}}\mathit{=}\mathit{\sum }_{\mathit{n}\mathit{=}\mathit{0}}^{\mathit{\infty }}{x}^{\mathit{n}} \\ \mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{\Rightarrow }\mathit{}\frac{\mathit{x}}{\mathit{1}\mathit{-}\mathit{x}}\mathit{=}\mathit{\sum }_{\mathit{n}\mathit{=}\mathit{0}}^{\mathit{\infty }}{x}^{\mathit{n}\mathit{+}\mathit{1}} \end{gathered}$$

Now,

$$\begin{gathered} \Rightarrow \frac{d}{dx}\left(\frac{x}{1-x}\right)=\sum _{n=0}^{\infty }\left(n+1\right)x^n \\ \Rightarrow \frac{x}{{\left(1-x\right)}^2}=\sum _{n=0}^{\infty }\left(n+1\right)x^n \\ \sum _{n=0}^{\infty }\left(n+1\right)\left(n+2\right)x^n=\frac{2}{{\left(1-x\right)}^3}.N=e^{\mu /k_{B}T}\frac{1}{{\left(1-e^{-\sout{h}\omega /k_{B}T}\right)}^3} \\ \mu =k_{B}T\left[InN+3\mathrm{In}\left(1-{\mathrm{e}}^{-\sout{\mathrm{h}}\mathrm{\omega }/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\right)+\frac{3}{2}h\omega \mathit{/}{k}_{\mathit{B}}T\right] \end{gathered}$$

In order to find energy, we have to calculate:

$E=\sum _{n=0}^{\infty }{N}_n{\epsilon }_n=\frac{\sout{h}\omega e^{\left(\mu -3\sout{h}\omega /2\right)/k_{B}T}}{2}\sum _{n=0}^{\infty }\left(n+\frac{3}{2}\right)\left(n+1\right)\left(n+2\right)x^n,x=e^{\sout{h}\omega /2/k_{B}T}$

We use the same trick as before:

$$\begin{gathered} \sum _{n=0}^{\infty }\left(n+1\right)\left(n+2\right)x^n=\frac{2}{{\left(1-x\right)}^3}/.x^{3/2}\sum _{n=0}^{\infty }\left(n+1\right)\left(n+2\right)x^{n+3/2} \\ =\frac{2x^{3/2}}{\left(1-x\right)}/\frac{d}{dx}\sum _{n=0}^{\infty }\left(n+\frac{3}{2}\right)\left(n+1\right)\left(n+2\right)x^{n+1/2} \\ =\frac{3x^{1/2}\left(1+x\right)}{{\left(1-x\right)}^4}\Rightarrow \sum _{n=0}^{\infty }\left(n+\frac{3}{2}\right)\left(n+1\right)\left(n+2\right)x^n \\ =\frac{3\left(1+x\right)}{{\left(1-x\right)}^4} \end{gathered}$$

So, energy is:

$$\begin{gathered} E=\frac{\sout{h}\omega e^{\left(\mu -3\sout{h}\omega /2\right)k_{B}T}}{2}\frac{3\left(1+e^{\sout{h}\omega /2k_{B}T}\right)}{\left(1-e^{\sout{h}\omega /2k_{B}T}\right)},e^{\left(\mu -3\sout{h}\omega /2\right)k_{B}T}=N{\left(1-e^{\sout{h}\omega /2k_{B}T}\right)}^3 \\ \Rightarrow E=\frac{3}{2}N\sout{h}\omega \frac{1+e^{\sout{h}\omega /2k_{B}T}}{1-e^{\sout{h}\omega /2k_{B}T}} \end{gathered}$$

(b)

For$k_{\mathrm{B}}T\ll \sout{\mathrm{h}}$we have$e^{\sout{-h}\omega /2k_{B}T}\to 0$

So energy is $E=\frac{3}{2}\sout{h}\omega N$. . All particles are in the ground state.

(c)

For$k_{\mathrm{B}}T\gg \sout{\mathrm{h}}$we have$e^{\sout{h}\omega /2k_{B}T}\to 0\approx 1-\frac{\sout{h}\omega }{k_{B}T}$. So energy is$E=3NkBT$

Equipartition theorem says that energy of a system in thermal equilibrium is equal to:

$E=n_f\frac{N{k}_{B}T}{2}$

Where$n_f$ is number of degrees of freedom. In our case, particle has 6 degrees of freedom (3 kinetic and 3 potential).