Consider a gas of $n$identical spin-0 bosons confined by an isotropic three-dimensional harmonic oscillator potential. (In the rubidium experiment discussed above, the confining potential was actually harmonic, though not isotropic.) The energy levels in this potential are $\epsilon =nhf$, where $n$is any nonnegative integer and $f$is the classical oscillation frequency. The degeneracy of level $n\text{is}(n+1)(n+2)/2$.

(a) Find a formula for the density of states, $g\left(\epsilon \right)$, for an atom confined by this potential. (You may assume $n>>1$.)

(b) Find a formula for the condensation temperature of this system, in terms of the oscillation frequency $f$.

(c) This potential effectively confines particles inside a volume of roughly the cube of the oscillation amplitude. The oscillation amplitude, in turn, can be estimated by setting the particle's total energy (of order $kT$) equal to the potential energy of the "spring." Making these associations, and neglecting all factors of 2 and $\mathrm{\pi }$and so on, show that your answer to part (b) is roughly equivalent to the formula derived in the text for the condensation temperature of bosons confined inside a box with rigid walls.

Number of particles =

Here,

$g\left(\epsilon \right)$=density of states,

$k$= Boltzmann's constant,

$T$= temperature.

The energy levels in the 3-dimensional harmonic oscillator,

$\epsilon =nhf$

Here,

$n$is any nonnegative integer and $f$is the classical oscillation frequency, and $h$is the Planck's constant

$n=\frac{\epsilon }{hf}$

The degeneracy level of $n$=

$g\left(n\right)dn=\frac{1}{2}(n+1)(n+2)$

Assuming $n>>1$$g\left(n\right)dn=\frac{n·n}{2}dn$,

$g\left(n\right)dn=\frac{n^2}{2}dn$

Differentiating the equation $\epsilon =nhf$on both side

$d\epsilon =dnhf$

$dn=\frac{d\epsilon }{h{f}^{\text{'}}}$

Substituting the value of $dn=\frac{d\epsilon }{h{f}^{\text{'}}}$and $n=\epsilon /hf$in the equation $g\left(n\right)dn=\frac{n^2}{2}dn$,

$g\left(\epsilon \right)d\epsilon =\frac{1}{2}{\left(\frac{\epsilon }{hf}\right)}^2\frac{d\epsilon }{hf}$

$=\frac{1}{2}\frac{{\epsilon }^2}{(hf{)}^3}d\epsilon$

Thus, Number of density states,$g\left(\epsilon \right)=\frac{1}{2}\frac{{\epsilon }^2}{(hf{)}^3}$

Substituting the value of $\mu =0,N_0=0,g\left(\epsilon \right)=\frac{1}{2}\frac{{\epsilon }^2}{{\left(h{f}^{\text{'}}\right)}^3}$

$N={\int }_0^{\infty }\frac{1}{2}\frac{{\epsilon }^2}{(hf{)}^3}\frac{d\epsilon }{e^{\frac{\epsilon }{kT}}-1}$

Let ,$x=\frac{\epsilon }{kT}\text{and}dx=\frac{d\epsilon }{kT}$

$=\frac{1}{2}{\left(\frac{k{T}_C}{hf}\right)}^3{\int }_0^{\infty }\frac{x^{2}dx}{e^x-1}$

${\left(\frac{k{T}_C}{hf}\right)}^3=\frac{2N}{{\int }_0^{\infty }\frac{x^{2}dx}{e^x-1}}$ (Equation-2)

As we know,

${\int }_0^{\infty }\frac{x^2}{e^x-1}dx=\Gamma \left(3\right)\zeta \left(3\right)$

$=2!(1.202)$

$=2.404$

Substituting the value of ${\int }_0^{\infty }\frac{x^2}{e^x-1}dx=2.404$in Equation-2

${\left(\frac{k{T}_C}{hf}\right)}^3=\left(\frac{2N}{2.404}\right)$

$T_C=\frac{hf}{k}{\left(\frac{N}{1.202}\right)}^{\frac{1}{3}}$

Thus, the condensate temperature for this system=$\frac{hf}{k}{\left(\frac{N}{1.202}\right)}^{\frac{1}{3}}$

We have,

The expression for potential energy

$E_{\mathrm{pot}}=\frac{C{L}^2}{2}$

$k{T}_C=\frac{C{L}^2}{2}$

Here,

$C$= spring constant

$L$= distance from the equilibrium position.

Angular frequency of system, $\omega =\sqrt{\frac{C}{m}}$

$C=m{\omega }^2$

Substituting the value of C in equation $k{T}_C=\frac{C{L}^2}{2}$

$k{T}_c=\frac{m{\omega }^2{L}^2}{2}$

$\frac{k{T}_C}{{\omega }^2}=\frac{m{L}^2}{2}$

$\frac{{\omega }^2}{k{T}_c}=\frac{2}{m{L}^2}$

Multiplying and dividing left side with $h^2$,

$\frac{{\hslash }^2{\omega }^2}{{\hslash }^{2}k{T}_C}=\frac{2}{m{L}^2}$

$\frac{(hf{)}^2}{{\hslash }^{2}k{T}_c}=\frac{2}{m{L}^2}$

Applying square on both sides

${\left(k{T}_C\right)}^2=(hf{)}^2{\left(\frac{N}{1.202}\right)}^{\frac{2}{3}}$

$k{T}_C=\frac{(hf{)}^2}{k{T}_C}{\left(\frac{N}{1.202}\right)}^{\frac{2}{3}}$

$={\hslash }^2\left(\frac{(hf{)}^2}{{\hslash }^{2}k{T}_c}\right){\left(\frac{N}{1.202}\right)}^{\frac{2}{3}}$

Substituting the value of $\frac{2}{m{L}^2}\text{=}\frac{(hf{)}^2}{{\hslash }^{2}k{T}_C}$

$k{T}_c={\hslash }^2\frac{2}{m{L}^2}{\left(\frac{N}{1.202}\right)}^{\frac{2}{3}}$

$={\left(\frac{h}{2\pi }\right)}^2\frac{2}{m{L}^2}{\left(\frac{N}{1.202}\right)}^{\frac{2}{3}}$

Substituting $V^{1/3}\text{=}L$

$k{T}_C=\frac{1}{\pi }\left(\frac{h^2}{2\pi m}\right){\left(\frac{N}{V}\right)}^{\frac{2}{3}}{\left(\frac{1}{1.202}\right)}^{\frac{2}{3}}$

$=0.318\left(\frac{h^2}{2\pi m}\right){\left(\frac{N}{V}\right)}^{\frac{2}{3}}$

Therefore, the expression obtained was roughly equal to the condensate temperature of bosons confined inside a box with rigid walls.