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Q. 7.73

Expert-verifiedFound in: Page 325

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

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.

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

(b) The condensation temperature for this system =$\frac{hf}{k}{\left(\frac{N}{1.202}\right)}^{\frac{1}{3}}$

(c) The expression obtained was roughly equal to the condensate temperature of bosons confined inside a box with rigid walls.

Number of particles =

$N={\int}_{0}^{\infty}g\left(\epsilon \right)\frac{d\epsilon}{{e}^{\frac{(\epsilon -\mu )}{kT}-1}}$ (Equation-1)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\xb7n}{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 ,$x=\frac{\epsilon}{kT}\text{and}dx=\frac{d\epsilon}{kT}$

$N={\int}_{0}^{\infty}\frac{1}{2}\frac{{\left(xk{T}_{c}\right)}^{2}}{(hf{)}^{3}}\frac{dx\left(k{T}_{c}\right)}{{e}^{x}-1}$$=\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.

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