Consider a Bose gas confined in an isotropic harmonic trap, as in the previous problem. For this system, because the energy level structure is much simpler than that of a three-dimensional box, it is feasible to carry out the sum in equation 7.121 numerically, without approximating it as an integral.*

(a) Write equation 7.121 for this system as a sum over energy levels, taking degeneracy into account. Replace $Tand\mu$with the dimensionless variables $t=kT/hf\text{and}c=\mu /hf$.

(b) Program a computer to calculate this sum for any given values of $tandc$. Show that, for $N=2000$, equation 7.121 is satisfied at $t=15$provided that $c=-10.534$. (Hint: You'll need to include approximately the first 200 energy levels in the sum.)

(c) For the same parameters as in part (b), plot the number of particles in each energy level as a function of energy.

(d) Now reduce $t$to 14 , and adjust the value of $c$until the sum again equals 2000. Plot the number of particles as a function of energy.

(e) Repeat part (d) for $t=13,12,11,\text{and}10$. You should find that the required value of $c$increases toward zero but never quite reaches it. Discuss the results in some detail.

Equation for Bose-Einstein condensation :

$N=\sum _{\text{all}s}\frac{1}{e^{\frac{\left({\epsilon }_1-\mu \right)}{kT}}-1}$ (Equation-1)

Sound waves behave almost like light waves. Each made of oscillation has a set of equally spaced energy levels, with the limit of energy equal to:

$\epsilon =nhf$

The relation between Fermi temperatures and Fermi energy is.

$t{\epsilon }_F=kT$

Here, $k$is the Boltzmann constant:

The relation between the chemical potential and Fermi energy is,

$\mu =c{\epsilon }_f$

we get,

$N=\sum _{\text{alls}}\frac{1}{e^{\frac{\left(nhf-c{c}_F\right)}{t{t}_F}}}\left(\text{here,}{\epsilon }_F=hf\right)$

$=\sum _{\text{alls}s}\frac{1}{e^{\frac{\left(n{e}_p-c_{r_r}\right)}{t{\epsilon }_{r_r}}-1}}$

$=\sum _{\text{alls}s}\frac{1}{e^{\frac{(n-c)}{t}-1}}$

Degeneracy of a level'$n\text{'}$is

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

Thus the Bose-Einstein distribution function:

$N=\sum _0^{\infty }\frac{1}{2}(n+1)(n+2)\frac{d\epsilon }{e^{\frac{(\epsilon -\mu )}{kT}-1}}$

we have,

$h\omega =1\left(\text{(}h\text{bar*omega *}\right)$

$kt=15\left(\text{(*Boltzmann constant*temperature *}\right)$

$N=2000(*\text{number of atoms}*)$

$\text{normsum}\left[N_{-}\right]:=sum\left[dist\left[c\right]*\frac{(n+1)*(n+2)}{2},\{m,0,sumlim\}\right]$

In an isotropic harmonic trap of 3-dimensional box, occupational number of particles can be attained by using this mathematic relation.

$\text{Listplot}\left[\text{occnum,plot range->}\right\{0,300\left\}\right];\left({}^{*}\text{plot occupation numbers}*\right)$

$\text{Listplot}\left[\text{Table}\right\{n,occ[n,-10.536,15]\},\{n,0,175\left\}\right]$

From the graph we can see that the occupancy peak is at height of. At this temperature the number of particles in the ground state is unity.

The following mathematical instruction accustomed plot the occupancy graph for $t=14$. At this temperature the occupancy within the state is tiny and roughly less than $1.5$.