In analogy with the previous problem, consider a system of identical spin$‐0bosons$trapped in a region where the energy levels are evenly spaced. Assume that $N$is a large number, and again let $q$be the number of energy units.

(a) Draw diagrams representing all allowed system states from $q=0$up to $q=6$.Instead of using dots as in the previous problem, use numbers to indicate the number of bosons occupying each level.

(b) Compute the occupancy of each energy level, for $q=6$. Draw a graph of the occupancy as a function of the energy at each level.

(c) Estimate values of $\mu$and $T$that you would have to plug into the Bose-Einstein distribution to best fit the graph of part(b).

(d) As in part (d) of the previous problem, draw a graph of entropy vs energy and estimate the temperature at $q=6$from this graph.

We need to draw the diagram representing all allowed system states from$q=0$up to$q=6$Instead of using dots.

Suppose we have a spin-$0boson$trapped in a region where the energy level is evenly spaced, ,the representation of the states from zero energy units $q=0$to six energy unit $q=6$is shown unit $q=0$is shown in the following figure, for $q=1$there is one state, for $q=1$there is one state, for $q=2$ there are two states, for $q=3$there are three states, for $q=4$there is five states, for $q=5$there are seven states and finally for $q=6$there are eleven states.

We need to draw a graph of the occupancy as a function of the energy at each level.

To find the occupancy of each level, we add the number in each level and divide it over 11 states, in the lowest level the sum is $19$, in the second-lowest level the sum is $8$, in the third-lowest the sum is $4$, in the fourth-lowest level the sum is $2$, in the fifth-lowest level the sum is$1$, in the sixth-lowest level the sum is $1$,and the sum of the seventh level is zero, so the occupancies are:

$\frac{19}{11}\frac{8}{11}\frac{4}{11}\frac{2}{11}\frac{1}{11}\frac{1}{11}00....$

the energy of the level equals the spacing between the levels $\eta$multiplied by the order of the level, so when we plot a graph between the dimensionless energy and the probability, we draw this between $\raisebox{1ex}{$\epsilon $}\!\left/ \!\raisebox{-1ex}{$\eta $}\right.$and $\overline{\eta }$, the graph will looks like this

We need to find Estimate values of $\mu$and $T$that you would have to plug into the Bose-Einstein distribution to best fit the graph of part(b).

For Bose Einstein distribution, the chemical potential equals the energy level when the occupancy goes to infinity, from the graph we see that the occupancy goes to zero at $\epsilon =0$,so the chemical potential is zero.Bose-Einstein distribution is given by:

role="math" localid="1649939548278" ${\overline{n}}_{BE}=\frac{1}{e^{{\displaystyle \raisebox{1ex}{$\left(\epsilon -\mu \right)$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}-1}$

set $\mu =0$then,

${\overline{n}}_{BE}=\frac{1}{e^{{\displaystyle \raisebox{1ex}{$e$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}}-1}$

I got a good match when $kT=2.2\eta$, therefore the Bose-Einstein distribution will be

${\overline{n}}_{BE}=\frac{1}{e^{{\displaystyle \raisebox{1ex}{$e$}\!\left/ \!\raisebox{-1ex}{$2.2\eta $}\right.}}-1}$

I plot this function and the points on the same graph,

We need to draw draw a graph of entropy vs energy and estimate the temperature at$q=6$ from this graph.

The entropy equals Boltzmann constant multiplies by the natural logarithm of the multiplicity, that is:

$\frac{S}{k}=\ln \left(\Omega \right)$

the multiplicities that correspond each energy units are shown in the following table:

$$\begin{gathered} \frac{q\Omega {\displaystyle \raisebox{1ex}{$s$}\!\left/ \!\raisebox{-1ex}{$k$}\right.}}{010} \\ 110 \\ 220.693 \\ 331.098 \\ 451.609 \\ 571.946 \\ 6112.397 \end{gathered}$$

a plot between $qand\raisebox{1ex}{$S$}\!\left/ \!\raisebox{-1ex}{$k$}\right.$looks like this:

The slope of the graph is :

$\frac{1∆S}{k∆q}=\frac{2.397-0.693}{6-2}=0.426$

but,

$q=\frac{U}{\eta }∆q=\frac{∆U}{\eta }$

thus,

$\frac{\eta ∆S}{k∆U}=0.426$

the temperature equals the difference in the energy $U$over the difference in the entropy that is

$T=\frac{∆U}{∆S}$

thus,

$\frac{\eta }{k}\times \frac{1}{T}=0.426$

$kT=2.35\eta$

and which is a rough agreement with the result of part (c).