# SM:Parastatistics, grand canonical ensemble function

## Homework Statement

Hi
I am looking at the question attached.

Parts c and d, see below

## The Attempt at a Solution

First of all showing that $<N>$ and $<n_r>$ agree

I have $Z=\Pi_r z_r$, where $Z$ here denotes the grand canonical ensemble.
So therefore we have $\log(Z) = \sum\limits_r log (z)$ and since $<N>=\frac{1}{\beta}\frac{\partial}{\partial \mu} \log Z$ the calculation seems pretty trivial unless there are some subtleties I am missing. The only subtlety I can think of however is the requirement of convergence such that the derivative can be taken inside the summation and then the derivative of the summation= sum of the derivative of each term, please see below.

So I have:
$z_r= \sum\limits_{n_r=0}^{n_r=p} x_r^{n_r }$
so $<n_r> =\frac{1}{\beta}\frac{\partial}{\partial \mu} \log ( \sum\limits_{n_r=0}^{n_r=p} x_r^{n_r } ) = \frac{1}{\beta} \frac{\sum\limits_{n_r=0}^{n_r=p} n_r \beta x_r^{n_r }}{\sum\limits_{n_r=0}^{n_r=p} x_r^{n_r } }$ (1)

And

$<N>= \frac{1}{\beta}\frac{\partial}{\partial \mu} \sum\limits_r \log ( \sum\limits_{n_r=0}^{n_r=p} x_r^{n_r } ) = \frac{1}{\beta} \sum\limits_r \frac{\partial}{\partial \mu} \log ( \sum\limits_{n_r=0}^{n_r=p} x_r^{n_r } ) = \frac{1}{\beta} \sum\limits_r \frac{\sum\limits_{n_r=0}^{n_r=p} n_r \beta x_r^{n_r }}{\sum\limits_{n_r=0}^{n_r=p} x_r^{n_r } }$ (2)

Where the 2nd equality has assumed convergence over the summation $\sum\limits_r$.

QUESTION 1- 2ND EQUALITY USING CONVERGENCE
So the question asks to use (1) and $<N>= \sum\limits_r <n_r>$ to show that (2) is consistent with the latter. It is clear that the two agree. HOWEVER, I am unsure how to justify convergence over the summation of $r$. Here $r$ is a single particle state, as far as I'm aware, (although most likely physically unlikely) this can be infinite, an $n_r$ is the number of particles occupying a single $r$ state. And so the summation over $r$ is infinite. If I $E_r > \mu$ then this is an infinite summation of the logarithm of a summation which I know converges...but I don't think this really helps me out? (since i have $\sum\limits_r \log (\sum\limits_{n_r} x_r^{n_r })$ where $x_r = \exp^{-\beta(E_r-\mu)}$).

QUESTION 2- UNABLE TO USE GENERAL EXPRESSION TO RECOVER BOSON DISTRUBUTION
so for fermions $p=1$ for bosons $p=\infty$.
Looking at (1), the numerator multiplies $n_r$ (dropped down from the exponentiial derivative) and so clearly when $n_r=\infty$ this expression is dodgy/invalid- I am unsure what is wrong, however, with my above working.
The best I can seem to do is to recover the boson distribution from the expression at an earlier step, more generalised, which was:

$\frac{1}{\beta} \frac{\partial}{\partial \mu}( \log (\frac{1}{1-x_r}))$ it its easy to show that I attain $\frac{1}{x_r^{-1}-1}$ as expected. (where I have use infinite summation formula for this and the assumption $E_r > \mu$
However I imagine the question wants you to be verifying your final, simplified expression. I am unsure what is wrong with (1)...

the fermion distribution recovery is ok:
$<n_r> = \frac{1}{\beta} \frac{\sum\limits_{n_r=0}^{n_r=1} n_r \beta x_r^{n_r }}{\sum\limits_{n_r=0}^{n_r=1} x_r^{n_r } }$
this gives:
$\frac{1}{\beta}( 0 + \frac{\beta x_r}{1+x_r} )= \frac{1}{x_r^{-1}+1}$ as expected.

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I can't say for sure, but I suspect that for part (b) you are expected to evaluate the sum in $z_r= \sum\limits_{n_r=0}^{n_r=p} x_r^{n_r }$ and get a fairly simple result for $z_r$ in terms of $x_r$ and $p$. Then you can use this in later parts of the problem.