Details regarding the high temperature limit of the partition function

In summary, the steps leading towards converting the sum to an integral require that we assume that the energies of the system are polynomial in the quantum number $n$, that the degeneracies are polynomial in $n$, and that the energies are indexed by a quantum number $q$. This works well for large $n$, but might not be accurate for small $n$. Approximating the sum by an integral allows us to use a well-approximated approximation for $Z$.
  • #1
EE18
112
13
Homework Statement
Consider a quantum system where the energy of the ##n##th level, ##\epsilon_n \geq 0##, is some polynomial function of ##n## of degree ##q## (##q > 0##), and the degeneracy of this level, ##g_n##, is some other polynomial function of n of degree ##r## (##r \geq 0##). In this limit one may approximate the sum in the partition function by an integral. Furthermore, the asymptotic behaviour is dominated by the highest-order terms in the polynomials. Show this.
Relevant Equations
See work below.
My main question here is about how we actually justify, hopefully fairly rigorously, the steps leading towards converting the sum to an integral.

My work is below:
If we consider the canonical ensemble then, after tracing over the corresponding exponential we get:
$$Z = \sum_{n=0}^\infty g_ne^{-\beta \epsilon_n}$$
where we have assumed, per the question, that we have a quantum system where the eigenenergies are indexed by some quantum number ##n## which takes on nonnegative integer values. Further, we are told to assume that the energies ##\epsilon_n## are polynomials in ##n## of degree ##q## and the degeneracies ##g_n## are polynomials in ##n## of degree ##r##:
$$\epsilon_n = a_qn^q+...+a_0$$
and
$$g = b_rn^r+...+b_0.$$
Now define, on looking at the arguments of the exponentials in ##Z## [Question, why does this work for ##n=0## case? Do we have to separate that from the rest of the integral if we are being rigorous? In that case, should the integral below start at 1?],
$$x_n := ({\beta\epsilon_n})^{1/q} = (\beta a_qn^q)^{1/q}\left(1+\frac{1}{q}(a_{q-1}n^{-1}+...+a_0n^{-q}) + ...\right)$$
where in the last step we've used a binomial expansion [Why is this expansion justified in this limit? I see why it works for large ##n##, but shouldn't this not be good for small ##n##? E.g. ##n = 1##?. For large ##n## this becomes
$$x_n \approx (\beta a_qn^q)^{1/q} \sim \beta^{1/q}n$$
where ##\sim## denotes an asymptotic scaling relation (i.e. the proportionality is independent of $n$).

From our work above, it's apparent that (on multiplying and dividing terms by ##\beta^{1/r}## and using ##g_n \sim n^r## for large ##n##) that (NB that ##x_n^q## means ##(x_n)^q## where ##x_n## is related to ##n## as above)
$$Z = \sum_{n=0}^\infty g_ne^{-\beta \epsilon_n} \sim \sum_{n=0}^\infty \frac{\beta^{r/q}}{\beta^{r/q}}n^re^{-x^q_n} = \frac{1}{\beta^{r/q}}\sum_{n=0}^\infty x_n^re^{-x^q_n}.$$
We now move to approximate this sum by an integral. Now define the increments
$$\Delta x := x_{n+1} - x_n = \beta^{1/q}.$$
For ##\beta## very small these becomes very small, so that the Riemann sum becomes well-approximated by the corresponding integral:
$$Z \sim \frac{1}{\beta^{r/q}}\sum_{n=0}^\infty x_n^re^{-x^q_n} = \frac{1}{\beta^{r/q}}\sum_{n=0}^\infty \frac{\Delta x}{\beta^{1/q}} x_n^re^{-x^q_n} \approx {\beta^{-\frac{r+1}{q}}}\int_{0}^\infty dx \, x^re^{-x^q} \sim \beta^{-s}$$
for ##s = \frac{r+1}{q}## and where in the last step we've noted that the integral is just some number (constant with respect to temperature or $\beta$).Any help with justifying the bolded steps would be much appreciated!
 
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  • #2
Just bumping this if possible
 

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