- #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!
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!
