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Suppose I have a single particle on a cubic lattice with lattice site length $a$ and $N$ lattice cells. The particle is feeling a potential $v(\boldsymbol{r})$ which is a continuous spatially dependent potential and a kinetic energy $t(\boldsymbol{p})=-\frac{p^2}{2m}$. One can then formulate the partition function as

$Z=\frac{1}{h^3}\int\mathrm{d}\boldsymbol{p} \mathrm{e}^{-\beta t(\boldsymbol{p})} \int \mathrm{d}\boldsymbol{r}\mathrm{e}^{-\beta v(\boldsymbol{r})}=\frac{1}{\Lambda^3}\int \mathrm{d}\boldsymbol{r}\mathrm{e}^{-\beta v(\boldsymbol{r})}=\frac{a^3}{\Lambda^3} \sum_i^N \mathrm{e}^{-\beta v(\boldsymbol{r_i})}$

Is there something wrong with the last step? Taking the limit $a\rightarrow0$ should give the partition function of a free particle feeling the potential $v$,i.e.

$Z=\frac{1}{\Lambda^3}\int \mathrm{d}\boldsymbol{r}\mathrm{e}^{-\beta v(\boldsymbol{r})}$

but here $a\rightarrow0$ gives just zero...

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  • $\begingroup$ Hint: How does $N$ vary with $a$? $\endgroup$ – lemon Jan 12 '17 at 15:39
  • $\begingroup$ $N$ should go to infinity then...is it possible to evaluate that limit? $\endgroup$ – Guiste Jan 12 '17 at 15:41
  • $\begingroup$ There's no need to. It is a standard result (it follows from the FTC) that those two terms converge as $a\to 0$. If you want to explicitly prove it then that becomes an exercise in real analysis and you'd be best asking it in the maths SE. $\endgroup$ – lemon Jan 12 '17 at 15:50
  • $\begingroup$ Just transform the sum into a Riemann sum, and there you go. $\endgroup$ – Ruslan Jan 12 '17 at 16:05
  • $\begingroup$ @lemon Ok that's totally right. Can you also help me how $Z$ would look like in the case of zero kinetic energy? $\endgroup$ – Guiste Jan 12 '17 at 16:09
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You are forgetting a -1, should be subtracted from your exponential and then evaluate the integral

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  • $\begingroup$ What do you mean by that? $\endgroup$ – Guiste Jan 16 '17 at 16:10

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