1
$\begingroup$

Consider a single particle on a cubic lattice with $N$ sites each with a site length of $a$. The particle has no kinetic energy but only feels a spatially varying continuous potential $v(\boldsymbol{r})$.

One can directly write down the partition function as sum over possible states (particle at different sites):

$$Z=\sum_i^N\mathrm{e}^{-\beta v(\boldsymbol{r_i})}$$

But what happens if we want to evaluate the partition function from the continuous representation, i.e.

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

Does that mean that $\frac{a^3}{h^3}\int\mathrm{d}\boldsymbol{p} = 1$?


Edit: As suggested by AccidentalFourierTransform we could integrate over the first Brilloin zone, this would exactly be in agreement with the above statement:

$\int\limits_{-\infty}^\infty \mathrm{d} \boldsymbol{p} = \hbar^3\int\limits_{-\infty}^\infty \mathrm{d} \boldsymbol{k} = \hbar^3\int\limits_{-\pi/a}^{\pi/a}\mathrm{d}\boldsymbol{k} = \frac{h^3}{a^3}$

Could someone comment on that or give an explanation? Is this a valid approach? Can e.g. atoms sitting on the lattice sites be considered as waves? And why do we not need to integrate over higher momenta as the Brilloin zone?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.