When calculating the canonical partition sum, we had the following: $$ Z_\text C = \sum_{\vec p} \sum_{\vec x} \exp(-\beta H(\vec p, \vec x)) $$

Now, since $\vec p$ and $\vec x$ are pretty much continuous when taking the limit $V \to \infty$, we can write it as integrals. However, there are certain prefactors now.

The following makes sense to me: Look at the wavenumber of the $x$-direction $k_x$. If we (without loss of generality) assume that the volume is a cube, the system has the volume $V = l^3$. The smallest wave number which is possible by quantum mechanics is $k_0 = 2\pi / l$. All other wavenumbers are $n k_0$ then. Then I can write: $$ \sum_{k_x} f(k_x) = \sum_{n = 1}^\infty f(n k_0) $$

Now I can write the sum over $n$ with an integral, like so: $$ \int \mathrm dn \, f(n k_0) $$

To get back an integration over $k_x$, I do a substitution, giving me a prefactor: $$ \frac{l}{2\pi} \int \mathrm dk_x \, f(k_x) $$

With momentum $p = \hbar k$, it will be $l/(2\pi\hbar)$. Doing this for all three dimensions, I get $V / (2\pi\hbar)^3$, which is the same in my textbooks.

But what happens when I do another sum/integration in space? With the same quantisation, $n x_0$, with $x_0 = l / N$, where $N$ is just some large number would give me another prefactor $N / l$. For three dimensions, this is $N^3 / V$.

Now I do have this huge number $N$ in the formula, which does not belong there. If I omit this, and just integrate over $x$, $Z_\text C$ will have the dimensions of a volume, but it should have no dimension.

Where is the mistake?


1 Answer 1


In QM, the sum over discrete state is always equivalent in the classical limit to the integral over $x,p$ divided by the volume of the basic "cell" which is $(2\pi\hbar)^N$, regardless of the basis you choose etc. It may be shown in many ways.

If $N$ means the number of lattice sites per length of the physical system, such an $N$ always cancels. If the number of states along the $x$ axis increases by some factor, the number of states along the momentum $p$ axis decreases by the same factor.

To actually see that, you must understand that a quantized (in some units integer) $x$ means that $p$ takes values in some interval only (because $p$ is effectively periodic) and vice versa. The length of the periodicity is always inversely proportional to the spacing of the dual variable.


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