Continuous Approximation

The goal of this subsection is to show that the error introduced by approximating Eq. (206) according to Eq. (207) is negligible when L y sufficiently large. For simplicty, we show this for a 1-dimensional problem, where

$$\sum_{K_x} \frac{1}{e^{\beta( \frac{\hbar^2}{2m}K_x^2 -\mu )} +1} = \sum_{K_x} f(K_x) \Delta K,$$ (225)

with

$$f(K_x)=\frac{L_x}{\pi} \frac{1}{e^{\beta (\epsilon(K_x) -\mu)} +1},$$ (226)

a decreasing function of $K_x$ and

$$\Delta K = \frac{\pi}{L_x},$$ (227)

Remember, that $K_x=K_x(n_x)$ is a function of the quantum number $n_x$, as defined by Eq. (204), where $n_x=1,2,...$. The discrete sum, introduced by Eq. (225), can be represented by the following diagram,

The diagram shows that,

$$\sum_{n_x=1}^{\infty} f(K_x(n_x)) \Delta K \leq \int_0^{\infty} d... ...x)) \Delta K = \sum_{n_x=1}^{\infty} f(K_x(n_x)) \Delta K + f(K_x(0)) \Delta K,$$ (228)

since $f(K_x(n_x)$ is a decreasing function of $K_x$. Therefore,

$$0 \leq \int_0^{\infty} d K f(K) - \sum_{n_x=1}^{\infty} f(K_x(n_x... ...f(K_x(n_x)) - \sum_{n_x=1}^{\infty} f(K_x(n_x)) }_{f(K_x(0))} \right] \Delta K.$$ (229)

Eq. (229) shows that the discrete sum becomes equal to the integral in the limit when $\Delta K \rightarrow 0$. This is in fact the case when $L \rightarrow \infty$, since

$$\Delta K=K_x(n_x) -K_x(n_x-1)= n_x \frac{\pi}{L} - (n_x -1) \frac{\pi}{L}$$

and therefore $\Delta K \rightarrow 0$ when $L \rightarrow \infty$.