# Limit evaluation on Kronig Penney Model

I am a beginner to QM and was studying about the Kronig Penney model and understood the derivation of the following equation, $$\cos(ka) = \frac{P}{\lambda a} \sin(\lambda a) + \cos(\lambda a)$$ where $$\lambda = \sqrt{\frac{2mE}{\hbar^2}}$$ and $$k= \frac{2n\pi}{Na}$$ where $$'N'$$ is the no.of lattices considered and $$'a'$$ being the distance between adjacent lattice points

Now, I want to evaluate its limits explicitly when $$\hspace{1 mm} P \to 0 \hspace{2 mm}$$ and $$\hspace{1 mm} P \to \infty \hspace{1 mm}$$ to calculate the energy at these conditions by obtaining possible values for $$\lambda a$$. I understand that $$\hspace{1 mm} P \to 0 \hspace{1 mm}$$ would be similar to the situation of a free particle and $$\hspace{1 mm} P \to \infty \hspace{1 mm}$$ would be similar to that of a particle in a box. I thought of approaching this using Taylor series, expanding $$\frac{\sin x}{x}$$ at $$0$$.

$$\cos (ka) = P\sum_{i=0}^\infty (-1)^i \frac{(\lambda a)^{2i}}{(2i+1)!} + \sum_{i=0}^\infty (-1)^i \frac{(\lambda a)^{2i}}{(2i)!}$$ $$\cos (ka) = \sum_{i=0}^\infty(-1)^i (\lambda a)^{2i} \left[\frac{P}{(2i+1)!}+\frac{1}{(2i)!} \right]$$ The LHS is constrained i.e. $$\cos (ka) \in [-1,1]$$. So as $$P\to\infty$$, $$\lambda a$$ must converge to satisfy the constraint on LHS. So, does this mean $$(\lambda a)\to n\pi \hspace{1 mm}?$$ where $$n=0,1,2,3...$$ which implies $$\frac{\sin (\lambda a)}{\lambda a} \to 0$$. I am not so sure of how to apply these conditions. And I do not know what to do for $$P\to 0$$.

I will need some guidance regarding this

• Note that you have already taken the limit $V_0 \to \infty$ and $b \to 0$ to obtain the form of the dispersion relation you have. Moreover, since $P = M V_0 b a/\hbar$, the $P \to \infty$ limit would be equivalent to taking $a \to \infty$, and the limit would be a particle in a single delta-function potential, not in a finite-width box. May 21 '20 at 15:17
• I see. I get your point. But in the graph of $E(k)$, I observed that as $P\to\infty$, the energies tend to colapse horizontally flat giving discrete energy levels like that of a particle in a box. May 21 '20 at 15:46