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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

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  • $\begingroup$ 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. $\endgroup$ May 21 '20 at 15:17
  • $\begingroup$ 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. $\endgroup$
    – Suraj S
    May 21 '20 at 15:46

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