I'm solving a quantum mechanics problem for a certain set of potentials and arrived to the following bound state condition for even solutions: $\tan[k(a-b)]=\cot(kb)$, which turns out to be $\tan[k(a-b)]=\tan(\frac{\pi}{2}-kb)$. My question is very simple and i'm kind of embarrassed to ask: Why is $k(a-b)=\frac{\pi}{2}-kb+(n-1)\pi$? Other example, for $\tan[k(a-b)]=\tan(-kb)$: $k(a-b)=-kb+n\pi$, where $n=1,2,....$ Thank you in advance!

  • $\begingroup$ It depends on the physical nature of $a$ and $b$. For example in bound states, the value of $k$ is not arbitrary and in turns governs the first quantum state and so on. Plot the graphs for specific values of $a$ and $b$ to see how the solutions behave. $\endgroup$ Jan 14 '17 at 20:56
  • $\begingroup$ Im assuming that k(a-b) is a variable y. Therefore I only have to plot tan(y) $\endgroup$
    – RicardoP
    Jan 14 '17 at 21:21

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