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In the book Quantum Mechanics by N. Zettili (page 224, 2nd edittion), the potential $V(x)$ is defined as: $$V(x) = \begin{cases} 0 & ; \,x \lt 0 \\ V_0 & ; \, 0 \le x \le a \\ 0 & ; \, x \gt a \\ \end{cases}$$ For the case $E \gt V_0$, the wave functions in the regions: $$\psi(x) = \begin{cases} \psi_1(x)=Ae^{ik_1x}+Be^{-ik_1x} & ; \,x \le 0 \\ \psi_2(x)=Ce^{ik_2x}+De^{-ik_2x} & ; \, 0 \lt x \lt a \\ \psi_3(x)=Ee^{ik_1x} & ; \, x \ge a \\ \end{cases}$$

How does the case ($0 \le x \le a)$ of $V(x)$ correspond to the case ($0 \lt x \lt a$) of $\psi(x)$? How is the equality sign moves to the first and third regions? Why and how does the domain condition change?

Some books like Griffiths and also on Wikipedia, I don't find any equal sign there.
The three cases are given for: (i) $x \lt 0$ (ii) $0 \lt x \lt a$ (iii) $x>a$.

Don't we need to include $x=0$ and $x=a$ ? If not, then how can be the four boundary conditions applied on the wavefunction $\psi(x)$?
What am I missing? I am really confused.
TIA

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    $\begingroup$ In general one requires the wave function to be continuous everywhere. Hence by gluing the solutions in the different regions together you effectively remove this problem. $\endgroup$ – NDewolf Mar 12 at 14:56
  • $\begingroup$ Yes, I know it. But it doesn't clear my confusion. $\endgroup$ – raf Mar 12 at 15:12
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    $\begingroup$ It sounds like people were careless about the boundaries. Nothing deep here. Because the solution must be continuous, the functions on both sides of a boundary point give the same value. So it doesn't matter which function you use at the boundary. So people didn't worry about it. $\endgroup$ – mmesser314 Mar 12 at 15:21
  • $\begingroup$ Can it be written as: $$V(x) = \begin{cases} 0 & ; \,x \le 0 \\ V_0 & ; \, 0 \le x \le a \\ 0 & ; \, x \ge a \\ \end{cases} $$ $$\psi(x) = \begin{cases} \psi_1(x)=Ae^{ik_1x}+Be^{-ik_1x} & ; \,x \le 0 \\ \psi_2(x)=Ce^{ik_2x}+De^{-ik_2x} & ; \, 0 \le x \le a \\ \psi_3(x)=Ee^{ik_1x} & ; \, x \ge a \\ \end{cases}$$ ? $\endgroup$ – raf Mar 12 at 15:38
  • $\begingroup$ @raf That's not correct. You're assigning two different values to $V(x)$ both at $x=0$ and $x=a$, in that way it's not a function. You can't have all of the domains be closed, if one is closed the next one should be open. $\endgroup$ – user137661 Mar 12 at 15:46
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The boundary conditions demand continuity of the wave function and its derivative at the boundary. For example, for $x=0$ we need to have \begin{equation} \psi_1(x) = \psi_2(x), \psi_1'(x) = \psi_2'(x). \end{equation} This means that both(!!) $\psi_1(x)$ and $\psi_2(x)$ are defined at this point (and equal)!

This is a rather fine point, and the books seem rather sloppy on it, since it doesn't change much in practice.

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  • $\begingroup$ So, can it be written as: $$V(x) = \begin{cases} 0 & ; \,x \le 0 \\ V_0 & ; \, 0 \le x \le a \\ 0 & ; \, x \ge a \\ \end{cases} $$ $$\psi(x) = \begin{cases} \psi_1(x)=Ae^{ik_1x}+Be^{-ik_1x} & ; \,x \le 0 \\ \psi_2(x)=Ce^{ik_2x}+De^{-ik_2x} & ; \, 0 \le x \le a \\ \psi_3(x)=Ee^{ik_1x} & ; \, x \ge a \\ \end{cases}$$ ? $\endgroup$ – raf Mar 12 at 15:36
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    $\begingroup$ The potential should not have the overlapping conditions (otherwise $V_0$ has to be 0). $\endgroup$ – NDewolf Mar 12 at 15:37
  • $\begingroup$ @NDewolf I agree. $\endgroup$ – Vadim Mar 12 at 15:39
  • $\begingroup$ @raf For piece-wise functions you really should only have one condition for each x value $\endgroup$ – BioPhysicist Mar 12 at 15:40
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    $\begingroup$ @raf It won't really matter for the final solution. Aaron has a point that, once you use strict inequalities in the definition of the potential you should have only one wave function there - which is the opposite of my answer. It is a bit of a paradox, but it has to do with the fact that such rectangular barriers do not exist in real world. $\endgroup$ – Vadim Mar 12 at 15:44

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