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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

raf
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    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. – NDewolf Mar 12 '20 at 14:56
  • Yes, I know it. But it doesn't clear my confusion. – raf Mar 12 '20 at 15:12
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    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. – mmesser314 Mar 12 '20 at 15:21
  • Can it be written as: $$V(x) = \begin{cases} 0 & ; ,x \le 0 \ V_0 & ; , 0 \le x \le a \ 0 & ; , x \ge a \ \end{cases} $$

    $$\psi(x) = \begin{cases} \psi_1(x)=Ae^{ik_1x}+Be^{-ik_1x} & ; ,x \le 0 \ \psi_2(x)=Ce^{ik_2x}+De^{-ik_2x} & ; , 0 \le x \le a \ \psi_3(x)=Ee^{ik_1x} & ; , x \ge a \ \end{cases}$$ ?

     – raf Mar 12 '20 at 15:38
  • @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. –  Mar 12 '20 at 15:46
  • In this case, the solution of the Schrödinger equation is not going to depend on the way you choose which of the domains of the definition of $V$ is closed and which is open, you just have to make sure you make it consistent. –  Mar 12 '20 at 15:50
  • so, these four situations are considered the same? @SV – raf Mar 13 '20 at 03:16
  • Strictly speaking no. Example #1 is not consistent in the domains of $\psi$ and $V$. But since $\psi$ is continuous examples 1-3 are physically equivalent. Example #4 is just plain wrong, you discarded the values of the functions at the discontinuity (i.e., the union of the domains is not $\mathbb R$) –  Mar 14 '20 at 22:03
  • Example #1 was taken from Zettili's book and example #4 from Wikipedia. How about this way: $$V(x) = \begin{cases} 0 & ; ,x \lt 0 \ V_0 & ; , 0 \lt x \lt a \ 0 & ; , x \gt a \ \end{cases} \ \psi(x) = \begin{cases} \psi_1(x)=Ae^{ik_1x}+Be^{-ik_1x} & ; ,x \le 0 \ \psi_2(x)=Ce^{ik_2x}+De^{-ik_2x} & ; , 0 \le x \le a \ \psi_3(x)=Ee^{ik_1x} & ; , x \ge a \ \end{cases}$$? – raf Mar 15 '20 at 03:07
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    #6 $$V(x) = \begin{cases} 0 & ; ,x \lt 0 \ V_0 & ; , 0 \le x \le a \ 0 & ; , x \gt a \ \end{cases} \ \psi(x) = \begin{cases} \psi_1(x)=Ae^{ik_1x}+Be^{-ik_1x} & ; ,x \le 0 \ \psi_2(x)=Ce^{ik_2x}+De^{-ik_2x} & ; , 0 \le x \le a \
    \psi_3(x)=Ee^{ik_1x} & ; , x \ge a \ \end{cases}$$?     – raf Mar 15 '20 at 03:16
  • Related: https://physics.stackexchange.com/a/754354/247642, https://physics.stackexchange.com/a/715950/247642 – Roger V. Mar 21 '23 at 13:12

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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.

Roger V.
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  • So, can it be written as: $$V(x) = \begin{cases} 0 & ; ,x \le 0 \ V_0 & ; , 0 \le x \le a \ 0 & ; , x \ge a \ \end{cases} $$

    $$\psi(x) = \begin{cases} \psi_1(x)=Ae^{ik_1x}+Be^{-ik_1x} & ; ,x \le 0 \ \psi_2(x)=Ce^{ik_2x}+De^{-ik_2x} & ; , 0 \le x \le a \ \psi_3(x)=Ee^{ik_1x} & ; , x \ge a \ \end{cases}$$ ?

     – raf Mar 12 '20 at 15:36
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    The potential should not have the overlapping conditions (otherwise $V_0$ has to be 0). – NDewolf Mar 12 '20 at 15:37
  • @NDewolf I agree. – Roger V. Mar 12 '20 at 15:39
  • @raf For piece-wise functions you really should only have one condition for each x value – BioPhysicist Mar 12 '20 at 15:40
  • At which region/s should be the x=0 and x=a included for $V(x)$ and $\psi(x)$? – raf Mar 12 '20 at 15:41
  • @AaronStevens, then how we write $\psi_1(0)=\psi_2(0)$ ? – raf Mar 12 '20 at 15:42
  • @raf $\psi_1$ and $\psi_2$ can be defined for all $x$ technically as functions. But when you explicitly write out a piece-wise function you need to make sure only one choice goes to each $x$ value – BioPhysicist Mar 12 '20 at 15:44
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    @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. – Roger V. Mar 12 '20 at 15:44