# Boundaries of finite potential well

I'm currently studying the simplest models of potentials in QM, and I've run across an apparent inconsistency in my textbook: when describing a finite potential as a piecewise function, I've noticed the author defines the value of the potential twice in the boundaries, as follows:

$$V(x) = \begin{cases}V_0, \ \ |x|\leq a/2\\0, \ \ |x|\geq a/2\end{cases}$$

where $$a$$ is the width of the well and $$V_0<0$$. Does this make sense? Intuitively, you'd think you need to remove one of the two equalities. Otherwise, the particle is subject to two different values of the potential at the same point (I know this is incorrect wording in QM but mind me, this is not a formal question). If this is the case, which one of the two equalities should be removed? As usual, thank you for your time.

EDIT: Specified the sign of $$V_0$$

• You are correct that the question is mistakenly doubly defining those points inconsistently. However, it will be of no consequence, so you are free to choose as you like, as long as you choose symmetrically, to allow symmetry analysis to simplify your solution. On an unrelated note, if it is a potential well, then it is customary to have the insides of the well be of zero potential and the outsides to have positive $V_0$ potential, extending all the way to infinity both ways. The expression you have seems to be for a square barrier. Commented Feb 25 at 18:38
• When an author is sloppy about something so simple, it makes you wonder about their ability to do the complicated stuff correctly. Commented Feb 25 at 18:43
• Comment to the post (v2): The value of the potential on a measure-zero set usually doesn't matter. Commented Feb 25 at 18:48

$$V(x)={\begin{cases} 0~{~~~~~~~~~~~~~~\text{if }}x\lt -a/2 ~\lor x \gt a/2 \\ V_{0}~~~~~~~~~~~~~{\text{if }}x \gt -a/2 ~\land x\lt a/2 \\ \text{undefined}~{\text{if }}x=-a/2 \lor x=a/2 \end{cases}}$$