# Issue with solving for the wavefunction of a simple infinite potential

For the potential given by $$V(x)=\left\{\begin{array}{ll}{\infty} & {x<0} \\ {-V_{0}} & {0

I am trying to solve for the wavefunction. However, I am stuck when trying to understand how to put some of the known information together. I know that the wavefunction must be $$0$$ for where the potential is infinite. An ansatz for the wavefunction between $$0$$ and $$a$$ is $$Ae^{ikx}+Be^{-ikx}$$. Therefore, because the wavefunction must be continuous at the $$0$$ boundary, I find that $$A=-B$$. The issue I have is that I thought the wavefunction needs to be continuously differentiable but setting the derivative of the wavefunction to be the same at $$0$$ gives me $$A=B$$ (meaning that the wavefunction is $$0$$ everywhere). I can see why this must be the case mathematically since sines and cosines never have a zero derivative where they also equal $$0$$. Is this just an artifact of approximating a potential with infinity? As in: do I not need the derivatives to match at the boundary here because it is just an idealised case?

• "I thought the wavefunction needs to be continuously differentiable" - Isn't the derivative of the wavefunction discontinuous at the infinite step in the potential? See, for example, section 2.5 in Griffiths "Quantum Mechanics". Also, see Discontinuity of wave function derivative – Alfred Centauri Aug 4 '19 at 11:39
• This is discussed in my Phys.SE answer here. – Qmechanic Aug 4 '19 at 11:41
• Check this (physics.stackexchange.com/q/490846) out too. – Paradoxy Aug 5 '19 at 10:12