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

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?

  • 3
    $\begingroup$ "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 $\endgroup$ – Alfred Centauri Aug 4 '19 at 11:39
  • 1
    $\begingroup$ This is discussed in my Phys.SE answer here. $\endgroup$ – Qmechanic Aug 4 '19 at 11:41
  • $\begingroup$ Check this (physics.stackexchange.com/q/490846) out too. $\endgroup$ – Paradoxy Aug 5 '19 at 10:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.