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What is the physical significance of having a wavefunction with a discontinuous derivative? Sometimes we impose continuity of the derivative in order to find values of unknown parameters and it's helpful. But sometimes it's not required and so it seems arbitrary.

Qmechanic
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Some Student
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2 Answers2

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The relation of the first derivative of the wave function is meant to keep the flux of propability constant across the interface. Therefore, if the effectice masses are different across a heterojunction ($m_1$ at left and $m_2$ at right), the contiuity of flux leads to the following relation across a heterojunction at $x = x_0$: $$ \frac{1}{m_1}\frac{\partial \Psi}{\partial x}\Big\vert_{x_0^-} = \frac{1}{m_2} \frac{\partial \Psi}{\partial x}\Big\vert_{x_0^+} $$

If there is a $\delta(x-x_0)$ potential at the boundary, it will involve in the continuity of flux throught the Shr$\ddot{\text{o}}$dinger equation.

ytlu
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Continuity of wave function is essential for finding momentum, momentum is the first derivative of wave function.

hat
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  • And? What then? – Physiker Apr 25 '21 at 13:13
  • Taking the first derivative of a wavefunction does not give you momentum. $\hat P|\psi\rangle\neq p$ – BioPhysicist Apr 25 '21 at 13:14
  • I meant, momentum operator, consists of derivative, and for that wave function should be continuous. – hat Apr 25 '21 at 14:32
  • Well... first the question is about the continuity of the derivative of the wavefunction, but your answer discusses continuity of the wavefunction. Regardless. the first derivative of the wavefunction need not be continuous. It's possible to take the Fourier transform of functions with a discontinuous first derivative -- see for instance Exercises 13-15 of https://web.stanford.edu/~hca/c276autumn2009/NA1_Fourier.pdf. – Andrew Apr 28 '21 at 17:17