I am reading an one-dimensional barrier problem. To evaluate the unknown constants $A,B,C,D$, the continuity of $\Psi(x)$ and $\frac{d\Psi}{dx}$ is used at $x=0$ where the potential has a discontinuity. I understand continuity of $\Psi$. But why is the derivative of $\Psi$ continuous at $x=0$? It has a step function which is not differentiable at $x=0$.
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$\begingroup$ physics.stackexchange.com/q/149001 This should help $\endgroup$– KenshinCommented Jun 26, 2017 at 8:04
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$\begingroup$ Shown here. quantummechanics.ucsd.edu/ph130a/130_notes/node141.html $\endgroup$– FarcherCommented Jun 26, 2017 at 8:05
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$\begingroup$ The question (v4) is essentially a duplicate of physics.stackexchange.com/q/19667/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Jun 26, 2017 at 9:53
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3$\begingroup$ Possible duplicate of Smoothness constraint of wave function $\endgroup$– Bill NCommented Jun 26, 2017 at 13:37
2 Answers
If the potential at some point is continuous or has finite discontinuty then from the Schrodinger equation one can show that the derivative of $\psi$ at that point is continuous.
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$\begingroup$ But it will cause infinite gradient. Which is a voilation of differentiation. $\endgroup$ Commented Jun 26, 2017 at 12:19
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$\begingroup$ @PrashantJoshi Intuitions always don't work. Please look at the proof in Griffith's book on quantum mechanics (2nd edition, page 83, Eqn. 2.121) for a mathematical proof of this. $\endgroup$– SRSCommented Jun 26, 2017 at 14:56
The short answer is that the first derivative of the wavefunction must be continuous in order for you to take the second derivative of the wavefunction. (Recall that the time independent, non-relativistic Schrödinger Equation involves two spatial derivatives.)
(I wrote the above because the links listed in the comments are nice and useful, but also very detailed; the above is a nice, non-rigorous physics-y argument that I believe captures the relevant point.)