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A particle is defined by a wave function, $Be^{-2x}$ for $x<0$ and $Ce^{4x}$ for $x>0$. For the wave function to be continuous at $x=0$, $B=C$. A wave function must be continuous for it to be valid.

However, another condition we were taught and I can find all over the internet, is that the first spatial derivatives of the wave function must also be continuous. For this to be true at $x=0$, $B$ cannot equal $C$. Therefore why is this a valid wave function?

Another problem: $\psi = iC/3 \times (x-2)$ from $x=2,5$ and $-iC/5 \times (x-10)$ from $x = 5,10$. else $\psi = 0$. Again, the derivative is discontinuous at $x=5$ since the lines have different slopes. Still, this example is considered a valid wave-function by the text. (Solid State Electronic Devices, 7th ed., 2.6(c) and 2.7)

Can we simply ignore isolated points of discontinuity?

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The derivative of $\psi(x)$ is continuous only where there is no infinite discontinuity in the potential. Examples of situations where $\psi'(x)$ is not continuous include a $\delta(x)$ potential and both ends of an infinite well.

The quick argument follows by integrating $\psi''(x)$ over a small region: \begin{align} -\frac{\hbar^2}{2m}\int_{-\epsilon}^\epsilon \psi^{''}(x)dx &=-\frac{\hbar^2}{2m}\left(\psi'(\epsilon)-\psi'(-\epsilon) \right)\\ &= \int_{-\epsilon}^\epsilon \,dx\, (V(x)-E)\psi(x)\, . \end{align} Thus, if the integrand on right hand side remains finite in the interval, the integral on the right goes to $0$ as $\epsilon\to 0$ and hence on the left hand side goes to $0$, implying continuity.

If as stated there is an infinite discontinuity in the integrand, then the integral on the right may give a non-zero value, which in turns gives a discontinuous $\psi'(x)$.

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I remember this exercise well, because the lecturer of my course in quantum mechanics gave us this homework assignment without us knowing anything about distributions.

The reason that the two cases are different cannot be understood properly from physics text books. The difference between $H = - \frac{\hbar^2}{2m} \Delta + V(x)$ for a “nice” potential (e. g. a smooth, bounded function with bounded derivatives) and, say, the case where $V(x) = \lambda \, \delta(x)$ is quite subtle.

The crucial notion here is that of a domain of an operator. One necessary condition for a vector $\varphi \in \mathcal{H}$ to be in the domain of an operator $H$ is that $H \varphi \in \mathcal{H}$ also needs to be in the domain. However, there may be additional conditions such as boundary conditions. So for example, you can have several mathematically distinct operators with the same operational prescription (say, $-\Delta$) but that differ on the domains they are defined on. One case that physicists are familiar with is the wave equation with Dirichlet and von Neumann boundary conditions (which could model a closed or semiopen pipe, for example) — the spectrum, i. e. the vibrational modes, will be different.

So the difference between $H$ with a “nice” and a $\delta$ potential lies in the domain of the respective Schrödinger operators: for a “nice” potential, the domain of $H$ is the domain of $-\Delta$. And in dimension one this domain consists of absolutely continuous functions whose derivative exists almost everywhere.

The mathematical definition of the Schrödinger operator with $\delta$-potential is more subtle, and what you actually do is define the free Schrödinger operator $H_0 = - \frac{\hbar^2}{2m} \Delta$ on a domain that differs from that of $-\Delta$. This domain contains the jump of derivative due to the $\delta$-potential in its definition. This justifies the computation by ZeroTheHero.

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