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Considering the time-independent Schrödinger equation, I can see for a finite potential, why the wavefunction has to be continuous, I can also see why the first derivative of the wavefunction is discontinuous when there is a Dirac delta potential, but I can't see what is forcing the wavefunction to be continuous in the case of an infinite potential (mathematically as well as physically).

Can someone explain this to me?

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    $\begingroup$ Continuous with respect to what? No wave function is required to be continuous with respect to spatial coordinates. In general, it is not defined on points (or other sets of measure zero). $\endgroup$ – yuggib Nov 13 '15 at 14:12
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    $\begingroup$ Related: physics.stackexchange.com/q/19667/2451 , physics.stackexchange.com/q/186323/2451 and links therein. $\endgroup$ – Qmechanic Nov 13 '15 at 14:42
  • $\begingroup$ @Qmechanic in your linked question the answer you gave, is similar to what I have seen before, can that be generalised to include the case of the Dirac delta? $\endgroup$ – Quantum spaghettification Nov 13 '15 at 16:40
  • $\begingroup$ @yuggib In some cases, wave function might not be continuous in space. But in infinite well case, since a derivate in space appears in the Schrodinger's equation, it is continuous in space. $\endgroup$ – user43796 Nov 14 '15 at 5:40
  • $\begingroup$ @user43796 A derivative in space in the equation does not necessarily yield continuity of the solution. There are plenty of non-smooth solutions to PDEs, especially when (as here) they are considered in the distributional sense (because $L^2$ is a subspace of $\mathscr{S}'$, the tempered distributions). $\endgroup$ – yuggib Nov 14 '15 at 9:01
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Without any attempt for rigorous arguments: If the wave function is not continuous across the delta potential, then evaluating the second derivative of it (because of the kinetic term in the Hamiltonian) would give you the derivative of a delta function, which is not a delta function, and the Schroedinger equation is not satisfied at the position of the potential.

Similarly, if the potential is a step function, not only the wave function is continuous at the step, but also so is its derivative, otherwise the Schroedinger equation is not satisfied at the step.

In short: it is important to see that the continuous conditions for the wave function and/or its derivative are required by the Schroedinger equation itself, and not something you take in from elsewhere.

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In any quantum mechanical setting, regardless of the type of the potential, one imposes the requirement that the wave-function $\Psi(x)$ (i.e., the solution to the Schrödinger equation) be $C^1$-smooth (continuous with well-defined 1st derivative) or at least, continuous, $C^0$ [see here for a definition], in position. Notice that a discontinuous wave-function will lead to a physically absurd situation, since the probability $P(x)$ of finding the particle in a position interval $[x, x + dx]$, is defined in terms of the absolute value of the wave-function $\Psi(x)$: $$P(x) = |\Psi(x)|^2 ~ ;$$ it is obvious that if the wave-function is not continuous at some point, the probability of finding the particle in the interval containing $x$ will not be defined! Therefore, to have a well-defined physical picture, the wave-function should be at least continuous in its domain.

Having a well-defined first derivative leads to a proper behaviour of other observable physical quantities like the particle current: $$ j = \frac{\hbar}{2m \mathrm{i}}\left(\Psi^* \frac{\partial \Psi }{\partial x}- \Psi \frac{\partial \Psi^* }{\partial x} \right) ~.$$ [Note that this is the simplest expression for the current, but that suffices for the current issue.]

For a detailed discussion, consult, e.g., Ballentine, L. E. “Quantum mechanics: a modern development”. World scientific (1998), sections 4.4 and 4.5.

Correction: The interpretation of $P(x)$ was corrected due to a comment by @ACuriousMind.

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    $\begingroup$ That is absolutely not true. The wavefunction can as well be discontinuous without any problem for the definition of probabilities. In general, wavefunctions are technically meaningless when evaluated at a point, as they are $L^2(\mathbb{R}^d)$ functions. $\endgroup$ – yuggib Nov 13 '15 at 14:44
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    $\begingroup$ Eigenstates of the momentum operator are not wavefunctions, and have no probabilistic/physical interpretation. They are just a convenient mathematical tool. And please, name me one experimentally measurable physical quantity that requires the evaluation of the wavefunction at a given position. And btw, the $j$ can be perfectly understood with derivatives in a weak (distributional) sense. If you want references, you may take a look at any book on the mathematical methods of quantum mechanics. $\endgroup$ – yuggib Nov 13 '15 at 15:36
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    $\begingroup$ In addition to what yuggib says, your $P(x)$ is not the probability, it is a probability density which has to be integrated over a region of space to give a probability. The probability to find any particle at a location $x$ is zero because integrating a density over a point - which is a zero measure set - is just zero. The requirement that this be a well-defined probability density is precisely the wavefunction being in $L^2$. $\endgroup$ – ACuriousMind Nov 13 '15 at 16:05
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    $\begingroup$ I think that yes, maybe your sources are not the best, for they have led up to misleading conclusions. When you say to use a "normalized eigenstate", you are actually using something that is not an eigenstate anymore (because you have to, roughly speaking, integrate it with a square integrable function). There are no eigenstates of the momentum (or position) operator that are also square integrable, i.e. that qualify as wavefunctions. Of course you may restrict just to continuous wavefunctions, but that restriction is unnecessary. $\endgroup$ – yuggib Nov 13 '15 at 16:58
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    $\begingroup$ I will say more, no one guarantees that your quantum evolution would preserve the space of continuous functions, i.e. even starting with a smooth function you may end up with a non-smooth one. An introduction to quantum mechanics with a sufficient level of rigor is given e.g. by Hall in his book "Quantum theory for mathematicians". $\endgroup$ – yuggib Nov 13 '15 at 17:01

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