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While analyzing stationary states and the infinite well potential, Griffiths claims in his book "Introduction to Quantum Mechanics" that $\Psi(x,0)$ and $\Psi(x,t)$ (i.e. the total wavefunctions) need not be continuous functions while the $\psi(x)$'s (i.e. the separable solutions) are continuous functions (pages 29 and 32 of 3rd. Edn). But if the former can be expanded linearly in terms of the latter, how is this possible? Where is the "discontinuity" allowed or how could it show up? Can anyone help me to clarify this?

I've found plenty of similar questions... so I apologize if this is repeated. However, I couldn't find any clear or concrete enough answer about the mathematical aspect of this claim.

Nicolás Budini
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  • This isn't a full answer, just commenting that mathematically it's quite possible to take the Fourier transform of a discontinuous function, which amounts to expanding a discontinuous function in a basis of continuous functions, in this case sinusoidal waves or complex exponentials. See, eg, https://math.stackexchange.com/q/2307392/. The key is that you need a superposition of an infinite number of basis functions. With a finite number you get the Gibbs phenomenon – Andrew Apr 13 '22 at 03:28
  • Possible duplicates: https://physics.stackexchange.com/q/262671/2451 and links therein. – Qmechanic Apr 13 '22 at 05:13
  • Thanks for the answers! As is evident there's some kind of discussion about whether or not a (total) quantum wavefunction should be (or not be) continuous. Anyway, which rules at last is how well quantum mechanics predicts the outcome of a real experiment... and it does quite well :) – Nicolás Budini Apr 13 '22 at 14:37

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