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In a infinite potential well of width $a$, an electron starts in the left half and at $t=0$; it is equally likely to be found at any point in that region.

To find the wavefunction at later times, we use the Fourier trick. But how could we proceed with this question when the initial wavefunction is discontinuous?

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  • $\begingroup$ Related: What is the spreading for rectangular wave packets? $\endgroup$ – Emilio Pisanty May 17 at 14:31
  • $\begingroup$ Contrary to popular belief induced by elementary books, continuity of wavefunction is never an issue, since a Lebesgue square-integrable function needn't be continuous. $\endgroup$ – DanielC May 17 at 15:22
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    $\begingroup$ It does not even make sense to ask whether a wave function is continuous, because a wave function is an element of $L^2$ and hence an equivalence class of functions, not a function. $\endgroup$ – WillO May 17 at 17:08
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No more than a potential well can be infinitely deep, or a pulley can be massless and frictionless. It is an idealized problem suitable for learning the subject.

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The minimum requirement for physically acceptable is that $$\int |\psi(x)|^2dx$$ be finite. The given function $$\psi(x,0)= \begin{cases} 2/L & 0\leq x\leq L/2 \\ 0& L/2< x\leq L \end{cases} $$ has a finite norm, Thus is acceptable.


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$$|\psi(t)\rangle=\mathcal{U}(t)|\psi(0)\rangle=e^{-iHt/\hbar}|\psi(0)\rangle$$ $$\psi(x,0)=\sum_n \phi_n(x) c_n$$ where $$\phi_n(x)=\sqrt{\frac{2}{L}}\sin\left(\frac{n\pi x}{L}\right)$$ You can find $c_n$ using $$c_n=\int \psi(x,0) \phi_n(x) dx$$ Once you have found this you can use the time evolution to find $\psi(x,t)$.

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    $\begingroup$ I think the OP is having issues with the SE depending on derivatives, but for a wavefunction as described above it has infinite / undefined derivatives. It might help to expand on that point. $\endgroup$ – BioPhysicist May 17 at 14:48
  • $\begingroup$ @BioPhysicist Is this correct now? $\endgroup$ – Young Kindaichi May 17 at 18:10
  • $\begingroup$ I never said anything was incorrect. And no, your edit did not address my comment at all. $\endgroup$ – BioPhysicist May 17 at 18:12
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The Hilbert space of square integral functions contains many functions on which the Hamiltonian is not defined, such as, indeed, discontinuous ones. However, when calculating the time evolution operator on those states where the Hamiltonian exists, one finds it has a unique extension to the entire Hilbert space. Look up stone's theorem: https://en.wikipedia.org/wiki/Stone%27s_theorem_on_one-parameter_unitary_groups although the mathematical details are quite hairy.

In essence, the above means that time evolution exists for all states, although not all states have an energy!

To make the connection with Fourier series explicit, sines and cosines are nice enough for the Hamiltonian to make sense on them, so one can calculate the time evolution by direct computation. Every function in the Hilbert space can be written as a Fourier series, even the discontinuous ones.

Any state $|\psi\rangle$ in the Hilbert space can be written as a Fourier sum: $|\psi\rangle=\sum_n a_n|f_n\rangle$, where the $a_n$ have to decay faster than $n^{-1/2}$, so that $\langle \psi|\psi\rangle = \sum_n |a_n|^2$ exists. However, for the Hamiltonian to make sense on a state, $\langle H\psi|H\psi\rangle = \sum_n n^4|a_n|^2$, which is clearly not true for all states in the Hilbert space. However, for the time evolution, one has $\langle U\psi|U\psi\rangle = \sum_n |a_n|^2$, so this exists for all states in the Hilbert space.

Note that the above also shows that "ugly" functions, such as discontinuous ones, have Fourier coefficients that decay slowly, and ones that have many derivatives have coefficients that decay faster.

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  • $\begingroup$ Of course, states without an energy cannot be created by any physically reasonable processen, hence the answer by mmesser314 $\endgroup$ – Anton Quelle May 17 at 17:34

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