You can't do that for the infinite square well! If you did, the well would be infinitely deep! So in that case we set the zero of energy at the bottom of the well, and all energies are positive.

May 17, 2020 · Infinite Square well: Abrupt change in well's length 7 Why is the wavefunction not equal to 0 at the spike of the Dirac delta potential, but it is 0 for the infinite square well boundaries?

Aug 28, 2015 · A well known exercise in basic quantum mechanics is the sudden (diabatic) increase of the length of an infinite square well. Now consider a particle in an eigenstate of an infinite well that is suddenly decreased in length. At first glance, this seems troublesome since the eigenstate of the initial well cannot be expanded in the eigenstates of ...

Sep 18, 2016 · I'm studying Quantum Mechanics for one week and I don't understand one thing about wave function harmonics. If we have a particle in an infinite square well with an initial wave function $ \psi(x,0),$ we can calculate the time evolution($ \psi(x,t)$), knowing all the $ \psi_n(x)$ do a Fourier series.

Oct 21, 2013 · The important point is that the infinite square well is a well-motivated idealization. What matters is that there are physically sensible reasons for considering it. In particular, there is no need to consider delta functions, especially an infinite number of them. $\endgroup$

Solutions for the wavefunction in an infinite square well with a delta function barrier in the middle are easily found online (see here for an example). I am wondering what the wavefunction is for an infinite square well with a delta function well in the middle. The setup is the bottom of the infinite square well is defined to be zero energy.

Mar 8, 2019 · Why is the wavefunction not equal to 0 at the spike of the Dirac delta potential, but it is 0 for the infinite square well boundaries? 1 Why isn't the delta-function-potential solution the same as the infinite square well solution?

Mar 4, 2021 · But I tried to find the odd and even wavefunctions by shifting the wavefunction of a particle in an infinite box (which is not symmetric), and got different result. I'll be glad if someone can explain the difference. Here's my calculation:

Imaginary constant in simplification of the Infinite square well problem. 0.

Mar 13, 2016 · Your limit b → a appears fine: The left-hand side blows up, and so therefore the tangent on the r.h.side must too, so the cosine in the denominator must vanish, and bK = n π/2 for some odd n, while K is the unique "inner" wavenumber, K=nπ/2b, familiar from the infinite square well. Might choose to flip the sign of your energy as Trimok ...