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The evanescent wave would be penetrating the box so why don't we account for that even if it decays, there might be some part protruding the box with walls of infinite potential.

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  • $\begingroup$ "there might be some part protruding the box with walls of infinite potential." No. If the potential rises from 0 to infinity at the wall, the force, i.e. negative gradient, is also infinite and directed into the box. $\endgroup$ Apr 30, 2020 at 8:46
  • $\begingroup$ More on Schr. eq. & boundary conditions: physics.stackexchange.com/q/177851/2451 and links therein. $\endgroup$
    – Qmechanic
    Apr 30, 2020 at 11:03

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The reason is that if the value of potential is infinite on the walls of the box (infinitely deep box), the penetration is actually 0. You can see this by considering the case of a finite valued potential on the box walls or rectangular barrier of potential, which behaves in the same way. When you reach a classically forbidden region the transmission coefficient $T$ is a decaying exponential whose exponent is proportional to the energy difference and the barrier width. In the case of an infinite box, both of these quantities tend to infinity and the exponential becomes essentially zero. So the conclusion is that for a problem with an extremely deep potential well the penetration is negligible for the scales of the problem.

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    $\begingroup$ I had a problem understanding this, thinking about infinitely large box. But then I realized "box is infinite" here is meant to say something about potential at the walls rather than the box size. $\endgroup$ Apr 30, 2020 at 13:37
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Ok, the answer from a physical perspective was already given by others, however it can also be reasond from the mathematical perspective. One of the axioms of Quantum mechanics says that observables are discribed by self-adjoint operators. Note that a self-adjoint operator $A$ satisfies $\langle \psi | A\varphi\rangle = \langle A \psi | \phi\rangle$, it is symmetric. Now considering the momentum operator $$P: \psi \mapsto (-i)\psi^\prime$$ of a particle living on the intervall $I=[a,b]$ we find that for $\psi$ and $\varphi$ in the domain of $P$ that $$\langle \psi | P\varphi\rangle = \int_a^b dx \bar\phi(x) (-i \varphi^\prime(x)) = - i\int_a^b dx \bar\phi^\prime(x)) \varphi(x) + \left[\bar\psi(x)(-i\varphi(x)) \right]_a^b$$ by partial integration. We obtain the equality $$\langle \psi | P\varphi\rangle = \langle P \psi | \varphi\rangle -i \left[\bar\psi(x)\varphi(x) \right]_a^b$$ and thus for the momentum operator to be symmetric, a necessary condition for being self adjoint, we have to restrict to wave functions such that $$\bar\psi(b)\varphi(b) - \bar\psi(a)\phi(a) = 0$$ is ensured. We see that for symmetry $\psi(a)=\psi(b)$ is sufficient, however, a further more technical analysis for the self-adjointness of $P$ will end in $\psi(a)=\psi(b)=0$. Frederic Schuller made a really good introduction lecture to mathematically more suffisticated approach to quantum mechanics. You can find it in YouTube.

Anyway, I hope this helps you understanding the reason for this assumption from a mathematical point of view!

Cheers!

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  • $\begingroup$ Your treatment a priori restricts the domain of $\psi$ to $[a,b]$, which is wrong. The initial domain is $\mathbb R$, and it's only due to the potential that $\psi$ vanishes outside $(a,b)$. $\endgroup$
    – Ruslan
    Apr 30, 2020 at 14:38
  • $\begingroup$ Depends on how you use "wrong". From a mathematical point of view, what is an infinite potential? It's a physicists way of saying that the particle is confined to the interval $I$. One way to discribe this mathematically is to restrict the wave function to $I$. $\endgroup$ Apr 30, 2020 at 14:44
  • $\begingroup$ Not quite. This approach is assuming too much. It will bite you if you e.g. try to apply it to Dirac's equation: you'll miss the Klein paradox. $\endgroup$
    – Ruslan
    Apr 30, 2020 at 18:30
  • $\begingroup$ Ok, I get your concern. However, we are talking about non-relativistic quantum mechanics. I do not dare to say that this way works for relativistic systems too. Actually, I don't think so but I didn't consider it. However, it works for the system of concern as some kind of limiting system of the finite box potential. Anyway, your concern is right, we have to be really warily with such limits and I'm aware of this. $\endgroup$ Apr 30, 2020 at 20:16
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In a way you are right. A particle with infinite energy can still escape the box, by tunneling or even by propagation. For a finite value of the wave function at boundary an infinite term occurs and the equation requires an infinite kinetic energy to match it. The potential is taken to be infinite for educational reasons. This forces the wave function to be zero at the boundary and outside the box, for finite energy solutions, and simplifies the problem of solving the wave equation considerably.

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