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I'm not sure if this is the right place to ask a question about the Schrödinger equation, but I'll take my chances anyway. Basically, I would like to know how one can set up a potential function that represents a double-slit barrier and then solve the Schrödinger equation for this potential. Of course, according to classical optics, we will obtain an interference pattern, but it would be nice to see a solution entirely within the quantum-mechanical framework. I see this as a problem in mathematical physics, so hopefully someone could kindly provide me with some references.

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    $\begingroup$ Young's experiment in QM is standard textbook material, possibly covered in your course. It is, predictably, best worked out in detail in  Ch 8 §61 of S I Tomonaga's classic ISBN-13: 978-0720401066  Quantum Mechanics, Vol. 2: New Quantum Theory . $\endgroup$ Aug 1, 2019 at 16:05
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    $\begingroup$ Try quantum Mechanics by EUGEN MERZBACHER , third edition which I found online, page 583, section 5. Coherence, Interference, and Statistical Properties of the Field.. "we apply these concepts to a quantum mechanical analysis of optical interference phenomena, and especially the familiar two-slit interference experiment." $\endgroup$ Aug 1, 2019 at 16:25

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The simplest Schrödinger's equation for this problem I can think of has an infinitely thin potential barrier with infinite height in the locations from which the electron should be reflected. Schrödinger's equation in this limit is equivalent to the one with constant potential, but with a tricky boundary condition:

$$\begin{align} \psi(x,y,z,t)&=0\text{ whenever }(x=0\text{ and }|y|\in[0,y_1]\cup[y_2,+\infty)),\\ |\psi(x,y,z,t)|&<\infty\text{ everywhere} \end{align}$$

where $y_1$ and $y_2$ are coordinates of one of the slits, and the other slit is the opposite to the first one with respect to the $x$ axis.

Then this problem can be solved as in the case of wave scattering by a set of elliptical cylinders (with minor axis set to zero). Such scattering problems for plane waves are solvable in terms of expansions in Mathieu functions. An example of such expansion see in [1] (paywalled).

As a way of somewhat simplifying this problem, if we are only interested in far field, we can consider two parallel strips instead of slits, and then apply Babinet's principle to find the solution for the original problem with slits. This is the way you want to go if you follow [1] directly (since it deals with two strips).

[1]: H. Ragheb and M. Hamid, "Scattering by two conducting strips with parallel edges", Can. J. Phys., vol. 66, pp. 376-383, May 1988.

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  • $\begingroup$ I am wondering, why should $\psi$ be zero at the slits? $\endgroup$
    – Tob Ernack
    Aug 1, 2019 at 15:34
  • $\begingroup$ @TobErnack you're right. The correction is to invert the region to exclude the slits from the line instead. Fixed now. $\endgroup$
    – Ruslan
    Aug 1, 2019 at 15:44
  • $\begingroup$ Try quantum Mechanics by EUGEN MERZBACHER , third edition which I found online, page 583, section 5. Coherence, Interference, and Statistical Properties of the Field.. "we apply these concepts to a quantum mechanical analysis of optical interference phenomena, and especially the familiar two-slit interference experiment." $\endgroup$ Aug 1, 2019 at 16:09
  • $\begingroup$ Sorry yes I'll move it. $\endgroup$ Aug 1, 2019 at 16:25
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I will only sketch out how one would arrive at the actual equation, as I am rather lazy at the moment.

First, you will have to decide how many spatial dimensions you want to include: Obviously, a double-slit experiment won’t work in one dimension, so we need at least two (which can then be generalised to three rather easily).

You then basically have a two-dimensional scattering problem, for which one would need a certain potential:

Assume waves to arrive from $x = -\infty$ parallel to the $x$ axis, to be scattered at a potential $U(x,y)$. $U$ should be $0$ nearly everywhere except for a certain barrier, possibly located around the $y$ axis. Something like this should do:

$$U_1(x,y) = \theta(y-y_1) \cdot \theta(x) \cdot \theta(1-x) $$ $$U_2(x,y) = \theta(y_0-y) \cdot \theta(-y_0 - y) \cdot \theta(x) \cdot \theta(1-x) $$ $$U_3(x,y) = \theta(-y-y_1) \cdot \theta(x) \cdot \theta(1-x) $$ $$U(x,y) = u_0 \cdot( U_1(x,y) + U_2(x,y) + U_3(x,y) \quad u_0 \textrm{ large or }\infty, \quad y_{0,1} > 0 $$

$U_1$ is meant to describe a potential of height $u_0$ in the area where $0 < x < 1$ and $y > y_1$ (that is, the upper part of the barrier). $U_2$ describes a similar potential for $0 < x < 1$ and $- y_0 < y < y_0$, $U_3$ is similar to $U_1$ in the lower half-plane for $y < -y_0$.

As you can see, this potential lacks basically any symmetry one could remotely hope for. I would try a planar wave ansatz, but it is really not nice.

Maybe someone else has a better idea? :)

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I suggest that you consult the next reference about quantum potential, in which De Broglie–Bohm theory considers that famous experiment according to its pseudo classic point of view, in which the path of the particle is followed meaningly, explaining its intrinsical behaviour due to the fact that the derived potential $Q(\vec{x},t)$, has to do directly with the quantic conduct (since the center of mass of the particle), independently of the usual $U(\vec{x},t)$. In other words, Bohm considers that wave function can be written like $\Psi(\vec{x},t)=R(\vec{x},t)\cdot\exp{[\frac{i}{\hbar}\cdot S(\vec{x},t)]}$, where $R\doteq{\rho}^{0.5}$ represents the probability density and $S$ is the already known action. After introducing $\Psi$ into the Schrödinger equation, it is obtained that $Q(\vec{x},t)\propto\dfrac{\nabla^2 R}{R}$. At references of the first link, you can find a lot of articles, explaining us how considering Quantum Mechanics in this way, we can appreciate/follow almost deterministically the trajectory of the particle.

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