Mathematically, what happens at the screen of the 2-slit experiment?

Question

Mathematically, what happens at the screen of the 2-slit experiment? How would one represent the detecting screen at the receiving end? What boundary condition should be used there to represent the detecting screen?

When one solves the Schrodinger Equation, one needs to specify the boundary conditions. For the 2-slit experiment, there should be a potential with two small gaps representing the screen with the two slits, and there should be a potential representing the detector screen at the receiving end. In what way should one write these potentials, especially the one for the detector screen. That is, what would the boundary condition at the screen be?

There is a related question and answer at What is the wavefunction of the Young Double Slit experiment? but it does not address mathematically what happens at the detecting screen.

Answer 1

In modeling a quantum experiment, Schrodinger's equation is not the end of the story. We use Schrodinger's equation (with an appropriate choice of potential and boundary conditions) to represent the evolving state of the system, however the measurement process (in this case a detection screen) is completely absent from this part of the model.

The detection screen is represented instead by an appropriate choice of self-adjoint operator, which in this case would be a multiplication operator corresponding to the axis along which the screen is aligned.

In other words, we model the double-slit screen by an appropriate choice of potential and boundary conditions on the wavefunctions, and we model the detection screen by a separate self-adjoint operator. The two components of the model unite at the point where we calculate probabilities using the Born rule.

As an interesting aside, if you're wondering how to model a detection screen itself as a quantum system (rather than as an 'external' observable), join the club! This is an unsolved problem in quantum mechanics which goes under the name of the Measurement Problem. Its successful resolution would be worthy of a Nobel Prize, so you are unlikely to get that question answered here I'm afraid ;-)

Answer 2

The double slit result is a simplified version of the much more complicated set ups of particle physics experiments.In these experiments, the primary interaction region is modeled using quantum field theory, of point particles interacting in a summed series of feynman diagrams.

The outgoing particles leaving the interaction region towards the detectors are treated essentially as wavepackets, within an uncertainty region consistent with the Heisenberg uncertainty relations,HUP, i.e. modeled as a classical particle. The hits in the detectors are also treated macroscopically, again because the interaction region of the outgoing particle with the detector is way larger than the quantum mechanical constraints of the HUP.

The equivalent interaction region for the two slit experiments, where a quantum mechanical solution is necessary is the "electron scattering off two slits", with the specific geometrical bounds. The screen is the detector.

The interactions of the electrons on the screen , the measured points, are way over the bounds of the HUP and the electron wavepacket is well approximated by the point in (x,y) space.

If you are worrying that one should in principle write out one mathematical expression for the whole experiment, i.e.that there is an entanglement of the initial electron wavepacket impinging and going through the slit plane with the same wavepacket hitting the screen, the density matrix formalism helps clearing up this: when dimensions become macroscopic with respect to the HUP the off diagonal elements of the entanglement of the screen particles with the incoming electron wavepacket are zero, the phases are lost, exactly because the involved dimensions are macroscopic, as gauged by using the HUP.

So viewed as one experimental setup, the quantum mechanical interaction is at the level of the two slits where the distances have been chosen so as to be commensurate with the de broglie wavelength of the electron. The screen is just a detector.

Of course there is a quantum mechanical interaction where the electron hits the screen: atoms become ionized and the energy is distributed in a many body way. The density matrix for this part has nondiagonal elements, i.e. quantum mechanical phases, only with the wavefunctions of the local scattering. The two slit wavefunction has no off diagonal element in the density matrix with the wavefunctions of the local screen scatters.

Answer 3

Mathematically, what happens at the screen of the 2-slit experiment?

Well, an electron has a wave nature, that's why we can diffract electrons. It isn't a point particle. But when we detect it on the screen, we see a dot. So, mathematically what happens at the screen is comething akin to a Fourier transform. See Steven Lehar's web page on that. Pay special attention to the optical Fourier Transform where the incident wave is converted into something pointlike:

How would one represent the detecting screen at the receiving end? What boundary condition should be used there to represent the detecting screen?

Sorry, I don't know.

When one solves the Schrodinger Equation, one needs to specify the boundary conditions. For the 2-slit experiment, there should be a potential with two small gaps representing the screen with the two slits...

Take a look at Ehrenberg and Siday's 1949 paper The Refractive Index in Electron Optics and the Principles of Dynamics*. An electron is depicted as plane waves going past a solenoid in figure 2. (Edit : not unlike the Gaussian beam Neurofuzzy referred to on the other question.) So I imagine one could emulate this and represent one electron as a plane waves going through two slits, much like water waves.

and there should be a potential representing the detector screen at the receiving end. In what way should one write these potentials, especially the one for the detector screen. That is, what would the boundary condition at the screen be?

Sorry, I don't know. But note that at the screen we have an interaction between two electromagnetic extended entities, and the combination looks pointlike.

There is a related question and answer at What is the wavefunction of the Young Double Slit experiment? but it does not address mathematically what happens at the detecting screen.

I took a look. Note that you send one electron through the apparatus, and you get one dot on the screen. See the Wikipedia article. So you might think the electron is pointlike. However you know the light going through a lens isn't pointlike. And when you see the pattern build up, you know the electron isn't pointlike either. After all, it's the Schrödinger wave equation, not the Schrödinger point-particle equation.

• I've got a non-paywall link to the full paper somewhere. I'll dig it out tonight.