2
$\begingroup$

Okay,so I am learning about the double slit experiment done with electrons. I saw this picture, which shows the interference pattern being built up slowly with increasing number of electrons: enter image description here

I just wanted to confirm whether I have the correct understanding. The fact that the first image has a random distribution, shows that each electron interferes with itself and strikes a point on on the screen which would be dictated by the probability function. The interference pattern is the result of the same interference of many electrons and is a statistical property of many electrons.

Also, does this mean the electron travels as a wave, but then it obviously must strike as a particle since it hits a well defined spot on the screen?

$\endgroup$
3
$\begingroup$

The fact that the first image has a random distribution, shows that each electron interferes with itself and strikes a point on on the screen which would be dictated by the probability function.

Yes.

The interference pattern is the result of the same interference of many electrons and is a statistical property of many electrons.

Sort of. Each electron impact obeys (technically, samples) the probability distribution, which contains the interference. You need many hits for the probability distribution to become evident, but saying that the interference is exclusively a statistical phenomenon is slightly contentious.

Also, does this mean the electron travels as a wave, but then it obviously must strike as a particle since it hits a well defined spot on the screen?

Yes. There is a disparity in the evolution of quantum systems: wavelike, continuous, and linear ("unitary") when they're left 'by themselves' and discrete, particle-like, discontinuous, nonlinear, when they're 'measured'. The current state of affairs is not really satisfactory, as there isn't an ironclad rule to say which situations are 'systems by themselves' and which situations are 'measurements', so there's still much to understand here. The overall problem is known as the measurement problem, and while there's been some impressive progress recently, we're still far from anything like a satisfactory understanding of these matters.

$\endgroup$
  • $\begingroup$ I'm surprised you agreed with OP's statement that electron interferes with itself. Feynman repeatedly cautioned in his lectures to think like that. $\endgroup$ – user36790 Sep 24 '15 at 13:45
  • $\begingroup$ @user36790 : see the Aharonov-Bohm Wikipedia article where there's a Feynman quote re understanding. The "Aharonov-Bohm effect" was actually predicted by Ehrenberg and Siday in their 1949 semi-classical paper The Refractive Index in Electron Optics and the Principles of Dynamics. See figure 2 where the electrons are shown as plane waves going past the solenoid, retarded on one side and advanced on the other. Electron refraction and diffraction is for real. $\endgroup$ – John Duffield Sep 24 '15 at 16:33
  • 1
    $\begingroup$ @user36790 My own attitude is closer to Anna's last paragraph. I work with interfering electronic wavefunctions on a daily basis, so for me the electron and the wavefunction are pretty much the same thing (within reason), but maybe I should be a bit clearer on that. Note, however, that just because Feynman said something doesn't automatically make it true! $\endgroup$ – Emilio Pisanty Sep 24 '15 at 16:48
  • $\begingroup$ @Emilio Pisanty: Even I know that you can't be wrong! I just behaved like a bit pedantic; sorry for that:) Also, as Feynman explained it rigorously in his lectures in vol.III; that's why I referenced him & nothing else. If it is wrong to cite one, than I apologise for my action. $\endgroup$ – user36790 Sep 24 '15 at 17:25
  • $\begingroup$ @user36790 It's the phrasing "I'm surprised you disagree with Feynman's admonishments" that's a bit... I dunno. Anyways. $\endgroup$ – Emilio Pisanty Sep 24 '15 at 19:01
5
$\begingroup$

The fact that the first image has a random distribution, shows that each electron interferes with itself and strikes a point on on the screen which would be dictated by the probability function.

What a) tells us is that a single electron was fired at two slits and was deflected to a point at an angle from a straight projections from the slits. The same would happen if one threw a billiard ball at two slits with the analogous sizes to the diameter of the ball.

b) and c) tells us that the shooter kept mostly hitting edges.

It is d) that shows a clear interference pattern in a distribution that answers to the question "what is the probability if I throw electrons against a double slit of appropriate dimensions that it will hit (x,y) on the screen."

The conclusion is that an electron does not behave like a billiard ball, i.e. classical mechanics, it does not have the behavior of a classical billiard ball when scattered.

This behavior is described accurately by solutions of the quantum mechanical equation with the boundary problem "electron scattering off two slits". The square of these solutions, called wave functions, give us the probability distribution .

The statement "each electron interferes with itself" is misleading/confusing as far as the behavior of matter in dimensions where quantum mechanics prevails ( i.e. commensurate with h_bar). "The wave function describing the electron has interference terms passing the boundary of the two slits" is more correct. It is not a mass wave, nor an energy wave.

$\endgroup$
  • 1
    $\begingroup$ It might be worth emphasizing that the OP's image is from an actual double-slit electron diffraction experiment, reported in New J. Phys. 15 033018 (2013). $\endgroup$ – Emilio Pisanty Sep 24 '15 at 16:45
1
$\begingroup$

I just wanted to confirm whether I have the correct understanding. The fact that the first image has a random distribution, shows that each electron interferes with itself and strikes a point on on the screen which would be dictated by the probability function.

More or less. It isn't quite random, and I'd say it's dictated by the nature of electrons which is modelled using a probability function. But yeah, sounds to me like you've got it.

The interference pattern is the result of the same interference of many electrons and is a statistical property of many electrons.

As above. You could maybe chuck in the fact that the interference pattern is there as a result of each electron interfering with itself, and that you only see the pattern emerge when you send a lot of electrons through one after the other.

Also, does this mean the electron travels as a wave, but then it obviously must strike as a particle since it hits a well defined spot on the screen?

Yep. That's the crux of it. You can see something similar in the optical Fourier transform, see Steven Lehar's web page:

enter image description here

The electron wave goes through both slits, but when it's detected on the screen it gets converted into a dot. Then if you try to detect the electron at one of the slits, it gets converted into a dot and goes through that slit only, so the interference pattern disappears. You don't need any many-worlds multiverses to explain the dual slit experiment.

$\endgroup$
0
$\begingroup$

From the theory of light waves we know that for similiar experiment an interference pattern occurs when light quanta interacts with the system. Now with electrons there is an unique "wave" for that particular experiment that guides those electrons that hit the screen. Initially electrons must have equal speed and direction for that clear pattern to emerge.

We can assosiate that "wave" $\psi$ with any electron in an abstract sense, and just like with light we have $|\psi|^2$ that stands for intensity of the hits on the screen (probability distribution). So for every electron there is a wave that depends on the system (e.g. double slit experiment, hydrogen atom, ...)

$\endgroup$

protected by Qmechanic Sep 24 '15 at 13:12

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.