3
$\begingroup$

I'm really confused about the fact that there seems to be two types of waves at play: the EM wave, which I understand to be an actual fluctuation of EM fields in space, and this other type of bulk "wave" that's referred to in explanations of the double slit experiment, whereby the bulk light emanates as a circular, water-like wave from the two slits. The circular path of the bulk wave seems to be what results in the fringe pattern, depending on the phase (the EM phase?) of the two waves at the point where they interfere at the detector. These bulk-like waves turn out to be a consequence of quantum mechanics and even occur with single photons, but at the time the debate seemed to be between particles and these circularly-emanating waves. However, is the only reason they can interfere at all because they are fundamentally EM waves that undergo destructive interference when in opposite phase? There is a difference between the waves in question and EM waves, right?

If the only reason there are dark fringes is because the quantum mechanical distribution causes physical EM waves to interfere, how can electrons generate a fringe pattern as well? I thought the de Broglie waves were complex mathematical abstractions and not representative of physical electron waves. Are de Broglie waves in any way analogous to EM waves? Does the fringe pattern fro electrons occur because of destructive interference of de Broglie waves? Is this fundamentally different from the photon interference?

$\endgroup$

1 Answer 1

3
$\begingroup$

An electromagnetic wave is a field, it has a (possibly zero) value at every point in space at every time.

A wavefunction is not a field. It does not have a value at every point in space. For $n$ particles it is a function from $\mathbb R^{3n}$ into the joint spin state of the system. For the simplest case of one particle with spin zero it looks like a function from $\mathbb R^3$ into $\mathbb C$ so you might think it is a complex scalar field. It isn't and that's going to bite you later. And then everything will be confusing and mysterious. If you want to think quantum mechanics is confusing and mysterious then a good way to do that is to think a wavefunction is a complex scakar values field in space.

other type of bulk "wave" that's referred to in explanations of the double slit experiment, whereby the bulk light emanates as a circular, water-like wave from the two slits.

That does not happen. But you also need to be clear, there is a classical double slit experiment with light where the electromagnetic field does spread through space and pass through slits and form interference patterns that have nothing to do with quantum mechanics.

The circular path of the bulk wave seems to be what results in the fringe pattern, depending on the phase (the EM phase?) of the two waves at the point where they interfere at the detector.

Even uncharged particles experience interference when you take quantum effects into account. So it's just the phase of the (possibly zero) spin (which still has a phase even when it's zero spin).

These bulk-like waves turn out to be a consequence of quantum mechanics and even occur with single photons, but at the time the debate seemed to be between particles and these circularly-emanating waves.

If you are going to study history, you have to ask about a specific moment of history and a specific person since each person's understanding changes at specific moments in history.

However, is the only reason they can interfere at all because they are fundamentally EM waves that undergo destructive interference when in opposite phase?

No. Again, uncharged objects can still interfere. And the interference doesn't happen in physical space like $\mathbb R^3$ it happens in configuration space like $\mathbb R^{3n}$.

There is a difference between the waves in question and EM waves, right?

Every possible difference imaginable. An EM wave is a field in space and has electromagnetic fields (with six components) as its values. A Quabtum wavefunction goes from configuration space and is spin valued.

I thought the de Broglie waves were complex mathematical abstractions and not representative of physical electron waves.

It's a gauge theory so a wavefunction does have too many degrees of freedom. But it's pretty close.

Are de Broglie waves in any way analogous to EM waves?

You can add two together, just like electromagnetic waves.

Does the fringe pattern fro electrons occur because of destructive interference of de Broglie waves?

Yes. But keep in mind the interference fringes happen in regions of configuration space. And what you see when you look at the screen are residuals of that.

Is this fundamentally different from the photon interference?

I don't know what you are saying there.

You mentioned a classical double slit experiment. Are you saying the experiment was fundamentally different?

You could actually pass water waves through some slits and you notice an interference. The interference happens in a similar way for every other kind of interference. But since the wave has more space to move through sometimes when one particle hits the screen another particle has been deflected a different direction, so the wave in configuration space doesn't overlap and so no interference happens. The interference happened in the overlap. This is why interference fringes can be "destroyed" really its like deflecting the beam going through the right slit down and the beam through the left slit up so they don't overlap any more. But if the up and down in the the position of a different particle the residual of just that particle looks lined up when the waves are not overlapping.

Running the particles through one at a time is designed to control for the possibility that the many objects going through are just pushing against each other (like water does). If it's just one particle at a time going through then it has to be something else.

$\endgroup$
6
  • $\begingroup$ Wow thanks for all your help that clears up a lot. So all interference that generates fringe patterns happen in real 3n, whereas the classical EM wave is in real 3 and is irrelevant. I got confused because I have seen the fringe patterns explained classically with water like waves emanating curcularly from the slits and with the interference occurring in real 3. You mentioned a classical double slit experiment. Are you saying the experiment was fundamentally different? Or that they simply interpreted it classically at the time, but the fringes actually result from the quantum effect? $\endgroup$
    – Jory
    Commented Mar 6, 2016 at 20:26
  • $\begingroup$ The way I see it, when they did the double slit with individual photons and still got the fringes, that sort of hinted at the fact that the interference was not occurring in real 3. Sorry if I'm not thinking about this in the right way. I am a mere biologist. $\endgroup$
    – Jory
    Commented Mar 6, 2016 at 20:28
  • $\begingroup$ So if the de Broglie wave is oscillating in configurational space and not in real 3 (sorry if I misunderstood again), why does the de Broglie wavelength define the resolution limit of electron microscopes, which is defined as a length in R. advanced-microscopy.utah.edu/education/electron-micro $\endgroup$
    – Jory
    Commented Mar 6, 2016 at 22:02
  • $\begingroup$ I mean I understand that phenomena in configurational space have implications for real space, but the resolution limit in microscopy is supposedly defined as being approximately half the wavelength of incident particles. This makes sense to me for light when thought of as an EM wave and in relation to this answer physics.stackexchange.com/questions/40850/… , as light is an EM wave in real space. How can electrons with wavelengths in configurational space abide by the same wavelength/resolution rule as light in real space? $\endgroup$
    – Jory
    Commented Mar 6, 2016 at 22:07
  • 1
    $\begingroup$ A wavefunction doesn't have to have a wavelength. And in general the wave vector $k$ can have components in each of the $3n$ directions. And the same optics ideas apply. I don't know why you think anything would be different. $\endgroup$
    – Timaeus
    Commented Mar 6, 2016 at 22:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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