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First off, sorry to throw in another question from someone who hasn't studied the maths.

I'd like to see if I have a correct (if very basic and non-mathematical) understanding of the wave and statistical nature of Quantum Mechanics, and to avoid any complexity I'd like to look at the Double Slit Experiment without the slit.

So we have a macro object (which is itself a very complex Quantum object that for the most part can be modeled with "classical physics") that emits electrons and can be switched to emit one electron at a time. Off to one side of the electron emitter is a flat screen that acts as an electron detector.

When we tell the emitter to produce a single electron what we get is a (spherical in 3-d space?) wave of some specific energy that travels from the emitter at the speed of light. As the wave intersects with objects (other waves, fields, "classical objects") with which it is capable of interacting with there is a percentage chance (that can be modeled through QM maths) that the wave will interact with it. Not each electron emission will interact with the screen, which is expected, and the wave will continue until it reacts with an object that absorbs its energy and becomes part of the Quantum state of another object.

So, is my understanding as presented in the previous paragraph correct?

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Okay, I promise I won't make any assumptions about this stuff until such time (or if such time) as I take up the maths. –  David Rouse Dec 5 '11 at 23:07
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When we tell the emitter to produce a single electron what we get is a (spherical in 3-d space?) wave of some specific energy that travels from the emitter at the speed of light.

We get an electron, with a particle trajectory, that travels at the speed defined by the electric and magnetic potentials that make the gun that produces the electron. If the screen is a TV screen the energy it has will be given up in interacting with the phosphor screen. That is why images can be formed.

As the wave intersects with objects (other waves, fields, "classical objects") with which it is capable of interacting with there is a percentage chance (that can be modeled through QM maths) that the wave will interact with it.

The quantum mechanical "wave" is not a physical wave. It is a probability distribution, i.e. a mathematical probability that the particle is at a specific (x,y,z) that sometimes conforms with a wave description and sometimes a particle one, within the dimensions of the little cube where the electron is "localized", which are very small, and have to do with limits given by the Heisenberg uncertainty principle; hbar is a very small number. The two slit experiment works when the slits are separated by distances compatible with the Heisenberg uncertainty principle. To summarize: the electron coming out of the gun will have a specific classical trajectory and momentum; the QM probabilities which sometimes lead to wave like traces enter in the very small dimensions about the electron trajectory. In the large dimensions of your thought experiment the electron is a classical particle.

Not each electron emission will interact with the screen,

the wave will continue until it reacts with an object that absorbs its energy and becomes part of the Quantum state of another object.

It is only a probability "wave", not a water wave. The electron will stop on the screen.

All electrons will end up on the screen.

Now if your screen were a very small TV , within the delta(p)*delta(x) dimensions where quantum mechanical effects are dominant, then the wave nature of the path would become apparent in the interaction of the electron with individual atoms of the screen ( that is the size where QM rules), but that is a different problem.

Edit: the above treats the general description you have given in your question. Now if one looks only on the part of the trajectory where delta(p)*delta(x) is of the order of hbar, then there does exist a a probability distribution that describes the possible path of the electron that gives wave like interactions . This means that spatially the probability of finding the electron is not constant, there are peaks and valleys within the hbar limits of its existence. It will hit the screen at some specific x,y ( if z is the direction of motion) within those limits. If you keep throwing electrons on the screen on the same macroscopic path, microscopically each electron is incoherent with the previous one so there will be no pattern in the dispersion of the hits.

It will look like error and it looks like that with the one slit experiment. What the second slit does within the hbar limits is to bring out the phase of the probability distribution for each individual electron, it changes the wave function as it passes the slits so the probability wave function interferes with itself.

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"It is a probability wave" is misleading. It is not a probability wave. It is only similar to a probability wave. It does not have a probability interpretation away from the large systems and decoherence limit at best. –  Ron Maimon Dec 5 '11 at 18:50
    
@RonMaimon OK, tried to edit it –  anna v Dec 5 '11 at 20:46
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There are multiple ways to understand it, but the first is pretty simple:

When the electron or photon is shot from the gun, what is shot is a wave. It is similar to a water wave or sound wave, which can be described as being a complex number, of which the real and imaginary parts vary sinusoidally as a function of time and position. Going through the two slits, you get an interference pattern. That's just like any wave.

We don't really know what medium the wave is "in", but it has a certain energy at every point, which is proportional to the square of its complex magnitude. Every wave acts like that, but the way this energy manifests is as the probability of the particle being observed.

Just as in electrical engineering, where the voltage of a signal is its amplitude, and its power is proportional to amplitude squared, what we have is a wave whose "size" is its amplitude, which can be positive or negative or complex, but its power, that is its probability, is always just a positive real number.

(EDIT: I'm only using the terms "power" and "energy" because they form a good analogy to probability. They are not actual physical power, although in the case of particles that carry energy, the resulting actual power is proportional to the probability, if you sum over many particles.)

When you have an interference pattern, the wave reinforces in some places, and cancels out in other places, meaning it has more power (= probability) in the places where it reinforces, and less in the places where it cancels.

So if you shoot a particle, the probability forms an interference pattern, and if you shoot a lot of them, that's where they accumulate. Notice, that means one of the places they pile up is exactly in the middle between the two slits, which isn't in line with either slit.

P.S. When you shoot the electron or photon, the wave you shoot is not just a simple sine wave, but a weighted sum of sine waves of different frequencies whose composite forms a wave packet. It's just like if you build a speaker in the right way, you can make it emit pulses of sound that are very directed. They don't necessarily fan out much in space or time, but can be almost like "particles" of sound. Drawings of photons are often depicted as little bunches or packets of waves, and that's trying to capture the idea. So if you have a gun shooting particles, if you try to represent them as actual sine waves, they have to be composed of a lot of closely related waves (as in a fourier transform) to get them to have the localization that you want.

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Saying that the wavefunction has "a certain energy at every point" is misleading at best. It isn't energy in the physics sense of the word. –  Harry Johnston Dec 5 '11 at 23:34
    
@HarryJohnston: I'm trying to find a simple way to say it, that captures the idea of amplitude, by reference to everyday concepts. Have you got a suggestion? –  Mike Dunlavey Dec 6 '11 at 1:15
    
Anna's answer was pretty good. –  Harry Johnston Dec 6 '11 at 1:34
    
@HarryJohnston: Her answers are always good. I'm just trying to take a slant that might not scare off a young student, while not hand-waving too much about the concept of "amplitude", which is the only way the concept of "probability wave" makes any sense, to me at least. –  Mike Dunlavey Dec 6 '11 at 1:56
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This is a really good description of what people say about quantum mechanics, and I'm going to throw a wrench in the works. I don't believe we have pea shooters for electrons. I know people will say that we do, but I don't believe that we have pea-shooters for photons either. If we really did have pea shooters, then the situation you describe would be very very difficult to explain. There is plenty of stuff that is indeed hard to explain, and people like to say it's exactly equivalent to the way you describe it. It is arguably very similar to what you describe, but it's not quite the same in any experiment I'm aware of. The difference between the great paradoxes that everyone talks about and the actual reality is that in fact, we do not yet have a pea shooter for electrons.

EDIT: This edit is posted to deal with some of the many objections to what I wrote above. There is a pretty big difference between an emitter whose intensity can be turned down very low, so that in some arbitrary time interval there is a very low probability of two electrons being emitted, and an emitter that can be switched to produce single electrons at will. The former exists and the latter does not, notwithstanding Harry Johnson's claim. The fact that Anna sees single electron tracks in a cloud chamber does not mean she is able to produce such tracks at will. In fact, I believe she can activate a source and wait for it to produce tracks in its own good time. This is very different than what the OP asks for: a source that will produce single electrons on demand.

Similarly, the fact that she can buy a detector which claims to detect single photons does not mean she is able to produce such photons at will. There is no pea-shooter for photons. If there were, then we would not have to resort to the much-talked about experiment by Thorne et al to demonstrate anti-bunching. We would simply fire single photons at a half-silvered mirror with a pea shooter and watch them click in individual detectors, 50% of the time on the left, and 50% of the time on the right. This simple experiment has never been done. Why not? Because we have no pea-shooters for photons.

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What is a "pea shooter for electrons"? –  Harry Johnston Dec 5 '11 at 3:41
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Sure we do. Confine a single electron in an electron trap, then release it. Bingo. –  Harry Johnston Dec 5 '11 at 4:22
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Photons are trickier, of course, but if you dial down the intensity enough, the odds are good that at any given time only one (relevant) photon is present in the system. AFAIK, a sufficiently low probability of a second photon doesn't have a disproportionate effect on the maths. –  Harry Johnston Dec 5 '11 at 4:25
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en.wikipedia.org/wiki/Electron_gun –  anna v Dec 5 '11 at 5:14
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@Marty, I don't see any fundamental problem with using an electron trap as a "pea-shooter". You'd have to prepare the trap ahead of time, and it will only produce one electron, but neither of these requirements run contrary to the experiment the OP describes. –  Harry Johnston Dec 5 '11 at 23:25
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