# Question; double-slit and wavefunction relationship

1. Jun 1, 2012

### James S Saint

I am having troubles finding anything online that reveals the conceptual concerns relating to the probability wavefunction in regards to the travel of a particle and then how that plays into the double-slit experiment. No math is required. I merely need to know the concepts involved.

1) exactly what is being implied concerning the path of a particle by the wavefunction?
2) exactly how is that notion being applied such as to explain the double-slit phenonema?

Thks.

Last edited: Jun 1, 2012
2. Jun 3, 2012

No takers?

3. Jun 3, 2012

### HallsofIvy

Staff Emeritus
It's hard to understand exactly what you are asking. What do you mean by "what is being implied"? And what do you mean by "how is that notion being applied". When you say "I am having troubles finding anything online that reveals the conceptual concerns relating to the probability wavefunction". Every site or book I have seen is very clear as to exactly what a "probability wavefunction" is. What exactly do you mean by "conceptual concerns"?

4. Jun 3, 2012

### YAHA

You want to look at Feynman Lectures on Physics vol. 3. Double slit experiment is best explained through the Path Integral formulation of Quantum Mechanics. Feynman gives basics of that in first couple of chapters there. Essentially, wave function squared is a probability amplitude of an event happening. If one slit is open, you get a typical bell-curve distribution behind it. Same happens when the other slit is open. However, when both of them are open at the same time, there is interference on the detector screen. This can be deduced mathematically from the fact that probability amplitudes of complex(!) wavefunctions do not just add algebraically, but rather, they create a cross term. The cross term is precisely the thing that creates interference. Well, actually, nature creates interference and cross term is just our crude way of explaining it

5. Jun 4, 2012

### James S Saint

Well thank you.

But now, I agree that immediately behind the slit, there would be a Gaussian type of probability concerning the presence of a particle and also one concerning whether that particle's location could be measured at that point (two distinctly different concerns).

But wouldn't that also be true for any and every point along the proposed path of that particle, both before and after the slit? It seems that the path itself must be consistent in any probability calculations until there is actual interference involved.

6. Jun 4, 2012

### YAHA

I am not sure what you mean by "presence of a particle and also one concerning whether that particle's location could be measured at that point." The assumptions of the experiment state that the detector screen allows the measurement (detection) of the particle's position upon its arrival.

Once again, see Feynman lectures volume 3. If you measure the particle's path before the slit (say, measure right at the slit) you disturb it and no interference is observed. Hence, you will see the Gaussian behind the slit at which you performed the measurement. As soon as you stop measuring, you will begin seeing interference again.

If you clarify your language a bit, we could explain it better :)

Last edited: Jun 4, 2012
7. Jun 5, 2012

### James S Saint

Well, I suspect that I would be better at picking up your language use than the other way around, but it seems we need to start somewhere. The book that you mentioned is around \$113.00 and the Youtube lectures and various online papers that I can find don't seem to cover much detail or they merely express the unassociated math without any clear indication of what they are really saying regarding the actual experiment. If I could find something sufficient online, I wouldn't be bothering you guys.

So starting from what I understand of the experiment (maybe you can then step in with QM concepts involved);

A particle leaves a source toward an obstruction. If the particle manages a near-miss situation, it continues onward to strike a screen. As it turns out if the obstruction was such that the particle could only possibly get around it on one side, the pattern on the screen displays what would be expected. But if the obstruction is such that the particle could possibly get around it on either side, the pattern on the screen becomes similar to an "interference pattern". The mystery seems to be about why an interference type of pattern would show up when only one particle was shot out at a time.

Quantum Physics gets into it (thank you De Broglie) explaining that there is a probability function involved in locating where the particle would be at any one time. When the second part of the obstruction (the "second slit") is closed off, that probability function reduces to zero and thus no interference pattern should be expected.

Now my question relates to exactly what is it that the probability function is describing (besides merely some probability). Is it describing the probability of the particle being at any one point along the possible paths?

I understand that the very seed thought to the whole probability function originated from the uncertainty principle. That would indicate that the function is actually a probability of being able to measure the position of the particle, not the probability of whether the particle was actually there. But I don't know what the intention in the explanation given by De Broglie really was.

I am not trying to clear up what De Broglie necessarily meant as much as how it applies to the double slit experiment. How does the probability function work into the experiment? Is it concerned with the location of the particles? If so, how so? Is it concerned with the probability of a particle being detected? If so, how does that relate to the experiment?

There seems to be a disconnect between the idea that QM solves the riddle and exactly what that solution is. Or at least I can't seem to find anyone fessing up to it.

8. Jun 6, 2012

### James S Saint

Well, I don't know how I missed that post... sorry.

Hopefully I have answered the questions that you asked. Perhaps since those lectures seem clear to you, you could perhaps address my more exact concern? ...please :shy:

"Conceptual Concerns" == Exactly how are the concepts involved applied to the experiment?

9. Jun 6, 2012

### DrewD

There is no riddle. Just the way things are and then the question as to whether or not we can describe what is happening. I'll do my best, but don't expect that it will necessarily make perfect sense. Imagine trying to describe gravity to somebody that had never been on a planet.

NOTE: a particle being in a particular place and measuring the particle in that particular place are the same thing.

The first step is to make things as simple as possible. Let it take place in a large room that is two dimensional, not just thin, we are only considering two dimensions because visualizing more than that will be tough, but mathematically it will work fine. There is a divider in the room and a light that is so dim that only one photon at a time is emitted on one side pointed at the center of the divider. On the other side of the divider there a photo-detector.

The center of the divider has slit A and slit B. As you know, when either slit is the only slit open, there is a Gaussian distribution. What does that exactly mean? It means that after a long time, the number of photons measured is greatest right in front of the slit and tapers of as a bell curve. If both are open, there is an interference pattern. This is the same as the interference pattern that you would see if you had water waves.

Now we can imagine looking down and "seeing" the probability wave move across the room. As soon as the photon is released it would spread out similar to the way that you would expect any other wave to spread out and it would interfere just like water waves. But this wave is in fact only the probability that the photon would be measured at any particular point.

So if we could see this wavefunction spread out, it would indeed tell us the probability that the particle is at any given point in the room. But we can't actually see the whole wave. It really is a probability of measuring something. All you can do pick some points and measure them over and over and eventually find that the result is what you would find from a wave. The most obvious measurement is on the opposite wall using the photo-detectors.

You may still be annoyed that I haven't exactly answered you first two questions, but, I think they are misguided. The particle does not have a well defined path when the interference pattern is observed! If it has a well defined path, the interference pattern is not observed. That is a fairly shocking idea, but it is true.

if you want a more eloquent explanation of this experiment, check a library for the Feynman lectures. Any college library will have them and I bet a number of public libraries do too.

10. Jun 7, 2012

### James S Saint

Thank you DrewD,

Actually, you did in fact answer the questions that I had asked. I still have to wonder if what you have said is really what "they" intended (whoever "they" are).

I have an issue with that particular understanding (besides the measurement = existence issue). Let me restate what I believe that you have said and then explain my concern.

Proceeding from a particle source would be a time dependent, radiating probability distribution wave. This, for any one particle, would be a single wave pulse traveling through time, fanning out in much the same way that a single water wave would leave an object inserted into the surface of a pool. The wave, through time, travels toward the slitted shield.

When that wave reaches the open slits, it becomes 2 waves of similar nature as before, merely less intense. Because those waves are spatially separated, through time, they fan out behind the shield and cross each other before they get to the detector screen.

Now my concern is that unlike water, probability wave pulses do not interfere directly with each other such as to produce secondary waves. They merely pass through each other without causing added ripples. So when they reach the detector screen, they are still merely 2 waves that reach the screen and stop. They don't reflect around or resonate in any way.

Such waves do not produce any interference pattern at all. And the next particle would produce the exact same wave distribution as would every one following. That screen would produce nothing but a Gaussian buildup of hits behind the 2 slits and nothing interesting at all.

So what is being left out here?

Last edited: Jun 7, 2012
11. Jun 7, 2012

### Sonderval

That is incorrect. The two "probability waves" in your picture do "interfere" in a sense.

You have to distinguish between the wave function (which is the propability amplitude) and the probability (which is the amplitude squared).

If you add up two wave functions (which you can since they can directly be added), you now have a new wave function which is the sum. However, the probability is the square of the wavefunction, and the sum of two squares is not the square of the sum:
$(a+b)^2 = a^2 + b^2 + 2ab$
The additional 2ab-Term is what you could call the "interference" of the probabilty.

12. Jun 7, 2012

### James S Saint

I think you misunderstood my point.

As the probability wave "travels" (just mentally. It doesn't actually travel at all), multiple waves (existing only in the mind) do not partially reflect off of each other, producing secondary turbulence waves as physical water waves would.

With a physical wave running across another physical wave, new waves are formed that head off in new directions. That offers the opportunity to eventually create an interference pattern. But probability waves have no physical substance. They are just mental constructs. Thus wavefunctions cannot produce secondary reflection waves merely off of the idea of them since they really weren't traveling in the first place.

13. Jun 7, 2012

### Sonderval

Sorry, I thought your argument was physical, not (pseudo-)philosophical. To me, as long as wave function can be used to successfully describe reality, they are "real". The fact that they have no "substance" is irrelevant, it feels like people in the 19th century claiming that electromagnetic waves need some ether to travel in because fields have no "substance".

Physically, what I explained was exactly the reason why you have linearly additvie wave functions but still interference effects.

14. Jun 7, 2012

### DrewD

Interference is not caused by waves reflecting off of each other. Interference is caused but when a point on a wave with a certain negative amplitude is found at the same point as another wave with the same amplitude, but positive. (actually it happens no matter what the amplitudes, but when they are almost equal and opposite one finds the most dramatic situation where the amplitudes become zero).

There is no need for the waves to be made from physical particles fluctuating. The physics of interference do not require such a wave. True, there are phenomena that will only happen in fluids, but interference is not one of them.

15. Jun 7, 2012

### James S Saint

Well, I guess that I'm still not being clear. I am not talking about philosophical issues.

If I imagine that there is a "4" written on my desktop and that number is traveling across the desktop, when it runs across another number 4 that I have imagined traveling across the desk, I can't claim that at that moment there is an "8" when they pass over each other. And even if I did, I can't then claim that "8" necessarily travels slower than a "4" so the numbers hesitate and that is why the edge of the desk isn't perfectly straight.

The wavefunction isn't a causative entity until it actually encounters something physical. The wavefunction wasn't actually traveling through space. I merely imagine it traveling so as to understand what things might look like IF that wavefunction were at the point in space. Whether the wavefunctions add is only relevant if there is some physical entity at the location where they came together. In the midst of mental traveling, there is no adding going on any more than the two "4"s adding as they crossed my desk.

For interference patterns to be generated, there must be more than merely 2 waves striking the screen. How did you get more than 2 waves of any nature?

16. Jun 7, 2012

### James S Saint

I guess this is the real question;
"For interference patterns to be generated, there must be more than merely 2 waves striking the screen. How did you get more than 2 waves of any nature?"

Last edited: Jun 7, 2012
17. Jun 7, 2012

### DrewD

That is incorrect. Interference only requires two waves. They do not need to create any reflections or turbulence.

You should learn more about simple waves before venturing too far into quantum mechanics. If the amplitudes are 4 and they cross, they indeed add and the amplitude is 8. I don't care what you claim. Learn more about the general theories of waves and then come back to quantum mechanics.

18. Jun 7, 2012

### YAHA

That, my friend, no body knows. Neither Feynman, or Einstein or any other poster here. It is just the way the nature is. And 2 waves is a minimum to get interference classically. In fact, quantum mechanics describes a free particle as a superposition of infinite number of waves. They can interfere very well with each other to form something called a wave packet. It is therefore not unreasonable to think that they can interfere as they pass through 2(!) slits.

19. Jun 7, 2012

### James S Saint

I'm afraid that one needs some support. Can you show me any logic at all that displays how merely 2 waves can create an interference pattern? We are not talking about repeating sinusoidal waves, but merely two lone wave humps. It is logically impossible for any interference pattern to be created from them.

That tells me that QM has not actually revealed any solution at all.
Is that true? If so, why is the double slit experiment always being represented as a QM topic?

Last edited: Jun 7, 2012