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The quote below is from Matt Strassler's blog:

a particle is a ripple with many crests and troughs; its amplitude, relative to its overall length, is what tells you that it is a single particle.

If I understand correctly, what he calls "ripple" is "probability wave". Why is it that the amplitude of a probability wave is the sign of "a single particle"?

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You should accept anwers. –  OmnipresentAbsence Feb 23 '13 at 19:59
    
None of the answers are definitive enough to accept. –  Zeynel Feb 24 '13 at 20:36
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You've asked 12 question and you haven't accepted one answer yet. Most of your questions (including this one) have fantastic answers. None of the answers are definite enough? Either you don't understand the answers or you don't understand your own questions. How can you possibly consider Terry's not definitive enough? –  OmnipresentAbsence Feb 24 '13 at 22:03
    
@OmnipresentAbsence: While I agree that the OP should probably accept a few answers, it's better to refrain from nay hints of rude language; thanks :) –  Manishearth Feb 26 '13 at 15:53
    
@Manishearth Yep, got a little carried away there –  OmnipresentAbsence Feb 26 '13 at 19:41
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2 Answers

In the bit of his article you quote Prof Strassler is specifically talking about photons, so the wave is an electromagnetic wave. Any EM wave can be interpreted as a particle, though this interpretation isn't always useful. An infinitely long wave corresponds to a photon with a well defined momentum but a very poorly defined position. If you pin down the wave so it has non-zero amplitude in a small area this corresponds to a photon with a well defined position but poorly defined momentum. (Incidentally such a localised wave is usually called a wave packet.)

If you look at fig 4 in the article (this is what the comment you link to is about) you'll see it shows a wave packet, and this is what we intuitively think off as a photon. However a continuous wave is a photon too, just not one with a well defined position.

Incidentally, the above trade-off between position and momentum might make you think of Heisenberg's uncertainty principle. That's exactly what it is!

Note that Matt Strassler's comment isn't just about the amplitude, it's the amplitude relative to the length i.e. how localised the wave packet is.

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I must admit I don't quite get his last point about the amplitude relative to the length: surely the only thing that can tell you how many particles are present is the number operator? What is amplitude even referring to - the amplitude of the classical mode function used for the photon basis? The ratio of the amplitude to length would have to be dimensionless in order to represent a number of photons. How is that done? –  twistor59 Jan 27 '13 at 21:14
    
You may be reading to much into a throwaway comment. I suspect he means a single photon and just means the amplitude and packet length are roughly inversely proportional. –  John Rennie Jan 28 '13 at 7:35
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Maybe, but earlier on the page he says The real world is quantum mechanical, so in fact (as described in this article) the amplitude A cannot be just anything; it takes discrete values, and those values are proportional to the square root of n, a positive integer (or zero), which is the number of quanta of oscillation in the wave. so I thought maybe it was the quantized amplitude he was talking about, because that's what's determined by the number of photons present (at least for number eigenstates). But then I don't get the "ratio of amplitude to length" thing! –  twistor59 Jan 28 '13 at 8:07
    
@JohnRennie wrote: "If you look at fig 4 in the article"... When I look at fig 4 I only see a wave. Where is the particle in that animation? –  Zeynel Jan 29 '13 at 0:32
    
@JohnRennie wrote: "Any EM wave can be interpreted as a particle" How is the word particle defined here? –  Zeynel Jan 29 '13 at 1:12
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Why is it that the amplitude of a probability wave is the sign of "a single particle"?

Take a spring or coil of any kind. Look at it from the side, or even better, project a shadow of it onto a piece of paper. In both cases you will see what looks like a sinusoidal wave complete with peaks, troughs, and zero points. But the spring has a smooth, constant radius that doesn't show any such peaks and troughs.

A quantum amplitude is very much like the radius of a spring, and the sinusoidal representations of a quantum wave function that you see in most text books are very similar to the shadows projected by such springs.

When Strassler said that the amplitude represents "a single particle," he was trying to emphasize that a sequence of peaks and troughs separated by zero points is really a single entity (like the spring) with a single smoothly changing amplitude (the radius of the spring). The peaks and troughs are just illusions acquired by projecting the wave function onto an overly simple 2D screen or page of paper.


(I added the above on 2013-03-04 My original answer, sightly edited, is below.)

I looked at Matt Strassler's blog, and I'm pretty sure his real intent was just to keep the questioner from thinking that a particle is always a single peak in a probability wave function. That is just wrong, and Professor Strassler was trying to make sure that readers didn't get into the habit of thinking in such terms.

Here's a slightly different way of looking at probability wave functions that may help.

In an earlier answer about Fourier transforms I argued that a much better way to think of probability wave functions is to use a complex plane perpendicular to real space, and then visualize how the amplitude or height of the wave function maps into that space. The problem is that such an approach requires thinking in a five-dimensional space, which is tough! However, you can cheat a bit by looking at only one XYZ dimension at a time (e.g. the length X of a box), and "borrowing" YZ to represent the complex plane.

Now, if you do that for say an electron bouncing back and forth between two ends of a box with length along X, the idea of "peaks" and "troughs" in the wave function pretty much disappears. Instead, you get various sorts of moving helical coils (moving electrons) and skip-rope-like stable states (the resonant or "stationary" wave function solutions) along a rope (representing amplitude) that stretches from one end of the X box to the other.

I should mention that it never ceases to amaze me just how close the differential equations that control rotating loops in an ordinary string or rope are to the equations that control this composite real-and-complex representation of wave functions. For example, if you take a hose laying in the yard and give it a quick circular jerk at one end, you will see a short helical wave move from your hand and travel down the length of the rope. What's remarkable is that both the helix and the way it moves have almost exact mappings into the wave function of an electron wave packet moving through space. Moreover, the "skip rope" loop solutions represent the resonant states in which you in effect have the electron helices going both directions at once (a "quantum superposition" of left and right moving states of the same electron).

In this rotating-rope model you only get peaks and troughs when you project a shadow of these coils onto a piece of paper. Think for a moment about how a spring or Slinky looks from the side and you can see how the usual sinusoidal curves with peaks and troughs can emerge from by limiting your perspective two only two dimensions (the projected shadow of the amplitude rope).

I'm pretty sure in fact that that was the message Professor Strassler was trying to get over in his comment: There is just one more-or-less continuous amplitude (the coils of the helix) where the particle is located, with those peaks and trough literally just being shadows of the underlying reality of the wave function.

Incidentally, I have to mention it: The rope analogy becomes even more powerful if you "standardize" the total volume enclosed by the various rotating coils and loops along the X length of the rope. If you do that, then each the volume enclosed along any segment of the rope becomes the probability of finding the particle along that part of the X rope. So, if you have two resonant loops (think of expert skip-ropers with double loops) along X, then the probability of finding the electron becomes 50% in either loop -- and 0% in the center! The electron sort of magically "tunnels through" that part of real space to get to either side. If you look carefully, however, the rope itself is moving like crazy at that same central location, so it's not quite as simple as saying that the electron is "not there" at the center. Its wave function is very active there, but just does not allow the particle to be found at that spot if you look for it.

I should point out also that by intent I just undid my whole argument! That is, while I just argued that the projected peaks and troughs of a wave function do not accurately convey the continuity their underlying complex amplitudes, it is not correct to say that the wave functions amplitudes for a single particle are contiguous. It is actually very common for them not to be, since for example the various lobes ("skip rope loops") of electron orbitals in atoms are examples of discontinuous electron wave function amplitudes. Such electrons can be found in certain disconnected regions of space, but not in the regions in between.

So, bottom line questions and answers:

  1. Is the existence of an amplitude a sign of a particle? Yes, pretty much by definition.

  2. Do the peaks and troughs of the real part of the wave function mean anything about where the particle really is? No; you must use the complex wave function for that.

  3. If you find a single stretch of complex amplitude that surrounded entirely by zero amplitudes in XYZ space, does that stretch necessarily represent an entire particle? No, definitely not, since that very situation happens all the time in atoms. The wave function may be broken up into many pieces, some conceivably quite far apart in XYZ space.

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Terry Bollinger: Is there a contradiction in your answer? If I read it correctly you answer "it is" and "it is not" to the same question: [Why is it that the amplitude of a probability wave is the sign of "a single particle"?] It isn't. [Is the existence of an amplitude a sign of a particle?] Yes, pretty much by definition. –  Zeynel Feb 26 '13 at 23:06
    
Ah, got it: When I said [the amplitude] "isn't" [a single particle], what I meant is that it may be only part of a single particle. Oddly, Strassler seemed to have thought in his blog entry that a particle wave function is always in one continuous piece, like a little blob of dough. That is emphatically not the case, so I was really critiquing Strassler with that line for giving an incomplete desciption. On the other hand, literally by definition, a non-zero amplitude does always imply that there is a particle around somewhere, even if you can only seeing some tiny part of it. –  Terry Bollinger Feb 27 '13 at 0:37
    
Terry Bollinger: Is it possible to write a shorter answer including the ideas of these comments so that I can accept it? Thanks. –  Zeynel Mar 2 '13 at 19:47
    
Zeynel, thanks. I'll add this as a preface in front so people don't get confused; let me know if it's sufficient after I write it. –  Terry Bollinger Mar 5 '13 at 0:43
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