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Particle/wave duality concerning light

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First off I would like to say I have just recently become emamored by theoretical physics and the like, don't know much about the subject, and am overwhelmed by the amount of information that I had no idea existed until a few days ago. But the only way to get answers is to ask right?

Perhaps I have misunderstoodsomething. To my understanding light is a particle without rest mass that exhibits particle/wave duality. On the other handhow thencan light have an electromagnetic frequencyif it is a particle? I think what happened is I don't fully understand the particle/wave duality orI missed something or I don't know. Somebody help please.

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kerplunk: To my understanding light is a particle without rest mass that exhibits particle/wave duality.

Or, one could just as well say that light is an electromagnetic wave and so has frequency, wavelength, velocity, and amplitude.

It is often said that which "face" light presents depends upon what we expect. If we do an experiment meant to show that light is a particle, it will oblige and act like a particle; on the other hand, if we do an experiment to show that light is a wave, it will happily play along. Wave or particle...we'll never catch it acting as both at the same time (however, IIRC, this last part has been shown to be wrong under certain conditions).

Another point is that the higher the frequency, the more particle-like light can behave. For example, Compton scattering is done with X-rays because the light has to have a sufficiently high frequency to act as little "billiar balls".

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Thanks that helps.

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Something I forgot to mention. There's a symmetry that helps address your concern: waves have particle aspects and particles have wave aspects - matter waves. That is, even particles - such as electrons - have wave aspects such as frequency, wavelength, etc. (and it's not limited to just electrons or protons - all matter has wave aspects, even a bowling ball).

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Yes that does help clear up a few of my muddled thoughts. Thanks.

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And thanks for the fish

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OK Telmad. If we assume that the photon wave mode is at a freq we would relate to "green" (say 510nm wavelength), as a "wave" it would have to have a "length" (wavelength) which would have peaks and valleys. Thus the wave's amplitude would increase or decrease along the wave cyclically. But we would not "measure" this variation in amplitude with a detector, would we? If the "wave" has a given discrete freq value, where does "white light" come from? Would it be represented by a wave with a continuosly varying phase? What is the freq/ wavelength of white light?

My background is electronics, specifically audio/ video/ RF. So I think of waves as displayed on an o'scope. I am having problems "visualizing" the correlation.

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FreeThinker: If we assume that the photon wave mode is at a freq we would relate to "green" (say 510nm wavelength), as a "wave" it would have to have a "length" (wavelength) which would have peaks and valleys. Thus the wave's amplitude would increase or decrease along the wave cyclically. But we would not "measure" this variation in amplitude with a detector, would we?

I'm not familiar with using detectors so quite frankly, I don't know. My guess is that we would detect the cyclic changes in the magnitude of the electric and magnetic fields associated with such a light wave.

(I'm not sure that calling it a variation in amplitude is correct: amplitude is the maximum diplacement from the equilibrium position - the height of a crest or depth of a trough. Even though the "height" (magnitude) of a sine wave continuously varies between 1 and -1, it still has only one amplitude: 1).

FreeThinker: If the "wave" has a given discrete freq value, where does "white light" come from? Would it be represented by a wave with a continuosly varying phase? What is the freq/ wavelength of white light?

When someone says white light they usually mean something like sunlight. The white light of sunlight is not a single color/wavelength but rather a composite of multiple colors/wavelengths. That's why a prism (and even droplets of water when forming a rainbow) can disperse sunlight into its component colors.

This is starting to touch a bit on how we perceive color. We have color-sensitive photoreceptors (cones) that each has a pigment that responds to red, blue, or green light. A photon of the appropriate color (frequency/wavelength) triggers a conformational change in a protein. This is like throwing a switch and sets off a whole cascade of reactions within the cell, culiminating in the closing of an ion channel in the cell membrane. This goes on to generate an action potential ("nerve impulse" or "electrical signal") that propagates down the axon (neuronal process - it comes together with others to form the optic nerve) and to the brain where the stimulus is interpreted. If red, blue, and green light are all separately striking our retina, we will perceive white light because all three cone types are stimulated.

FreeThinker: My background is electronics, specifically audio/ video/ RF. So I think of waves as displayed on an o'scope. I am having problems "visualizing" the correlation.

My "background" is in computer information systems and biology (majored in CIS and minored in biology). Physics, cosmology, and what have you are just "hobbies" of mine. I may not know the subject matter thoroughly enough myself to resolve whatever problems you are having (I may have them too).

For anyone interested....

If was actually Einstein who really got the "photons are particles" thing going, when he published his 1905 paper on the photoelectric effect. (The following – just like the rest of this post - is off the top of my head…I don’t claim it’s 100% accurate). In the photoelectric effect, a open circuit is created but putting some sort of material that is not a good conductor in the circuit. Thus, even with a voltage applied, no current flows (okay FreeThinker, I know current doesn’t flow). But when light strikes the material, electrons are knocked loose and then net movement of electric charge occurs: current. The physics of the day held that light was a wave, with the energy spread out across a wave front. An electron is a very miniscule thing and so would intersect only a very tiny fraction of the energy of a given wave. Thus, it would take a long time for a given electron to build up enough energy to become freed. However, experiments showed that a current was produced immediately. Also, theory held that the frequency of the light should be immaterial and that amplitude should be key: experiment showed otherwise. If

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So I think of waves as displayed on an o'scope.

Hey, FreeT - thanks. I usually feel like I'm way over my head in the discussions in some of these threads. I felt a little lost when I wanderred into this one. But after reading the first 6 posts, I finally got to yours. When I saw the o'scope remark, i went back and re-read the rest of them and it helped me put things into perspective. Relating things to o'scopes helped for me, as I spent many years using the o'scope to help with my Navy job. Not much use for one in my current job (MOM), but it's nice to be able to relate to things. So thanks for putting things into perspective for me.

Just one question though... What does this mean?????

And thanks for the fish

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One more thing that might help. Picture a sine wave on an oscilloscope. Each point that makes up the curve corresponds to a solution to an equation: here, simply y = sin x. The same general principle applies to the wave equations used in quantum mechanics (of course, those solutions are three dimensional and form a contour, though). Now, looking at a 2D wave pattern on an oscilloscope as an analogy, where along the wave would one be most likely to find a photon? At the crests and troughs. This is because the probability of finding a photon at a particular location is the square of the amplitude* of the wave at that point. So where sin x = 1/2 (for example, when x = 30 degrees or pi/6 radians) the probability of finding of photon there would be (1/2)^2 = 1/4. Where sin x = (2^-2)/2 (for example, when x = 45 degrees or pi/4 radians) the probability of finding a photon there would be [(2^-2)/2]^2 = 1/2. Where sin x = (3^-2)/2 (when x = 60 degrees or pi/3 radians) the probability of finding a photon there would be [(3^-2)/2]^2 = 3/4. And so on. So the higher the graph of the wave function, the more likely one is to find a photon there if a measurement were done.

Note that the above is just an analogy. For example, when x = 90 degrees (pi / 2 radians) then sin x = 1, which would mean that the probability of finding the photon at that point on the wave would be 100%; and when x = 0 degrees, sin x = 0 so the probability would be 0. Quantum mechanics does not let us know position of particles with 100% accuracy.

*Hhmmm, that kind of contradicts what I said about amplitude not changing along a wave pattern. Too lazy to look up if the word amplitude is the one actually used here or if its magnitude or something.

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Originally posted by: IrishEyes

So I think of waves as displayed on an o'scope.

Not much use for one in my current job (MOM),

Are you kidding? No use for an electron trace modulating the display on a phosphoresent screen as a parent?

Ever heard of TV? :-)

Just one question though... What does this mean?????

And thanks for the fish

kerplunk's tag is:

"The story so far:

In the beginning the Universe was created.

This has made a lot of people very angry and been widely regarded as a bad move."

"The knack of flying is learning how to throw yourself at the ground and miss."

All Douglas Adams (Hitchhikers Guide to the Universe) stuff. As is "And thanks for the fish".

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*Hhmmm, that kind of contradicts what I said about amplitude not changing along a wave pattern. Too lazy to look up if the word amplitude is the one actually used here or if its magnitude or something.

Ya it does. And that is part of where I am having trouble. Along with a photon, especially if is not a particle or perhaps has not collapsed to one yet, having frequency info.

Again, white light consists of a SPREAD of freq.

In some visible light production systems (long name for light bulbs?), they use tri-color stimulous, such as floresent tubes. They actually produce 3 distinct colors that the human eye merges into the color tone reflected off of the object. It would be easy to costruct the representation of 3 sets of photons, each with their relative "amplitude" (magnetude,... ) and freq. The selected gasses each produce specific freq (color) photons

If a "red" LED produce one photon we could theoretically "measure" the freq and strength of the single photon and it would spec out as a red photon of a given strength.

But in incandesent blubs, they are wideband black body radiators. They produce a continous spectrum of light. Thus we can not use the same photon model.

If a photon from a "white light source" was "measured", what freq (color) would it be? There is no "WHITE" freq. Would the single photon contain a BAND of freqs? ALL freqs? In a single one? Or does a white light produce "colored" photons like a floresent tube, just across a wider range of freqs?

Is there a "white" photon, the way there woud be a "red" photon? If a white LED produced a single photon, what freq would it be?

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I have a video that discusses the "white light" from a gas filled, electric discharge tube and it states that the light is composed of multiple colors. When a photon absorbs energy it jumps to a higher-energy level, then falls back down, (re)emitting a photon. The 'height' of the drop determines what frequency/wavelength/color the emitted photon will be. Even if there is a single type of gas in the tube, such as helium, there will be different possible energy-level jumps/falls possible (for example, look at the Lyman series for hydrogen). So even a single gas absorbing energy and then emitting light can produce red, green, yellow, or what have you photons. For us humans, these colors are all mixed into what we perceive to be white light. If we were to pick a single photon at the source and measure it, we would find it to be red...or green...or what have you. I assume that the distinctness is retained as it travels to our retinas, but since waves can be superposed, I am not positive.

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When a photon absorbs energy it jumps to a higher-energy level, then falls back down, (re)emitting a photon.

First let me correct your statement. My guess being that you made the error inadvertantly. It should be

"When an electron absorbs energy".

The electron "absorbs energy" by absorbing the photon.

Now RE the rest of the info:

Even if there is a single type of gas in the tube, such as helium, there will be different possible energy-level jumps/falls possible (for example, look at the Lyman series for hydrogen). So even a single gas absorbing energy and then emitting light can produce red, green, yellow, or what have you photons.

This is not accurate. In fact the Lyman series for hydrogen shows the spectrum of radiation emitted by hydrogen is non-continuous. It CAN NOT "produce red, green, yellow, or what have you photons". It can only produce freq which are discrete natural number multiple of the primary wavelength.

Where n= the principal quantum number referring to the energy level of the electron, we get n = 2 to n = 1 is called Lyman-alpha, 3 to 1 is Lyman-beta, 4 to 1 is Lyman-gamma, etc. Further these wavelengths are in the UV region, and they are not visible.

This is why when gases are used, discrete wavelengths result, monochromatic. As I described in floresent tubes. So we are back to black body radiators being the only real source of continuous "white light". Where does this put us RE a "white light" photon?

If we were to pick a single photon at the source and measure it, we would find it to be red...or green...or what have you. I assume that the distinctness is retained as it travels to our retinas, but since waves can be superposed, I am not positive.

So would a "white light" consist of an infinite spread of wavelengths simultaneously? Does a black body radiator generate a random series of discrete freq photons with it then being impossible for a resultant white light without some critical number of individual discrete freq photons?

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FreeThinker: First let me correct your statement. My guess being that you made the error inadvertantly. It should be

"When an electron absorbs energy".

The electron "absorbs energy" by absorbing the photon.

Yep, a careless oversight. But your other "gotchas" don't only fail, some of them are just flat out wrong. I'll demonstate in my next post below.

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TeleMad: Even if there is a single type of gas in the tube, such as helium, there will be different possible energy-level jumps/falls possible (for example, look at the Lyman series for hydrogen). So even a single gas absorbing energy and then emitting light can produce red, green, yellow, or what have you photons.

FreeThinker: This is not accurate. In fact the Lyman series for hydrogen shows the spectrum of radiation emitted by hydrogen is non-continuous. It CAN NOT "produce red, green, yellow, or what have you photons". It can only produce freq which are discrete natural number multiple of the primary wavelength.

Hmnmm, seems you’ve gone a bit out your way to try to find fault with statements. Let me set you straight on a few things.

First, note that the reference to hydrogen is only a parenthetical, set apart from the main statements. I did NOT say that hydrogen produces red, green, yellow, or what have you photons. Yet that misconception is one of the things that your “gotcha” relies upon.

Second, I did NOT state or imply that the possible wavelengths for any single gas are continuous. I listed a few discrete examples of colors and then said “or what have you”. This was meant to convey that the particular single gas used would determine what colors were produced (helium would produce red, yellow, and green, whereas chlorine would produce a different set of colors). This is another misconception of yours that your “gotcha” relies upon.

Third, the reason I picked red, green, and yellow specifically, and mentioned helium - both in the main, non-parenthetical statements - is that those are the colors the video I referenced explicitly mention that helium can produce.

Fourth, I mentioned the Lyman series of hydrogen parenthetically to show the principle being alluded to: multiple possible energy-level “jumps” and “drops” for a single gas/element. After all, most people familiar with the basics of atomic structure have heard of the Lyman series, but they are not familiar with any He series.

So your “counters” are all invalid.

FreeThinker: Where n= the principal quantum number referring to the energy level of the electron, we get n = 2 to n = 1 is called Lyman-alpha, 3 to 1 is Lyman-beta, 4 to 1 is Lyman-gamma, etc. Further these wavelengths are in the UV region, and they are not visible.

If you are claiming that hydrogen gas doesn’t produce any visible light, you are wrong….

”Now consider light from a hydrogen lamp, essentially a cathode ray tube with hydrogen at low pressure as the residual gas. When an electric discharge passes through the tube, some of the hydrogen atoms are energized or excited through collisions with the cathode rays [electrons]. These excited hydrogen atoms give off energy as light. If was pass light from a hydrogen lamp through a slit and prism, we observe only a few images of the slit These images appear as narrow colored lines separated by dark regions; this is a discontinuous spectrum.” (General Chemistry: An Integrated Approach: Second Edition, John W Hill & Ralph H Petrucci, Prentice Hall, 1999, p289)

So hydrogen gas does produce visible light, multiple wavelengths of visible light, in fact.

FreeThinker: This is why when gases are used, discrete wavelengths result, monochromatic.

Wrong. I pointed out just above that hydrogen produces multiple wavelengths of visible light. And, as I alluded to originally, helium gas will produce multiple, visible colors…

”Figure 7.15 The emission spectrum of helium

The light source is a helium lamp. The visible spectrum of helium consists of six colored lines in the visible portion of the electromagnetic spectrum.” (General Chemistry: An Integrated Approach: Second Edition, John W Hill &