Sciencemadness Discussion Board

light waves - photon amplitude

quantumchromodynamics - 23-1-2014 at 04:38

I am confused by the literature regarding the amplitude of light waves.

As is generally accepted, Einsteins 1910 paper on the photoelectric effect, demonstrates that light behaves as particles. With this theory the common idea of amplitude does not apply to photons. Light is emitted and absorbed in mass-less quantum packets called photons. It is the bulk number of photons emitted or absorbed that is perceived as the brilliance of light or is measured as spectral power density of light. In other words, there are more or less bullets at a particular energy, but not bigger or smaller bullets.

The double slit experiment, and others, shows that photons behave as waves. They show interference patterns with adjacent nodes and anti-nodes appearing at integral wavelengths. OK, a photon is a quantum particle and therefore exhibits a dual nature as a wave. Makes sense to me. The shorter the wavelength, a higher frequency, and a higher energy is assigned to the photon. Photons with very high energy, such as gamma rays, have very high penetrating and destructive power.

Normally, one associates the amplitude of a wave with the energy of a wave. At a quantum level it is the wavelength of the photon that is associated with energy. I can intuitively feel how a tightly bound "angry" photon carries more energy than a lazy photon. This also makes sense to me.

However, if a photon is a wave, and a photon has a frequency, then where did the amplitude go? How can a wave exist that has no amplitude?

DraconicAcid - 23-1-2014 at 09:35

I'm sure it has an amplitude, but I'm not sure that it's measurable.

phlogiston - 23-1-2014 at 12:46

In the quantum interpretation, the photon (or any other particle) wave is a probability wave. The (squared) amplitude of this wave represents the likelyhood of finding that photon in a certain position. With large numbers of photons, it would be proportional to the number of photons at any given instant.
The classical amplitude of the electromagnetic wave at that point is then the sum of the contribution of all those photons interacting with a test charge at that location.

[Edited on 23-1-2014 by phlogiston]

a photon has no amplitude

quantumchromodynamics - 26-1-2014 at 02:03

After some careful study I believe my question is invalid. I was confusing classical and quantum waves. A photon transports a packet of momentum from place A to place B as the quantum variable called frequency. There is no quantum variable called amplitude that is a property of a photon. Amplitude is an emergent field property, classical field property, and only makes sense after wave function collapse.

In case any future scientist may care...

IrC - 26-1-2014 at 02:41

Interesting question, if amplitude is a scalar it seems one could term it's momentum an 'amplitude'.

"For example, the pressure of electromagnetic radiation on an object derives from the transfer of photon momentum per unit time and unit area to that object"

http://en.wikipedia.org/wiki/Photon

Just thinking out loud.

Marvin - 26-1-2014 at 05:31

The amplitude of the photon is the strength of the electric vector (and magnetic vector). Photons are not fermions, as many photons as you like can occupy the same volume, the amplitudes just add. Lasers, for example giant pulse ruby lasers can produce sparks in mid air, when the strength of the electric vector exceeds air's dielectric strength at the focus.

quantumchromodynamics - 26-1-2014 at 05:56

Cool Marvin! Is there a quantum of EM vector assigned to a photon? I thought a photon only had one quantum state variable. That variable is energy, thought about as wavelength, and hence frequency. I guess a Bose/Einstein condensate makes sense for a boson. Is a classic EM vector then assigned to a macro pile of photons, because photons are bosons, and a whole macro pile of them can occupy the same position in space?

stoichiometric_steve - 26-1-2014 at 08:13

Noob question:

Doesn't the amplitude correlate with the number of impacting photons at a certain wavelength, the amplitude being directly proportional to the number?

[Edited on 26-1-2014 by stoichiometric_steve]

phlogiston - 26-1-2014 at 12:22

Yes it is, as I wrote above as well.

Marvin, you are (perhaps deliberately) mixing quantum and classical views of that phenomenon.

In the spirit of Einsteins famous quote "Everything Should Be Made as Simple as Possible, But Not Simpler":
It would be oversimplifying to say that each photon is a little electromagnetic wave and these tiny waves add up to give a macroscopically observable wave with a larger amplitude proportional to the number of photons. The problem is that the photon wave is not a small electromagnetic wave. The photon wave is a probability wave, a pure quantum concept, and the macroscopically observable electromagnetic forces are caused by interactions of photons (absorption/emission) with charged particles.

In the example of a powerful laser ionising the air:

Classically: The EM vector exceeds the dielectric strength.

Quantum: Atoms are excited so rapidly by the large numbers of photons that they become ionised before they can decay to ground state. Incidentally, with a powerful enough laser, this process actually produces roentgen if electron vacancies can form in the inner shells.

[Edited on 27-1-2014 by phlogiston]

quantumchromodynamics - 26-1-2014 at 16:32

Quote: Originally posted by phlogiston  
The problem is that the photon wave is not a small electromagnetic wave. The photon wave is a probability wave, a pure quantum concept,


Right

phlogiston - 26-1-2014 at 16:50

As in "yeah right, dude" or as in "i agree"?

[Edited on 27-1-2014 by phlogiston]

i agree

quantumchromodynamics - 26-1-2014 at 22:58

phlogistion: Sorry for the brief quote and comment. I am a literal person, and I am confused by sarcasm. Your statement regarding the contextual meaning of the word "wave", as it pertains to either classical or quantum mechanics, helps sort this particular question, as well as other questions I am working on.

I believe the quotation "answers" the question.

I took your meaning as follows. A photon is a probability wave, (wave packet??), which is not the same thing as a tiny piece of a classical EM wave. Just because both classical and quantum waves have frequency, does not imply that both classical and quantum waves have amplitude. Frequency in a quantum sense, is not the same thing as frequency in classical sense. The word frequency is used because the two cases are similar, but they are not the same phenomenon.

This is just one example of a problem I am having. Many terms used in quantum mechanics are the same words used in classical mechanics. Concerning quantum mechanics, a subtle differences in a detail can lead to a radically different and unexpected conclusion. Also, depending on the age of article or publication, as concepts flesh themselves out, these subtle details change.

I continue to find assertions that seem both true and false at the same time.
This makes me crazy.

Gearhead_Shem_Tov - 27-1-2014 at 15:40

Quote: Originally posted by quantumchromodynamics  

... Many terms used in quantum mechanics are the same words used in classical mechanics. Concerning quantum mechanics, a subtle differences in a detail can lead to a radically different and unexpected conclusion. Also, depending on the age of article or publication, as concepts flesh themselves out, these subtle details change.

I continue to find assertions that seem both true and false at the same time.
This makes me crazy.


Well, lad, you aren't the first, and you won't be the last. Folks who claim to "understand" QM really mean they understand QM's formalism -- the equations. Nobody really understands what it means in a gut-level, intuitional way, though. Perhaps if humans had evolved in a universe where quantum phenomena were readily apparent at the macro level then our brains might be wired to "get" it the same way we "get" Newtonian mechanics.

Failing that, we have only looking at the equations and trying to ferret out predictions; many of them are bizarre, most of them utterly are counterintuitive, and, yet, QM is still the most comprehensively tested and verified theory we have ever had.

-Bobby

a possible insight about waves

quantumchromodynamics - 27-1-2014 at 21:44

QM: Quantum mechanical
NM: Newtonian mechanical

I woke up thinking about mechanics with a possible insight. Maybe the essential term "wave" has a different meaning in QM than it does in NM.

Possibly:

A QM wave is a set theoretic vector of states that evolves according to a wave equation. It is only in the sense that sets of states get repeated as the state vector moves along, does a QM wave seem to "wiggle". Since waves wiggle, the word "wave" got used.

The wiggle of a QM wave still has time dependent frequency, but not any definite amplitude. A QM wave is quite different from an NM wave in this regard.

phlogiston - 28-1-2014 at 03:08

A probability function does have amplitude. To make it meaningful it is normalized, such that its integral over all of the universe equals 1. In other words, the particle has a 100% chance of being somewhere. Then, the integral of the square of the amplitude over a smaller region of space represents the chance of detecting the particle in that region.

stoichiometric_steve - 30-1-2014 at 16:51

Quote: Originally posted by phlogiston  
In other words, the particle has a 100% chance of being somewhere.


you made my day. now i'm really happy for that particle and its 100% chance of being somewhere! someone should make a motivational poster out of this.

forgottenpassword - 31-1-2014 at 03:11

Mathematics and reality do not necessarily coincide.

phlogiston - 31-1-2014 at 05:18

youtube: Feynman: Mathematicians versus Physicists

quantumchromodynamics - 1-2-2014 at 01:17

Respectfully, a probability wave function does not have an amplitude. A probability wave function describes a distribution of possibilities. In the same way, a photon is not a probability wave function. The former is a property of nature, while the latter is a mathematical abstraction.

My intention is to clarify the meanings of certain (subtle, confusing) terminology used in the description of QM.

QM is a battlefield.

phlogiston - 1-2-2014 at 03:59

I am not sure I fully understand your reasoning, please explain a bit more.

Indeed the probability wave function describes a distribution of possibilities, but that seems compatible to me with the wave function having an amplitude. It is even sometimes literally called a probability amplitude .

[Edited on 1-2-2014 by phlogiston]

neptunium - 1-2-2014 at 12:27

i think the probability amplitude is just a term in probability to determine the likelyhood of presence it has not much to do with a photon amplitude.
this whole debate is interesting because any electromagnetic wave (as described before) has 2 components
one electric
one magnetic
the frequency of wich determine the energy (as in E=Hv)
because the speed of light is the same regardless , the amplitude (i assume) must be the same for any photon on any part of the electromagnetic spectrum.
because the number of vibrations per unit of time only determine the energy ,the amplitude must be the same if we reason by classical physic. otherwise the photon (or quanta) spend too much time reaching high amplitude and lows and looses energy therefore frequency wich does not support observations .

back again to this topic

quantumchromodynamics - 19-3-2014 at 01:09

Gentlemen, for the past month I have been indirectly researching this topic by going over a friends PhD work at the University of Oregon in QM. Like a bad music song, or an "ear worm", that you can't stop playing in your head, this issue has become a "science worm", and keeps playing over and over again in my head. Is this science "madness?"

I have a greater familiarity with the formal mathematics of QM as it is currently taught at university. My study further underscores there is generally a problem with terminology used to teach and to describe QM. As one example, the term "amplitude" as related to a QM wave function is not fundamental or intuitive, and it is the exception rather than the rule, that an author takes the time to even indicate a difference exists. In QM a photon amplitude is often defined to be a complex unit vector, without a classical magnitude, and only a spin direction, so the photon can participate in wave like superposition. A question. What does this definition of amplitude have to do with the energy of the photon? The answer. Nothing. These are two different and valid applications of the term amplitude. Could this subject possibly be more ironic or confusing?

Even more central to the issue, a quantum state vector, or a quantum basis vector, or a quantum ket, certainly implies properties of a classical system state vector. However the actual definition of a ket is generally skipped over. This happens because the differences are hard to understand and chasing them down and figuring out what they mean is hard work. I think the situation is similar to the subtle difference between the terms "uncertainty" and "in-determinism". The proper choice of terms is hugely important, especially here, as it changes ones' entire relationship with the universe. Is probability intrinsic to reality, or is probability a tool for computing reasonable answers to questions we don't know enough stuff about? Are my actions determined before hand, or do I have free will? Shall I go and kill that Arab on the beach?

As an aside, this is what I think about kets. A quantum ket represents knowledge about a chunk of a quantum state vector. Broken off chunks of state vectors make sense inside of QM calculations and there just isn't an analogy in a classical sense.

With clarity, Neptunium hits my original thoughts right on the nose, and expresses the ambiguity using good terminology. In some important sense, the term amplitude is inherently vague, and this is not good.