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Ray McDavis
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Waves are energy moving. Light is moving packets of fixed amounts of energy. Why must we invoke particles to understand light? How is a wave (energy moving) antithetical to packets of energy?
In what way does light behave that is incompatible with waves?
In summary, waves are energy moving, while photons are particles. We must invoke particles to understand light because low intensity light always detects discrete events, while waves only describe the average behavior of a huge number of individual detections.
- #2
PeterDonis
Mentor
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Because when we detect low intensity light, we always detect discrete events. For example, in a double slit experiment, if you run it with the light intensity fairly high, an interference pattern will appear on the detector screen, but as you turn the light intensity down further and further, you don't see the complete pattern all at once but fainter and fainter. You see the pattern resolve into individual dots that hit different points on the screen at random times, and the interference pattern gets built up over time by the dot impacts.Ray McDavis said:
So the wave patterns that you get from the classical theory of light don't describe what we actually detect when we detect low intensity light. They only describe the average behavior of a huge number of individual detections (and the reason high intensity light looks like a continuous wave pattern is that there are too many detections happening in too short a time interval for us to resolve them into individual events).
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It was Einstein's explanation for the photoelectric effect that began the quantised theory of light:Ray McDavis said:
https://en.wikipedia.org/wiki/Photoelectric_effect
- #4
DrChinese
Science Advisor
Gold Member
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Ray McDavis said:
To add to the correct answers PeterDonis and PeroK: The answer to this gets complicated fairly quickly. When the number of light waves is a fixed number (such a 1, 2, 3, etc - a/k/a a Fock state), light can present different properties than when the particle number can vary. In the below paper, which is intended for an undergraduate physics audience, the particle number is exactly 1. Note that the experimental detection of a single particle of light (photon) at a specific expected place is itself a strong indicator that light is not simply a wave.
http://people.whitman.edu/~beckmk/QM/grangier/Thorn_ajp.pdf
Abstract: "While the classical, wavelike behavior of light interference and diffraction has been easily observed in undergraduate laboratories for many years, explicit observation of the quantum nature of light (i.e., photons) is much more difficult. For example, while well-known phenomena such as the photoelectric effect and Compton scattering strongly suggest the existence of photons, they are not definitive proof of their existence. Here we present an experiment, suitable for an undergraduate laboratory, that unequivocally demonstrates the quantum nature of light. Spontaneously downconverted light is incident on a beamsplitter and the outputs are monitored with single-photon counting detectors. We observe a near absence of coincidence counts between the two detectors—a result inconsistent with a classical wave model of light, but consistent with a quantum description in which individual photons are incident on the beamsplitter. More explicitly, we measured the degree of second-order coherence between the outputs to be g(2)(0)=0.0177+/-0.0026, which violates the classical inequality g(2)(0)>=1 by 377 standard deviations."
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- #5
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Indeed, one should abandon the teaching of photons as being "particles" at all. Photons are very special, because they are the quanta of a spin-1 massless quantum field, and as a detailed mathematical analysis shows, there is no way to even define a position observable for photons. All that is defined in the sense of a possible "localization of photons" are probabilities for the detection of a photon at a given place, where the detector is located. E.g., you can put a photoplate (or a CCD cam as its modern equivalent) somewhere and detect electromagnetic radiation. If the electromagnetic radiation is of sufficiently low intensity you can observe the statistical nature of its detection in single small spots on the screen. It's pretty likely in this case that the detection event represents the absorption of a single photon by the material of the detector.
On the other hand, as is stressed in the above quoted abstract, whether you really have single photons before these detection events took place depends rather on the preparation of the em. radiation, i.e., it's source. In fact what's usually quoted as demonstrating the existence of photons in introductory sections even of physics textbooks at the university level, is not accurate. Indeed, the photoelectric effect and the Compton effect are entirely describable at the level quoted in the textbooks in the semiclassical approximation, i.e., treating the electromagnetic radiation as a classical field. The quantum aspects and ##\hbar## enter only through the quantization of the matter (for the photoelectric effect bound electrons for the Compton effect (quasi-) free electrons).
On the other hand, as is stressed in the above quoted abstract, whether you really have single photons before these detection events took place depends rather on the preparation of the em. radiation, i.e., it's source. In fact what's usually quoted as demonstrating the existence of photons in introductory sections even of physics textbooks at the university level, is not accurate. Indeed, the photoelectric effect and the Compton effect are entirely describable at the level quoted in the textbooks in the semiclassical approximation, i.e., treating the electromagnetic radiation as a classical field. The quantum aspects and ##\hbar## enter only through the quantization of the matter (for the photoelectric effect bound electrons for the Compton effect (quasi-) free electrons).
- #6
MRMMRM
- 15
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"In particular, black body radiation, photoluminescence, generation of cathode rays from ultraviolet light and other phenomena associated with the generation and transformation of light seem better modeled by assuming that the energy of light is distributed discontinuously in space. According to this picture, the energy of a light wave emitted from a point source is not spread continuously over ever larger volumes, but consists of a finite number of energy quanta that are spatially localized at points of space, move without dividing and are absorbed or generated only as a whole."
https://sites.pitt.edu/~jdnorton/le...05_docs/Einstein_Light_Quantum_WikiSource.pdf
You must invoke paticles to understand the different energy properties of light, because light is not packets of fixed energy. Photons are a fixed delta in space or time, "C". Like L*W*H determines volume, not density. The Frequency of the EMF determines their individual energy, not total light energy. Our eyes detects different frequencies as different colors. The brightness of a single color can vary. This is not because the energy of each individual packet varies, but the amount of packets entering our eyes does. Of course the EM spectrum extends out much further on both ends than what our eyes can perceive.
If you want to think of it as a fixed energy point of view, then you would have to think you could emit more red photons than violet photons, if you had sufficent energy to make them. You would possibly also have a remainder lower frequency EMF from the energy left over, unable to make a light photon.
Imagine seeing two ocean waves the same height hit a break wall, but one gets reflected by the break wall, and the other pushes the wall and everything behind it inland 100ft. It would be a strange sight, but that's what light waves are like. Not exactly like other observable waves in how they carry energy. Based on observed water level, one wave you would say that behaved wave like, and the other, you would say it behaved particle like.
https://sites.pitt.edu/~jdnorton/le...05_docs/Einstein_Light_Quantum_WikiSource.pdf
Ray McDavis said:
You must invoke paticles to understand the different energy properties of light, because light is not packets of fixed energy. Photons are a fixed delta in space or time, "C". Like L*W*H determines volume, not density. The Frequency of the EMF determines their individual energy, not total light energy. Our eyes detects different frequencies as different colors. The brightness of a single color can vary. This is not because the energy of each individual packet varies, but the amount of packets entering our eyes does. Of course the EM spectrum extends out much further on both ends than what our eyes can perceive.
If you want to think of it as a fixed energy point of view, then you would have to think you could emit more red photons than violet photons, if you had sufficent energy to make them. You would possibly also have a remainder lower frequency EMF from the energy left over, unable to make a light photon.
Imagine seeing two ocean waves the same height hit a break wall, but one gets reflected by the break wall, and the other pushes the wall and everything behind it inland 100ft. It would be a strange sight, but that's what light waves are like. Not exactly like other observable waves in how they carry energy. Based on observed water level, one wave you would say that behaved wave like, and the other, you would say it behaved particle like.
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MRMMRM said:
Unfortunately, what you say is not correct. Photons are not localized in space, although they are the quanta of the quantized Electromagnetic field. See, for example:
https://www.physicsforums.com/threads/why-no-position-operator-for-photon.906932/
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MRMMRM
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PeroK said:
Unfortunately, what you say is not correct. Photons are not localized in space, although they are the quanta of the quantized Electromagnetic field. See, for example:
https://www.physicsforums.com/threads/why-no-position-operator-for-photon.906932/
Plancks law, sure does descibe black body radiation nicely. It doesn't ignore position, but ensures that it's defined.
Frequency and momentum are the same thing.
Photons are moving at a constant speed, and you can't give them a single defined position, other wise they would cease to exist. Just like if a moving mass stopped, its momentum would cease to exist. That doesn't mean the energy or mass are not localized in space. Thinking of them as a particle occuping a volume of space is ok, just know that, that energy is never stationary or accelerating while it exist as a photon.
Then if you wanted to know a photons "inertia" or "moment of inertia" if you think of them as spinning, well I would say that is Plancks constant.
- #9
Nugatory
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These two statements are mutually inconsistent. Speed is defined as the magnitude of the velocity and velocity is defined as the change of position with time, so if we can’t assign a definite position to a photon we can’t define its speed either (although the momentum of a photon is OK because we have a definition other than the the classical ##p=mv##).MRMMRM said:
But if the two statements are mutually inconsistent they can’t both be right…. And the resolution is that photons don’t have a position in the usual sense of the word, so “photon speed” is not an especially meaningful concept. One way of seeing this is to consider a standing electromagnetic wave in a cavity: where are the photons and in what sense could we say that they are moving? To properly understand the role of photons in this case we need to do something along the lines of these class notes (which are no substitute for a real quantum electrodynamics course, but get the flavor across).
- #10
aguarneri
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Talking about the experiments on the nature of light (waves or particles) does it make a difference if the light is polarized (meaning, if I understand correctly, that the waves "vibrate" all in the same plane) or not polarized ? I am not sure if my question is relevant as experiments are on individual photons and not about a beam of light.
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- #11
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Planck's law is about thermal radiation in a cavity. In a cavity the photons, i.e., the normal modes, are not localized at all, because they are standing waves.MRMMRM said:
Also free photons, moving in free space are not localizable at all, at least not in direction of their propagation. You can't even formally define a position operator fulfilling all the properties it must have as such.
- #12
HomogenousCow
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The canonical example for why a classical treatment of electromagnetism fails is the divergence of the field energy density in a thermalized cavity. To simplify the algebra let's look at a scalar wave equation in 1-D, the qualitative results are the same.
In this case the Hamiltonian is
$$H = \int d^3x~ \frac{1}{2}\pi^2+\frac{1}{2}c^2(\nabla \phi)^2$$ where ##\pi## is the conjugate momentum field. A general solution which obeys the boundary conditions ##\phi(0)=\phi(L)=0## is given by
$$\phi = \sum^\infty_{n=1}\left[A_n \cos(\frac{n\pi c}{L}t) + B_n \sin(\frac{n\pi c}{L}t)\right] \sin(\frac{n\pi}{L}x)$$ Using ##\pi = \dot \phi##, the Hamiltonian simplifies greatly to
$$H = \frac{\pi^2c^2}{4L}\sum^{\infty}_{n=1} n\left(A_n^2+B_n^2\right)$$ which looks like an infinite sum of harmonic oscillators with frequencies increasing as ##\sqrt n##. When we plug this into the boltzman distribution and calculate the partition function we get$$Z = \int \left(\prod^{\infty}_{n=1}dA_n dB_n\right) e^{-\beta H} \propto \prod^{\infty}_{n=1} \frac{1}{ {\beta n}}$$ which doesn't look very promising. Indeed the average energy is then $$\langle E \rangle = -\frac{\partial}{\partial \beta} \log Z = \sum^{\infty}_{n=1} k T \to \infty$$ A similar result follows if we consider ##\mathbf{E}## and ##\mathbf{B}## instead of ##\phi##; the energy of the field in thermal equilibrium appears to be infinite. The problem is that the system is equivalent to an infinite number of harmonic oscillators each requiring ##kT## of energy to reach thermal equilibrium.
If we look at this system in the framework of quantum mechanics something interesting happens and this divergence goes away. The above discussion is actually valid in quantum mechanics up to the third equation
$$H = \frac{\pi^2c^2}{4L}\sum^{\infty}_{n=1} n\left(A_n^2+B_n^2\right)$$ All we have to do is promote ##A_n## and ##B_n## to operators, in fact they simply correspond to ##\hat{x}## and ##\hat{p}## for the harmonic oscillator of mode ##n##. Now that we have quantized harmonic oscillators, the possible energies become discrete and in fact each mode can only have energy $$E_{n} = \frac{\pi^2 c^2 }{2L}n \times \hbar m = \gamma nm$$ where ##m## is an integer and ##\gamma = \pi^2 c^2 \hbar / 2L##. The partition function for this quantized system is then
$$Z = \left(\prod^{\infty}_{n=1}\sum^{\infty}_{m=0}\right) \exp \left(-\beta \sum_{n=1}^{\infty} \gamma n m\right) = \prod^{\infty}_{n=1} \frac{1}{1-\exp(-\beta \gamma n)} $$ The average energy of the system is then $$\langle E \rangle = \sum^{\infty}_{n=1} \frac{n \gamma}{\exp(\beta \gamma n)-1}$$ What this means is that each mode ##n## of the system has an average value of ##m## given by $$\langle m \rangle = \frac{1}{\exp(\beta \gamma n)-1}$$ which is the famous Bose-Einstein distribution.
In this case the Hamiltonian is
$$H = \int d^3x~ \frac{1}{2}\pi^2+\frac{1}{2}c^2(\nabla \phi)^2$$ where ##\pi## is the conjugate momentum field. A general solution which obeys the boundary conditions ##\phi(0)=\phi(L)=0## is given by
$$\phi = \sum^\infty_{n=1}\left[A_n \cos(\frac{n\pi c}{L}t) + B_n \sin(\frac{n\pi c}{L}t)\right] \sin(\frac{n\pi}{L}x)$$ Using ##\pi = \dot \phi##, the Hamiltonian simplifies greatly to
$$H = \frac{\pi^2c^2}{4L}\sum^{\infty}_{n=1} n\left(A_n^2+B_n^2\right)$$ which looks like an infinite sum of harmonic oscillators with frequencies increasing as ##\sqrt n##. When we plug this into the boltzman distribution and calculate the partition function we get$$Z = \int \left(\prod^{\infty}_{n=1}dA_n dB_n\right) e^{-\beta H} \propto \prod^{\infty}_{n=1} \frac{1}{ {\beta n}}$$ which doesn't look very promising. Indeed the average energy is then $$\langle E \rangle = -\frac{\partial}{\partial \beta} \log Z = \sum^{\infty}_{n=1} k T \to \infty$$ A similar result follows if we consider ##\mathbf{E}## and ##\mathbf{B}## instead of ##\phi##; the energy of the field in thermal equilibrium appears to be infinite. The problem is that the system is equivalent to an infinite number of harmonic oscillators each requiring ##kT## of energy to reach thermal equilibrium.
If we look at this system in the framework of quantum mechanics something interesting happens and this divergence goes away. The above discussion is actually valid in quantum mechanics up to the third equation
$$H = \frac{\pi^2c^2}{4L}\sum^{\infty}_{n=1} n\left(A_n^2+B_n^2\right)$$ All we have to do is promote ##A_n## and ##B_n## to operators, in fact they simply correspond to ##\hat{x}## and ##\hat{p}## for the harmonic oscillator of mode ##n##. Now that we have quantized harmonic oscillators, the possible energies become discrete and in fact each mode can only have energy $$E_{n} = \frac{\pi^2 c^2 }{2L}n \times \hbar m = \gamma nm$$ where ##m## is an integer and ##\gamma = \pi^2 c^2 \hbar / 2L##. The partition function for this quantized system is then
$$Z = \left(\prod^{\infty}_{n=1}\sum^{\infty}_{m=0}\right) \exp \left(-\beta \sum_{n=1}^{\infty} \gamma n m\right) = \prod^{\infty}_{n=1} \frac{1}{1-\exp(-\beta \gamma n)} $$ The average energy of the system is then $$\langle E \rangle = \sum^{\infty}_{n=1} \frac{n \gamma}{\exp(\beta \gamma n)-1}$$ What this means is that each mode ##n## of the system has an average value of ##m## given by $$\langle m \rangle = \frac{1}{\exp(\beta \gamma n)-1}$$ which is the famous Bose-Einstein distribution.
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- #13
MRMMRM
- 15
- 3
https://gilles.montambaux.com/files/histoire-physique/Bose-1924.pdf
Bose applied Plancks law to the classical electromagnetism, Quantizing the field down to the photon.
Dirac went on to refine this for electrons appling relativity to them as waves.
but,
Really Feynman said it best.
"A photon goes from one place and time to another place and time.
An electron goes from one place and time to another place and time.
An electron emits or absorbs a photon at a certain place and time."
So you can't really say a photon (light) is a particle, but it has helped tremendously advance physics.
Bose applied Plancks law to the classical electromagnetism, Quantizing the field down to the photon.
Dirac went on to refine this for electrons appling relativity to them as waves.
but,
Really Feynman said it best.
"A photon goes from one place and time to another place and time.
An electron goes from one place and time to another place and time.
An electron emits or absorbs a photon at a certain place and time."
So you can't really say a photon (light) is a particle, but it has helped tremendously advance physics.
1. What is the wave-particle duality of light?
The wave-particle duality of light refers to the fact that light exhibits both wave-like and particle-like behavior. This means that in some experiments, light behaves like a wave, while in others it behaves like a stream of particles called photons.
2. How does light exhibit wave-like behavior?
Light exhibits wave-like behavior through phenomena such as diffraction, interference, and polarization. These behaviors can be explained by the wave nature of light, as described by the electromagnetic wave theory.
3. What are some characteristics of light waves?
Light waves have properties such as wavelength, frequency, and amplitude. They can also travel through a vacuum at the speed of light, and can be reflected, refracted, and diffracted.
4. In what ways does light behave like a particle?
Light behaves like a particle in that it can be emitted and absorbed in discrete packets of energy called photons. These photons have a specific energy and momentum, and can interact with matter in a particle-like manner.
5. What are some experiments that demonstrate the particle nature of light?
The photoelectric effect, Compton scattering, and the double-slit experiment are all experiments that demonstrate the particle nature of light. These experiments show that light can behave like a stream of particles, rather than a continuous wave.
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