I What is a Photon? - Physics Basics Explained

[Fooality]

But geometric optics is already a shortcut to quantum optics and the Maxwell equations - so for further shortcuts you need to go into the other direction - simplifying geometric optics. This has no longer anything to do with quantum physics!

Yes, and there are many attempts to do just that people are working on. My curiosity about QM came from the fact that the concept of rays is awkward for computation. I wondered if there was some other view, maybe waves or something else as far as a simplified approximation that might be out there. Its really not important unless its interesting to you. Maybe I'm curious because of quantum computers, or quantum limits in chip manufacture, or anything else. QM is and will be a big part of the computer world. My only real point is I for one like the simplified views people offer...

 

[vanhees71]

Well, to any complicated problem there's always a simple answer, which, however, is usually wrong ;-)).

 

[lox_and_whiskey]

Great article. No better way to introduce photons then talking about harmonic oscillators. A simple harmonic oscillator is anything with a linear restoring potential. Simple things like a spring, or a string with tension, or a wave. Erwin Schrödinger described mathematically how a harmonic oscillator stores energy and how to calculate how much energy it stores.

For the photon, the value of this restoring potential is the Planck constant (h). Planck’s constant, relates the amount of energy stored in a photon to its wavelength (λ). Planck’s constant tells you the amount of time it takes the photon to undergo one cycle of whatever its doing given that the photon has a specific amount of energy. The equation E for energy = h / λ, tells us that a photon with low energy will take much longer to complete one cycle of the wave then a photon with high energy.

My favorite model of a photon as a harmonic oscillator is as an expanding and contracting ball of energy flying though the air. It immediately lends understanding to the particle and wave nature of the photon. A photon of a specific wavelength has a specific energy. The properties of a photon change periodically over time and distance depending on the wavelength. High energy photons oscillate very fast and store a lot of energy, low energy photons oscillate very slowly. Photons can be in the same place at the same time, sometimes reinforcing each other, sometimes cancelling each other out and it looks like there are no photons at all. The uncertainty principle: ΔxΔp ≥ h/4π falls from this. The photon is either big affecting a wide area, or it’s tiny and only affecting one small area, it cannot be both at the same time. The importance of visualizing the photon as a harmonic oscillator cannot be overstated.

That's a very nice explanation.

As a general question i'd be curious if something were elaborated on, is there any difference between vacuum and two photons canceling each other out?

 

[vanhees71]

Two photons cannot simply cancel out in the vacuum due to energy-momentum conservation. You can have (theoretically) processes like inverse pair annihilation, i.e., $\gamma \gamma \rightarrow \mathrm{e}^+ \mathrm{e}^-$.

 

[A. Neumaier]

waves or something else as far as a simplified approximation

waves are not a simplification but a computational burden as they must be computed at every point in space and not only (as rays) where they meet a surface. Thus it is far more expensive to work with waves than with rays. Forget quantum mechanics for image rendering. Also forget quantum computing (at least for the next 10 years) - at present, they cannot even sort a list of 1000 items in an acceptable time, even in a very generous view.

 

[A. Neumaier]

You can have (theoretically) processes like inverse pair annihilation,

But only at energies high enough to convert it into the mass of two electrons. This requires a very large intensity.

 

[vanhees71]

Indeed, I've somewhere read that people are after an experimental verification for this process (in lowest order perturbation theory a pure QED process), but as far as I know, it's not yet observed.

 

[maline]

My favorite model of a photon as a harmonic oscillator is as an expanding and contracting ball of energy flying though the air. It immediately lends understanding to the particle and wave nature of the photon

Sorry, but this is just wrong- not in the sense of "oversimplified" like the models George Box referred to, but simply unrelated to reality. The conception of a "photon" that "flies through the air" is presumably a classical electromagnetic wave packet. This is, I think, acceptable as a (very) oversimplified model. But waves do not "expand and contract as they move"! The motion of a wave consists precisely of "following" a peak of the wave, while the field & energy values at the peak are constant (or diminishing if the wave spreads out).
A good way to think of a wave is as a chain of oscillators (say springs, connected end to end, with small frictionless masses between each pair). When one is "energized" (compressed) it can relax by passing the energy on to the next spring. This makes a "ripple" that passes along the chain. If one spring is moving harmonically, the next one will also oscillate with a slight delay, and you get a moving sine wave. There is no one oscillator that moves.

 

[Fooality]

waves are not a simplification but a computational burden as they must be computed at every point in space and not only (as rays) where they meet a surface. Thus it is far more expensive to work with waves than rays...

[Mentor's note: A digression on quantum computing has been moved into its own thread: https://www.physicsforums.com/threads/status-of-quantum-computing.880521/]

I don't know what you mean here. With classical waves, like audio, its pretty straightforward to compute their value along a surface some distance from the source.

You're shooing me away from QM, but I don't see the problem with looking into it. For instance, in the ray tracing model tells me a photon moves in a straight line, and if it hits, say, a mirror, it bounces off at the angle of incidence every time. But when I listen to Feynman's talks, he says that point it hit in the mirror only emerges as a probability, given by his path integral. So 'rays' only probably exist, right? If they were a computationally graceful lie, I'd run with it. But they're truly not. Read the second paragraph of the wiki article on angle of incidence for one of the many reasons. Its not the fact its a lie that bothers me, its the bizarre idea that its the ONLY lie which approximates the truth that does.

Referring to to the best scientific model for a system, when seeking to simulate it, just doesn't strike me as a far out idea.

 

[A. Neumaier]

Referring to to the best scientific model for a system, when seeking to simulate it

Why don't you then start with the standard model? This is the best scientific modle for reality on Earth that we currently have. You'll find that, to caclulate anything of interest to you, you need to climb up the standard ladder of approximations until you reach ray optics and even that - the highest rung of the ladder, simpler than the Maxwell equations - is not fast enough, as you complained. There is no use at all starting at very accurate but expensive descriptions when the simpler (and still fairly accurate) models are already too slow.

 

[Fooality]

Why don't you then start with the standard model? This is the best scientific modle for reality on Earth that we currently have. You'll find that, to caclulate anything of interest to you, you need to climb up the standard ladder of approximations until you reach ray optics and even that - the highest rung of the ladder, simpler than the Maxwell equations - is not fast enough, as you complained. There is no use at all starting at very accurate but expensive descriptions when the simpler (and still fairly accurate) models are already too slow.

I would love to learn the standard model in depth. Even if none of it pans out in terms of computation, the more I learn, the more I realize its a worthwhile thing just to know.

Why, after all, are you a physicist? Done you feel a certain thrill at understanding this universe we live in at a deeper level? Don't blame others for feeling the same thrill, even if they know less about it.

 

[A. Neumaier]

Don't blame others for feeling the same thrill

I don't blame you for wanting to learn quantum mechanics. I just wanted to warn you that it will not give you a faster way of creating photorealistic images. By the way, a PhD student of mine wrote his thesis on accelerating ray tracing. It was long ago, though. See http://www.sbras.ru/interval/Library/Thematic/CompGraph/DivideConquer.pdf

 

[exponent137]

Bill,
Your transfer from the last equations in page 6 in your link
http://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf
is fine. This is a direct transfer from classical to quantum harmonic oscillator. Who is a author of this text, you?

But, I think that calculation of quantum harmonic oscillator can be still simpler. Namely, in appropriate units, classical harmonic oscillator can be described as a circling in circle in a plane $x$ and $p_x$:
$2H=x^2+p^2$
Because of this circle, without solving this differential equation we can easily guess that
$x = x_0 e^{-i\omega}$
Classical creation and destruction operators $a^+ ,a$ are in agreement with this way of thinking:
$a=x+ip_x$,
because this means the radius of this circle. Thus it is not necessary to calculate quantum $\hat{a}$, but we can guess it.

Additionally, when we go into QM oscillator, we obtain additional part $1/2$ in $n+1/2$, which is a consequence of uncertainty principle.

Because of this, I think, that QM oscillator can be much easier derived. But, for instance, if we concentrate on equations:
$\hat{a} u_n= n^{1/2} u_{n-1}$ (1)
$\hat{a}^+ u_n= (n+1)^{1/2} u_{n+1}$ (2)
I think that derivation of these equations is too long. Why to use complicated Hermitian polinoms and so on, if the result is so short?
Do you know any other derivations of (1) or (2)? Maybe, if we use a wave function which is mix of $x$ and $p$ representations, it can be easier to obtain these simple results?p.s. I hope that you all understand what the above simbols mean?

 

[nikkkom]

My favorite model of a photon as a harmonic oscillator is as an expanding and contracting ball of energy flying though the air. ... The uncertainty principle: ΔxΔp ≥ h/4π falls from this. The photon is either big affecting a wide area, or it’s tiny and only affecting one small area, it cannot be both at the same time.

This can't be correct. A photon with fixed energy has a completely undetermined position. Hence, it's totally unlike "ball flying though the air" - when one visualizes a ball, it has a known position.

 

[nikkkom]

I'm looking for a description of photons in the language of U(1) gauge group.
My understanding is that the explanation of electromagnetism as U(1) gauge group works as follows:

Every point in space has a U(1) value (a complex number with absolute value 1) "attached" to it. These values are generally not the same everywhere.

Multiplying all these values everywhere by constant U(1) value is unobservable. (This multiplication is often termed "rotation" because multiplying complex numbers with absolute value 1 move them around the unit cicrle in the complex plane).

However, multiplying them by non-constant (varying in space) U(1) values is observed as existence of electromagnetic field.

This far it's clear. Now, *how exactly these values vary through space* in a few typical electromagnetic setups? This is where I don't have a clear picture.

(1) a constant electric field between charged plates?
(2) a constant magnetic field (say, inside a solenoid)?
(3) a photon?

 

[maline]

A photon with fixed energy has a completely undetermined position

True, but you can define a photon that has a spread of energies. See Vanhees' post #19 in this thread.

If I understood correctly, this is a superposition of the various pure energy-momentum-spin one-photon modes, with coefficients that square-integrate to unity.

 

[vanhees71]

This can't be correct. A photon with fixed energy has a completely undetermined position. Hence, it's totally unlike "ball flying though the air" - when one visualizes a ball, it has a known position.

No photon has any kind of position, because there's no position observable for a photon. I can't count, how often I've mentioned this only in this thread! :-(

 

[maline]

No photon has any kind of position, because there's no position observable for a photon

Well there must be some form of "spatial" difference between a pure momentum one-photon mode and a photon immediately after emission from an atom.

How about defining "position" as follows: If I understand correctly, the electric and magnetic field strengths at each point are observables. Thus we can calculate the Hamiltonian at each point and get an expectation value for the energy density there. So we have a description of an energy distribution over space. We can refer to the center of this distribution as the photon's position, and the second moment will describe how "spread out" it is.

 

[vanhees71]

But this doesn't define a position operator. Of course, it's all you can observe, namely the detection probability with the detector placed at a certain location.

 

[A. Neumaier]

We can refer to the center of this distribution as the photon's position

This defines the field's position, not the photon's position, as there is only a single position of this kind for any N-photon state!

 

[maline]

But this doesn't define a position operator. Of course, it's all you can observe, namely the detection probability with the detector placed at a certain location.

Okay, I am just looking for a meaningful way of describing "where the photon is". The shape of this distribution will depend on the momentum distribution, in a way that is very similar to the position probability distribution for a massive particle. Correct?

This defines the field's position, not the photon's position, as there is only a single position of this kind for any N-photon state!

Right, I was describing one-photon states. But thank you for pointing this out. In general, is there any good way to describe a two- photon state as the sum of two one-photon states? How about as a member of the tensor product of two "one-photon" state eigenspaces?

 

[A. Neumaier]

"where the photon is".

The photon is somewhere where the e/m field is. Thus in typical experiments, it is somewhere in the beam. If the beam goes in z-direction, the photon has a fairly well-defined x, y, but no definable z, while it has a fairly well-defined $p_z$, but very uncertain $p_x$ and $p_y$. This is consistent with the uncertainty relations.
The operators that are simultaneously measured are the momentum in the direction of the direction of flight and two transversal position coordinates. These are given by well-defined operators. Their preparation or measurement is enough to tell where the photons are destroyed at the screen, but not the moment in time (which is well known to be random).

 

[maline]

If the beam goes in z-direction, the photon has a fairly well-defined x, y, but no well-defined z,

In what sense of "position" are the x and y coordinates well-defined?

 

[A. Neumaier]

In what sense of "position" are the x and y coordinates well-defined?

In the sense that transverse operators exist for them. The obstruction to the existence of a complete position operator with 3 commuting components is gauge invariance, which eliminates the longitudinal degree of freedom that other particles have. But for a photon in a beam (idealized as a state with exact $p_z$), one can choose a frame in the direction of motion, and in this particular coordinate system one can construct two commuting operators giving the transversal position. This is indeed what we do observe, and explains everything naturally.

 

[edguy99]

That's a very nice explanation.

As a general question i'd be curious if something were elaborated on, is there any difference between vacuum and two photons canceling each other out?

Thanks, WRT difference between vacuum and photon cancellation, they look the same, but I think you see the difference over time. If two streams of photons are flying together with slightly different wavelengths, you will see at certain times and locations where there appear to be no photons (due to cancellation) - and it looks like a vacuum, and other locations with lots of photons (reinforcement).

 

[edguy99]

Two photons cannot simply cancel out in the vacuum due to energy-momentum conservation. You can have (theoretically) processes like inverse pair annihilation, i.e., $\gamma \gamma \rightarrow \mathrm{e}^+ \mathrm{e}^-$.

The photon cancellation in interference is in the range of 1 to 5 eVolts where say a stream of 1.97 eVolt red photons. Pair production would be at much higher energy levels.

 

[edguy99]

Sorry, but this is just wrong- not in the sense of "oversimplified" like the models George Box referred to, but simply unrelated to reality. The conception of a "photon" that "flies through the air" is presumably a classical electromagnetic wave packet. This is, I think, acceptable as a (very) oversimplified model. But waves do not "expand and contract as they move"! The motion of a wave consists precisely of "following" a peak of the wave, while the field & energy values at the peak are constant (or diminishing if the wave spreads out).
A good way to think of a wave is as a chain of oscillators (say springs, connected end to end, with small frictionless masses between each pair). When one is "energized" (compressed) it can relax by passing the energy on to the next spring. This makes a "ripple" that passes along the chain. If one spring is moving harmonically, the next one will also oscillate with a slight delay, and you get a moving sine wave. There is no one oscillator that moves.

I disagree. A stream of expanding and contracting balls will look exactly like a wave. You can see models of a stream of 50 photons, each 1.97 evolts over a period of 4.5 femtoseconds. The animation starts with a single 630nm photon trapped in a 2520nm cavity and grows to a photon stream looking just like a wave and interfering just like a wave. http://www.animatedphysics.com/games/photon_cavity.htm

All visual representations have value and I do see value in modelling this type of motion as oscillations in your chains. I prefer to think of the oscillation of the photon it self.

 

[edguy99]

This can't be correct. A photon with fixed energy has a completely undetermined position. Hence, it's totally unlike "ball flying though the air" - when one visualizes a ball, it has a known position.

The ball has a fixed location, only if you know where its center is. If that ball has expanded quite large, and your one measurement told you that you were within the ball, you would not know where the center is. So no, the ball does not have a known position.

 

[maline]

I disagree. A stream of expanding and contracting balls will look exactly like a wave. You can see models of a stream of 50 photons, each 1.97 evolts over a period of 4.5 femtoseconds. The animation starts with a single 630nm photon trapped in a 2520nm cavity and grows to a photon stream looking just like a wave and interfering just like a wave. http://www.animatedphysics.com/games/photon_cavity.htm

I repeat, this has no resemblance to reality. If you think it does, please post a reference to a peer- reviewed publication that describes such a model.

 

[vanhees71]

The photon cancellation in interference is in the range of 1 to 5 eVolts where say a stream of 1.97 eVolt red photons. Pair production would be at much higher energy levels.

I do not understand what you mean. Can you describe any (real physics) setup, where two photons exactly annihilate into the vacuum? If so, something's wrong in your calculation, because energy-momentum conservation holds, and the vacuum state is the state of lowest possible energy, i.e., the ground state, and any two-photon energy eigenstate has a larger energy; so the two photons cannot simply annihilate.

Also in classical electromagnetism (and it's much better to think in terms of classical electromagnetism as the classical analogue of photons than to think about them in terms of classical particles) there's no way to make two waves cancel everywhere and at any time completely since this also would violate energy conservation. So your idea is a misconception already from the classical point of view. Before approaching photons, you have to master classical electrodynamics (i.e., the Maxwell equations) and then relativistic quantum field theory.

 

[edguy99]

@vanhees71, the original poster linked to an article talking about modelling a photon as a harmonic oscillator. Within the context of "no model is perfect, but some are illuminating", this model has some very important aspects. A harmonic oscillator with 2 states conveys the visual information contained in the concept of wave reinforcement and interference. Interference in this context has nothing to do with the destruction of photons, or the creation of electron pairs from a high energy photon. One of the simplest ways to see this is to delay a coherent stream of photons by half a wavelength and then send that wave together with the original wave. The interference causes the 2 waves to "appear" to disappear. They don't really of course disappear, but the point it makes, and the principle it teaches is very effective visually.

Modelling the photon as an oscillator as the original article did, has many important ideas built in. It introduces the idea of the relationship between wavelength and energy (the faster the oscillation, the more the energy). It allows for visual representation of reinforcement and interference (a photon at a peak state cancels a photon at a trough state, two photons at a peak state look twice as strong although they are still 2 independent photons). It allows an understanding of diffraction where something that follows a wave equation will not always go in straight line around corners or always reflect off a smooth surface - the result is dependent on the "phase" of the photon.

 

[A. Neumaier]

Modelling the photon as an oscillator

You are completely missing the point of the criticism. You are not at all modeling photons but modes of the free, classical electromagnetic field - as understood already before quantum mechanics existed. No quantum mechanics is involved; therefore all your talk about photons is spurious.

 

[maline]

@vanhees71, the original poster linked to an article talking about modelling a photon as a harmonic oscillator.

Let me summarize that article. Note that I am not at all an expert- I am learning these ideas here on this thread! But I think I get it; someone will hopefully correct me if not.

The harmonic oscillator discussed there is the quantum equivalent of the magnitude of a spin component of a spatial frequency in the Fourier decomposition of the classical vector potential (in the radiation gauge).
The point to take is that frequency components are not "located in space" at all- they are found by the Fourier transform, which is an integral over all of space!

Classically, such a component represents a plane wave with circular polarization. See http://en.wikipedia.org/wiki/Circular_polarization for diagrams, but remember that these diagrams only show the field values on the Z axis. The values are constant throughout each X-Y plane.

Now classically, there is not much reason to think of such a mode as a (single) "oscillator". Although the field values in a particular direction at any point do oscillate, the wave as a whole simply moves forward. In fact, the magnitudes of the fields, including the vector potential, are constant everywhere- only the direction changes.

In the quantum case, the field values of each frequency mode are quantum observables that do not commute with energy or momentum. A pure momentum (spatial frequency) state must be represented by a function that gives a complex "quantum amplitude" for each of the possible field values for (the corresponding Fourier component of) one of the fields.

If we look for such a function that represents a pure energy state -an eigenstate of the Hamiltonian -it turns out that the solutions, in terms of magnitude of the vector potential OR of the electric field, are the same as the solutions for energy eigenstates of the simple quantum harmonic oscillator, in terms of position. This is because the two Hamiltonians (can be made to) have the same form. See http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator for these solutions.

The amplitude for each (X-Y) vector value will also have a complex phase depending on the direction- the field is in all directions (perpendicular to the Z axis) at once. This allows the wave propagation (rotation in space) to consist of nothing but a complex phase rotation! Of course, this is required for energy eigenstates by Schrodinger's equation.

Other pure momentum states can be constructed as (discrete) sums of the pure energy states. These in fact oscillate in time, just as the corresponding states of he simple quantum harmonic oscillator do. The oscillations consist of energy transfer between the electric field and the magnetic field, at all points in space as one.

A photon is not an oscillator; it doesn't oscillate at all. It is an excitation of one of these oscillators from one (stationary) energy eigenstate to the next energy state. As in the case of the simple quantum harmonic oscillator, the energy difference between two neighboring states is the frequency times Planck's constant.

A "one-photon state" can also refer to a general quantum electromagnetic state with the property that in its Fourier decomposition, each mode is in the first excited state. This allows for photons with localized spatial distributions, like ones that were emitted from atoms. However, once more than one photon is involved, it is complicated if not impossible to define a position for each one separately.

 

[edguy99]

I believe the original idea of the photon as a harmonic oscillator started as an effort to explain the photons relationship between energy and wavelength. A simple harmonic oscillator is anything with a linear restoring potential. The simplest examples are things like a spring, or a string with tension, or a wave. In 1926, an Austrian physicist Erwin Schrödinger, described mathematically how a harmonic oscillator stores energy and how to calculate how much energy it stores. The Schrödinger equation describes the system’s “wave function” or “state”, a condition that changes over time on a periodic basis.

For the photon, the value of this restoring potential is known as the Planck constant (h). Planck’s constant, relates the amount of energy stored in a photon to its wavelength (λ). Put another way, Planck’s constant tells you the amount of time it takes the photon to undergo one cycle of whatever its doing given that the photon has a specific amount of energy. The equation E for energy = h / λ, tells us that a photon with low energy will take much longer to complete one cycle of the wave then a photon with high energy.

@maline, this modelling of the photon leads to an understanding of the energy stored in a photon and the concept of "phase", where some photons have a matching phase (i.e.. they are coherent) and some do not. The concept of spin relates to the orientation of the electrical axis (or the orientation of the Jones vector in qm) independent of the periodicity of the wavelength. If you look at the Maxwell equation, the electrical component "oscillates" on a periodic basis perpendicular to the "oscillation" of the magnetic axis. The change in direction of the electrical axis over time is what we call spin.

 

[vanhees71]

Please read the first few chapters of Landau&Lifshitz vol. II, where the free em. field is quantized canonically in radiation gauge. There it becomes very clear, in which correct sense you can treat photons as an infinite set of uncoupled harmonic oscillators.

 

[maline]

I believe the original idea of the photon as a harmonic oscillator started as an effort to explain the photons relationship between energy and wavelength. A simple harmonic oscillator is anything with a linear restoring potential. The simplest examples are things like a spring, or a string with tension, or a wave. In 1926, an Austrian physicist Erwin Schrödinger, described mathematically how a harmonic oscillator stores energy and how to calculate how much energy it stores. The Schrödinger equation describes the system’s “wave function” or “state”, a condition that changes over time on a periodic basis.

For the photon, the value of this restoring potential is known as the Planck constant (h). Planck’s constant, relates the amount of energy stored in a photon to its wavelength (λ). Put another way, Planck’s constant tells you the amount of time it takes the photon to undergo one cycle of whatever its doing given that the photon has a specific amount of energy. The equation E for energy = h / λ, tells us that a photon with low energy will take much longer to complete one cycle of the wave then a photon with high energy.

@maline, this modelling of the photon leads to an understanding of the energy stored in a photon and the concept of "phase", where some photons have a matching phase (i.e.. they are coherent) and some do not. The concept of spin relates to the orientation of the electrical axis (or the orientation of the Jones vector in qm) independent of the periodicity of the wavelength. If you look at the Maxwell equation, the electrical component "oscillates" on a periodic basis perpendicular to the "oscillation" of the magnetic axis. The change in direction of the electrical axis over time is what we call spin.

There a many points that are wrong or confused in this account, but I am out of patience to discuss them. For the last time, if you want this "flying harmonic oscillator" idea to be taken seriously, you must provide references. Otherwise it has the status of a "personal theory", which are not allowed on PF.

 

[Nugatory]

@vanhees71, the original poster linked to an article talking about modelling a photon as a harmonic oscillator.

That is a complete misdescription of the relationship between photons and harmonic oscillators, as described in Bhobba's original post. The number of oscillators is unchanging (and infinite) and completely unrelated to the number of photons present, so the photons are not being modeled as oscillators.

 

[pr3dator]

Yes the photon just a quanta of field. The field made of quantum harmonic oscillator.
The real problem here: QM say that the field is not element of reality.
Then element of what?
of a dream?

 

[vanhees71]

Well, isn't light an "element of reality"? Admittedly it's the only fundamental quantum field we are able to perceive with our senses directly, but I'd say it's an example for a quantum field which for sure is an "element of reality" in the sense that it is observable, and in this case even directly with our senses.

In my opinion all this philosophical uttering about "reality" is just nonsense. There's not even a common definition of what the word reality means. In physics reality is what is observable.

 

[edguy99]

Please read the first few chapters of Landau&Lifshitz vol. II, where the free em. field is quantized canonically in radiation gauge. There it becomes very clear, in which correct sense you can treat photons as an infinite set of uncoupled harmonic oscillators.

Thank you for the reference. On page 108, in The wave equation section:

First of all we note that such fields must necessarily be time-varying.
f = f₁*(t-(x/c)) + f₂*(t+x/c) (47 - 2)
Suppose, for example, f₂ = 0, so that
f = f₁*(t-(x/c))
Let us clarify the meaning of this solution. In each plane x = const, the field changes with the time; at each given moment the field is different for different x. It is clear that the field has the same values for coordinates x and times t which satisfy the relation t—(x/c) = const, that is,
x = const +ct.

This means that if, at some time t = 0, the field at a certain point x in space had some definite value, then after an interval of time t the field has that same value at a distance ct along the Z axis from the original place. We can say that all the values of the electromagnetic field are propagated in space along the X axis with a velocity equal to the velocity of light, c.

Thus, f₁*(t-(x/c)) represents a plane wave moving in the positive direction along the X axis.

I feel there is nothing new or different here. The animation at http://www.animatedphysics.com/games/photon_cavity.htm does this exactly. You are looking sideways on a 2d, grid of oscillators. The first part of the animation is the representation of one photon. At each step, the size of the photon is transferred to the next grid location to the right and made either slightly larger or slightly smaller, depending on the phase of the photon. The time interval used here is 0.1 femtoseconds (10^-15), if you use the STEP button, it advances frame by frame.

 

[maline]

"It is clear that the field has the same values for coordinates x and times t which satisfy the relation t—(x/c) = const, that is,
x = const +ct."
This is the most basic description of a wave, and it is very, very different from "the size of the photon is transferred to the next grid location to the right and made either slightly larger or slightly smaller, depending on the phase of the photon".
For one thing, there is no "made slightly larger or smaller". Also, plane waves, by definition, fill (at least) the entire YZ plane at each moment! They do not resemble "balls of energy".

You are looking sideways on a 2d, grid of oscillators

Ah, we're getting somewhere! This is a major improvement over calling "a photon" one "harmonic oscillator".

 

[vanhees71]

Thank you for the reference. On page 108, in The wave equation section:
I feel there is nothing new or different here. The animation at http://www.animatedphysics.com/games/photon_cavity.htm does this exactly. You are looking sideways on a 2d, grid of oscillators. The first part of the animation is the representation of one photon. At each step, the size of the photon is transferred to the next grid location to the right and made either slightly larger or slightly smaller, depending on the phase of the photon. The time interval used here is 0.1 femtoseconds (10^-15), if you use the STEP button, it advances frame by frame.

Argh. Of course, I meant Landau/Lifshitz vol. IV about quantum electrodynamics. Sorry for the confusion.

 

[edguy99]

Argh. Of course, I meant Landau/Lifshitz vol. IV about quantum electrodynamics. Sorry for the confusion.

On page 5 of Landau/Lifshitz vol. IV, Talking about Quantization of the Free Electromagnetic Field:

With the purpose of treating the electromagnetic field as a quantum object, it is convenient to begin from a classical description of the field in which it is represented by an infinite but discrete set of variables. This description permits the immediate application of the customary formalism of quantum mechanics. The representation of the field by means of potentials specified at every point in space is essentially a description by means of a continuous set of variables. ...

On page 11, Introducing Photons:

These formulae enable us to introduce the concept of radiation quanta or photons, which is fundamental throughout quantum electrodynamics. We may regard the free electromagnetic field as an ensemble of particles each with energy ω (= ħω) and momentum k (=nħω/c). The relationship between the photon energy and momentum is as it should be in relativistic mechanics for particles having zero rest-mass and moving with the velocity of light. ... The polarization of the photon is analogous to the spin of other particles; ... It is easily seen that the whole of the mathematical formalism developed in §2 is fully in accordance with the representation of the electromagnetic field as an ensemble of photons; it is just the second quantization formalism, applied to the system of photons. ...

And Continues:

The plane waves ... may be treated as the wave functions of photons having given momenta k and polarizations e^α.

I agree completely. A photon is a plane wave traveling through space at the speed of light.

Also, today is the birthday of Erwin Schrödinger (ca. 1887–1961). A great quote:

The world extended in space and time is but our representation.

 

[Nugatory]

The plane waves ... may be treated as the wave functions of photons having given momenta k and polarizations e^α.

I agree completely. A photon is a plane wave traveling through space at the speed of light.

"The plane wave may be treated as the wave function of a photon" and "A photon is a plane wave" are different statements. If you completely agree with the first, you're rejecting the second.

 

[Giulio Prisco]

Hi Guys and Gal's

In answering a question in general physics I came across the following which explains at a reasonably basic level what a photon is, spontaneous emission etc at the level of basic QM with a bit of math:
http://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf

IMHO its much better than the usual misleading hand-wavey stuff and even if you don't follow the math would allow a general gist to be had.

Thanks
Bill

What is a Photon? Introduction to Quantum Field Theory
https://works.bepress.com/charles_torre/92/
Now this course is published in book form, open access CC.
It seems a good first intro, with all and only the detail that the newcomer really needs, I wish I had this book when I was studying these things the first time.

 

[vanhees71]

Looks indeed great!