Are photons identical? As they are considered fundamental like electrons, it would seem the answer would have to be yes. The mathematics of quantum mechanics may even require this, I don’t recall, but it does for electrons.
But if they are identical how is it they have different energies? The energy of a photon can be essentially any value (at least to some vague upper limit imposed by the energy available in the universe). But how do they “hold” this energy. An electron can “store” more energy by going faster, or being in a more energetic orbital in an atom. Photons can’t do that.
The standard answer would be, they have a different frequency or wave length. But that isn’t an inherent property as I can change what I observe by moving towards or away from the oncoming photon. There has to be something more fundamental which “tells” two photons traveling along “next to each other”, “You, photon A, you will always have more energy than photon B no matter the frame of reference of the observer.”
I suppose this is all coupled up with a two photon system can have mass even though a single photon does not, but then who gets to decide which photons are in a system together?
Wavelength is not an inherent property of a particle, and electrons moving at different speeds have different wavelengths, too, governed by the same rules. You can change the wavelength of any particle just by changing your motion relative to it.
Chronos thanks for your reply. I guess I worded it badly as that’s what I was trying to say in part. Frequency/wave length cannot be an inherent property of a photon becasue different observers disagree about the absolute levels. But when we see two photons coming at us, we can see them with different wavelengths. There is no way to go to the rest frame to say “See they’re identical in their rest frames.” as we can do for electrons.
So this is the thought experiment. Consider a set of observers in different inertial frames moving along the parallel lines. Two monochromatic lasers at rest relative to each other pump out photons towards them. One observer at rest relative to the lasers sees orange and yellow. Another observer moving towards the lasers sees yellow and green, another moving away sees red and orange, etc. The observers don’t agree on the amount of energy in the photons, but they do all agree that laser A emits more energetic photons than laser B.
What is it about the photons that makes sure that they observe this? If it were electron pumps shooting electrons at the observers, it would be that the more energetic electrons were coming towards all the observers faster (though each observer would disagree about the speed just as they do the energy). But this can’t be what makes it so for the photons obviously. What is it about the A photons that makes all observers agree they are more energetic.
The photon has more momentum energy. That energy is conserved. If you give more energy to a baseball, or an electron, then it increases its velocity, which increases it’s momentum. If you give more energy to a photon, it can’t go any faster, so that momentum energy ends up as higher frequency/shorter wavelength.
A conceptual way to think about it that probably does not translate well to the math is that the photon is actually “going faster”. If you trace out its route, with its wavelength moving it up and down along its path, it is following a longer route than a photon of lower energy. Since the propagation of the wave is at C, then the particle that is following along the wave, as well as traveling cyclically perpendicular to the wave, is traveling, in a naive way, “faster than C”. That “velocity” is the momentum of the photon, and is how you tell them apart.
Given that photons are really waves, and not particles, there is obviously not an actual particle traveling along at greater than C tracing the crests and troughs of the wave, but that might give you some idea of how that energy is transmitted.
There are particles, and there are the states they are in. In your example, you have two identical photons in different states. How you describe those states is frame-dependent. For you electron case, it so happens that you can find a frame where the energies line up, but their (vector) momenta are still different. The electrons are still in different states.
Boosting into different reference frames is a red herring, though, made seemingly relevant by your choice of states. Take your two electrons instead to be the electrons in a helium atom. One is spin-up and one is spin-down, and again those terms are describing the states that the indistinguishable electrons are in.
Which part, the part that I said does not translate to the math, or that I said it does not translate to the math?
I was giving a conceptual idea, which I agree is incorrect on the face of it, as there is no such thing as a photon particle in the first place. The photon as a particle is already a conceptualized version of the underlying math, so yeah, I was stretching the analogy to give a better idea as to how the momentum is bound up into the wave using familiar concepts from classical mechanics.
If you have a better conceptual analogy, go for it, I’d like to hear it too.
Photons also have the familiar polarization property. Two otherwise identical photons with different polarizations behave different. E.g., when passing thru some media, when reflecting at certain angles, etc.
My reply was about the conceptual idea itself. The idea that the “wave” of a photon has any transverse spatial size is very dangerous and will lead to many more incorrect conclusions than correct ones. There is nothing that can be measured in units of distance transverse to the direction of motion. The wave doesn’t physically move “side to side” in any spatial sense, math or concept.
The concept of identical particles as mentioned in the OP is a fundamental one. It shouldn’t be confused with how those (identical) particles are behaving, are situated, are interacting, etc.
As a non-quantum example, imagine you have two identical solid blue marbles, perfectly uniform and indistinguishable to the naked eye. Place one marble in a bowl marked “Bowl 1” and another in a bowl marked “Bowl 2”. If you leave the room and come back later, you have no way of telling whether someone came in and switched the marbles around. The two marbles are different in that they are in different states, but you just can’t say “which” marble is in which state, because that question doesn’t make sense if the marbles are indistinguishable.
In quantum and statistical mechanics, true indistinguishablility has important observable consequences.
That’s fine, and I don’t disagree. It was not my analogy, but one I saw on a documentary some years back. I’m pretty sure it had Michio Kaku in it, so take it for what that’s worth.
As analogies go, it seems as valid as using a trampoline and bowling ball to conceptualize gravity, or molasses as the higg’s field (both of which are entirely wrong, both of which I hear all the time). The question of the OP was what is the intrinsic difference between two photons of different energies, and giving an example of two things that have a different kinetic energy, even though they propagate at the same speed seemed appropriate. The OP could understand how electrons of differing velocities were different, but could not see how photons that travel at the same velocity could be. Adding the idea of kinetic energy in a direction perpendicular to the propagation is flawed, but still seems useful in translating that extra degree of freedom that photons have to a classical mechanics analogy, even if that is not actually what is happening.
The last bit though, I did not know. You are saying that there is no width to a photon’s wave, that it can only be defined as a line?
Well, the relative wrongness of wrong things is a bit subjective. I would rate treating a photon’s wave as wiggling the photon back and forth as particularly egregious, whether or not there may be other egregious analogies about other things out in the wild.
Energy and momentum are the things that make the particles (well, the states they are in) distinguishable in both the photon and electron case. The “missing degree of freedom” you are trying to find is just that the electrons’ states are still distinguishable because of their momentum, just like the photons’ states. That is, the misapprehension in the OP was thinking that the photons are always different-looking whereas the electrons can be made same-looking, but that’s not true if you look at momentum and energy together, as you must. (And you must because it is the energy-momentum four-vector that is relativistically invariant. Cherry picking one piece of that four-vector is leaving part of the story out.)
No, that’s a different can of worms (opened below). I was noting that the wiggling of a photon wave isn’t the photon moving side to side in space. It’s only moving forward. The thing that’s varying sinusoidally along the photon’s path is the magnitude and direction of the electromagnetic fields. The fields are located on the path, but their vector values are pointing sideways. Nothing is physically moving sideways.
To let the unrelated worms out: a photon of definite energy has infinite transverse extent. The term of art is “plane wave”. This is an idealization and can’t be achieved in reality, but it’s an invaluable (and usually fantastically good) approximation for lots of physical systems. In scenarios where you do care about the width of a wave or wave packet, the run-of-the-mill approach is to think of the wave as a superposition of waves of different momenta. The result is that you have a wave with a finite transverse physical extent but not a definite momentum. This transverse width, though, has nothing to do with any physical displacement of the photon back and forth in step with its frequency(ies).
There is no such thing as “momentum energy”. Not sure if that was a typo or not.
Also, photons are not waves. Photons are photons. We sometimes use a wave model to describe their behavior, but that does not make them waves. I’ve not had my head stuck in a physics book for decades, but I don’t recall any of them using your analogy. Do you have a cite of one doing so? If not, there is probably a reason for that.