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  • When an electron jumps from a high to a lower energy level and a photon is created, can the photon be detected from all vantage points or only from a discrete position? Does the atom nucleus cast a shadow on the disturbance of the em field that the photon creation causes? What I find difficult to understand about a photon being detectable everywhere due its wave nature is with so many photon creations causing ripples (or is it disturbances) in the em field at any moment, how is it possible to see (detect) anything discrete at all if waves are the only explanation of visibility as I would expect so much cross interference with all the different disturbances being created at every atom where an electron changes energy that the em field would just be chaos (I'm a Cs major, not physics though which might explain my confusion)
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closed as too broad by David Z May 5 at 21:27

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Why do you think photons have a “length”? Mainstream physics treats them and all other elementary particles as point particles. $\endgroup$ – G. Smith May 5 at 16:25
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    $\begingroup$ partial duplicate of physics.stackexchange.com/q/94049 $\endgroup$ – Ben Crowell May 5 at 16:35
  • $\begingroup$ refined question title, just to see if that works $\endgroup$ – The_Sympathizer May 5 at 22:33
  • $\begingroup$ in making less "broad". $\endgroup$ – The_Sympathizer May 5 at 22:51
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    $\begingroup$ @Waslap This is an interesting case-study. Questions on Physics SE should be specific and focused on conceptual issues. The question was put on hold because it appears to be a list of questions ("too broad"), but I wonder if you really meant it to be a single specific question about a conceptual issue that you weren't sure how to express. If the "list" of questions was really just an attempt to bracket that one core conceptual issue that was difficult to express directly, then maybe explaining this situation in the question would lead to a re-open vote. Don't know, but maybe worth a try. $\endgroup$ – Chiral Anomaly May 5 at 23:54
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is the photon immediately travelling at the speed of light?

Yes, in the sense that if you immediately measure its energy and momentum, they are related in the way that is required if and only if a particle is moving at the speed of light ($E=pc$). But it doesn't have a well-defined velocity vector in the sense you would imagine for a classical particle, because it doesn't have a well-defined trajectory.

Does the photon immediately assume its final length?

"Length" could mean the wavelength or the length of the wavetrain, which are different things. The main issue here is the wavelength. If it did immediately assume its final wavelength, we would have a paradox, since it would have to travel a distance $>\lambda$ in zero time, but that would give it an infinite velocity. At a short time $t$ after emission, the energy-time uncertainty principle says that there must be an uncertainty in the photon's energy $\gtrsim h/t$. This causes its wavelength to be uncertain by a corresponding amount, and if you work out the algebra, this is enough to make the paradox go away.

Can the photon be detected from all vantage points or only from a position that happens to be directly in its path?

A photon doesn't have a path. A path is a property of a classical particle, not of a quantum-mechanical wave-particle. The probability of detecting it in a given direction depends on the radiation pattern, just as it would for a classical radiation pattern. (This is an example of the correspondence principle.)

How is the photon's initial trajectory decided?

It doesn't have a trajectory.

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For answering questions like this, the quantum superposition principle is essential.

Quantum processes are smooth. The name "quantum" comes from the fact that some measurement outcomes are limited to a finite list of outcomes rather than a continuum of possible outcomes, but processes are still smooth. In fact, processes in quantum theory are more smooth than processes in classical theory! This is possible because of the quantum superposition principle. I'll illustrate this in a simpler toy model first, and then I'll explain how it relates to the question about photons.

Toy model: Moving smoothly in a discrete space

To illustrate the claim that processes in quantum theory are more smooth than processes in classical theory, and to explain how this relates to the quantum superposition principle, consider a toy model of a particle (not a photon) in a discrete space where the coordinates can only be integers instead of real numbers. In a classical model, the particle would be unable to move smoothly. In order to move at all, it would need to suddenly disappear from one point and reappear at a different point, because the coordinates can only change by integer amounts. But in a quantum model, the particle can move smoothly even though the coordinates can still only have integer values. This is possible because of the superposition principle: If $|A\rangle$ represents the state where the particle is located at point $A$, and if $|B\rangle$ represents the state where the particle is located at point $B$, then states like $$ \alpha|A\rangle+\beta|B\rangle $$ are also possible, for any complex numbers $\alpha$ and $\beta$ with $|\alpha|^2+|\beta|^2=1$. The points $A$ and $B$ still have integer coordinates, but the particle's location is indefinite in that weird way that makes quantum theory so much fun. In loose and inadequate words, the particle is partly at $A$ and partly at $B$. Even though the coordinates of those points are integers, the coefficients $\alpha$ and $\beta$ can still be varied continuously, so the degree to which the particle is located at each of the two discrete points can be varied continuously. In this way, the particle can "move" smoothly from being entirely at $A$ to being entirely at $B$, without ever entering any of the space in between.

One moral of this story is that quantum theory cannot be treated as "classical theory with some randomness." The quantum superposition principle is essential, and classical theory (with or without randomness) has no good analog for this.

What happens when a photon is emitted?

When an electron jumps from a high to a lower energy level and a photon is created, is the photon immediately travelling at the speed of light?

Even if the energy levels available to the electron are discrete, the electron still transition smoothly from one to the other, thanks to the superposition principle. This works just like it did in the preceding toy model, but now instead of talking about points in a discrete space, we're talking about discrete energy levels. This transition may occur very quickly, so calling it a "jump" may be okay; but when we use that language, we need to remember that the transition (the "jump") occurs smoothly, even if it's relatively quick.

The total energy of the system is conserved, and as the electron smoothly transitions from the higher energy-level to the lower level, light is smoothly emitted. But wait: how can the emission of a photon be a continuous process? Isn't a photon a discrete all-or-nothing kind of thing? Once again, ambiguous language is the culprit. Yes, there is a sense in which light consists of discrete all-or-nothing things ("photons"). For example, when we measure the amount of energy in light with a given frequency, the answer is always an integer multiple of Planck's constant times that frequency. In this sense, photons are things that can be counted. But when we're not measuring the energy, the quantum superposition principle allows states involving a superposition of different numbers of photons. That's exactly what happens when light is being emitted while an electron transitions smoothly from one energy-level to another: we have a quantum superposition of the form $$ \alpha|A\rangle+\beta|B\rangle \tag{1} $$ where $A$ means "the electron is in the higher energy level and no photon is present," and $B$ means "the electron is in the lower energy level and a photon is present." Even though $A$ and $B$ differ from each other in a discrete way, the coefficients in the superposition, the $\alpha$ and the $\beta$, can still vary continuously in time. That's what happens when an atom emits a photon. The process is smooth, even though measuring the number of photons would give a discrete answer.

By the way, the act of measuring the number of photons would modify the emission process that we're trying to study, because we can't measure something unless we interact with it.

Also by the way, if both coefficients are non-zero in the state (1), then the energy-level of the electron and the presence/absence of the photon are entangled with each other. During the transition, neither the electron nor the photon has a state of its own. Only the pair has a well-defined state, namely a superposition of options $A$ and $B$. This kind of entanglement is essential for understanding how the emission process can be smooth.

...is the photon immediately travelling at the speed of light? Does the photon immediately assume its final length?

To answer this, another issue needs to be addressed. When we say that an photon is a "point" particle, what we mean is that the interaction between photons and (say) electrons is local, with no action-at-a-distance. It would be just as accurate to say that Maxwell's equations describe the electromagnetic field as a "point" field — it's not the field itself that's localized, but rather the way the field interacts with charged matter. The interaction is point-like, even if the field and the matter are not. The important message is that the photon itself can be spread out in space as an extended wavepacket, even though it's a "point" particle in the sense that its interactions are local.

As the electron is transitioning, the disturbance in the quantum elecromagnetic field is propagating outward at the speed of light in the $B$ part of the state (1). The whole process is smooth: the length of the disturbance grows while the transition is occurring, and the superposition principle is again what makes this possible. In the end, after the transition is complete, we will have a photon with some finite length (the length of the wavepacket) travelling away from the atom at the speed of light.

Can the photon be detected from all vantage points or only from a position that happens to be directly in its path?

Which path? The superposition principle strikes again! The photon is not emitted in any single well-defined direction, just like it doesn't have any single well-defined pointlike location. If we surround the atom with a large spherical shell of photon-detectors, then only one of those detectors will "click" — but that doesn't mean that the photon was moving exclusively in that direction prior to the measurement. The double-slit experiment is the prototypical way of demonstrating this, and there are many others.

How is the photon's initial trajectory decided?

In the superposition over all directions, some directions might be weighted more heavily than others (in the sense that those coefficients in the superposition have larger magnitudes), but a natural emission process won't result in any one trajectory for the photon. If we try to measure the photon's location at any given time, then we'll get a specific location as the outcome of that measurement, but again: that's a consequence of the act of measurement.

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  • $\begingroup$ damn, you took almost all the wind out of what I was going to put down as it would have been more or less the same but with some added details :( $\endgroup$ – The_Sympathizer May 5 at 20:35
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    $\begingroup$ This is a nice presentation that does a good job of highlighting the fundamental structure of quantum mechanics. But I can't help feeling that this sidesteps or fails to address some of the questions the OP asks, or that maybe it needs to be developed more. You're treating this as a two-state system, but it isn't a two-state system. The generic quantum-mechanical two-state system oscillates, it doesn't decay exponentially. To get exponential decay, you need to couple to the continuum, which isn't a single state. In a two-state model, you also can't address the OP's questions about propagation. $\endgroup$ – Ben Crowell May 5 at 21:41
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    $\begingroup$ @BenCrowell You're right that photon emission cannot be described by a two-state system, and I didn't mean to suggest that it can. I'm glad you posted the comment, though, because it hadn't occurred to me that the two-state warm-up might be mistaken for the whole answer. I meant to use the two-state illustration as a bridge between the "toy model" and the photon-emission case, just to introduce the superposition idea as simply as possible. The propagation-oriented wording later in the answer is meant to convey that there is more going on than just transitioning between two states. $\endgroup$ – Chiral Anomaly May 5 at 22:55
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    $\begingroup$ @ChiralAnomaly: I see what you mean. Yeah, I guess what I'm finding unsatisfying is that I'd like to see the whole thing gel more. Of course we're basically trying to explain QFT without the math, which is hard. $\endgroup$ – Ben Crowell May 6 at 0:21
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The photon is immediately moving at the speed of light. Photons in vacuum always move at c. They don’t accelerate or decelerate.

Its wavefunction immediately has a wavelength that depends on the difference in the atomic energy levels, but the photon does not have a length because it is a point particle.

The photon can be detected anywhere, but, depending on the atomic transition, it may be more likely to be detected in some directions than others.

There is no trajectory to decide; the probability of detecting the photon at any position is described by a quantum-mechanical wavefunction, not by a classical path.

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    $\begingroup$ Its wavefunction immediately has a wavelength that depends on the difference in the atomic energy levels, This would violate the energy-time uncertainty relation. There is no trajectory to decide; the probability of detecting the photon at any position is described by a quantum-mechanical wavefunction, not by a classical path. It's actually not technically true that a photon has a wavefunction. The system has a wavefunction, but the photon doesn't have a wavefunction that can be expressed in the position basis. $\endgroup$ – Ben Crowell May 5 at 16:40
  • $\begingroup$ Agreed. Your answer is the better one. $\endgroup$ – G. Smith May 5 at 16:48
  • $\begingroup$ @Ben Crowell does the atom nucleus cast a shadow on the disturbance of the wave that the photon creation causes? What I find difficult to understand about a photon being detectable everywhere due the wave nature is with so many photon creations causing ripples in the em wave at any moment, how is it possible to see (detect) anything discrete at all if waves are the only explanation as I would expect so much interference with all the different disturbances being created at every atom where an electron changes energy that the em wave would just be chaos (I'm a Cs major, not physics though) $\endgroup$ – Waslap May 6 at 13:10

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