# Thread: Width of a photon

1. Hello everyone,

Electromagnetic waves are usually depicted with 2 orthogonal waves. Most drawings are usually identical.

But do the two wave have a width? How could they have no width at all?

Nic.

2.

3. The vector arrows point in the directions of the electric and magnetic field. The lengths of the vector arrows are the magnitudes of electric and magnetic field and do not represent a size in space.

4. Originally Posted by Nic321
Hello everyone,

Electromagnetic waves are usually depicted with 2 orthogonal waves. Most drawings are usually identical.

But do the two wave have a width? How could they have no width at all?

Nic.
I think this may be quite a deep question. If you think of the famous 2 slit experiment in which photons are passed sequentially through a pair of slits and you get, dot by dot, an interference fringe pattern on the screen, does one consider the "width" of a photon to be that of the individual dots, i.e. the "particle-like" quanta that are individually detected? Or does one consider that the wavefunction must pass through both slits, in order for an interference fringe pattern to appear? Clearly the area of space explored by the latter is greater than the size of the former. So what does "width" really mean, then? It's tricky, I think.

5. Originally Posted by KJW
The vector arrows point in the directions of the electric and magnetic field. The lengths of the vector arrows are the magnitudes of electric and magnetic field and do not represent a size in space.
Ok, but the electric and magnetic fields are in space. At one point in space ( along the axis of propagation of the photon ), the electric and magnetic fields oscillate, but they must also change in the neighbouring points of space. No?

6. Originally Posted by exchemist
Originally Posted by Nic321
Hello everyone,

Electromagnetic waves are usually depicted with 2 orthogonal waves. Most drawings are usually identical.

But do the two wave have a width? How could they have no width at all?

Nic.
I think this may be quite a deep question. If you think of the famous 2 slit experiment in which photons are passed sequentially through a pair of slits and you get, dot by dot, an interference fringe pattern on the screen, does one consider the "width" of a photon to be that of the individual dots, i.e. the "particle-like" quanta that are individually detected? Or does one consider that the wavefunction must pass through both slits, in order for an interference fringe pattern to appear? Clearly the area of space explored by the latter is greater than the size of the former. So what does "width" really mean, then? It's tricky, I think.
I am not sure about the proper interpretation of the double slit experiment, but the interference pattern occurs when the photons are emitted one by one, so I imagine the photon goes through both slits and interfers with itself to produce the interference pattern.

I am not sure at all.

7. Originally Posted by Nic321
Originally Posted by exchemist
Originally Posted by Nic321
Hello everyone,

Electromagnetic waves are usually depicted with 2 orthogonal waves. Most drawings are usually identical.

But do the two wave have a width? How could they have no width at all?

Nic.
I think this may be quite a deep question. If you think of the famous 2 slit experiment in which photons are passed sequentially through a pair of slits and you get, dot by dot, an interference fringe pattern on the screen, does one consider the "width" of a photon to be that of the individual dots, i.e. the "particle-like" quanta that are individually detected? Or does one consider that the wavefunction must pass through both slits, in order for an interference fringe pattern to appear? Clearly the area of space explored by the latter is greater than the size of the former. So what does "width" really mean, then? It's tricky, I think.
I am not sure about the proper interpretation of the double slit experiment, but the interference pattern occurs when the photons are emitted one by one, so I imagine the photon goes through both slits and interfers with itself to produce the interference pattern.

I am not sure at all.
Exactly, that's what I'm getting at. So the wavefunction goes through both, but each quantum is detected at a particular spot. I found the following on the internet which may help a bit: it is NOT a trivial question at all : quantum mechanics - How fat is Feynman

Maybe a real physicist wil interject at this point to take things further - I shall shortly be out of my depth.

8. Originally Posted by Nic321
Originally Posted by KJW
The vector arrows point in the directions of the electric and magnetic field. The lengths of the vector arrows are the magnitudes of electric and magnetic field and do not represent a size in space.
Ok, but the electric and magnetic fields are in space. At one point in space ( along the axis of propagation of the photon ), the electric and magnetic fields oscillate, but they must also change in the neighbouring points of space. No?
Yes, but the diagrams are not showing this. The diagrams are only showing a single line without width in the direction of propagation.

As for the "width of a photon", this has no meaning in quantum mechanics unless the width has actually been observed. It is a characteristic feature of quantum mechanics that objects do not possess definite properties unless those properties have been observed directly or indirectly. This is known as counterfactual indefiniteness.

9. Originally Posted by KJW
Originally Posted by Nic321
Originally Posted by KJW
The vector arrows point in the directions of the electric and magnetic field. The lengths of the vector arrows are the magnitudes of electric and magnetic field and do not represent a size in space.
Ok, but the electric and magnetic fields are in space. At one point in space ( along the axis of propagation of the photon ), the electric and magnetic fields oscillate, but they must also change in the neighbouring points of space. No?
Yes, but the diagrams are not showing this. The diagrams are only showing a single line without width in the direction of propagation.

As for the "width of a photon", this has no meaning in quantum mechanics unless the width has actually been observed. It is a characteristic feature of quantum mechanics that objects do not possess definite properties unless those properties have been observed directly or indirectly. This is known as counterfactual indefiniteness.
Nevertheless, if you point a stream of photons at, or not quite accurately at, an atom capable of absorbing them, the probability of absorption must depend on how close by to the atom the photons pass, must it not? It will not be a yes/no hit or miss phenomenon, clearly, but there will be some effective, probabilistically defined, "width" will there not?

10. Originally Posted by exchemist
Originally Posted by Nic321
Originally Posted by exchemist
Originally Posted by Nic321
Hello everyone,

Electromagnetic waves are usually depicted with 2 orthogonal waves. Most drawings are usually identical.

But do the two wave have a width? How could they have no width at all?

Nic.
I think this may be quite a deep question. If you think of the famous 2 slit experiment in which photons are passed sequentially through a pair of slits and you get, dot by dot, an interference fringe pattern on the screen, does one consider the "width" of a photon to be that of the individual dots, i.e. the "particle-like" quanta that are individually detected? Or does one consider that the wavefunction must pass through both slits, in order for an interference fringe pattern to appear? Clearly the area of space explored by the latter is greater than the size of the former. So what does "width" really mean, then? It's tricky, I think.
I am not sure about the proper interpretation of the double slit experiment, but the interference pattern occurs when the photons are emitted one by one, so I imagine the photon goes through both slits and interfers with itself to produce the interference pattern.

I am not sure at all.
Exactly, that's what I'm getting at. So the wavefunction goes through both, but each quantum is detected at a particular spot. I found the following on the internet which may help a bit: it is NOT a trivial question at all : quantum mechanics - How fat is Feynman

Maybe a real physicist wil interject at this point to take things further - I shall shortly be out of my depth.
Thank you for the link, that's interesting. So in fact the photon occupies a certain volume. The electric and magnetic fields really oscillate in space.

11. Originally Posted by exchemist
Originally Posted by KJW
Originally Posted by Nic321
Originally Posted by KJW
The vector arrows point in the directions of the electric and magnetic field. The lengths of the vector arrows are the magnitudes of electric and magnetic field and do not represent a size in space.
Ok, but the electric and magnetic fields are in space. At one point in space ( along the axis of propagation of the photon ), the electric and magnetic fields oscillate, but they must also change in the neighbouring points of space. No?
Yes, but the diagrams are not showing this. The diagrams are only showing a single line without width in the direction of propagation.

As for the "width of a photon", this has no meaning in quantum mechanics unless the width has actually been observed. It is a characteristic feature of quantum mechanics that objects do not possess definite properties unless those properties have been observed directly or indirectly. This is known as counterfactual indefiniteness.
Nevertheless, if you point a stream of photons at, or not quite accurately at, an atom capable of absorbing them, the probability of absorption must depend on how close by to the atom the photons pass, must it not? It will not be a yes/no hit or miss phenomenon, clearly, but there will be some effective, probabilistically defined, "width" will there not?
The width of the photon will essentially be the width of the photon source (plus any divergence and diffraction effects). The reason one gets interference in the double-slit experiment is because the photon has as much freedom as it is permitted to have. It doesn't constrain itself to a single slit when it is free to pass through both. When one measures which slit the photon passed through, even if this measurement was after the photon passed through the slits, one has constrained the photon to pass through only a single slit and the double-slit interference is destroyed.

12. Originally Posted by KJW
Originally Posted by Nic321
Originally Posted by KJW
The vector arrows point in the directions of the electric and magnetic field. The lengths of the vector arrows are the magnitudes of electric and magnetic field and do not represent a size in space.
Ok, but the electric and magnetic fields are in space. At one point in space ( along the axis of propagation of the photon ), the electric and magnetic fields oscillate, but they must also change in the neighbouring points of space. No?
Yes, but the diagrams are not showing this. The diagrams are only showing a single line without width in the direction of propagation.
Ok, so the electric field and the magnetic field really oscillate in space.

As for the "width of a photon", this has no meaning in quantum mechanics unless the width has actually been observed. It is a characteristic feature of quantum mechanics that objects do not possess definite properties unless those properties have been observed directly or indirectly. This is known as counterfactual indefiniteness.

But isn't the width of the photon related to the amplitude of the electric and magnetic waves? And we know these 2 amplitudes, so we should know the width of the photon? So why would we know it only once it has been observed?

The probability of finding a photon at one place is proportional to the wave function squared. On the drawing on the lower left here , the blurred circles represent the probability of finding the photon are on the x axis. Will the photon be observed exactly on the x axis or in a volume around it?

I am sorry but I am confused between the amplitude of the electric and magnetic waves and the width of the photon itself.

13. Originally Posted by Nic321

I am sorry but I am confused between the amplitude of the electric and magnetic waves and the width of the photon itself.
The photon (very much like the electron) is a pointlike particle (dimensionless). You are confusing its dimensions with its position, I have explained that to you in a prior thread, the one about fitting a photon inside a black hole of very small Schwarzschild radius. It is the position that is "smeared" probabilistically, as per QM, not its dimension. Bottom line, your question about "Width of a photon" has no answer since it has no meaning.

14. Originally Posted by KJW
The width of the photon will essentially be the width of the photon source (plus any divergence and diffraction effects).
I should also mention coherence length.

15. When one measures which slit the photon passed through, even if this measurement was after the photon passed through the slits, one has constrained the photon to pass through only a single slit and the double-slit interference is destroyed.
Could it be that the act of measuring the photon after it has passed one slit forces the photon which in fact passed through the 2 slits to rejoin, and continue as if it had passed through only one?

Otherwise it would mean that the part of the photon which passed through the other slit goes back in time back through the slit it went through and goes through the slit where the photon was measured.

16. Originally Posted by Howard Roark
The photon (very much like the electron) is a pointlike particle (dimensionless).
I think it should be noted (and the point that I'm making elsewhere) that this is only true if the observable being measured is a pointlike observable.

17. Originally Posted by Howard Roark
Originally Posted by Nic321

I am sorry but I am confused between the amplitude of the electric and magnetic waves and the width of the photon itself.
The photon (very much like the electron) is a pointlike particle (dimensionless). You are confusing its dimensions with its position, I have explained that to you in a prior thread, the one about fitting a photon inside a black hole of very small Schwarzschild radius. It is the position that is "smeared" probabilistically, as per QM, not its dimension. Bottom line, your question about "Width of a photon" has no answer since it has no meaning.
I understand that it is smeared probabilistically. When I am talking about the 'width' of the photon, I am not referring to the fact that it is a pointlike particle ( I agree with that ), I am referring to the fact that it occupies a certain volume ( probabilistically ). The pointless particle is everywhere in that volume at the same time with a certain probability of being at each point of that volume.

18. Originally Posted by Nic321
Ok, so the electric field and the magnetic field really oscillate in space.
The directions of the vector fields are directions in space.

Originally Posted by Nic321
But isn't the width of the photon related to the amplitude of the electric and magnetic waves?
No, not at all. The amplitude of the electromagnetic field relates to the number density of photons.

Originally Posted by Nic321
So why would we know it only once it has been observed?
How else would we know it if it is never observed?

19. Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321

I am sorry but I am confused between the amplitude of the electric and magnetic waves and the width of the photon itself.
The photon (very much like the electron) is a pointlike particle (dimensionless). You are confusing its dimensions with its position, I have explained that to you in a prior thread, the one about fitting a photon inside a black hole of very small Schwarzschild radius. It is the position that is "smeared" probabilistically, as per QM, not its dimension. Bottom line, your question about "Width of a photon" has no answer since it has no meaning.
I understand that it is smeared probabilistically. When I am talking about the 'width' of the photon, I am not referring to the fact that it is a pointlike particle ( I agree with that ), I am referring to the fact that it occupies a certain volume ( probabilistically ). The pointless particle is everywhere in that volume at the same time with a certain probability of being at each point of that volume.
So, stop using the term "width". It does not apply.

20. Originally Posted by Howard Roark
Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321

I am sorry but I am confused between the amplitude of the electric and magnetic waves and the width of the photon itself.
The photon (very much like the electron) is a pointlike particle (dimensionless). You are confusing its dimensions with its position, I have explained that to you in a prior thread, the one about fitting a photon inside a black hole of very small Schwarzschild radius. It is the position that is "smeared" probabilistically, as per QM, not its dimension. Bottom line, your question about "Width of a photon" has no answer since it has no meaning.
I understand that it is smeared probabilistically. When I am talking about the 'width' of the photon, I am not referring to the fact that it is a pointlike particle ( I agree with that ), I am referring to the fact that it occupies a certain volume ( probabilistically ). The pointless particle is everywhere in that volume at the same time with a certain probability of being at each point of that volume.
So, stop using the term "width". It does not apply.
OK Howard, but I'm now curious as well. Tell me, how can one determine how close a photon has to pass, to an atom capable of absorbing it, in order for the absorption probability to be significant? If it flies by at a distance of 0.1nm from the nucleus, for example, Id have thought the probability must be higher than if it passes 10nm away.

Is there not some probability distribution analogue of a "width" here, similar to the fuzzy but useful idea of the "size" of an atomic orbital, say?

21. Originally Posted by exchemist
Tell me, how can one determine how close a photon has to pass, to an atom capable of absorbing it, in order for the absorption probability to be significant? If it flies by at a distance of 0.1nm from the nucleus, for example, Id have thought the probability must be higher than if it passes 10nm away.

Is there not some probability distribution analogue of a "width" here, similar to the fuzzy but useful idea of the "size" of an atomic orbital, say?
The probability is determined by the wavefunction. The wavefunction is determined by how it was initially prepared and its subsequent evolution.

22. Originally Posted by exchemist
Originally Posted by Howard Roark
Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321

I am sorry but I am confused between the amplitude of the electric and magnetic waves and the width of the photon itself.
The photon (very much like the electron) is a pointlike particle (dimensionless). You are confusing its dimensions with its position, I have explained that to you in a prior thread, the one about fitting a photon inside a black hole of very small Schwarzschild radius. It is the position that is "smeared" probabilistically, as per QM, not its dimension. Bottom line, your question about "Width of a photon" has no answer since it has no meaning.
I understand that it is smeared probabilistically. When I am talking about the 'width' of the photon, I am not referring to the fact that it is a pointlike particle ( I agree with that ), I am referring to the fact that it occupies a certain volume ( probabilistically ). The pointless particle is everywhere in that volume at the same time with a certain probability of being at each point of that volume.
So, stop using the term "width". It does not apply.
OK Howard, but I'm now curious as well. Tell me, how can one determine how close a photon has to pass, to an atom capable of absorbing it, in order for the absorption probability to be significant? If it flies by at a distance of 0.1nm from the nucleus, for example, Id have thought the probability must be higher than if it passes 10nm away.

Is there not some probability distribution analogue of a "width" here, similar to the fuzzy but useful idea of the "size" of an atomic orbital, say?

KJW ninja'd me :-)

23. Originally Posted by Howard Roark
Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321

I am sorry but I am confused between the amplitude of the electric and magnetic waves and the width of the photon itself.
The photon (very much like the electron) is a pointlike particle (dimensionless). You are confusing its dimensions with its position, I have explained that to you in a prior thread, the one about fitting a photon inside a black hole of very small Schwarzschild radius. It is the position that is "smeared" probabilistically, as per QM, not its dimension. Bottom line, your question about "Width of a photon" has no answer since it has no meaning.
I understand that it is smeared probabilistically. When I am talking about the 'width' of the photon, I am not referring to the fact that it is a pointlike particle ( I agree with that ), I am referring to the fact that it occupies a certain volume ( probabilistically ). The pointless particle is everywhere in that volume at the same time with a certain probability of being at each point of that volume.
So, stop using the term "width". It does not apply.
The photon is not just a pointless particle, it is also a wave. So one can wonder about the width of the wave.

And in the original post, I was really talking about the width of the electric and magnetic waves.

24. Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321

I am sorry but I am confused between the amplitude of the electric and magnetic waves and the width of the photon itself.
The photon (very much like the electron) is a pointlike particle (dimensionless). You are confusing its dimensions with its position, I have explained that to you in a prior thread, the one about fitting a photon inside a black hole of very small Schwarzschild radius. It is the position that is "smeared" probabilistically, as per QM, not its dimension. Bottom line, your question about "Width of a photon" has no answer since it has no meaning.
I understand that it is smeared probabilistically. When I am talking about the 'width' of the photon, I am not referring to the fact that it is a pointlike particle ( I agree with that ), I am referring to the fact that it occupies a certain volume ( probabilistically ). The pointless particle is everywhere in that volume at the same time with a certain probability of being at each point of that volume.
So, stop using the term "width". It does not apply.
The photon is not just a pointless particle, it is also a wave. So one can wonder about the width of the wave.
Then you can use :

The more precision you have in measuring the photon frequency, the less precision you have in measuring its "occupied volume".

25. Originally Posted by Howard Roark
Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321

I am sorry but I am confused between the amplitude of the electric and magnetic waves and the width of the photon itself.
The photon (very much like the electron) is a pointlike particle (dimensionless). You are confusing its dimensions with its position, I have explained that to you in a prior thread, the one about fitting a photon inside a black hole of very small Schwarzschild radius. It is the position that is "smeared" probabilistically, as per QM, not its dimension. Bottom line, your question about "Width of a photon" has no answer since it has no meaning.
I understand that it is smeared probabilistically. When I am talking about the 'width' of the photon, I am not referring to the fact that it is a pointlike particle ( I agree with that ), I am referring to the fact that it occupies a certain volume ( probabilistically ). The pointless particle is everywhere in that volume at the same time with a certain probability of being at each point of that volume.
So, stop using the term "width". It does not apply.
The photon is not just a pointless particle, it is also a wave. So one can wonder about the width of the wave.
Then you can use :

The more precision you have in measuring the photon frequency, the less precision you have in measuring its "occupied volume".
I take it and are the momenta in the y and z directions. How can the photon have a momentum along these directions? Do you mean it is moving within the wave up and down and left and right with a momentum in each direction?

26. Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321

I am sorry but I am confused between the amplitude of the electric and magnetic waves and the width of the photon itself.
The photon (very much like the electron) is a pointlike particle (dimensionless). You are confusing its dimensions with its position, I have explained that to you in a prior thread, the one about fitting a photon inside a black hole of very small Schwarzschild radius. It is the position that is "smeared" probabilistically, as per QM, not its dimension. Bottom line, your question about "Width of a photon" has no answer since it has no meaning.
I understand that it is smeared probabilistically. When I am talking about the 'width' of the photon, I am not referring to the fact that it is a pointlike particle ( I agree with that ), I am referring to the fact that it occupies a certain volume ( probabilistically ). The pointless particle is everywhere in that volume at the same time with a certain probability of being at each point of that volume.
So, stop using the term "width". It does not apply.
The photon is not just a pointless particle, it is also a wave. So one can wonder about the width of the wave.
Then you can use :

The more precision you have in measuring the photon frequency, the less precision you have in measuring its "occupied volume".
I take it and are the momenta in the y and z directions.
Nope they are all the ERRORS (the incertitude) in measuring the momentum in the respective direction.

Do you mean it is moving within the wave up and down and left and right with a momentum in each direction?
No, I do not mean that, I mean exactly what I wrote.

27. Nope they are all the ERRORS (the incertitude) in measuring the momentum in the respective direction.
Ok, then let's say it is moving along the x axis. What would delta py and delta pz be?

28. Originally Posted by Nic321
Nope they are all the ERRORS (the incertitude) in measuring the momentum in the respective direction.
Ok, then let's say it is moving along the x axis. What would delta py and delta pz be?
Whatever your measurement error is. Check your measuring devices.

29. Originally Posted by Howard Roark
Originally Posted by Nic321
Nope they are all the ERRORS (the incertitude) in measuring the momentum in the respective direction.
Ok, then let's say it is moving along the x axis. What would delta py and delta pz be?
Whatever your measurement error is. Check your measuring devices.
Ok I can imagine to some extent that there could be a momentum in the y and z direction. And the larger the energy, the larger the amplitude of the wave, so the larger the square of the wave function which gives the probability distribution, right? So the larger the momentum uncertainty will be in the y and z axis, no?

30. Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321
Nope they are all the ERRORS (the incertitude) in measuring the momentum in the respective direction.
Ok, then let's say it is moving along the x axis. What would delta py and delta pz be?
Whatever your measurement error is. Check your measuring devices.
Ok I can imagine to some extent that there could be a momentum in the y and z direction. And the larger the energy, the larger the amplitude of the wave, so the larger the square of the wave function which gives the probability distribution, right? So the larger the momentum uncertainty will be in the y and z axis, no?
The momentum uncertainty has nothing to do with the momentum value, it is a function of your instruments, the interaction with the particles and your skills as an experimentalist.

31. Originally Posted by Howard Roark
Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321
Nope they are all the ERRORS (the incertitude) in measuring the momentum in the respective direction.
Ok, then let's say it is moving along the x axis. What would delta py and delta pz be?
Whatever your measurement error is. Check your measuring devices.
Ok I can imagine to some extent that there could be a momentum in the y and z direction. And the larger the energy, the larger the amplitude of the wave, so the larger the square of the wave function which gives the probability distribution, right? So the larger the momentum uncertainty will be in the y and z axis, no?
The momentum uncertainty has nothing to do with the momentum value, it is a function of your instruments, the interaction with the particles and your skills as an experimentalist.
So it is only a experimental problem, the uncertainty is not due in any way to the wave function of the photon?

32. Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321
Nope they are all the ERRORS (the incertitude) in measuring the momentum in the respective direction.
Ok, then let's say it is moving along the x axis. What would delta py and delta pz be?
Whatever your measurement error is. Check your measuring devices.
Ok I can imagine to some extent that there could be a momentum in the y and z direction. And the larger the energy, the larger the amplitude of the wave, so the larger the square of the wave function which gives the probability distribution, right? So the larger the momentum uncertainty will be in the y and z axis, no?
The momentum uncertainty has nothing to do with the momentum value, it is a function of your instruments, the interaction with the particles and your skills as an experimentalist.
So it is only a experimental problem, the uncertainty is not due in any way to the wave function of the photon?
Well not exactly. After all, the Uncertainty Principle itself is a direct consequence of particles having a wavefunction. But the fact that the expectation value of the momentum in the y and z directions happens to be zero in this case doesn't prevent there being an uncertainty around that average zero value.

I suspect what we are really saying is that any question about the "width" of a photon needs to be asked in the context of a particular physical arrangement, in which the photon may interact with something (a "measurement" if you like). Perhaps it is worth a thought experiment as to what sort of arrangment would potentially interact with a photon in away that depended on its "width".

33. Originally Posted by exchemist
Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321
Originally Posted by Howard Roark
Originally Posted by Nic321
Nope they are all the ERRORS (the incertitude) in measuring the momentum in the respective direction.
Ok, then let's say it is moving along the x axis. What would delta py and delta pz be?
Whatever your measurement error is. Check your measuring devices.
Ok I can imagine to some extent that there could be a momentum in the y and z direction. And the larger the energy, the larger the amplitude of the wave, so the larger the square of the wave function which gives the probability distribution, right? So the larger the momentum uncertainty will be in the y and z axis, no?
The momentum uncertainty has nothing to do with the momentum value, it is a function of your instruments, the interaction with the particles and your skills as an experimentalist.
So it is only a experimental problem, the uncertainty is not due in any way to the wave function of the photon?
Well not exactly. After all, the Uncertainty Principle itself is a direct consequence of particles having a wavefunction. But the fact that the expectation value of the momentum in the y and z directions happens to be zero in this case doesn't prevent there being an uncertainty around that average zero value.
That's what I thought, the fact that the particle has a probability distribution is important here.

I suspect what we are really saying is that any question about the "width" of a photon needs to be asked in the context of a particular physical arrangement, in which the photon may interact with something (a "measurement" if you like). Perhaps it is worth a thought experiment as to what sort of arrangment would potentially interact with a photon in away that depended on its "width".
From what I understand, the width depends on the observer. For instance, 2 observers observing one photon with different wavelength ( depending on how the observer is moving) will not see the same probability distribution, so probably won't see the same width.

34. This all depends on what you call photon. What theorists call photon is photon of single wavelenght is ideal case but impossible to create or be created as it would be infinite in time and infinite in space. What experimentators are calling photon is in fact coherent pulse that has expectation value of photon number operator 1. It has smeared probabilistic distribution in energy/ therefore its finite in time and also probabilistic distribution in k-vector space therefore its finite object in space. Real photon is superposition of infinitely many plane-wave-single-wavelenght "idealistic" photons. The distribution functions specificaly in energy are given by broadened spectral lines (which depend on time of life of level) of atoms or objects that created the photon in first place.

35. Originally Posted by Gere
This all depends on what you call photon. What theorists call photon is photon of single wavelenght is ideal case but impossible to create or be created as it would be infinite in time and infinite in space. What experimentators are calling photon is in fact coherent pulse that has expectation value of photon number operator 1. It has smeared probabilistic distribution in energy/ therefore its finite in time and also probabilistic distribution in k-vector space therefore its finite object in space. Real photon is superposition of infinitely many plane-wave-single-wavelenght "idealistic" photons. The distribution functions specificaly in energy are given by broadened spectral lines (which depend on time of life of level) of atoms or objects that created the photon in first place.
What about a photon produced by a natural process, say a photon created by the sun? Each photon received from the sun has a precise wavelength, doesn't it? what is the probabilistic distribution of such a photon?

36. Originally Posted by Nic321
Originally Posted by Gere
This all depends on what you call photon. What theorists call photon is photon of single wavelenght is ideal case but impossible to create or be created as it would be infinite in time and infinite in space. What experimentators are calling photon is in fact coherent pulse that has expectation value of photon number operator 1. It has smeared probabilistic distribution in energy/ therefore its finite in time and also probabilistic distribution in k-vector space therefore its finite object in space. Real photon is superposition of infinitely many plane-wave-single-wavelenght "idealistic" photons. The distribution functions specificaly in energy are given by broadened spectral lines (which depend on time of life of level) of atoms or objects that created the photon in first place.
What about a photon produced by a natural process, say a photon created by the sun? Each photon received from the sun has a precise wavelength, doesn't it? what is the probabilistic distribution of such a photon?
The question you are asking ignores the information that Gere gave you, so perhaps it needs to be repeated, in slightly different terms.

Forget about photons for a second and just consider a sinusoidal wave. In order for that wave to consist truly of a single frequency, it must conform to sin(t) for all t. Not just for some t, but all t from the infinite past to the infinite future.

If the sine function applies only to a finite interval -- as it must, since the universe has existed for a finite time -- then its spectrum no longer consists of a single frequency. This result has nothing to do with QM; it comes from 18th century mathematics. And it applies to all photons, whether they were created by natural or artificial processes (whatever that's supposed to mean).

37. Originally Posted by tk421
Originally Posted by Nic321
Originally Posted by Gere
This all depends on what you call photon. What theorists call photon is photon of single wavelenght is ideal case but impossible to create or be created as it would be infinite in time and infinite in space. What experimentators are calling photon is in fact coherent pulse that has expectation value of photon number operator 1. It has smeared probabilistic distribution in energy/ therefore its finite in time and also probabilistic distribution in k-vector space therefore its finite object in space. Real photon is superposition of infinitely many plane-wave-single-wavelenght "idealistic" photons. The distribution functions specificaly in energy are given by broadened spectral lines (which depend on time of life of level) of atoms or objects that created the photon in first place.
What about a photon produced by a natural process, say a photon created by the sun? Each photon received from the sun has a precise wavelength, doesn't it? what is the probabilistic distribution of such a photon?
The question you are asking ignores the information that Gere gave you, so perhaps it needs to be repeated, in slightly different terms.

Forget about photons for a second and just consider a sinusoidal wave. In order for that wave to consist truly of a single frequency, it must conform to sin(t) for all t. Not just for some t, but all t from the infinite past to the infinite future.

If the sine function applies only to a finite interval -- as it must, since the universe has existed for a finite time -- then its spectrum no longer consists of a single frequency. This result has nothing to do with QM; it comes from 18th century mathematics. And it applies to all photons, whether they were created by natural or artificial processes (whatever that's supposed to mean).
Let's be concrete... In the drawing on the upper left here , what you say means that the frequency at the start of the wave front is not the same as the frequency at the middle of the pulse. When we say that a photon has a given frequency, we are refering to the frequency in the middle of the pulse. Is that correct or did I get it completely wrong?

38. Thats not photon, thats wavefunction of electron probably. Frequency and position isnt really connected like this. You just have probability distribution of frequency and probability distribution of k-vector/position of photon. These things are connected as tk421 tried to explain. Key word is Fourier transform.

39. Originally Posted by Gere
Thats not photon, thats wavefunction of electron probably. Frequency and position isnt really connected like this. You just have probability distribution of frequency and probability distribution of k-vector/position of photon. These things are connected as tk421 tried to explain. Key word is Fourier transform.
To be honest, it seems to me neither you nor tk421 is really dealing with what seems to me to be Nic's question. All this you both describe is the well-known linear momentum and position relations for a QM wave-particle, along the direction of propagation, in terms of a wave packet versus a monochromatic sine wave. Classic Fourier series/uncertainty principle stuff, in fact.

But, as I understand it, Nic's question is about the degree of dispersion or "width" of a photon perpendicular to the direction of propagation. That is interesting, it seems to me, in that it is not generally talked about in the textbooks. Earlier I posed the question of how far away from an atom that is able to absorb it, can a photon pass, and still have an appreciable probability of being absorbed?

The expectation value of its momentum normal to the direction of propagation is of course zero, but there will be an uncertainty around this value. Conversely the uncertainty in its position on the axes normal to the axis of travel will define the "width" of the photon. I think this is what he was trying to get at originally.

40. Originally Posted by exchemist

But, as I understand it, Nic's question is about the degree of dispersion or "width" of a photon perpendicular to the direction of propagation.
Had you paid attention, you would have seen that I answered that.

41. Oh, true exchemist. A bit sloppy on my part. The thing is that the distribution function (Wigners) is in fact function in whole phase space. Therefore not only is it smeared in say but also at and . This smearing (its usually 3D gaussian) through FT tels you the "size" of photon perpendicular to momentum. Such size or width is not sharp but one can "define" it to be say full-width-at-half-maxima of distribution.

42. Originally Posted by Howard Roark
Originally Posted by exchemist

But, as I understand it, Nic's question is about the degree of dispersion or "width" of a photon perpendicular to the direction of propagation.
Had you paid attention, you would have seen that I answered that.

Originally Posted by Howard Roark
So, stop using the term "width". It does not apply.
If this is your answer I recommend R. Loundon: The quantum theory of light.

43. Originally Posted by Gere
Originally Posted by Howard Roark
Originally Posted by exchemist

But, as I understand it, Nic's question is about the degree of dispersion or "width" of a photon perpendicular to the direction of propagation.
Had you paid attention, you would have seen that I answered that.

Originally Posted by Howard Roark
So, stop using the term "width". It does not apply.
If this is your answer I recommend R. Loundon: The quantum theory of light.
This is not the answer I was referring to. But I would be interested in a quotation from the above book where the author introduces the notion of "photon width".

44. Originally Posted by exchemist
To be honest, it seems to me neither you nor tk421 is really dealing with what seems to me to be Nic's question.
That's because I was responding narrowly only to the sub-part of his question that I quoted in my reply. Sorry if I had implied that my answer applied more generally. I'm still thinking of the best way to respond to what I think is his question, because the answer depends a bit (actually, quite a bit) on the precise nature of the question (e.g., whether one is asking about absorption cross-sections, say, which is the nature of your follow-up). His is a thoughtful question and deserves an equally thoughtful response.

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