Is a photon an EM wave packet?

In summary: One can also say they have a random phase.In summary, the similarities between the classical wave packet and the QM photon are striking, as they both move at the group velocity and are subjected to an uncertainty principle. However, the concept of a photon is a strange notion in quantum mechanics and it has nothing to do with the uncertainty principle. The energy of a photon is given by E^2 = E1^2 + E2^2 in the region of superposition, and this explains why the linear superposition of two wavepackets does not conserve energy. In quantum mechanics, the addition of two states always results in a squared value, which explains why we often see squared terms in calculations.
  • #1
da_willem
599
1
The similarities between the classical wave packet and the QM photon are striking. They both move at the group velocity. They both are subjected to an uncertainty principle. A classical wave packet is like a photon a region of concentrated energy, ...

So is a photon something like the quantised version of a wavepacket?
 
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  • #2
da_willem said:
The similarities between the classical wave packet and the QM photon are striking. They both move at the group velocity. They both are subjected to an uncertainty principle. A classical wave packet is like a photon a region of concentrated energy, ...

So is a photon something like the quantised version of a wavepacket?

Quantum mechanics deals with classical electromagnetic field and therefore "photon" is a strange notion to it.The photon has nothing to do with the uncertainty principle,as his velocity,momentum,energy,wavelength are given.Trying to figure out where is a photon is a nonsense...It travels with the speed of light.
No,a photon is a particle just like the electron...Wave behavior of light is similar to the one of particles...So,their particle behaviors are the same as well.
 
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  • #3
I have an interesting observation here.Suppose you have two photons(represented by wavepackets) close by.In the region of superposition how should the fields add? Suppose you have E1(x) and E2(x) as the two wavepackets at a given instant.One may naively expect the field to be E1(x) + E2(x) in the region of superposition--but this is not correct.Energy goes as E^2, so the linear superposition does not conserve energy.So one should have E in the region of superposition given by E^2 =E1^2 + E2^2.Comments please.
 
  • #4
gptejms,

If the two wave E1 and E2 are totally uncorrelated (in phase), then the power term is also the sum of their respective power, the cross-term E1*E2*cos(phase) is statistically zero.

Seen in QM, two statistically uncorrelated waves represent in fact two different photons. Think to the two-holes-in-a-screen experiment: two different photons emitted from each hole have uncorrelated phases, and their energy will also add-up like their amplitudes.

On the contrary, if their phase are correlated, then the two waves are actually only one photon. Think to the two-holes-in-a-screen experiment: two correlated waves add-up after the screen and correspond to only photon to be detected.
 
  • #5
More simply (at least for me :yuck: ),
In QM you add states |psi>=|psi1>+|psi2>, but you always compute the results (energy, momentum, etc...) with the square of these states: there is no surprise (think on the probability conservation law).
For example, when you have a doubt, just think with the density matrix rho=|psi><psi| and <a>=tr(rho.A) (where A is any observable, ie for example the projector of an energy value measurement).
Therefore you have the simple result: rho=|psi><psi|<>|psi1><psi1|+|psi2><psi2|=rho1+rho2.

This is the mathematical consistency of QM and it explains somewhat why we find a lot of squares in the results!

Seratend.
 
  • #6
lalbatros said:
gptejms,

If the two wave E1 and E2 are totally uncorrelated (in phase), then the power term is also the sum of their respective power, the cross-term E1*E2*cos(phase) is statistically zero.

Seen in QM, two statistically uncorrelated waves represent in fact two different photons. Think to the two-holes-in-a-screen experiment: two different photons emitted from each hole have uncorrelated phases, and their energy will also add-up like their amplitudes.

On the contrary, if their phase are correlated, then the two waves are actually only one photon. Think to the two-holes-in-a-screen experiment: two correlated waves add-up after the screen and correspond to only photon to be detected.

In a laser you have photons in phase(this does not amount to saying there is just one photon as you say) and the cross-term is not zero.If E1 ~ E2 = E then you get 4E^2 as the answer for energy if you superpose the amplitudes of the 'wavepackets' in a linear way(whereas the correct answer is 2E^2).
 
  • #7
gptejms,

You are right.
We should go to the limit of very low intensity, single photon counting.
Then, of course, two photons have uncorrelated waves.
 

1. What is a photon?

A photon is a fundamental particle of light that carries energy and momentum. It is considered both a particle and a wave, exhibiting properties of both.

2. Is a photon an EM wave packet or a particle?

A photon is both an EM wave packet and a particle. It behaves as a wave in certain experiments, such as the double-slit experiment, and as a particle in others, such as the photoelectric effect.

3. How does a photon travel?

A photon travels through space at the speed of light, which is approximately 299,792,458 meters per second. It does not require a medium to travel through, unlike other types of waves.

4. Can a photon be created or destroyed?

A photon cannot be created or destroyed, but it can be converted into other forms of energy. For example, when a photon is absorbed by an atom, it can cause an electron to jump to a higher energy level.

5. How does the energy of a photon relate to its frequency and wavelength?

The energy of a photon is directly proportional to its frequency and inversely proportional to its wavelength. This relationship is described by the equation E = hf, where E is energy, h is Planck's constant, and f is frequency.

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