Quantization of E/M field

[JK423]

I'd like to ask one 'newbie' question on the quantization of the E/M field.
I know that when we quantize the E/M field we get infinite in number harmonic oscillators.

I just want to know, what's the 'physical meaning' of these harmonic oscillators?
How do we interpret this result?

Thanks in advance

 

[JK423]

Anyone?

 

[bapowell]

I'd like to ask one 'newbie' question on the quantization of the E/M field.
I know that when we quantize the E/M field we get infinite in number harmonic oscillators.

I just want to know, what's the 'physical meaning' of these harmonic oscillators?
How do we interpret this result?

Thanks in advance

In ordinary QM, the states of the harmonic oscillator are energy eigenstates: we move up the ladder of states by adding a bit of energy, $$\hbar \omega$$, each time. In QFT, the states of the harmonic oscillator are not energy eigenstates, but states of definite particle number. The creation and annihilation operators that show up in the Fourier decomposition of a quantum field create and annihilate particles, rather than raise and lower energies in the harmonic oscillator.

So the infinite number of oscillator states correspond to the quantum field's potential to create an infinite number of particles.

 

[JK423]

Thanks a lot!

 

[ytuab]

So the conclusion of the "physical meaning" (not only the "mathematical" meaning) proves to be (photon) particles? Is it really OK?

Even in QFT photons, the photon satisfies the Maxwell's equation like the electromagnetic waves. What's the relation with this?

The ordinary particle has the definite figure(volume) and the character, and of course the ordinary rigid particle can not be divided. But the "photons" are not known well about this?
If one photon can't be divided , it can't interfere with itself by splitting.
How does this harmonic oscillator explain about this phenomenon?

If we can clearly say the photon is a rigid particle and can't be divided,
we can clearly say the strange nonlocal phenomena such as the entanglement of the two photons is "really" occurring.
But if the photon can be divided like the electromagnetic waves, the entanglement is not needed?

The photon is a particle(the entanglement really occur) or a wave(the entaglement is not needed?)
Which case (partcle or wave) is more natural?

Of course, the relativistic QFT (QED) is a local theory. So how can we explain about the nonlocal phenomena?
See also this thread.

 

[bapowell]

So the conclusion of the "physical meaning" (not only the "mathematical" meaning) proves to be (photon) particles? Is it really OK?

Yup.

Even in QFT photons, the photon satisfies the Maxwell's equation like the electromagnetic waves. What's the relation with this?

Right. The photon is associated with the eigenmodes of the Fourier decomposition of the vector potential, $$A_\mu$$. And so the vector potential assumes the role of the photon field in QED. Of course, it satisfies Maxwell's Eqs.

The ordinary particle has the definite figure(volume) and the character, and of course the ordinary rigid particle can not be divided. But the "photons" are not known well about this?
If one photon can't be divided , it can't interfere with itself by splitting.
How does this harmonic oscillator explain about this phenomenon?

I'm not sure I follow you here. It sounds like you are referring to the double slit experiment and the quantum mechanical nature of the photon's wavefunction. All of this can be understood with ordinary, non-relativistic QM.

I'm not sure I can answer the remaining questions about entanglement, although I'm sure someone else can. These are more fundamentally quantum mechanical phenomena, and I don't know if QFT offers anything beyond what non-relativistic QM can on these matters.

 

[ytuab]

Thanks for reply, bapowell.

The photon has the two natures of a particle and wave.
The interference of one photon can be caused only by the wave nature , not by the particle.

So if one photon is a wave, it can be detected as a "photon" only when the strength of the EM waves is above some threshold?

Suppose two pair photons with the opposite polarization axes are flying in the opposite directions.
The photon A bumps into the polarizing filter A, while photon B bumps into the filter B.
When the angle between the photon A and the filter A is $$\theta_{1}$$, the probability photon A passes the filter A is $$\cos^2 \theta_{1}$$. (the reflection is $$\sin^2 \theta_{1}$$).
While the angle between the photon B and the filter B is $$\theta_{2}$$, the pass is $$\cos^2 \theta_{2}$$, and the reflection is $$\sin^2 \theta_{2}$$.
But this case satisfies the Bell inequality and is different from the experimental results.

The experimental results shows that the instant the photon A has passed the filter A, the polarized angle of the photon B becomes the same as the filter A. So the probability that both the two photons pass each filter is $$\cos^2 (\theta_{1}-\theta_{2})$$. This case violates the Bell inequality in some angle($$\theta_{1}-\theta_{2}$$). Right?

But if the photon is a wave, the passing photon A can be detected only when the polarized angle of the photon A is near the angle of the filter A due to the detecting threshold?

If the photon is a particle, it can't be divided, so there are only two patterns (pass or reflect) as shown in the Bell inequality.
But in case of a wave, there are three patterns (pass (detected as a photon) or reflect or not detected as a photon because the strength is under the threshold)?
If so, the wave nature of the photon causes the illusion that the entanglement is occurring?

 

[Cthugha]

So if one photon is a wave, it can be detected as a "photon" only when the strength of the EM waves is above some threshold?

No, this assumption is ruled out by coincidence counting experiments with single photon sources. The observed antibunching cannot be explained by any classical probability distribution of the amplitude/photon number, which would be the basis for a wave-only view. You need a nonclassical probability distribution which is used in the quantized approach.

If the photon is a particle, it can't be divided, so there are only two patterns (pass or reflect) as shown in the Bell inequality.

That is nonsense. The photon does not need to split. It is sufficient that there are indistinguishable probability amplitudes leading from the same initial to the same final state. These do interfere. The notion of interference being a property of particles is a semantic error introduced back in the days of the famous quote from Fermi. Read Glauber's Nobel speech for an easy-to-follow introduction into what interference and quantized excitations of the em field indeed mean.

 

[ytuab]

No, this assumption is ruled out by coincidence counting experiments with single photon sources. The observed antibunching cannot be explained by any classical probability distribution of the amplitude/photon number, which would be the basis for a wave-only view. You need a nonclassical probability distribution which is used in the quantized approach.
That is nonsense. The photon does not need to split. It is sufficient that there are indistinguishable probability amplitudes leading from the same initial to the same final state. These do interfere. The notion of interference being a property of particles is a semantic error introduced back in the days of the famous quote from Fermi. Read Glauber's Nobel speech for an easy-to-follow introduction into what interference and quantized excitations of the em field indeed mean.

Ok. So if one photon is (splitted) by the splitter into the two different paths, we can detect one photon only in one of the two paths at the same time?

For example, the first light strength is 10, then it was splitted by the splitter, suppose the detector can detect it when the strength is 6 as a single photon. So 10 is detected as one photon not 2 photons(=12).
But the splitted photon(6 and 4)(EM-waves) can interfere with itself?
And the 4 side is not detected, only 6 side is detected as a photon?

Is this case possible?

 

[SpectraCat]

Ok. So if one photon is (splitted) by the splitter into the two different paths, we can detect one photon only in one of the two paths at the same time?

For example, the first light strength is 10, then it was splitted by the splitter, suppose the detector can detect it when the strength is 6 as a single photon. So 10 is detected as one photon not 2 photons(=12).
But the splitted photon(6 and 4)(EM-waves) can interfere with itself?
And the 4 side is not detected, only 6 side is detected as a photon?

Is this case possible?

No, that is not what happens in the experiment. As Cthuga said, the photon amplitude is not split in the manner you describe. If a photon with "strength" 10 goes in, then a photon with "strength" 10 is detected ... no matter what you do. Keep in mind that *exactly* the same experiment works with massive particles ... no one ever detects half an electron ... they detect the intact particle.

The two slit experiment is not about "splitting photons", it is about entanglement of the wavefunction of the particle with the physical paths in the detector.

 

[bapowell]

ytuab,

I've always felt that the 'wave-particle duality' concept was unnecessary and only led to confusion. How can something be both a wave and a particle, but behave like one or the other depending on its mood? Obviously that's impossible. In reality, light is a particle. It is comprised of photons. Always. The wavelike properties of light (and electrons, etc) derive from the quantum mechanical wavefunction of the particle. So photons aren't splitting in double slit experiments anymore than electrons do in the same setup. It's purely a result of the physical manifestation of the probability amplitude of the particle.