Why do photons come in quanta?

[mavc]

Hi,

Why do photons come in quanta? Why are the quanta not bigger or smaller or continuous?

Thanks,
Anna.

 

[Dickfore]

Because the creation and annihilation operators of the electromagnetic field obey the commutation relations:

$$\left[b_{\mathbf{k} \lambda}, b_{\mathbf{k}' \lambda'}\right] = \left[b^{\dagger}_{\mathbf{k} \lambda}, b^{\dagger}_{\mathbf{k}' \lambda'}\right] = 0$$

$$\left[b_{\mathbf{k} \lambda}, b^{\dagger}_{\mathbf{k}' \lambda'}\right] = \delta_{\mathbf{k}, \mathbf{k}'} \, \delta_{\lambda, \lambda'}$$

Then, the number operator obeys the following commutation relations:

$$\left[b^{\dagger}_{\mathbf{k} \lambda} b_{\mathbf{k} \lambda}, b_{\mathbf{k}' \lambda'}\right] = \left[b^{\dagger}_{\mathbf{k} \lambda}, b_{\mathbf{k}' \lambda'}\right] \, b_{\mathbf{k}' \lambda'} = - \delta_{\mathbf{k}, \mathbf{k}'} \, \delta_{\lambda, \lambda'} \, b_{\mathbf{k}, \lambda}$$

This means that the eigenvalues of the number operator can be changed in unit amounts. Since the vacuum corresponds to zero photons, these eigenvalues are non-negative integers.

 

[Brainguy]

The quantum is something I don't understand enough. But atoms always have electrons flying around them right? They fly in certain paths called shells. The shell closest to the nucleus us the shortest path right? It is called the base state. Electrons with the least energy stay at the base state because they don't have enough energy to go the longer ways. An electron with high energy will fly on he shell farthest from the nucleus simply because it can. When an electron gains or loses energy (emitting or absorbing a photon) it leaps to either a farther out shell, or a farther in shell. If it lost energy, it goes closer to the nucleus to complete it's rounds faster. An electron can not jump to a space in between the shells. It can only jump to a shell farther or closer to the nucleus. So electrons can only be on a shell. Sorta like being on a ladder or a slide. On the slide you can stand anywhere. On a ladder u can only stand on the rungs(shells) you can move up and down the ladder but can never put your foot on a space with no rungs.

 

[mavc]

Because the creation and annihilation operators of the electromagnetic field obey the commutation relations:

$$\left[b_{\mathbf{k} \lambda}, b_{\mathbf{k}' \lambda'}\right] = \left[b^{\dagger}_{\mathbf{k} \lambda}, b^{\dagger}_{\mathbf{k}' \lambda'}\right] = 0$$

$$\left[b_{\mathbf{k} \lambda}, b^{\dagger}_{\mathbf{k}' \lambda'}\right] = \delta_{\mathbf{k}, \mathbf{k}'} \, \delta_{\lambda, \lambda'}$$

Then, the number operator obeys the following commutation relations:

$$\left[b^{\dagger}_{\mathbf{k} \lambda} b_{\mathbf{k} \lambda}, b_{\mathbf{k}' \lambda'}\right] = \left[b^{\dagger}_{\mathbf{k} \lambda}, b_{\mathbf{k}' \lambda'}\right] \, b_{\mathbf{k}' \lambda'} = - \delta_{\mathbf{k}, \mathbf{k}'} \, \delta_{\lambda, \lambda'} \, b_{\mathbf{k}, \lambda}$$

This means that the eigenvalues of the number operator can be changed in unit amounts. Since the vacuum corresponds to zero photons, these eigenvalues are non-negative integers.

Hi,

Thanks for your reply - however my maths is pretty rusty. Could you please explain in layman's terms, or define what the symbols mean?

Regards,
Anna.

 

[mavc]

The quantum is something I don't understand enough. But atoms always have electrons flying around them right? They fly in certain paths called shells. The shell closest to the nucleus us the shortest path right? It is called the base state. Electrons with the least energy stay at the base state because they don't have enough energy to go the longer ways. An electron with high energy will fly on he shell farthest from the nucleus simply because it can. When an electron gains or loses energy (emitting or absorbing a photon) it leaps to either a farther out shell, or a farther in shell. If it lost energy, it goes closer to the nucleus to complete it's rounds faster. An electron can not jump to a space in between the shells. It can only jump to a shell farther or closer to the nucleus. So electrons can only be on a shell. Sorta like being on a ladder or a slide. On the slide you can stand anywhere. On a ladder u can only stand on the rungs(shells) you can move up and down the ladder but can never put your foot on a space with no rungs.

Hi,

Is this the same process that happens in nuclear fusion or antimatter annihilation, where gamma radiation is generated?

Thanks,
Anna.

 

[Dickfore]

Hi,

Thanks for your reply - however my maths is pretty rusty. Could you please explain in layman's terms, or define what the symbols mean?

Regards,
Anna.

The creation/annihilation operator formalism in Quantum Mechanics is known as Second Quantization. There are so called canonical commutation (for bosons) and anti-commutation (for fermions) relations between them, just like the famous $[x, p] = i \, \hbar$ commutation relation for the position and momentum in the First Quantization. I wrote them above for the creation/annihilation operators of the electromagnetic field.

 

[Polyrhythmic]

The quantum is something I don't understand enough. But atoms always have electrons flying around them right? They fly in certain paths called shells. The shell closest to the nucleus us the shortest path right? It is called the base state. Electrons with the least energy stay at the base state because they don't have enough energy to go the longer ways. An electron with high energy will fly on he shell farthest from the nucleus simply because it can. When an electron gains or loses energy (emitting or absorbing a photon) it leaps to either a farther out shell, or a farther in shell. If it lost energy, it goes closer to the nucleus to complete it's rounds faster. An electron can not jump to a space in between the shells. It can only jump to a shell farther or closer to the nucleus. So electrons can only be on a shell. Sorta like being on a ladder or a slide. On the slide you can stand anywhere. On a ladder u can only stand on the rungs(shells) you can move up and down the ladder but can never put your foot on a space with no rungs.

What you are describing here is basically the Bohr model, which is nice to picture but actually not accurate. In quantum mechanics, electrons and other particles are described by the so-called wave function. The essence is that an electron is no point particle which orbits the nucleus. The wave function is something which spreads through space and allows one to define a probability density for measuring an electron somewhere. You should read some basic text on quantum mechanics.

 

[cragar]

Maybe reading about the photo-electric effect would be a good place to start. At least it gives experimental evidence.

 

[vanhees71]

The photo-electric effect is totally explainable with the interaction of the electrons with a classical em. field. It does not demonstrate the existence of photons but the quantum theory of (non-relativistic) electrons in the metal.

A photon is by definition a one-particle Fock state of the electromagnetic quantum field!

 

[cragar]

really, In the photo-electric effect you need a minimum energy particle to liberate an electron. Then classically why couldn't I shoot a bunch of low-energy EM waves until I liberate it?

 

[Naty1]

No one really knows why some phenomena are quantum like and others particle like...it's like asking why an electron is a "quanta", a particle of mass...nobody really knows.
Like other elementary particles, photons behavior is best explained by quantum mechanics and will exhibit wave–particle duality, exhibiting properties of both waves and particles.The math posted above, if correct, simply describes WHAT happens not WHY...

Is this the same process that happens in nuclear fusion or antimatter annihilation, where gamma radiation is generated?

no, not the same process; fusion and annihilation are NUCLEAR reactions...Chemical reactions typically involve electrons outside the nucleus, like hydrogen and oxygen combining to form water.

In fusion, certain smaller nuclei combine to form a heaver nucleus; in fission, large nuclei breakdown into smaller...and binding energies within the nucleus accounts fior the energy produced.

Check Wikipedia for fission and fusion for some additional information.

 

[mavc]

Thanks for all of your replies.

Dickfore, I don't understand why the eigenvalues would be integer values... in particular I don't understand your statement: "This means that the eigenvalues of the number operator can be changed in unit amounts. Since the vacuum corresponds to zero photons, these eigenvalues are non-negative integers". Do bosons or fermions interacting with one another generate/annihilate an electromagnetic field? Is this an empirical or theoretical relation?

cragar, I'm aware of the photoelectric effect. I was wondering about the why behind the quantised nature of em waves rather than the experimental observations..

vanhees71, what did you mean by "one-particle Fock state of the electromagnetic quantum field"?

Naty1, I'm aware that nuclear fusion & fission release energy, but why is some of it in the form of gamma em waves? Is it to do with the interacting particles during the fission/fusion process?

Regards,
Anna.

 

[TriTertButoxy]

Hi,

Why do photons come in quanta? Why are the quanta not bigger or smaller or continuous?

Thanks,
Anna.

What do you want your answer in terms of?

Are you asking, why must we quantize the electromagnetic field? I could just say 'because that's the way it is' but I'm sure that's not what you want.

Or are you asking, why does the formal quantization of a field lead to discrete particle-like objects? but this requires some mathematical understanding...

The bottom line is quantum mechanics is what causes the photons to come in quanta, but why the universe operates under the Rules of Quantum Mechanics is unknown. Scientists generally take that as a given.

 

[mavc]

What do you want your answer in terms of?

Are you asking, why must we quantize the electromagnetic field? I could just say 'because that's the way it is' but I'm sure that's not what you want.

Or are you asking, why does the formal quantization of a field lead to discrete particle-like objects? but this requires some mathematical understanding...

The bottom line is quantum mechanics is what causes the photons to come in quanta, but why the universe operates under the Rules of Quantum Mechanics is unknown. Scientists generally take that as a given.

I'm asking a form of the 2nd q - i.e. how come em waves come in discrete packets? This may sound like a silly question... but is it possible that space-time itself is divided into discrete pockets? and the em waves must therefore divide themselves up into these pockets, making them into discrete packets of energy?

Thanks,
Anna.

 

[Dickfore]

Dickfore, I don't understand why the eigenvalues would be integer values... in particular I don't understand your statement: "This means that the eigenvalues of the number operator can be changed in unit amounts. Since the vacuum corresponds to zero photons, these eigenvalues are non-negative integers".

The hamiltonian of the electromagnetic field is:

$$H = \hbar \, \sum_{\mathbf{k}, \lambda}{\omega_{\mathbf{k} \lambda} \, b^{\dagger}_{\mathbf{k}, \lambda} \, b_{\mathbf{k}, \lambda}}, \ \omega_{\mathbf{k} \, \lambda} = c k$$

It simply counts the number of photons in each mode, multiplying it with the corresponding energy of the photons. That is why:

$$n_{\mathbf{k}, \lambda} = b^{\dagger}_{\mathbf{k} \lambda} \, b_{\mathbf{k}, \lambda}$$

is identified as the operator that counts the number of photons in each mode. The above commutation relations (or, algebra) determines the possible eigenvalues of $n_{\mathbf{k}, \lambda}$ as the non-negative integer, the ground state being identified as the state with zero photons in each mode.

Do bosons or fermions interacting with one another generate/annihilate an electromagnetic field? Is this an empirical or theoretical relation?

Division of particles into Bosons and Fermions is simply according to what statistics they obey (or whether their creation/annihilation operators obey commutation of anti-commutation relations). Photons are the quanta of the electromagnetic field. They have spin 1 and are therefore bosons. Electrons are spin-1/2 particles and are fermions. The interaction between photons and fermions is the electromagnetic interaction. It comprises the essence of Quantum Electrodynamics (QED), the first successful field theory which reproduces all experiments in this field.

 

[TriTertButoxy]

I'm asking a form of the 2nd q - i.e. how come em waves come in discrete packets? This may sound like a silly question... but is it possible that space-time itself is divided into discrete pockets? and the em waves must therefore divide themselves up into these pockets, making them into discrete packets of energy?

Thanks,
Anna.

Within the framework of quantum field theory, the reason photons come in discrete quanta is not because space-time is dividing into discrete pockets. The discreteness of the photons is coming directly from an uncertainty between the value of the electromagnetic field and the rate at which it is changing. Mathematics then shows that this uncertainty leads to the amplitude of electromagnetic field coming in discrete packets. These discrete packets are then interpreted to be particles, named the photons.

I'm fairly certain this answer is not very enlightening, but this is how quantum field theory works. The uncertainty relation is what is making them come in discrete packets of energy.

 

[Brainguy]

You are all hurting my brain. Apparently this isn't as easy as i thought. Do u guys have like, a second keyboard with quantum symbols?

 

[mavc]

Me too, Brainguy. I will have to read up on quantum mechanics to understand the answers, I'm afraid. I feel like I'm missing too much information. And it probably shows through my questions. However I do feel like I'm coming closer to understanding the "rules", even though I don't know why the rules are there in the first place, which was my initial question.

Dickfore, what is a mode? What do the eigenvalues represent? I know that it's the solution to $$A x = \lambda x$$but what does it mean?

TriTertButoxy, do you have a link or further details on the maths that goes between the uncertainty and the amplitude coming in discrete packets? How does an amplitude come in discrete packets?

Thanks for your patience,
Anna.

 

[TriTertButoxy]

TriTertButoxy, do you have a link or further details on the maths that goes between the uncertainty and the amplitude coming in discrete packets? How does an amplitude come in discrete packets?

Thanks for your patience,
Anna.

This is a very basic result of quantum field theory, and every introductory text on the subject will have it under 'canonical quantization of a field.' (Peskin and Schroeder, Srednicki, Bjorken and Drell, Itzykson and Zuber, to name a few)

But before you pick up a quantum field theory book, I'd do two things
1. Brush up on your classical mechanics skills (learn what a mode is) and make sure you understand the Lagrangian/Hamiltonian formalism
2. Review the basic elements of quantum mechanics, so you know how the eigenvalue problem to various physical systems.

Wikipedia articles may suffice :)

I hope that helps. I'm sorry to have to differ you to texts since there's a formalism that goes into showing that the uncertainty relation is what leads to photons coming in discrete packets.

 

[edguy99]

Hi,

Why do photons come in quanta? Why are the quanta not bigger or smaller or continuous?

Thanks,
Anna.

One way to think of this is by an example. Every element has a specific amount of energy that is released when an electron falls into its first energy level. This photon is called the K-alpha emission and helps to uniquely identify elements. If an electron already exists in this first energy level, then a slightly higher energy photon is emitted called the K-beta photon. This type of photon example is an x-ray for your teeth. Electrons are dropped into the first energy level of copper atoms and x-rays shoot out. The property of the element is determining the energy of the photon, hence it cannot be a bit bigger or smaller.

If an electron is substantially turned or accelerated due to coulomb force, it may emit a Bremsstrahlung (braking radiation) photon at various wavelengths. These photons or quanta are indeed a varity of sizes.

 

[Dickfore]

Dickfore, what is a mode? What do the eigenvalues represent? I know that it's the solution to $$A x = \lambda x$$but what does it mean?

Modes are the frequencies of standing electromagnetic waves, where $\lambda$ labels the two possible polarization for a given k-vector. To avoid certain divergencies arrising from the infinite volume of space, it is simplest to consider periodic boundary conditions, where each component of the electromagnetic field is periodic:

$$f(x + L_{x}, y, z) = f(x, y + L_{y}, z) = f(x, y, z + L_{z}) = f(x, y, z)$$

and can be expanded in a Fourier series:

$$f(\mathbf{x}) = \frac{1}{V} \, \sum_{\mathbf{k}}{f_{\mathbf{k}} \, \exp{(i \mathbf{k} \cdot \mathbf{x})}}, \ f_{\mathbf{k}} = \int{d\mathbf{x} \, f(\mathbf{x}) \, \exp{(-i \, \mathbf{k} \cdot \mathbf{x})}}$$

where the possible k-vectors are of the form:

$$\mathbf{k} = 2 \pi \, \left(\frac{n_{x}}{L_{x}} \, \mathbf{e}_{x} + \frac{n_{y}}{L_{y}} \, \mathbf{e}_{y} + \frac{n_{z}}{L_{z}} \, \mathbf{e}_{z}\right), \; n_{x}, n_{y}, n_{z} = 0, 1, \ldots$$

Then, the wave equation:

$$\frac{1}{c^{2}} \, \frac{\partial^{2} f}{\partial t^{2}} - \nabla^{2} \, f = 0$$

becomes:

$$\left(\frac{\omega^{2}}{c^{2}} - \mathbf{k}^{2} \right) \, f_{\mathbf{k}}(\omega) = 0$$

The only non-zero Fourier components are for $\omega = \pm \omega_{\mathbf{k}}, \; \omega_{\mathbf{k}} = c |\mathbf{k}|$.

These are the classical modes of the electromagnetic field. They correspond to the degrees of freedom of the field and are equivalent to a system of independent linear harmonic oscillators and are quantized accordingly.

 

[A. Neumaier]

I will have to read up on quantum mechanics to understand the answers, I'm afraid. I feel like I'm missing too much information. And it probably shows through my questions.

Answering in an understandable way would be easier if you'd tell us what kind of background you have, so that one can target the answer to your level.


For general background needed for quantum mechnaics (and to see what you need to brush up),
look at my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html

what is a mode? What do the eigenvalues represent? I know that it's the solution to $$A x = \lambda x$$but what does it mean?

Typically, A is an observable, lambda one of the values it can take, and x the state vector, describing a state in which A is guaranteed to have the value lambda.

 

[mavc]

Hi,

Thank you all for your kind answers. For my background, I've done most of the first year science subjects (physics, chemistry, biology) but I went on to do computer science with a minor in maths for the rest of my bachelor's degree. I'm also at the end of my PhD, in microanalysis and simulation.

I've always wondered about why things are the way they are, and the computer scientist in me tells me to understand how the code works before I try to manipulate it. I guess my objective here is that I am trying to manipulate my understanding of these concepts, in order to understand our current technology, as well as to try to come up with new technology.

I realize that I'm just at the start of it, and that in recent history there has been at least 500 years or so worth of geniuses whose shoulders I must stand upon. And a seemingly short question has in actual fact lots of thinking behind it in order to understand the answer. So I really appreciate your patience in answering my questions.

Thanks once again.

Regards,
Anna.

 

[mavc]

Hi Dickfore,

I think I can follow those equations. Is it possible, though, that the quantisation result comes through because the initial assumption/representation of a particle in a box with periodic boundary conditions is only valid for certain solutions, as laid out by the maths? If we had a different representation, would we get a different result?

Or does this mathematical result give us predictions that are verified experimentally, so the initial assumption is therefore proven to be correct?

Regards,
Anna.

 

[Dickfore]

Hi Dickfore,

I think I can follow those equations. Is it possible, though, that the quantisation result comes through because the initial assumption/representation of a particle in a box with periodic boundary conditions is only valid for certain solutions, as laid out by the maths? If we had a different representation, would we get a different result?

Or does this mathematical result give us predictions that are verified experimentally, so the initial assumption is therefore proven to be correct?

Regards,
Anna.

What quantization? What particle in a box?

 

[mavc]

Hmm I must be thinking of the completely wrong thing...

 

[San K]

Hi,

Why do photons come in quanta? Why are the quanta not bigger or smaller or continuous?

Thanks,
Anna.

quanta is thought to be the smallest form/packet of energy.

in my opinion: things are discrete in timespace (hence you have quanta), however outside timespace they are spreadout/continuous

 

[Polyrhythmic]

quanta is thought to be the smallest form/packet of energy.

in my opinion: things are discrete in timespace (hence you have quanta), however outside timespace they are spreadout/continuous

How could anything be outside of spacetime?

 

[San K]

How could anything be outside of spacetime?

good question. Please note that the idea/discussion is just a hypothesis

i will compose it, when i have more clarity and post it later

 

[edguy99]

quanta is thought to be the smallest form/packet of energy.

in my opinion: things are discrete in timespace (hence you have quanta), however outside timespace they are spreadout/continuous

I am not sure I agree with the definition. Quanta technically is the plural of http://en.wikipedia.org/wiki/Quantum" and I think the term gets sometimes very confused.

Consider the energy of individual photons that are produced from copper atoms when electrons are shooting around (Cu line, ignore the Mo line).

Notice that although there are 2 peeks around 1.5Angstrom, the ranges around 1 angstrom are continuous. In other words, photons can come in any size, not just particular multiples of some number. Particular photons with particular energies are sometimes created because of the structure of the atom that created them, ie. in the copper atom you can see when electrons are dropping into particular orbitals by the energy of the photon that is emitted. I don't think it is correct to think of a photon as made up of a number of quanta.