What is a Photon

• I
Mentor
Hi Guys and Gal's

In answering a question in general physics I came across the following which explains at a reasonably basic level what a photon is, spontaneous emission etc at the level of basic QM with a bit of math:
http://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf

IMHO its much better than the usual misleading hand-wavey stuff and even if you don't follow the math would allow a general gist to be had.

Thanks
Bill

• OmneBonum, fluidistic, exponent137 and 4 others

A. Neumaier
For those who want to really know, including all the fine points that the question includes, a collection of relevant articles on the topic can be found here:

The Nature of Light: What Is a Photon?
Optics and Photonics News, October 2003
http://www.osa-opn.org/Content/ViewFile.aspx?Id=3185

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• kith and vanhees71
tionis
Gold Member
Isn't true that a photon doesn't exist until we make a measurement?

Delta2
Homework Helper
Gold Member
Isn't true that a photon doesn't exist until we make a measurement?

I think it exists but it exists in a superposition of states, that is a state that is beyond our classical understanding.

tionis
Gold Member
I think it exists but it exists in a superposition of states, that is a state that is beyond our classical understanding.

Yes, that is my understanding too. But can we distinguish between a photon field and an electron field before a measurement is done? IOW, if all of these fields are in a superposition, how can we tell that an electron field and a photon field are not the same field manifesting as a particular object according to what we wish to measure?

A. Neumaier
how can we tell that an electron field and a photon field are not the same field
The strength of the electromagnetic photon field and the current density of the electron field are state dependent. But the fields exist independent of the particular state, and are know to be distinct because of the way they appear in QED. For example they differ in spin, and hence in the form the basic observable field values take (field strength resp. current density).

The e/m field excitations manifest themselves as observable photons only in the moments they are detected by a counter or screen.

• ComplexVar89 and tionis
tionis
Gold Member
The strength of the electromagnetic photon field and the current density of the electron field are state dependent. But the fields exist independent of the particular state, and are know to be distinct because of the way they appear in QED. For example they differ in spin, and hence in the form the basic observable field values take (field strength resp. current density).

The e/m field excitations manifest themselves as observable photons only in the moments they are detected by a counter or screen.

Thanks, A. Neumaier. A few questions, if I may: what is the difference between a wavefunction and a field? Could I say that in the universe there is only one wavefunction that manifests as different fields, or do different fields have their own wavefunction, and if so, what is/are the difference between them other than exhibiting different values at the moment of measurement? Also, you said that the fields are ''known to be distinct because of the way they appear in QED,'' but is this a prediction of the theory before measurement?

A. Neumaier
Thanks, A. Neumaier. A few questions, if I may: what is the difference between a wavefunction and a field?
It is precisely the same difference in quantum field theory as between a wave function and position in case of single particle quantum mechanics. The wave function defines in both cases a pure state, whereas the components of position resp. the fields averaged over a region of observation define the primary observables.

do different fields have their own wavefunction
No. Different fields are like different particles in a molecule. They define different observables but there is only one wave function for the molecule, not one for every particle.

tionis
Gold Member
It is precisely the same difference in quantum field theory as between a wave function and position in case of single particle quantum mechanics. The wave function defines in both cases a pure state, whereas the components of position resp. the fields averaged over a region of observation define the primary observables.

Huh? I'm sorry I don't understand what you said. Are you saying the wavefunction is a pure state i.e, an undefined probabilistic state of the fields that doesn't have any properties until it is measured?

No. Different fields are like different particles in a molecule. They define different observables but there is only one wave function for the molecule, not one for every particle.

OK, so there is only one wavefunction for all the fields out there?

A. Neumaier
Huh? I'm sorry I don't understand what you said. Are you saying the wavefunction is a pure state i.e, an undefined probabilistic state of the fields that doesn't have any properties until it is measured?
A wavefunction describes a pure state, which is well-defined once the wave function is given. It determines the measurable field expectations. If the state is known, these field expectations are known, too, and can be checked by measurement.
OK, so there is only one wavefunction for all the fields out there?
Yes, if the state is pure. Otherwise there is only a density operator. We can never know enough about the state of a macroscopic system to decide which is the case, and effectively use always density operators. Pure states are useful only for very small systems.

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• tionis
tionis
Gold Member
A wavefunction describes a pure state, which is well-defined once the wave function is given. it determines the measurabler field expectations.

Yes, if the state is pure. Otherwise there is only a density operator. We can never know enough about the state of a macroscopic system to decide which is the case, and effectively use always density operators. Pure states are useful only for very small systems.

Thanks, Neumaier. I don't want to drag you into a never-ending series of questions and it's ok if you don't wish to answer anymore, but can you please break it down for me? In the hierarchy of all things quantum, what entity rules supreme? Is it the wavefunction, then the fields, then the particles/observables? I'm not still clear on that. Like for example, when physicists such as yourself search for a TOE, what exactly are you looking for? Is it an explanation of the physicality of the wavefunction, or whatever the wavefunction manifests into, meaning the observables?

Mentor
In the hierarchy of all things quantum, what entity rules supreme?

The field is a field of operators. It operates on something more general than the usual space in QM, called a Fock space:
https://en.wikipedia.org/wiki/Fock_space

Thanks
Bill

• tionis
A. Neumaier
In the hierarchy of all things quantum, what entity rules supreme?
The observables (i..e, the field operators) and the relations between them (commutation rules, field equations, equations of motion, operator product expansions) are the primary thing - without these one has no theoretical framework to speak about anything. They are independent of any state and therefore have a universal form. Having a TOE means knowing just this. From it one can determine (in principle) the particle content and the possible decays and reactions between particles, and scattering cross sections involving probabilities, which are the next important thing.

The expectations and correlation functions are the next important thing. They depend on the state, which is in general something very complex to which only coarse approximations can be determined experimentally. These determine what we actually find where in the world. The general laws for them (derivable without knowing more than the existence of a state) tell how macroscopic objects flow and deform.

• vortextor, tionis and bhobba
tionis
Gold Member
Thank you both. Until today, I thought the wavefunction was the most important and fundamental object in quantum mechanics, but how could it possibly be if all it is is an infinite number of probabilities that do not manifest until it's is acted upon by the field operators? At least that's what I gathered from your replies. If that is somewhat correct, then I have no further questions.

Mentor
Thank you both. Until today, I thought the wavefunction was the most important and fundamental object in quantum mechanics,

Have a look at Gleason's Theroem:
http://www.kiko.fysik.su.se/en/thesis/helena-master.pdf

That would tend to suggest operators are the fundamental thing.

Thanks
Bill

• tionis
tionis
Gold Member
Have a look at Gleason's Theroem:
http://www.kiko.fysik.su.se/en/thesis/helena-master.pdf

That would tend to suggest operators are the fundamental thing.

Thanks
Bill

Too advanced for me, but the consensus seems to be that operators are the fundamental objects, so yeah. Thanks.

edguy99
Gold Member
Great article. No better way to introduce photons then talking about harmonic oscillators. A simple harmonic oscillator is anything with a linear restoring potential. Simple things like a spring, or a string with tension, or a wave. Erwin Schrödinger described mathematically how a harmonic oscillator stores energy and how to calculate how much energy it stores.

For the photon, the value of this restoring potential is the Planck constant (h). Planck’s constant, relates the amount of energy stored in a photon to its wavelength (λ). Planck’s constant tells you the amount of time it takes the photon to undergo one cycle of whatever its doing given that the photon has a specific amount of energy. The equation E for energy = h / λ, tells us that a photon with low energy will take much longer to complete one cycle of the wave then a photon with high energy.

View attachment 194839

My favorite model of a photon as a harmonic oscillator is as an expanding and contracting ball of energy flying though the air. It immediately lends understanding to the particle and wave nature of the photon. A photon of a specific wavelength has a specific energy. The properties of a photon change periodically over time and distance depending on the wavelength. High energy photons oscillate very fast and store a lot of energy, low energy photons oscillate very slowly. Photons can be in the same place at the same time, sometimes reinforcing each other, sometimes cancelling each other out and it looks like there are no photons at all. The uncertainty principle: ΔxΔp ≥ h/4π falls from this. The photon is either big affecting a wide area, or it’s tiny and only affecting one small area, it cannot be both at the same time. The importance of visualizing the photon as a harmonic oscillator cannot be overstated.

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• Fooality
What determines if it big or tiny? The energy?

vanhees71
Gold Member
2021 Award
This is also not the entirely correct way to view photons, I'm sorry to say. First of all as massless quanta with spin ##1## there's no position operator. So the naive uncertainty relation, valid for massive particles, doesn't make sense to begin with. For a review, see

http://www.mat.univie.ac.at/~neum/physfaq/topics/position.html

The most easy way to introduce photons in a correct way is to look at the free classical electromagnetic field and fix the gauge completely, i.e., you choose the radiation gauge for the four potential ##A^{\mu}##, which consists in the two constraints
$$A^0=0, \quad \vec{\nabla} \cdot \vec{A}=0.$$
and then quantize the remaining two physical transverse components canonically. This leads to the mode decomposition
$$\hat{\vec{A}}(x)=\sum_{\lambda= \pm 1} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{\sqrt{(2 \pi)^3 2 |\vec{p}|}} [\vec{\epsilon}(\vec{k},\lambda) \hat{a}(\vec{p},\lambda) \exp(-\mathrm{i} x \cdot p) + \text{h.c.}]_{p^0=|\vec{p}|}.$$
The total energy and momentum, i.e., the Hamiltonian and the momentum of the em. field are defined as the normal ordered expressions
$$\hat{P}^{\mu} = \sum_{\lambda= \pm 1} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} p^{\mu} \hat{a}^{\dagger}(\vec{p},\lambda) \hat{a}(\vec{p},\lambda), \quad p^0=|\vec{p}|.$$
the annihilation and creation operators fulfill the commutator relations
$$[\hat{a}(\vec{p},\lambda),\hat{a}^{\dagger}(\vec{p}',\lambda')]= \delta^{(3)}(\vec{p}-\vec{p}') \delta_{\lambda \lambda'}.$$
Each single mode indeed fulfills the commutator relations of the harmonic oscillator, i.e., the electromagnetic field is equivalent to an infinite number of uncoupled harmonic oscillators. Thus the Fock states, i.e., the common occupation-number eigenstates of the number operators ##\hat{N}(\vec{p},\lambda)=\hat{a}^{\dagger}(\vec{p},\lambda) \hat{a}(\vec{p},\lambda)## (where only a finite number of occupation numbers is different from 0, ##N(\vec{p},\lambda)\in \{0,1,2,\ldots \}##.

Now we can unanimously define what a photon is: It's a single-photon state, i.e.,
$$|\psi \rangle=\sum_{\lambda=\pm 1} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \hat{a}^{\dagger}(\vec{p},\lambda) \phi(\vec{p},\lambda) |\Omega \rangle,$$
where ##|\Omega \rangle## is the vacuum state, for which all ##N(\vec{p},\lambda)=0## (ground state of the system).

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• fluidistic, tionis, Mentz114 and 2 others
edguy99
Gold Member
What determines if it big or tiny? The energy?

Yes, for sure. High energy photons have a tiny wavelength and store a lot of energy, low energy photons have a very large wavelenght.

vanhees71
Gold Member
2021 Award
Again, these statements are very misleading, not only but particularly for photons. One should emphasize that one cannot think about quanta, particularly massless quanta like the photon, in terms of classical fields ("waves") or particles. There is no wave-particle dualism, there is no position operator for photons and thus you cannot define in a reasonable way what a photon's position is. All these ideas are gone from modern physics for more than 90 years now!

The only way to describe the observable facts about photons is relativistic quantum field theory, i.e., in this case QED. It's the most accurate theory concerning the comparison between theory and experiment we have today. QED tells you that all you can measure are detection probabilities for photons, and this is expressed in terms of (gauge invariant) correlation functions. For a very good introduction to the quantum-optics aspects, see

M. O. Scully, M. S. Zubairy, Quantum Optics, Cambridge University Press

For the high-energy particle physics aspects I recommend

Schwartz, M. D.: Quantum field theory and the Standard Model, Cambridge University Press, 2014

• fluidistic, tionis and bhobba
Now we can unanimously define what a photon is: It's a single-photon state, i.e.,
|ψ⟩=∑λ=±1∫R3d3⃗p^a†(⃗p,λ)ϕ(⃗p,λ)|Ω⟩,|ψ⟩=∑λ=±1∫R3d3p→a^†(p→,λ)ϕ(p→,λ)|Ω⟩,​
|\psi \rangle=\sum_{\lambda=\pm 1} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \hat{a}^{\dagger}(\vec{p},\lambda) \phi(\vec{p},\lambda) |\Omega \rangle,
where |Ω⟩|Ω⟩|\Omega \rangle is the vacuum state, for which all N(⃗p,λ)=0N(p→,λ)=0N(\vec{p},\lambda)=0 (ground state of the system).
Would you mind describing this a bit in words? I see you are integrating over momentum, so clearly this photon will not have a definite frequency. What is it that defines this a a single photon? In a state with multiple photons, how would you define the number of photons? Is it in fact always a whole number?

tionis
Gold Member
Again, these statements are very misleading, not only but particularly for photons. One should emphasize that one cannot think about quanta, particularly massless quanta like the photon, in terms of classical fields ("waves") or particles. There is no wave-particle dualism, there is no position operator for photons and thus you cannot define in a reasonable way what a photon's position is. All these ideas are gone from modern physics for more than 90 years now!

This is PhD stuff. So everyone from a master's degree down is pretty much wrong or unaware (unless they visit PF or read the books) of what the correct explanation of a photon is?

This is PhD stuff.

No it's not :P In Poland you can learn it on your first year of masters.

tionis
Gold Member
What about in the US? When do you learn what's really going on? I mean, they teach you all the classical stuff then they tell you is not accurate enough or it's wrong?