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Could two photons with same frequency have different amplitudes and so different peak velocities of oscillations perpendicular to their direction? (to make a larger distance in same interval you need to be quicker).

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    $\begingroup$ No, photons are not oscillating in a direction perpendicular to their propagation direction. Their field amplitudes that have vectorial nature perpendicular to the direction of propagation oscillates in strength. Subtle but important difference. $\endgroup$ – flippiefanus Feb 21 at 13:58
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    $\begingroup$ This kind of thinking is great. Don’t let anyone discourage you from thinking of photons as individual particles. All photons travel at the same speed and photons with the same frequency would have the same amplitudes. They can be out of phase or incoherent. For instants, billions of coherent photons from a single source will create a light wave. $\endgroup$ – Bill Alsept Feb 21 at 15:45
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Could two photons with same frequency have different amplitudes

Photons do not have amplitude,i.e their energy is not distributed in space, let alone oscillate. They are elementary point particles in the standard model of physics. This is a quantum mechanical theory.

The particles in quantum mechanics can be represented by wave functions which are the solutions of the appropriate equations. For the photon the equation is quantized Maxwell's equation and you can read about the wave function $Ψ$ of a photon here.

This means there exists a probability for a photon to be at an (x,y,z,t) given by $Ψ^*Ψ$, the complex conjugate squared of the wavefunction. It is this probability that has a frequency given by the $E=hν$ where E is the energy of the photon and $ν$ the frequency of the classical light wave that zillions of such energy photons will build up. This can be shown with quantum field theory, for example here.

and so different peak velocities of oscillations perpendicular to their direction? (to make a larger distance in same interval you need to be quicker).

This has no meaning in the probabilistic quantum mechanical framework which models photons.

See this double slit experiment one photon at a time. The photons are all of the same energy and thus the light they build up is of one frequency.

enter image description here

Figure 1. Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

The little dots are the footprints of single photons, no oscillations. At left they look random, but at the rightmost frame they have built up the classical interference pattern of light of that frequency. The plot on the right is the quantum mechanical probability distribution for the experiment "photons impinging on given double slit" with its boundary conditions.

Keep in mind, each photon appears at a definite (x,y) on the screen which is a distance z. No photon oscillations.

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  • $\begingroup$ Of course I agree with 99.9% of this. But I thought I would add that the states of low photon number employed in such experiments are usually coherent states of low mean $n$. Those states do, just barely, give an electric field whose expectation value is non-zero and oscillates at the appropriate frequency. But what you wrote is correct for Fock states. $\endgroup$ – Andrew Steane Feb 21 at 19:29
  • $\begingroup$ If the photon has a path does peak meet peak along the whole path so the distance of the screen doesn't matter? $\endgroup$ – Krešimir Bradvica Feb 22 at 2:31
  • $\begingroup$ @KrešimirBradvica as I explain there is no peak, it is a point containing all the energy that is moving in the probable path $\endgroup$ – anna v Feb 22 at 5:22
  • $\begingroup$ @AndrewSteane This view fits experimental observations in the simplest way. That there are remarkable mathematical tools that give even more insights is wonderful, but I try to remind in my answers that physics is a discipline that observes nature and fits mathematical models to observations and experiments. I find too many people are inherently platonists, i.e. "mathematics creates reality" ( a metaphysical axiom) and not "mathematics models reality".. $\endgroup$ – anna v Feb 22 at 5:29
  • $\begingroup$ With detectors present they are free particles but with no detectors something else is directing them to make an interference pattern... $\endgroup$ – Krešimir Bradvica Feb 22 at 7:46
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The word "photon" refers to an aspect of the nature of electromagnetic radiation that arises in the treatment of the fields by quantum theory. The best way to get an idea of what the word means is to think of any given field as composed of many modes. A mode is a pattern of oscillation; it refers to an oscillation all at one frequency, but the spatial distribution of the oscillation will vary from case to case. When a given mode of frequency $\omega$ is in a state of well-defined energy with energy $(n + 1/2) \hbar \omega$ then we say there are $n$ photons present in that mode.

Now I used the word "oscillation" there, and this relates to your question. But it is not always straightforward to say what is oscillating. When $n$ is large then it is possible to have a configuration with several modes of close frequency all having many photons, and when they combine in a certain way, called coherent state, the total electric field and the magnetic field are oscillating. This is the situation in the kind of light waves that are treated by classical electromagnetism. Light from a laser is like this, and light from the thermal source such as the Sun can be expressed as a sum of many coherent states.

The relationship between the amplitude of the electric field and the average number of photons in a coherent state is roughly \begin{eqnarray*} \mbox{energy} &\simeq& n \hbar \omega \\ \Rightarrow \epsilon_0 E^2 V &\simeq& n \hbar \omega \end{eqnarray*} where $E$ is the electric field amplitude and $V$ is the volume occupied by the mode. The answer to your question is that each photon can be said to have a given energy, but it is a bit misleading to think of them as having amplitudes independently of one another. You could compare it to something more familiar, as follows. Suppose a classical wire were vibrating, such that the total energy in the vibration is $1$ joule. Asking about the oscillation amplitude for each photon would be like asking, what is the amplitude associated with each of the one thousand milli-joules which together form the total oscillation?

By the way, you can also have states where the idea of an oscillation does not apply in the ordinary sense of the word. When the state of a mode has a single precise value of $n$, as opposed to a superposition (for example, this often happens when $n$ is small), then the phase of the oscillation is spread out by quantum uncertainty, and then it is not correct to say the fields are oscillating in the ordinary sense of the word. There is a frequency but the electric field is not waving up and down. The situation is very similar to the energy eigenstates of a harmonic oscillator, if you know about those. The oscillator has a frequency $\omega$, but in a state of well-defined $n$ the modulus-squared wavefunction $(|\psi|^2)$ is completely static.

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    $\begingroup$ Please, give me an example so I can do some research about oscillators with frequency that have no evident oscillations..... $\endgroup$ – Krešimir Bradvica Feb 21 at 16:52
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    $\begingroup$ Photons are , nevertheless, particles. $\endgroup$ – my2cts Feb 21 at 18:31
  • $\begingroup$ @my2cts Well no, not quite. They are neither particles nor waves. They are what quantum physics might call 'particles' and that is fine if you are dealing with wavefunctions or Feynman diagrams and know what you are doing. But a field-state such as $\hat{a}^\dagger | 0 \rangle$ is not a particle in the ordinary sense of the word. For example, if it has well-defined momentum then the associated 'particle' is everywhere at once. $\endgroup$ – Andrew Steane Feb 21 at 19:39
  • $\begingroup$ @AndrewSteane You mean that the probability to find it is equal everywhere. $\endgroup$ – my2cts Feb 21 at 19:49
  • $\begingroup$ @my2cts True. But I don't think it best to interpret this in terms of a hidden variable such as in the Bohmian picture. $\endgroup$ – Andrew Steane Feb 21 at 21:21

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