What would happen if we had super fast and super sensitive oscilloscope that could probe and record into memory electric or magnetic field oscillation created by single photon that would fly near the probe/antenna?

  1. Can antenna detect change in electric or magnetic field of single photon without being directly hit by the photon? The photon just flies nearby but doesnt hit the solid conductor of the antenna.

  2. If 1. is indeed possible, what would the recorded waveform look like? How long will it be?

I know its going to be sinusoid, and I think the lenght should be 1 wavelenght but the thing is, having just one sinewave oscillation with instant start and stop is like doing FFT with rectangular window function, its going to create harmonics and distortion becose of the rapid start/stop.

That got me thinking that if the electromagnetic field oscillation of single photon was just one wavelenght long, it would mean there must be higher wavelenght photons riding alongside it to represent the upper harmonics caused by the rectangular window single cycle duration, its probably completly wrong but I thought about it.

Then I thought maybe the single photon waveform is more like gaussian window, longer than one wavelenght, with smooth increase as the photon gets closer and closer and equaly smooth "fade out" as it gets away.


2 Answers 2


With a few exceptions, photons are only known to exist when we observe then or detect them. During such observations the photon is destroyed. So the possibility to observe the effect of a photon without destroying it is minimal.

Yet, some clever guy figured out a way to detect a photon's existence without destroying it. This guy is Serge Haroche, and he earned a Nobel prize for his work. What he did was to design a cavity that can capture and hold a single (or a few) photons. Then he sends highly excited atoms through the cavity and afterward makes measurements on those atoms. The way these atoms evolve while passing through the cavity depends on whether the cavity contains a photon or not. This way he could detect the existence of photons in the cavity.

Also, what Haroche showed is that a single photon does have an electric field and a magnetic field (contrary to what anna v says in her answer), because the reason why the excited atom evolves as it does when it passes through the cavity is because of the effect that the electromagnetic field of the photon in the cavity has on the atom.

This is still very different from measuring the effects of a passing photon though. However, we can say a few things about a photon. Firstly, it is not like a little ball of light propagating in the beam of light. Instead one can think of it as a single excitation (one quantum of energy) of the entire beam. More photon means more excitations, but each photon effectively stretches throughout the entire beam. (This assumes perfect coherence. For partial coherence, the situation becomes more complicated.)

Hope this helps.


You have a basic misunderstanding. A photon does not have electric and magnetic fields. It is a quantum mechanical entity , a part of the basic elementary particles in the standard model of particle physics. It just has $energy=h*ν$, where ν is the frequency of the classical wave which will emerge when zillions of such photons superimpose their wave functions.

When it interacts, as here

enter image description here

it interacts as a particle, as these footprints show, the wave nature appearing when thousands follow the same boundary conditions, giving the classical interference which is the quantum mechanical probability distribution for these boundary conditions.

The wavefunction of the photon is a complex solution of quantized Maxwell's equation, and is connected with the E and B of the classical Maxwell equations, but the complex wavefunctions of quantum mechanics are unmeasurable. Only the $Ψ*Ψ$, the probability distribution of finding the photon at (x,y,z,t) , is measurable. One version of the wavefunction is here . One does not usually talk about the photon wavefunction because classical solutions do a very good job of describing all optics.

So a photon will just give a tick, as it interacts with the atoms of the oscilloscope, it cannot give an E field.

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    $\begingroup$ I always saw photon being shown as changing magnetic and electric field with sinewave waveform couple of cycles long travelling through space. I think those examples should not be used since they are wrong if you say we cant measure field waveform of single photon. $\endgroup$ Jul 8, 2019 at 20:34
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    $\begingroup$ @wavscientist Those pictures are definitely wrong. They display wave packets which are entirely distinct from photons. They are a futile attempt to apply concepts that we can understand to an entity (the photon) that is really beyond our ability to fully comprehend. $\endgroup$
    – garyp
    Jul 9, 2019 at 0:51
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    $\begingroup$ Yes, if people use such representations for photons, they are wrong. Any wave nature for photons are in the probability distribution, not the space and time of a single photon( except in the non measurable complex wavefunction) $\endgroup$
    – anna v
    Jul 9, 2019 at 5:40
  • $\begingroup$ By the way, can that probability distribution waveform of single photon be measured? $\endgroup$ Jul 9, 2019 at 16:41
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    $\begingroup$ look at the experment above. A single photo leaves a dot , on the left screen. It means that the photon hit an electron from the atom in the screen and the electron made enough other scatters to register on the screen. The probability distribution on the right, where hundreds of thousand of photons show a wave interference, which actually the interference expected from light at that frequency. Probability is a distribution of many events with the same boundary conditions. $\endgroup$
    – anna v
    Jul 9, 2019 at 17:48

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