This is not a duplicate. I am not asking about the connection between photons and EM waves nor wave particle duality.

I have read these questions:

What is the relation between electromagnetic wave and photon?

where annav says:

Conceptually watching the build up of interference fringes from single photons in a two slit experiment might give you an intuition of how even though light is composed of individual elementary particles, photons, the classical wave pattern emerges when the ensemble becomes large.

How many photons are needed to make a light wave?

where CuriusOne says in a comment:

Light doesn't ever behave like a particle or a wave. It behaves like a quantum field. People need to stop talking about it the way their great-grandfathers talked about it for a dozen years before Dirac wrote up with the correct explanation in the early 1930s! We have been over this wave-particle duality nonsense almost as long as we have been over the aether.

How does light oscillate?

where fffred says:

In light propagation, oscillation does not mean any movement in space. It is the value of the electromagnetic field, at one given point in space, that oscillates. For electromagnetic waves, there is no matter or photons that go up and down. Instead, you have to imagine that there is a little arrow associated to each point in space: this little arrow is the electric field direction. Another arrow, at the same point, is the magnetic field. These two arrows change size and direction with time, and in fact they oscillate.

How to imagine the electromagnetic waves?

where annav says:

The electromagnetic wave is described by the solution of classical maxwell's equation which has a sinusoidal dependence for the electric and magnetic fields perpendicular to the direction of motion of the wave. It is called a wave for this reason and the frequency is the repetition rate of the sinusoidal pattern. A single photon has only a detection probability distribution that "waves", as explained above. It is not a wave.

Can photons oscillate?

Why do electromagnetic waves oscillate?

where Bjornw says:

To clarify another part of your question - "what oscillates" - the answer is that the quantum amplitude for sending out the field correlations that build up the "photon" oscillates at the source, and this affects the destination. There is nothing "in between" that oscillates.

Is a single photon also a Maxwellian wave?

where WetSavannaAnimal says:

There is indeed a way wherein "one photon" can be thought of as a Maxwellian wave. So therefore one can construe the information contained in the Maxwellian fields as equivalent to the knowledge of the one photon state of the EM field. For every classical freespace solution to Maxwell's equations, there is a corresponding one-photon state and contrariwise.

So basically most of these answers say that photons are not waves, and nothing oscillates physically in space (3D) as the photon propagates in space. One says that there is a way where one photon can be thought of as a Maxwellian wave.

Now based on these, a photon should always propagate (in flat spacetime) in a straight (3D) line, no oscillation (physically in space), the photon itself does not oscillate as it propagates, just the field vectors.

Yet, we are talking about the photon traveling as a wave everywhere.

enter image description here

We are using these pictures to model the propagation of the EM field (that is coherently built up by photons) with this oscillating field.

How can we interpret polarization and frequency when we are dealing with one single photon?

where WetSavannaAnimal says:

Maxwell's equations exactly define the propagation of a lone photon in free space. The state of a photon can be defined by a vector valued state in Hilbert space and this vector valued state is a precise mathematical analogy of the E⃗ and H⃗ fields of a macroscopic, classical field.

Now as I understand, the Maxwell equations can describe a one photon state too, but it is not the photon itself physically that oscillates in space, it does not move up and down or in any direction other then the propagation. It is moving in a straight (3D) line.

Based on this, and the fact that photons are pointlike elementary particles, the propagation of the photon on the smallest scale is described by a straight 0 dimensional line, that has no thickness, and is completely straight (3D).

Thus when we say things like the photon is propagating as a wave, we use these pictures, we are using confusing statements, because the photon itself is just propagating in a straight line.

Do Photons Move in a Wave Like Pattern?

But this question has no answer basically stating yes or no.

Is light amplitude spatial?

where dmckee says:

So, no, nothing is moving off the line of the ray, but the because electric field is a vector the oscillation does have a direction associated with it (and therefore polarization makes sense).

So basically, nothing (not real photon) is moving off the straight (3D) line. It is just the static EM field components that oscillate, that we model with virtual photons.

Thus, the real photon moves in a straight line, and the virtual photons (that is just a model of the static field) oscillate.

So the photon is an excitation in the photon field, and that excitation propagates in a straight (3D) line, and the excitation does not oscillate itself.


  1. Do (real) photons oscillate or not and why do we still sometimes use the phrase the photon travels as a wave?
  • $\begingroup$ the propagation of the photon on the smallest scale is described by a straight 0 dimensional line Quantum particles don’t have classical trajectories. $\endgroup$
    – G. Smith
    Oct 27, 2019 at 19:46
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    $\begingroup$ The format with all the quoted comments upfront is really hard to read/understand. Before you have asked your question, it's completely unclear to the reader why these comments are being quoted and why one should read them. Comments are also ephemeral and can be deleted at any time, so they do not make good references at all. I would suggest cutting most, if not all, of these quotes, as they do not really contribute much to your actual question. $\endgroup$
    – ACuriousMind
    Oct 27, 2019 at 19:58
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    $\begingroup$ In addition to ACM's completely on-point observations, starting off with "This is not a duplicate" does very little beyond making sure that you have an antagonistic reader, particularly when you make it a standard formulaic kick-off used on all of your posts. It doesn't stop being a duplicate (if it is) just because you say it isn't. And the comment is utterly useless if you don't specify which other questions it's not a duplicate of. (If you don't know what those might be, that's a great sign that you shouldn't be writing that line.) $\endgroup$ Oct 27, 2019 at 20:02
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    $\begingroup$ why the downvote? $\endgroup$ Oct 28, 2019 at 0:50
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    $\begingroup$ "Thus, the real photon moves in a straight line, and the virtual photons (that is just a model of the static field) oscillate." No. Just no. There comes a point when you need to just for FSM's sake stop hoping to find enlightenment in studying many, many different explanations in words and learn the actual models, which means learning to do the math and slogging through all the hard, boring bits. Because the models are written about half in math. $\endgroup$ Oct 28, 2019 at 1:33

2 Answers 2


The crucial point here is completely unspecific to photons: You're trying to reason about quantum objects with classical thinking. Stop thinking of the photon as a little ball of light with a definite position and momentum.

Quantum particles do not "propagate in a straight line". This is classical thinking - "particles as billard balls", but it is not what happens. Every quantum object has a wavefunction that spreads according to the particular Schrödinger equation it obeys. There is no unique trajectory, the question "Which path did the particle take from A to B" is not a meaningful question unless you measured its position every step along the way.

The strange focus on "oscillation" in the question is also beside the point. The wavefunction that models a freely travelling particle is usually a Gaußian wavepacket. This moves, but it does not "oscillate". But this is the same as for classical electromagnetic waves - a beam of light is only an "oscillation" if it is eternal. If you're looking at actual pulses - i.e. a short "beam" moving, e.g. a light being turned on and off again - it's also just such a travelling wavepacket. Since the one-particle Schrödinger equation for the quanta of a free field is the same as the classical equation of motion for the field, this isn't particularly surprising - the classical EM field and the wavefunction of its photons are different things, but they obey equations of the same shape.

  • $\begingroup$ I was under the impression that the “wave function of a photon” was a controversial concept, as stated in this paper. You seem to imply that it is a widely-accepted concept (but it was not covered as part of my graduate studies). If so, can you point me to what you consider an authoritative treatment, with the appropriate equation and the correct Born rule? How many components does a photon wavefunction have? $\endgroup$
    – G. Smith
    Oct 27, 2019 at 20:19
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    $\begingroup$ @G.Smith You're right that for a massless photon, wavefunction and position are tricky concepts, but I think that's a technical problem distracting from the problems in this question. There's two ways out of this: 1. Since massive QED is also consistent (cf. Stückelberg formalism), you can endow the photon with a very small mass to be able to talk about its wavefunction. 2. Talk about the "probability of photon to be detected crossing a surface", which is close to the usual notion of a wavefunction being the "probability to be detected at a point" but still well-defined for massless particles. $\endgroup$
    – ACuriousMind
    Oct 27, 2019 at 20:24
  • $\begingroup$ Thank you so much! This is actually a good answer "The wavefunction that models a freely travelling particle is usually a Gaußian wavepacket. This moves, but it does not "oscillate".", "the classical EM field and the wavefunction of its photons are different things, but they obey equations of the same shape." Can you please elaborate on that? What is that shape? $\endgroup$ Oct 27, 2019 at 20:43
  • $\begingroup$ en.wikipedia.org/wiki/File:Wavepacket-a2k4-en.gif Are these the EM field vector that are moving or is this actually the wavefunction of the photon? Is this a completely different wave then the one in my question? $\endgroup$ Oct 27, 2019 at 20:51

Do (real) photons oscillate or not and why do we still sometimes use the phrase the photon travels as a wave?

Everything in physics we represent is based on models. And the existence of photons and the oscillation of their magnetic and an electric field component is a very useful model. The introduction of photons explains

  • the specific radiation curves for black body radiation,
  • the photoelectric effect,
  • the emission and absorption of energy in packages for subatomic particles,
  • the dots on a photo-emulsion film and the exposure of a CCD video chip.

The introduction of B and E field components for photons is based on the observation of radio waves, where these field components are direct measurable. In addition in the near field of an antenna it is common sense that the electric field induces a magnetic field and this field again induces an electric field.
If one accepts the model for accelerated charges that emit photons during their acceleration, one could imagine that the radio wave consists of zillions of photons. These photons form the radio wave and according the near field of the radio wave one could imagine that each photon in its propagation has an oscillating electric field and an oscillating magnetic field that induce each other.

By the way, the fact that photons receive E and B field components from their donor is not surprising when one realizes that the charged subatomic particles have both an (intrinsic) electric field and a magnetic field.

Such a model is applicable for a lot of phenomena:

  • The polarization of photons at polarizers can be explained by the interaction of the field components of the photons and the electrons on the surface of the polarizer grid.
  • The energy of the sum of both oscillating field components of the photon is a constant value and doesn’t vanish like in the model with synchronous E and B amplitudes.
  • The deflection of photons at a prism with different angle of deflection in dependency from the value of the field components (the energy content of the photon) again could be explained with the interaction of the surface electrons.
  • Last not least it would be an elegant way to explain at the end the interference pattern behind edges. Simply these interactions have discrete values and the fringes are the result of these discrete deflections.
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    $\begingroup$ This is dead wrong. QED is not a "model", and neither are photons. $\endgroup$ Dec 13, 2019 at 8:34
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    $\begingroup$ @EmilioPisanty I don’t know about these answers but you are wrong about the photon model. A particle (photon) model derives all light phenomenon but the wave model can only derive some. Besides you can’t even physically describe what the field is without resorting to billions of individual photons. $\endgroup$ Dec 13, 2019 at 9:00

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