A photon is associated with the equations $h\nu$ and $\frac{hc}{\lambda}$.

My book (Serway Modern Physics) says that Einstein explained the photoelectric effect by assuming that the classical wavefront had its energy distributed over bundles, with energy $h\nu$. (I've been told this is wrong elsewhere, but its not crucial to my question anyways).

With this picture I imagine that the Maxwellian wave is replete with photons, and its not too hard to digest that the energy of a photon could be given by $h\nu$ where $\nu$ is the frequency of the Maxwellian wave.

But what about a single photon produced, e.g., in an electronic transition in an atom? What does the frequency or wavelength variable in its energy a wavelength or frequency of? Is it also of a classical Maxwellian wave? Meaning that a single photon has a complete plane wave associated with it?

How is this Maxwellian wave distributed over space? Is the common picture of the photon as a little sperm correct then?

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    $\begingroup$ This is a good question (so upvote) and the short answer is no, the photon is not a Maxwellian wave but explaining this is non-trivial. $\endgroup$ – Alfred Centauri Apr 6 '15 at 2:52
  • $\begingroup$ @AlfredCentauri I disagree. A "photon" is a unit of excitation of an electromagnetic mode. If the electromagnetic field "has one photon in it" then I'm implicitly saying that there's some superposition of states (i.e. modes) each with one unit of excitation. This is still a wave, because each mode has a particular E and B field configuration. $\endgroup$ – DanielSank Apr 6 '15 at 6:52
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    $\begingroup$ Classical EM wave consists of brilions of photons. $\endgroup$ – Paul Apr 6 '15 at 7:31
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    $\begingroup$ See my answer to this question:physics.stackexchange.com/questions/66977/… $\endgroup$ – Virgo Apr 17 '15 at 4:52
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    $\begingroup$ Maxwellian waves and photons are from different theories. $\endgroup$ – user46925 Jun 12 '15 at 23:40

The maxwellian wave is an emergent phenomenon from a great multitude of photons with the frequency of the maxwellian wave. This is explained in this blog entry by Lubos Motl. I will give you my experimentalist's interpretation of this:

A photon as a quantum mechanical entity has a wavefunction. This wavefunction is a solution of a form of Maxwell's equations, where the derivatives in the equation are turned into quantum mechanical operators, operating on the psi of the photon. There are variables that describe the wavefunction including polarization and phases. It is not surprising that the frequency is the same for both uses of the Maxwell equations.

The phases coherently build up the macroscopic classical wave. I find this figure useful :


left and right handed circular polarization, and their associate angular momenta.

Note the direction of the spin of the photon, either +1 or -1 to its direction. It is the handedness of the macroscopic wave that reflects this. The phases in the solutions of the photon build up the fields and the circular polarization in this simple example.


There is indeed a way wherein "one photon" can be thought of as a Maxwellian wave.

We are dealing here with the quantized electromagnetic field. If the EM field is in a one photon state then one can compute two vector fields from the EM field's state that:

  1. Fulfill Maxwell's equations for freespace propagation and
  2. Uniquely define the one photon state.

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.

Strictly speaking, this idea only works with "free photons", i.e. between interaction events with other quantum fields, so that the photon number can be thought of as being conserved for the analysis.

The quantities which fulfill Maxwell's equations and which define the one-photon state are as discussed further in my answer here to the physics SE question "Relation between Wave equation of light and photon wave function?".

  • $\begingroup$ What is troubling to me about this view is that a Maxwellian plane wave of a single frequency exists over all space and time (else it would not be a single frequency). In the question, the OP asks if a photon emitted by an atomic transition has a complete plane wave associated with it. I do see that you address this concern somewhat in your penultimate paragraph but I'm not convinced. $\endgroup$ – Alfred Centauri Apr 6 '15 at 11:40
  • $\begingroup$ @AlfredCentauri All quantum fields have this same normalization problem with truly plane waves. Even so, a photon from a transition is never a lone plane wave, even without thermal broadening. What happens is that the excited emitter (with energy $E$ above ground state) is coupled to a broad range of frequencies in the EM field, and "tries" to couple to them all. In Wigner-Weisskopf theory, the frequency of driving term of the EM modes is of course $\omega_0 = E/\hbar$. So a mode at a greatly differing frequency but still coupled strongly is driven at this frequency, but is not resonant .... $\endgroup$ – WetSavannaAnimal Apr 6 '15 at 13:04
  • $\begingroup$ ...with it. Therefore, destructive interference prevents strong coupling. If you analyze this in equations, this detuning effect yields the Lorentzian lineshape. Without thermal broadening though it is very narrow: the Lorentzian width is inversely proportional to the transition lifetime. Have a look at the first chapter of Quantum Optics by Scully and Zubairy if you can get it for the Maxwellian interpretation of the photon. Also the works of Iwo Bialynicki-Birula - "photon wave function" and similar named papers. $\endgroup$ – WetSavannaAnimal Apr 6 '15 at 13:06
  • $\begingroup$ Wet, I'd like to continue this, perhaps in chat ( @DanielSank too ). However, I've got a busy day ahead so this will have to go on the back burner. $\endgroup$ – Alfred Centauri Apr 6 '15 at 13:27

Imagine that space-time is emergent, not a pre-existing framework which holds photons, be they waves or particles. If space-time emerges at the level of individual quanta or photons, you might say there's no excess space and time there for a wave to propagate through. But at larger distances, as the photon interacts with other elements of our universe, it exhibits complementary attributes of waves and particles, depending on the nature of the interaction or the measurement.

This fits with Niels Bohr's notion of complementarity, which says that quantum objects have characteristics which are complementary, which can not be simultaneously measured with 100% acccuracy. The total nature of a photon may need to be described as both a particle and a wave in order to capture all its possibilities and properties.

Serway (Physics for Scientists and Engineers, 1992, p. 1171) says that long wavelengths such as radio waves consist of photons with so little energy that it takes many photons striking a receiver for detection to occur. In such cases, it's difficult or impossible to detect the particle nature of each photon. Shorter wavelengths, however, are composed of high-energy photons, such that it takes fewer of them to elicit detection, and they are more likely to be detectable as particles.


Look into Wave-particle duality. It is a major part of Quantum Mechanics which answers your question.

A quick summary: light is not just a wave, not just a particle. In some situations it behaves like a wave; in others it behaves like a particle.

In some situations, neither wave nor particle sufficiently describe reality. A solid example of this is the Single Photon Double Slit Experiment. Quantum mechanics describes light differently (as a so called "waveform"), and QM's model matches what actually happens in that double slit experiment. Neither the classic wave nor classic photon models do a good job of modeling what happens.


It turns out that light can be thought of as a wave and a particle under different conditions. For example, as Cort Ammon described, the Double Slit experiment showed that light had properties of diffraction. On the other hand, the photoelectric effect considers light as packets of light called photons with certain energy $hf$. Another such experiment is the Compton scattering effect where light is said to impart momentum on an electron. A massless particle gives an electron momentum! I am assuming you are in your second, or third, year of physics (since that is when I used Serway), quantum mechanics is hard so don't be discouraged if you can't wrap your head around a concept. Many others are going through the same struggle.


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