I try to better understand how electromagnetic radiation works. So I have some questions. If an antenna emits at 100MHz (the charges on the antenna oscillate at 100MHz) what frequency will have the emitted photons? Will they have the same 100 MHz? In the case of the photons what will this frequency represent in fact?

Also what does the wavelength of the photons represent? I mean in the case of an electromagnetic wave , the wave represents the variation in intensity of the magnetic and the electric fields, and the wave-length is the distance between two crests of this variations (correct me if I am wrong). How about in the case of the photon wave-length?


3 Answers 3


Photon frequency and wavelength are the same as the corresponding classical mode.

When you quantize the electromagnetic field, you first treat the spatial dependence by decomposing the field into normal modes, which is a generalization of treating fields of the form $$E(t)=E_0(t)\sin(kx),$$ like you'd find in a conducting box of length $L$ such that $k=\pi n/L$ for some integer $n$. This is really no more than kinematics, or a restatetement of the problem if you will. The true dynamics of the field are then encoded in the temporal dependence. This dependence is through $E_0$, which obeys Maxwell's equations in the form $$\frac{d^2E_0}{dt^2}+\omega^2 E_0=0$$ for $\omega=ck$.

This equation describes a harmonic oscillator, and quantum mechanics says that harmonic-oscillator systems can only have a discrete set of possible energies, with an even spacing of $\hbar\omega$ between them. If the state of the field is such that there is, whenever you look, only one excitation present, then we say the field is in a single-mode, single photon state. This photon then has a well-defined frequency ($\nu=\omega/2\pi$) and wavelength ($\lambda=2\pi/k$).

There can then be another additional complication. In any given box, and more so in free space, there will always be more than one mode present. There is then the possibility that there be exactly one photon in the system, but that we don't know in which mode it's in - the photon is in a Schrödinger's-cat state. Usually the photon will be concentrated in modes within some bandwidth $\Delta\omega$ around some central frequency $\omega_0$, which means that the photon frequency is uncertain to some extent. The remarkable thing is that the electric field will then be nonzero in some spatially and temporally localized region: it will pass any given point in time of order $1/\Delta\omega$ and thus have a width of order $c/\Delta\omega$.

It is also important to realize that you can get this same effect with a classical field: a localized wavepacket will have a corresponding spread in its frequency and wavelength - exactly the same as it's hard to tell the pitch of short notes. The difference between a one-photon state and a weak classical field is a statistical one: for the same mean intensity, a one-photon wavepacket will always give one count on a photo-multiplier tube or avalanche photodiode, whereas a classical field will sometimes give none, sometimes several. This seems like a small difference but it makes things like single-photon interferometry possible.

  • $\begingroup$ Thank you for your explanation. Unfortunately I don't think I have the required knowledge of physics in order to understand it :(. Could you please explain it in laymen's terms? $\endgroup$ Jun 27, 2012 at 9:44

Maxwell's equations are to photons what the Schrödinger and Dirac equation are to electrons. Their solutions predict statistical observations of photons. Electrons do not have a wavelength, only electron wave functions have. The same is true for photons.


When people talk about the frequency or wavelength of a photon, they usually mean the frequency or wavelength of the electromagnetic field corresponding to that photon. Of course, in practice, a photon is made up of components with different frequencies spread over some range, but if that range is very small, you can treat it as pretty much a single frequency, and characterize the photon by saying it is of that frequency.


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