Does an electromagnetic wave necessarily contain many photons? I've often come across people saying from a quantum physics standpoint that an electromagnetic wave necessarily contains many photons. But doesn't the double-slit experiment conducted one photon at-a-time suggest that an electromagnetic wave corresponds with a single photon?
EDIT: This question is solely about the numerical relation between electromagnetic waves and photons.
 A: In quantum mechanical domain these type of question does not have meaning. Every single photon is associated with a wave and vice versa. But to talk whether an electromagnetic wave contains a single photon or not is an ambiguous statement. When people say 

an electromagnetic wave necessarily contains many photons

it only means that a incident beam of EM waves have many photons associated with it. That does not mean that it can't have more than one photon. In fact in can be proven that the number of photons in a coherent beam (such as Laser) is not a constant. Just like position-momentum uncertainty there is a number-phase uncertainty for the EM waves. So although we know what is the mean number of photons that constitutes the beam, we do not know the exact number.
On the other hand if your question is about whether one photon can be treated as an EM wave then of course it is yes.
A: It has to do with the total energy or power of the EM wave you're interested in, as well as the frequency of the wave. As a simple example, a 3mW laser at 500nm wavelength will produce roughly 7.55*10^15 photons per second. From how large this number is, it's not difficult to see how light will usually be made up of an extremely large number of photons.
For the double slit experiment happening one photon at a time, the experiment had to use very low intensity light to ensure only one photon on average was being released at a time, and the experiment had to occur over the course of months to capture enough photons for it to actually be measurable.
So to answer your question, EM waves like light or radio that we interact with will have huge numbers of photons, though waves with small numbers of photons are also possible. I also like Ari's answer for the detail on how the number of photons corresponds to an EM wave.
A: The very word, photon, belongs to the quantum mechanical regime. It is one of the elementary particles in the standard model of particle physics. Elementary particles are described with quantum mechanical wave functions, which are complex function. The complex conjugate square gives the probability of finding the particle at (x,y,z,t).
In the case of the double slit experiments single photons at a time and single electrons at a time display points on the screen which accumulated give a probability distribution that has wave, i.e. sinusoidal, behavior. The wavelength for the photon is the wavelength of the beam, and for the electron the de Broglie wavelength, and both particles leave points on the screen. It is the much discussed wave particle duality.
A beam of electrons is composed of electrons, and a beam of photons is composed of photons and is called light.  The wavefunction of the photon is complex and contains information of the classical potential A of Maxwell's equations. Thus, though the photon itself has only energy and spin to describe it,( spin +1 or -1,) it builds up the electric and magnetic fields in confluence with a huge number of other photons through the complex wavefunction. This illustration is useful:

The mathematical format of how classical electromagnetic fields emerge from a confluence of photons can be seen in this blog post by Lubos Motl.
See also this question and answers here.
