What is the length of a single photon? I don't mean the wavelength, I mean the full length in the direction of propagation. What is the total extent of a single photon in space? Does it even have a length?
Edit to clarify: Let's say a have a polar molecule, e.g. water. A standing EM wave (in this case our photon) will cause the molecule to oscillate (this is how a microwave works). So with a single photon forming the standing wave, how many water molecules can I put in a line, such that the same photon will be affecting them all at the same time? I'm not sure if that even makes sense, but that's what I'm thinking of when I refer to length.
 A: As knzhou says in a comment:

Depends on the photon. It’s like asking what the width of an ordinary particle’s wavefunction is. Sometimes it’s bigger and sometimes it’s smaller.

Some sources, such as lasers, emit photons with very long coherence lengths. For instance, the laser in your CD player probably emits wavetrains with lengths of ~10 cm.
Other sources emit photons with much shorter wavetrains. For example, if you look at thin film interference patterns made by light from a sodium discharge tube, the patterns never have more than ~100 fringes, which is because the sodium atoms emit wavetrains with lengths of ~100 wavelengths. This is determined by the properties of the atom. For example, if the half-life of the transition is ~100 periods of the light wave, then the wavetrain will have a length of $\lesssim100$ wavelengths.
A: The answer by @BenCrowell is closest to correct.  The OP clearly asks for the length of a photon in the direction of propagation.  As @safesphere commented, a single photon detection has no spatial extent (at least under ordinary circumstances).  However, the photon's wave nature is relatively easy to demonstrate, as in single-photon interference. 
An analogous experiment can be set up to demonstrate the spatial extent of a photon, both in the direction of propagation and in the directions at right angles to the direction of propagation: single-photon interference in a Michaelson interferometer, for example.  The pathlength difference at which fringe contrast drops to (and stays at) zero is equal to the coherence length of the photons.
A really graphic demonstration of the 3D extent of a photon can be done by setting up a "classical" single-photon 3D holography system, to make a hologram of a large 3D scene of known geometry, so that the path length difference for object and reference beam is known at a given point on the recording film, for every point in the scene.  Looking at the scene through the developed hologram at the given point amounts to a map of the spatial extent of the (identical) photons that recorded the hologram.
A: I want to extend knzhou's comment. In the quantum theory of the electromagnetic (EM) field the number of "photons" is the number of excitations of the EM field with a given wavevector $\mathbf k$  and a given polarization $\lambda$. In classical EM theory, a wave with definite $\mathbf k$ and $\lambda$ is a plane wave, which has infinite extent to begin with (so it would influence all the water molecules). Of course we may build a localized wave, but in order to do this we'll have to add many plane waves and therefore $\mathbf k$ will assume a range of values. So a standing wave does not have a definite $\mathbf k$ since it's a sum of two plane waves travelling in different directions! We already see that the notion of "extent" of a wave with definite values of $\mathbf k, \lambda$ is ill-defined.
Things carry out similarly in the quantum theory of the EM field, but some subtleties kick in. The first thing that changes is how we describe the field: we use an occupation number ket $|n_{\mathbf k, \lambda}\rangle$ that tells us how many photons (excitations) there are with a given $\mathbf k$ and $\lambda$ and the fields become field-operators that act on the ket. You may ask: "if the fields are operators now, how do we associate them with the classical fields (which are numbers)?". Well, we take the expectation value
$$
\langle n_{\mathbf k, \lambda}| \mathbf E |n_{\mathbf k, \lambda}\rangle
$$
which, if you carry out the calculations, will be zero for a state with a definite number of photons (but $\langle \mathbf E^2 \rangle$ will be infinite!). It'll be non-zero if our field-state is a superposition of different occupation numbers, for instance:
$$
|\psi\rangle = c_0|0_{\mathbf k, \lambda}\rangle + c_1|1_{\mathbf k, \lambda}\rangle.
$$
Indeed, the state that resembles a classical coherent wave is called a coherent state, which is a superposition of infinite photon number states. So even our notion of electric field depends on having an uncertainty in the number of photons.
So, in your example which water molecule would a single photon influence? The answer is still all of them, but only one will be excited by the photon! All molecules will feel the EM field of a single photon, but by conservation of energy only one can absorb the photon and excite. It's interesting to note that even in the absence of photons the field can influence the molecules and de-excite them, making them emit more photons.
In conclusion, both classical and quantum treatments of the EM field will not have a well-defined "length" of a wave/photon with definite $\mathbf k, \lambda$, and if you really want a "reasonable" length you'll have to give up the definiteness of $\mathbf k$, and hence have more than one photon.
A: According to our currently accepted theory, the Standard Model, photons are elementary particles.
This means that they do not have a spatial extension or an internal structure. We call them point particles.
That is why we cannot talk about the size of a single photon.
We would need a Theory Of Everything where elementary particles are made up of strings to talk about such a size.
But in the currently accepted theory, photons are point particles.
Though, sometimes we talk about the fact that photons do have wave characteristics, and they do have certain wavelengths where we could talk about the minimal size like Compton. Or we can talk about the Planck length, as a minimal size that is important in QM.
But in reality, the photon is a point particle.
please see here:
Why doesn't De Broglie's wave equation work for photons?
A: If you are talking about the spatial extension of a photon it is zero. Photons are elementary particles which have no size, in an atom size is created be a relationship between the strong force and electromagnetic force. But elementary particles ie. Photons, quarks  are sizeless.
A: You can only answer this question if you find an isolated photon source that spontaneously generates exactly ONE photon and is not disturbed. One of the few examples is the famous hydrogen line with a wavelength of 21 cm. The remarkably low transition probability A21 = 2.85e-15 1/s results in an extreme frequency accuracy. With this value and with Heisenberg's uncertainty principle, a coherence length of very many kilometers can be calculated. Experiments with many photons or even lasers are unsuitable for determining the properties of individual photons.
