How do a photon's two fields look like? The photon is a electromagnetic wave and the electric as well as the magnetic part of the wave can be written as follows (taken from http://en.wikipedia.org/wiki/Electromagnetic_wave_equation): [Broken] [itex]E(\vec r, t) = g(\omega t - \vec k \vec r)[/itex] where [itex]\vec r[/itex] is any three dimensional vector, [itex]\vec k[/itex] is a fixed 3D vector that basically denotes the direction of propagation of the field in dependence of time, and [tex]\vec k \vec r[/tex] denotes the scalar product of the two vectors. To simplify the question to come, I select [itex]\vec k=(1,0,0)[/itex] and with [itex]\vec r=(x,y,z)[/itex] we get [itex]E(\vec r, t) = g(\omega t - x)[/itex] This field propagates in the direction of the x-axis and intuitively (and given figures in books) one assumes that the field propagates on the x-axis. This, however, is not true, because for the complete 2D plane [itex](x_0, y, z)[/itex] where [itex]x_0[/itex] is a fixed value and y and z vary over the real values, we get: [itex]E((x_0, y, z), t_0) = g(\omega t - x_0)[/itex] That means that the value of E is identical over the whole y,z-plane for each x_0. So when looking only at the values of E (not at the equation), nothing would tell you in relation to which x-axis position within the y,z-plane you have to compute it. In fact you can choose any position of the x-axis. Now, if a photon hits my eye and I see light, I would again have thought that it travelled on a line right into my eye. On the contrary, the reasoning above seems to indicate that a whole y,z-plane hit my eye but somehow decided that my eye is "in the middle" and therefore produces an interaction to make me see light. Is this description generally true or am I missing something? Do the photon's fields indeed not have a "center line" along which they travel but rather just a direction? How come I can actually see a photon then? How is it decided that it interacts in my eye and not a meter left of me with the wall? Harald.