Since electromagnetic energy is carried by photons and moves in forms of waves, does it mean that a single photon when propagating through space doesn't follow the straight path but instead always moves up and down, up and down like a wave. If so another question arises the speed of propagation of light in vacuum is fixed meaning that it will always take the same amount of time for it to travel from point A to point B, but if a photon always moves up and down it will also mean that it travels longer distance than the distance between A and B and so it ill travel faster than light propagates, is it even possible, could you please clarify these concepts for me?
[Physics] the trajectory of a photon moving through a vacuum
electromagnetic-radiationphotonsvisible-lightwaves
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How could one single photon have only one direction? I mean at every point of its propagation the changing electric field creates a magnetic field in every direction (not just in the direction of the propagation). This would create an (elementary) wave source and it will propagate in every direction.
The photon is an elementary particle in the standard model of particle physics. As such it is described by a quantum mechanical wavefunction , in complex numbers, whose complex concugate square gives the probability density of finding the photon at (x,y,z,t). It is only characterized by its energy=h*nu, and its spin which is + or -1h towards its direction of motion. Thus the image you have of it as a source of radially propagating fields is wrong. The confluence of photons builds up the electric and magnetic fields of the emergent classical electromagnetic wave, but the classical wave format cannot be cut down to describe a photon.
Quoting from this answer by Motl (to a different question)
the wave function of a single photon has several components - much like the components of the Dirac field (or Dirac wave function) - and this wave function is pretty much isomorphic to the electromagnetic field, remembering the complexified values of E and B vectors at each point. The probability density that a photon is found at a particular point is proportional to the energy density (E2+B2)/2 at this point. But again, the interpretation of B,E for a single photon has to be changed.
Have a look at this answer of mine on a similar question.
This question is about the nature of the electromagnetic field. The electromagnetic field is a physical system that is most fully described by quantum field theory, and the results match those of classical field theory in certain limiting cases. The 'photon' is a physical picture which gives us a useful way to imagine certain aspects of this field. It is primarily a way to track energy movements.
The main thing you need to know is that energy is conserved, but photons are not. When energy moves from some other form to an electromagnetic form, then photons are created. When energy moves from an electromagnetic form to other forms, then photons are destroyed.
Another way of saying the same thing is to note that when an electron moves from a higher to a lower energy level in an atom, it does so through the way its charge pushes on the surrounding electromagnetic field, causing it to vibrate at a higher amplitude (the electric and magnetic parts both start to vibrate). This vibration, when it happens at a fixed frequency, can be conveniently modelled by saying it has a fixed amount of energy, equal to $h f$ where $h$ is Planck's constant and $f$ is the frequency. If this $h f$ is equal to the energy change $\Delta E$ in the atom, then we say one photon has been created. You can also find cases where two photons are produced, one at frequency $f_1$ and the other at $f_2$, and then $h f_1 + h f_2 = \Delta E$. This kind of process is much rarer but it illustrates that energy is conserved, but a given amount of energy can be expressed physically in more than one way.
Eventually a photon may arrive at some other atom and be absorbed. What happens then is that the oscillating electromagnetic field pushes on the electrons inside the atom, until one of them gains some more energy. The field vibration then falls away as the energy is transferred. We summarise the process by saying that the photon has been absorbed. Or, if you like, the photon 'dies'. This is just another way to say that the field has stopped vibrating.
Best Answer
The term photon belongs to the realm of quantum mechanics. The photon is a fundamental elementary particle in the standard model of particle physics. Electromagnetic energy is defined well in classical electrodynamics and it does move this energy as a wave in time and space.
A single elementary particle propagating through space is mathematically modeled by a wavefunction which is a solution of a quantum mechanical equation. This is a complex number function, it has a sinusoidal form but the only physically measurable effect is the probability of getting a "photon" signal at a specific (x,y,z,t). It is the probability that has a sinusoidal dependence in space time, not the photon, as can be seen in the answer here. The energy of the photon is h*nu, where nu is the frequency of the classical wave which will emerge from a large number of such energy photons.
So it is not possible to talk of a trajectory of a single photon at the microscopic quantum level. It is only macroscopically, when the atomic source is known, and the interaction footprint of the photon is detected on a screen or a camera that a straight line can be drawn which in effect is the optical ray of the classical em wave.
No, it is not possible in vacuum. The photon does not propagate as you imagine, and can only be described by its energy=h*nu and its spin direction. It always travels at c.
In the complicated quantized environment of a medium with an index of refraction the way the photon wavefunctions are related to the emergent classical wave, shows that the individual photon paths, which at the microscopic level are always in vacuum and travel with velocity c, can not be an optical ray. An individual photon impinging on a transparent medium will interact by elastic scatterings with the atoms of the lattice and certainly its path cannot be one straight line. In coherence with the zillions of photons in a classical em wave it is better to discuss the classical paths and let quantum mechanics take care of the individual interactions. A true analysis quantum mechanically needs quantum field theory and is unnecessarily complicated.