Recent content by Fwiffo

This is not as easy as it seems, It depends on which method you want to use to define the photon and which gauge you are working in. The simplest expression for a photon is |1> in fock space. If you want a "space-time" kind of wave you have a look at the paper "O. Keller / Physics Reports 411...

Good point. This by the way is nice proof that the commutation relation [A,B] = \alpha I is impossible in the discrete case where operators can be diagonalized.

The problem is not the continuous spectrum, although what you (and jostpuur) wrote is certainly right. The (bigger) problem is that two operators which don't commute don't have the same eigenvectors. The line above is impossible unless A and B commute. This whole exercise could be done with a...

<A><B>\ne<AB>

This is your basic wave/particle duality. The photon has wave like properties, these properties obey the maxwell equations (i.e the equations for electromagnetic waves). However it is quantized (a particle property) so two photons cannot interfere with each other in the same way two...

This thread has diverged into looking at 3 related topics. Photon position operators; Photon detection; and photon localization. All three are hard problems. I'll start with the easiest one which is detection/emission of (single) photons. Here the really big problem is that relativistic field...

Adding an index is not enough. This is not a Schroedinger equation but a second quantized approach, you have to use a whole different method. Normalization is not trivial, eqs 49 and 53 cannot be used with just an index added.

Like you said this works for spinless particles. I don't think including spin in this wavefunction is as simple as you make it out to be.

By cannonical I mean one which is used in most cases. You are right there are ways to define a position operator , but it usually has some drawbacks such as non commuting components, or being non relativistic, or not giving the position. One that seems to overcome a lot of problems is the one...

Where can I find this?

Regardless of interactions there is no position operator for photons. True localization is less of a problem for free photons, but the operator ( d/dp for example) does not work. Other choises can be constructed for spinless (massless) relativistic particles but a photon position operator is...

The position operator for a photon is a very big problem. A good review article on the subject is http://adsabs.harvard.edu/abs/2005PhR...411...1K by Ole Keller Another person who has done a lot of work on the subject is Margaret Hawton. It's hard to count the problems with this idea...

True, but the statistics after a measurement are very different from those you expect in classical physics which does not allow entanglement. The various Bell inequalities are ways of testing the theory. The simplest one is CHSH where the chances of winning a certain game should be 1/2 but are...

... sorry about braking this into two posts so when do we have to anti-symmetrize the state? The answer is, when it becomes meaningful for our purposes. You could say the state of all fermions in the universe is at all times antisymmetric, this should be true (as far as we know). If I look at...

Again the use of language is misleading. seemingly the state |AB>-|BA> is a singlet state, and in some cases it is entangled. But in the most general case there is no "real" entanglement, this is because the particles are identical so if I discover one particle in |A> I have no idea if it was...