Recent content by flippiefanus

OK so it depends on what state you have. The expectation value of ##a## for a single Fock state is zero. But you get a nonzero value when you have for example a coherent state. So now you expand the state in a Fock basis and then consider the phase of off-diagonal terms in its density operator...

Why do you say this experiment can be analysed in flat spacetime? Earth's gravity does not represent flat spacetime.

The Wikipedia statement probably intended to explain how the one-particle Hilbert space is used as a basic element to build up multiple particle states as tensor products. The Fock basis includes states with multiple particles, for which the occupation number is larger than 1.

Then we are in agreement. Personally I also prefer to work on phase space, but one needs to work through the basics to have a good understanding.

Yes, I agree that off-diagonal terms carry the phase information, but not in the Fock basis. One needs a basis involving the other degrees of freedom, like the spatiotemporal degrees of freedom.

That quote is also significant for what it does not say: it does not say that one cannot use quantum theory for such scenarios.

No, go back and look at #7, #9 and #16 again. It is not a strawman.

OK but what is the reason why we want such a probability interpretation? It is not used in classical theory where we measure intensity. The underlying reason is because we work with quanta.

True, I've expressed myself poorly. Let me clarify. If ##|\psi\rangle## and ##|\phi\rangle## are single photon states, then we can have a superposition ##|\psi\rangle a+|\phi\rangle b## that can lead to interference. However, the tensor product ##|\psi\rangle|\phi\rangle## does not give us...

No that is not the case. See #7, #9, and #16 for example. Those are the replies that I started making comments to.

A laser produces a multi-photon state. Not sure what you were trying to say here. The state of two lasers that are coupled cannot be represented by a tensor product state. If you had a tensor product state you would not see interference. So clearly it becomes a multi-photon process. There is no...

One needs to have some force that can affect neutrinos to slow them down. (Gravity also affects them but that is not a force.) The only force that affects them is the weak force, which we don't know how to use yet.

For what it is worth, much of the discussion that appear in the literature on this topic can be better understood by looking at the physical experiments. Many (if not most) of the EPR type experiments (to demonstrate the violation of the Bell inequality) are done with light. In such a case, the...

This is going to become a bit technical. If something is not clear, please let me know. All fields in quantum field theory are irreducible representations of the Poincare group, which combine Lorentz invariance (from special relativity) and translation invariance. That basically gives us...

Just for clarification, the excitations of fields that represent the quanta, which are (confusingly) often referred to as "particle" are not individual plane waves. Although a single quantum can theoretically be parameterised as a plane wave, such a representation would not be normalisable...