Recent content by PeterDonis

That's only correct for a free particle, i.e., zero potential. That's not the case in your example; you have a potential confining the particle to a finite region of space. You can't just wave your hands and arbitrarily say the wave function is zero outside the box. The wave function has to be...

That's true. I haven't computed the actual deltas from the energy eigenstates the OP is using, but of course that could be done to see where they actually fall. I would expect the ground state ##n = 0## to have the minimum uncertainty, so the OP's reasoning would apply there, but not necessarily...

Yes, it assumes that the only source of uncertainty is that inherent in the quantum wave function. That's the argument I understood the OP to be making in that case. In practice you're right, the actual uncertainties will often be much larger because of other factors that have nothing to do with QM.

You can't argue anything from an uncertainty relation where one of the factors is zero. That breaks the math.

And this invalidates the rest of your reasoning regarding ##\Delta p## being infinite. The correct conclusion from ##\langle H \rangle = 0## is not that ##\Delta p \to \infty##. It is that the uncertainty relations are undefined for this case. (Note that, unfortunately, many QM textbooks gloss...

No. The geodesic is a worldline. The 4-momentum is a 4-vector ##k^\mu## that is tangent to that worldline. And... No, this is not necessarily true. Suppose two light pulses (I'm using that term instead of "photon" for reasons explained in my post #10 just now) are both emitted at the same event...

Since we are talking about relativity, the term "photon" is out of place; that's a quantum term. A better term might be "light pulse" or "laser pulse". With that correction, the answer to your question is no. Each light pulse has a distinct worldline in spacetime and a distinct 4-momentum...

How are you obtaining these statements? Please show your work.

If the potential in the box walls has only a finite width, yes.

By analogy, "homogeneity of time" would mean invariance under time translations. And "homogeneity of space and time", which is what Landau and Lifshitz actually say, would mean invariance under any kind of translation, even one that combined "space" and "time" components.

What source? Again, please give a specific reference. I'm not sure there's actually a difference, but without seeing the actual other source you mentioned and what it says, I have no way of knowing.

Btw, I personally would argue that the statement I just quoted above, all by itself, is sufficient to establish that we must have ##ds = ds'##, because ##ds## is just the spacetime length of the interval, and that must be an invariant; it can't depend on which inertialf frame we choose. I think...

Ok, so their ##ds## and ##ds'## are describing the same infinitesimal spacetime interval in flat spacetime, just in two different inertial frames. The homoegeneity of space and time, i.e., spacetime. That just means that the way two different inertial frames are related to each other has to be...

Which book? Where? Please give a specific book/chapter/section/page reference.

@smaricic your latest post referencing Brian Greene's lecture on SR has been deleted, since its content makes clear that we all are wasting our time trying to help you. I'm going to give a summary of the key points below, even if it doesn't get through to you, for the benefit of others who might...