Recent content by PeterDonis

In which case it's certainly not zero. So either way the OP's expectation of being able to find a common diagonalization for ##H## and ##P## is false.

I don't see why the potential is "troublesome"; it's a function of ##x## whose derivative won't in general be zero, so it won't commute with ##p##.

It is. On an initial read this paper looks like nothing more than handwaving. That's the general opinion that Deur's work was trying to undermine. However, this paper doesn't appear to me, on an initial read, to be based on anything like the kind of field self-interaction that Deur's work is...

You're missing the point. I wasn't saying there was no reason to prefer the L-L pseudotensor to the Dirac pseudotensor; I think that's a matter of opinion. I was saying that "it's symmetric" is not a reason to prefer the L-L (2, 0) pseudotensor to the L-L (0, 2) pseudotensor.

As far as I know, the same would be true if we lowered both indexes, so again, this is not a reason to prefer the form with two upper indexes.

You can't--at least, not if you're going to take the viewpoint of Dirac (and L-L, for that matter) that it's perfectly okay to write curved spacetime equations using partial derivatives instead of covariant derivatives. In other textbooks, such as MTW, that viewpoint is not taken--everything in...

I know you are. And you have not responded at all to my argument in post #13 for why that is not correct--why ##T_{\mu \nu}##, with two lower indexes, is physically the correct expression for the density of stress-energy, the expression that does not intrinsically involve any factors of the...

Only if you think ##T^{\mu \nu}## represents what I've said it doesn't represent, and explained why. I can't remember if L-L use the same restricted class of coordinate charts that Dirac does. ##T_{\mu \nu}## with two lower indexes, which is what I have said is the correct expression for the...

Well, let's see. $$ M^{\mu \nu} = g^{\mu \alpha} M_\alpha{}^\nu $$ Therefore, $$ M^{\mu \nu}_{,\nu} = \left( g^{\mu \alpha} M_\alpha{}^\nu \right)_{,\nu} $$ which gives $$ M^{\mu \nu}_{,\nu} = g^{\mu \alpha} \left( M_\alpha{}^\nu{}_{,\nu} \right) + M_\alpha{}^\nu \left( g^{\mu...

Did you read the quote from your own post that my question was about? That's the claim I'm asking for a basis for. It was a claim you made, not someone else.

Yes, but now it's the state of atom plus environment, not just the state of the atom.

But unless and until someone finds such a theory and makes successful predictions from it, it's vaporware and we can't discuss it here. This subforum is for discussing interpretations of QM, not hypothetical new theories that don't even exist.

They are identical in the precise sense of that term in quantum mechanics. What is your basis for this claim? It has no basis whatever in standard QM.

You can--by creating an environment where the quantum states of the decay products are already occupied, or partially occupied. (And for the case of electron capture, you can change the half-life of the reaction by ionizing or partly ionizing the atom, so the electron availability for capture is...

This is a faulty analogy. The coins are tossed one at a time, and, as you say, the initial conditions can vary for each toss, so a deterministic model can still give different results for each toss. But you can prepare a sample of a huge number of identical radioactive atoms all at the same...