Recent content by gentzen

What is the relation of these axisymmetric static solutions of Einstein's equations to BSM physics?

I agree. Basically yes. The positions will only get entangled behind the Stern-Gerlach apparatuses. For describing the actual experiment, using momenta instead of positions would be more precise, i.e. ##\ket{\Psi}=\ket{\text{Bell state}}\ket{p_1}\ket{p_2}##. The momenta can be controlled more...

Those statements are well-defined as part of measurement protocols. (As you noticed, you cannot actually know the moment "just before the photon arrives", not just because you didn't measure "the position of the photon" or "the moment it was emitted", but because the photon would arrive at the...

You misunderstood: I would have pressed the "like" button below his post, not the one below yours. This was a direct and honest answer to your question to me. My initial comment was of course directed to both of you: To him in order to inform him about his mistake, and to you in order to...

Well, I would have "liked" that post, if that issue were absent. If you do see more issues with this thread, feel free to point them out explicitly.

You could have explicitly pointed out that it should have been |Ψ|² = |α|²⟨A|A⟩ + |β|²⟨B|B⟩ + αβ*⟨B|A⟩ + α*β⟨A|B⟩ where ⟨A|A⟩, ⟨B|B⟩, ⟨B|A⟩, and ⟨A|B⟩ are scalar products between the states |A⟩ and |B⟩.

and Cramer's version is very different from it: https://arxiv.org/abs/1503.00039

I hope that Kastner's version will be similar to it (that's what I meant by "Kastner's version seems closer to what I am looking for"). The Two-State Vector Formalism is fine for weak-measurement, but it is not obvious to me how to model normal "perturbing" measurements. The time-symmetric...

Because I wanted to better understand a sketch/drawing in chapter 4 of a "strange book" I was reading I started to dive a bit into the transactional interpretation (TI). I quickly learned enough to understand that sketch and finish chapter 4 (after which I decided to stop reading that book for...

OK, now I see where you are coming from. But your conclusion is not justified. The "weights" of the delta peaks in the higher order derivatives can converge to zero in the limit of an infinite sum of eigenstates, even if the individual eigenstates have non-zero weights for those peaks. (Of...

This argument does not convince me. Even if <p^6> and higher order ones diverge, why should that imply incompleteness of the eigenstates for L^2? In fact, I see no reason why the eigenstates for finite well potential (discrete and continuous together) should not be complete.

Yes, this is the implication. If a photon transfers energy somewhere, then either the photon vanishes completely, or its frequency goes down. But I answered, because I thought about the same question in the context of electrons in a crystal recently, and the eigenvalue problem one solves to get...

If you accept that "(almost) all functions can be decomposed" must be read as "all square integrable functions can be decomposed", then my impression is that the remaining part of your problem is related to the recently discussed issue that square integrable functions go to zero at infinity if...

If ##A## has eigenvalues ##\lambda_i## with corresponding eigenvectors ##v_i## and ##B## has eigenvalues ##\mu_j## with corresponding eigenvectors ##w_j##, then ##A\otimes B## has eigenvalues ##\lambda_i\mu_j## with corresponding eigenvectors ##v_i\otimes w_j##: $$(A\otimes B)\ v_i\otimes w_j =...

yes