Recent content by gentzen

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

Oh yes, indeed. Now I see your point. No, you are right. I was the one using looser terminology here, and didn't even notice.

The space of Hilbert-Schmidt operators on a Hilbert space ##H## is always naturally isometrically isomorphic to the tensor product of the Hilbert space ##H## with its dual ##H^*##. And in the finite dimensional case, all linear operators acting on the Hilbert space ##H## are Hilbert-Schmidt...

Because I am a mathematician, and being "naturally isometrically isomorphic" is sufficiently equal for me. On the other hand, the infinite dimensional case is important to me, and there the statement is just plain and simple wrong.

No idea where he got this from. It is correct in the finite dimensional case, but in the general case, you only get the Hilbert-Schmidt operators in this way:

If the "dyad picture" works nice for you, then you can think like that. However, the typical contexts where the "dyad picture" arises is in formulas from classical electrodynamics that are written with explicit scalar products (and vector products). In this context, the "dyad picture" is a...

Yes, it is. If you want a density matrix, then it is easiest to see it as a column vector in a specific basis. Otherwise, you should imagine a density operator. This is not difficult either, it is just a different picture. In that case, you "imagine" how it applies/operates on a given vector...