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(Small correction: We can take the observables to be the self-adjoint elements of $B(H)$, or any C${}^*$-algebra, and in your whole discussion $A$ should be assumed self-adjoint.) This can be reduced …

The C${}^*$-algebra $A$ is a red herring here. All the result is really saying is that if $T$ is a self-adjoint operator on a Hilbert space $H$ then we can find a family of measures $\mu_i$ on $\sigma …

The question is a little unclear --- you want something axiomatic but not rigorous? Anyway, if you don't care about rigor and you like Dirac deltas, I don't think there's any better place to start th …

Why are rigged Hilbert spaces a paper subject, not usually treated in rigorous textbooks: this is a good question. If you learn QM the way I did, you start by understanding that the physicists' Dirac …

Okay, there is a lot of confusion in this question. First, I'm not sure why you say ``it is common to begin the discussion with embedding $A$'' into $B(H)$. The point of the C${}^*$-algebra approach t …

The connection can be seen by taking $H = L^2(\mathbb{R}^3)$ in the first explanation. This is the Hilbert space of a nonrelativistic, spinless, three-dimensional particle. By direct summing the symme …

Question 1: Yes, if you take the von Neumann algebra morphisms to be normal $*$-homomorphisms. Restricting any such map to the projections will preserve sups and orthocomplements. Question 2: No, this …

Fine question. The first thing to notice is that your problem becomes simpler if you work in the $(1,1)$, $(0,1)$ basis of $\mathbb{Z}\times\mathbb{Z}$. I.e., we have an isomorphism from $\mathbb{Z}\t …

Here is a simple example that shows that the idea of spectral theory on pre-Hilbert spaces in the sense you are asking is hopeless. Consider the pre-Hilbert space consisting of the restrictions of all …

The question mentions topological vector spaces, but aren't Banach spaces a special case of sufficient interest to merit explicit mention? Every infinite dimensional Banach space is a non-locally comp …

I disagree with the previous comments. As Christian says, the sums all converge strongly if we interpret $|x\rangle \langle y|$ as a rank 1 operator. In particular, $\hat{n}$ is a coisometry which tak …

Another way to do it is to consider the bounded operators $(T_i + iI)^{-1}$, check that these are normal and commute so they can be simultaneously represented as multiplication operators on an $L^2$ s …

I'd like to try to give a more comprehensive answer. In the elementary formulation of quantum mechanics, pure states are represented by unit vectors in a complex Hilbert space $H$ and observables are …

This is not quite what you are asking, but I addressed time optimality in my paper "Time-optimal control of finite quantum systems", J. Math. Phys. 41 (2000), 5262–5269. We have results like: if any c …

To add to John Baez's answer, you can regard the symmetric Fock space over $L^2(X)$ as a measurable tensor product of the Hilbert spaces $l^2(\mathbb{N})$ over the index set $X$, and the antisymmetric …