/sci/ - Science & Math

>>15500960
Hello, Yukariposer here. I have had the opportunity to talk to Jake personally and have him explain his research to me. I was also present at his PhD defense.
The gist is this: the self-adjoint assumption is in general sufficient but not necessary for the real-valuedness of observables. You can control the failure with a unitary [math]\eta[/math] such that [math]\eta H = H^\dagger\eta[/math], and the modified sesquilinear form [math]\langle\eta-,-\rangle[/math] gets you the proper setting to talk about real-valued observables, including the energy. Moreover, the spectral theory of non-self adjoint operators is itself very interesting since it lies in some sort of intersection between complex algebraic geometry and operator algebra.
You are correct, of course, in that non-unitarity implies non-conservation of probability, and hence dissipation of information. Complex observables is a signature of this when you take the expectation of non-self-adjoint operators without this [math]\eta[/math]. In out-of-equilibrium dynamics, such non-self adjoint property of the Hamiltonian play a very significant role in dissipation-fluctuation. I in fact asked him if it's possible for him to give me an example of this [math]\eta[/math] in the context of OOE Lindblad master equation, but he was not able to give me an answer.

>>11486054
You're missing the point sweetie. I did not object to the fact that every vector can be written as a finite linear combination [math]in~the~direct~limit[/math] [math]\mathcal{V} =\lim\limits_{\rightarrow}V[/math], that's obvious to anyone that's already taken intro linear algebra. My point was that this endows the direct limit topology on [math]\mathcal{V}[/math], namely the weakest topology in which the projections [math]p:\mathcal{V}\rightarrow V= \operatorname{Span}\mathcal{B}[/math] are continuous; and indeed we must have a topology, as the moment you endow a norm the boundedness of [math]p[/math] becomes synonymous with continuity of [math]p[/math]. I'm saying this is a specific topology that absolutely does [math]not[/math] encompass "all" normed linear spaces. If you wanted to bring in topology in your objection you have to understand what it entails first.
>>11486053
>can I make sense of
Yeah, you can compute it with residues. Starting with a test function its Fourier transform also ends up a test function; using Faltung theorem [math]\mathcal{F}^{-1}(F(\xi)/\cos(k\xi)) = f(\xi) \ast \mathcal{F}^{-1}(\cos(j \xi))[/math] we know that this is also compactly supported and smooth. Though note that [math]\mathcal{F}(\cos(k\xi)) =\int\frac{d\xi}{2\pi}\cos(k\xi) = \delta(k)[/math] in the sense of distributions so you have to evaluate the convolution in [math]\mathcal{S}'[/math].
>what would happen if f is not a test function
Then you land outside of it, until you hit [math]L^2[/math]. By Parseval [math]\mathcal{F}[/math] is an automorphism and preserves the [math]L^2[/math]-inner product, so in a sense [math]\mathcal{S}'[/math] is the "biggest" space you can evaluate [math]\mathcal{S}[/math] and their Fourier transforms on. If you enlarge [math]\mathcal{S}[/math] you shrink your corresponding [math]\mathcal{S}'[/math].

>>11378930
>I tried to find the volume of my friend's dick
Nothing beats a good ol' hands-on (or mouth-on if you like) experiment in this case imo

>>11305741
Put Hermitian operators [math]a = A + A^*[/math] and [math]b = b + b^*[/math]. Let [math]c = a,b[/math] and [math]C=A,B[/math], then [math]c^2 = 2 + {C,C^*}[/math] whence [math]|c^2| = 2 (1 + |C|^2) [/math]. We now use two results:
1. the spectral theorem, which states that [math]\operatorname{im}C[/math] is dense in [math]\mathcal{H}[/math] for normal operators [math]C[/math], and
2. the spectral bound [math]|A| = \max |\operatorname{Spec}A|[/math], where the norm over the spectrum is taken with the essential Lebesuge measure.
Now we can bound [math]\langle \psi,c\psi\rangle \leq \sum_{\lambda_n \in\operatorname{Spec}c}|d_n|^2 \lambda_n \leq K (\operatorname{dim}\mathcal{H}) d_{n_\text{max}}\lambda_{n_\text{max}} = K (\operatorname{dim}\mathcal{H})|d_{n_\text{max}}|^2 |c|[/math], where [math]\psi = \sum_n d_n \psi_n[/math] is a decomposition into eigenfunctions of [math]c[/math], and [math]K[/math] counts the number of possible degeneracies in the maximal eigenspace.
Now since [math]\psi \in P\mathcal{H}^* [/math] is normalized, we have control over [math]\sum_n |d_n|^2 \leq 1[/math] hence [math]|d_{n_\text{max}}|^2 \leq 1[/math]; on the other hand, we can use Cauchy-Schwarz on the [math]Hilbert~space~of~operators[/math] whence [math]|(c,c)|^2 = |(c^2,I)|^2 \leq |c|^2[/math] where the inner product is defined as [math](A,B) = \operatorname{tr}_\mathcal{H}(A^*B)[/math]. Hence [eqn]\langle\psi,C\psi\rangle \leq \frac{1}{2}K(\operatorname{dim}\mathcal{H})\sqrt{2(1 + |C|)} \implies \langle\psi,A\psi\rangle + \langle\psi,B\psi\rangle \leq \frac{\sqrt{2}}{2}\operatorname{dim}\mathcal{H}\left(K_a\sqrt{1+|A|} + K_b\sqrt{1+|B|}\right).[/eqn]
You can already see the problem with your conjecture: you can't possibly get 1 on the RHS unless [math]A = B = 0[/math].
>>11306255
You need to take the non-relativistic limit before you evaluate the Feynman integral. Internal lines aren't on-shell.