So expressing that in coordinate space can be as trivial as $$<\psi|\psi>= \int <\psi|x> <x|\psi>d^3x= \int \psi^{\dagger}(x) \psi(x) d^3x$$
Excellent! So what is $$\langle \psi | \psi \rangle $$ physically?
So expressing that in coordinate space can be as trivial as $$<\psi|\psi>= \int <\psi|x> <x|\psi>d^3x= \int \psi^{\dagger}(x) \psi(x) d^3x$$
$$\ell^2$$ is infinite dimensional. Keep digging...Oh, I shouldn't have the infinity sign should I, in a finite $$\ell_2$$ space. Is that the spurious thing which was meant?
Congratulations once again - the spectrum of $$\hat{x}$$ is continuous (i.e. not discrete).Of course I've seen it before, it's used for a discrete basis state in this case for position so the completeness is given as $$\mathbb{1}=\int |x><x|d^3x$$.
You certainly did not make that clear until you had been caught out multiple times. And then you continued your intellectual impostures, instantly negating any sympathy or leeway might have been offered you on account of being a layman.I came here as a novice, made that clear, but it still fell on deaf ears, trying to state I made it out to be more.
You're being called out on not knowing the basics, while claiming that you know them. This happens all the time in any educational system, and life in general, and shouldn't be unfamiliar to you. When you got red marks on your maths exercises in school did you then feel that the grader was "out to get you"?As I said, I am sick to teeth about these lynching mobs.
You simply don't have the expertise to disagree. In my best assessment of your current competence level, based on the posts of yours I have read, you have years of material to cover before you could credibly claim to know Hilbert spaces.Even funkstar is trying to make it out you need to know a great deal more about Hilbert Space to say you know what it is - I disagree - I'm not going to start teaching on them, because yes, then I would need to know how the metric fit in to it, but to be honest, I don't think it would be very hard to find out how.
Congratulations once again - the spectrum of $$\hat{x}$$ is continuous (i.e. not discrete).
Ok Green Destiny, let's give you a nice easy test so you can at least show us you understand basic vector spaces, as you claim.
Consider the following four vectors:
$$\begin{pmatrix}1\\0\\0\\0\end{pmatrix} \begin{pmatrix}0\\1\\0\\1\end{pmatrix} \begin{pmatrix}0\\1\\0\\-1\end{pmatrix}\begin{pmatrix}3\\0\\5\\1\end{pmatrix}$$
Find a set of basis vectors for the vector space spanned by the vectors above, with the set of scalar multipliers being defined as the set of real numbers.
I suppose I would start by defining the matrices you gave in the correct fashion, by stating:
$$x,y,z,t= A \begin{pmatrix}1\\0\\0\\0\end{pmatrix} B \begin{pmatrix}0\\1\\0\\1\end{pmatrix}C \begin{pmatrix}0\\1\\0\\-1\end{pmatrix}D \begin{pmatrix}3\\0\\5\\1\end{pmatrix}$$
So
$$\begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix}= (AV_1,BV_2,CV_3,DV_3)$$
Would that be right? I am a bit weary.
You just need to find a set of linearly independent basis vectors for the vector space I defined above, that's all I'm asking for.