# Let $T$ be an operator on Hilbert space $H$...

Let $T$ be an operator on Hilbert space $H$. Let $A \in H$ and $\chi_A$ its characteristic function.

What does $\chi_A(T)$ mean?

for x in H

xhi_A (T)(x) = :

1 if T(x) is in A
0 otherwise.

Unfortunately no.

Here is a context btw

With your assumption, the dimension doesn't match

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then don't call chi a fucking characteristic function if it's not one. It's a projector.

It is a characteristic function

so what's wrong with the previous response?

T acts on H.

You take an element of H, you apply T to that element.
Then you check if T(x) is in A or not.
and that gives you the result
Maybe I'm misunderstanding something.

In the case of this picthe operator is A.
So you take Ax
you check if it's in the subset B_i
you multiply by ci
you sum over all i
etc

There is a big difference between $f(Tx)$ and $f(T)x$.

You are idiot. That's the problem.

Pleb here. What is a Hilbert space and how do I use it? I've been trying to understand the concept for a while now, but so far I haven't made any progress.

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Roughly speaking a Hilbert space is a (usually infinite dimensional) vector space with a geometric structure similar to that of normal Euclidean space. In fact, Euclidean spaces are exactly finite dimensional Hilbert spaces. Hilbert spaces are used in quantum mechanics. P

>that's the problem
not explaining the problem
k m8.
Come back when you understand that it doesn't matter here.

Nigga, you can't even into muh matching dimensions. What are you talking about? Clean out your eyes and look at the expression.

I really don't see which fucking dimensions are not matching, I explained how everything matches and you're just here spewing bullshit instead of actually giving information. If I'm wrong, correct me properly or don't say anything.

Ax is a vector. The summand here is supposed to be a vector. In fact, it's a vector from an invariant subspace.

there's a lot in here that doesn't make sense to my eyes, perhaps because of missing context. i'd guess though, that chi(A) is suposed to be the operator whose existence you get from the spectral theorem (assuming A to be normal). then you can apply any bounded, measurable complex function to your operator.

It's defined before introducing spectrum. It should be a limit of polynomial approximation I guess.

therefore the definition of a characteristic function doesn't fit the function used here. Which is my first point. Chi is not a characteristic function as defined in

It's always a good idea to consider a simple example.
Let $H = \mathbb{C}^n$ and let $T: H \to H$ be a self-adjoint operator with distinct eigenvalues $\lambda_1 , \ldots, \lambda_n$.

Then $T$ can be written as $T = \sum_{k=1}^n \lambda_k | \lambda_k \rangle \langle \lambda_k |$ and any element $| x \rangle \in H$ can be written as $| x \rangle = \sum_{k=1}^n c_k | \lambda_k \rangle$ for some $(c_k)$.

Now you can define a Functional Calculus $f(T)x := \sum_{k=1}^n f(\lambda_k) c_k | \lambda_k \rangle$. So if $f$ is the charactersitc function of some set $A$ you just put those $c_k$ that don't belong to an Eigenvalue in $A$ to $0$ while leaving those that belong to an Eigenvalue in $A$ alone. This makes it to a projection. If the set $A$ only includes a single Eigenvalue then it would just be the projection to the corresponding Eigenspace.

Yes it is. You can consider it as a limit of a polynomila function.

>Hilbert spaces are used in quantum mechanics. P
I find your kind of explanation confusing. Just because you explicitly use the word "Hilbert space" when you teach quantum mechanics, doesn't make it untrue that you use Hilbert spaces whenever you do something in R^2.

Same goes with
>usually infinite dimensional

Yes I know what you mean, but.

The problem is that it was introduced before introducing spectrum

alley cat, yee :)

Okay familia here's how it works. As stated before, Hilbert spaces are an abstraction of our usual Euclidean space into an arbitrary number of dimensions. This abstraction allows us two important extensions: defining the Hilbert space $\mathcal{H}$ over the complex field $\mathbf{C}$ and defining an inner product $\langle \cdot , \cdot \rangle$ in the same $\ell^2$ sense as in typical Euclidean space (basically, the Pythagorean theorem).

Point 1 is sorta trivial in the context of explaining the concept, so I'll skip that (you just add imaginary numbers to the real numbers, throw in some c.c.'s, and so forth).

>doesn't make it untrue that you use Hilbert spaces whenever you do something in R^2
Generally speaking, you're right. Here's the thing though -- when you speak just of $\mathbf{R}^2$, you're just talking about the Cartesian product between 2 real fields. There is no structure within it: no *metric* which tells me how vectors within $\mathbf{R}^2$ are additively measured from one another, no *inner product* to tell me how they can multiplicatively interact with one another. We take for granted that when we speak of the n-dimensional real field $\mathbf{R}^n$, we actually mean the metric space $E = (\mathbf{R}^n , d)$, where the metric $d : \mathbf{R}^n \times \mathbf{R}^n \to \mathbf{R}$ is defined by $d(x,y) = \sqrt{\sum_{j=1}^n (x_j-y_j)^2}$ for $x = (x_1, x_2, \dots , x_n),\ y = (y_1, y_2, \dots , y_n)$.

So it's not the field over which you define your space that matters -- it's really all about the metric, from which you have a sense of distance and length between elements of your space.


So let's bring that all back to the Hilbert space. We equip the Hilbert space, not with a metric, but an inner product: $\mathcal{H} = ( \mathbf{R}^n , \langle\cdot,\cdot\rangle )$ (I'm sticking with real Hilbert spaces for now). The inner product $\langle\cdot,\cdot\rangle : \mathbf{R}^n \times \mathbf{R}^n \to \mathbf{R}$ is defined as $\langle x , y \rangle = \sum_{j=1}^n x_jy_j$ (using the same vectors as in the previous post). Indeed, it is the usual inner/scalar/dot product we're so very familiar with in Euclidean space: that's because all Euclidean spaces are Hilbert spaces. (Note that you can define the metric/norm from the inner product and vice versa.)

The divergence, of course, happens when you consider $n \to \infty$ -- now, thinking of Euclidean space makes no sense! But what are infinite dimensional vectors but functions -- or rather, the sequence of values that a function takes over the domain of the space. We have to redefine some notions of length and direction with these infinite dimensional vectors, of course. We move from the finite vectors $x\in\mathbf{R}^n,\ y\in\mathbf{R}^n$ to functions $f\in L^2,\ g\in L^2$, where $L^2$ is the space of square-integrable functions on $\mathbf{R}$. We now define the inner product as $\langle f , g \rangle = \int f g$ with some measure of the integration (Riemannian is usually good enough for physics purposes, I believe). Just as when $x$ and $y$ orthogonal functions, we say $f$ and $g$ are orthogonal functions if $\langle f , g \rangle = 0$. Thus we have a sense of projection and direction between functions (infinite dimensional vectors), just as we did with the usual vectors.

In a sentence: (inf. dim.) Hilbert spaces allow us to extend the geometry of Euclidean spaces into function spaces.

Have you ever worked with fourier series? Congratulations, you've worked with the concept of an infinite dimensional Hilbert Space. Each function $\sin(\frac{n\pi x}{L}),\ \cos(\frac{n\pi x}{L})$ is orthogonal to all others over the range [math]-L

>They then form a basis in a Hilbert space
No, one of the first things you should learn about Hilbert spaces is that they never have a countable basis. This is usually proven with Baire category theorem.

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Whoah, thanks user. If only all posts on this shithole of a board could be this insightful.

So, it this actually a limit of a polynomial?

it is defined in terms of subspaces of H or measurable sets.

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Is it defined throught the spectral measure?

?