Redpill me on the bra-ket notation

Yes but the bras are just the dual vector space right? How hard can it be?

it gets fucked up when you realize that you can in a sense use both a countably infinite dimensional basis and uncountably infinite dimensional basis to represent the same things. That doesn't happen in finite dimensions, where the dualspace is always the same dimension as it's paired (non-dual) space.

For instance, consider the free particle and harmonic oscillator. For the free particle people usually pick the uncountably infinite Fourier transform basis functions which are not themselves normalizable, but you can form inner products with them that converge. In contrast, consider the harmonic oscillator with it's countably infinite series of basis functions. These are normalizable and can represent states themselves. The cuckoo thing is both bases can describe any normalizable wave function. It gets really interesting when you understand that even for something like the free particle, you can choose a countably infinite basis, so the difference between countably infinite and uncountably infinite basis vectors gets very subtle. The correct formulation is the Reize representation, and the correct way to think about these weird bases is how they are defined in terms of the a Gelfand triple. It's something that's really easy to overlook.

mathpages.com/home/kmath638/kmath638.htm

I don't know what a free particle or wave function is.

then why the fuck are you in a thread about bra-kets?

Because there has to be a rigorous mathematical definition for this dumbass notation.

of course. refer to gelfand triple, rigged hilbertspaces, and reize representation

Just some autistic notation made by autistic people making it more frustrating and taking the focus out of the question why the qm works

i just started studying quantum information a few months ago and it confuses me why physicists prefer this garbage notation. will it become more clear to me later on?

If it's not clear at the start, then it will never be clear. kys brainlet

>you can in a sense use both a countably infinite dimensional basis and uncountably infinite dimensional basis to represent the same things

Why would you go on the internet and tell lies? Dimension is always well-defined.

I am pointing out fucked up conclusions if you don't approach the math rigorously, which is what happened when I first learned quantum, and then thought about it more.

If you disagree with my assertion that the solutions to the harmonic oscillator (countably finite number of solutions) and the free particle (infinite number of solutions) can both represent elements in [math]L^2[/math], explain your reasoning.

The trick in resolving this is the Gelfand triple. You get hints of a problem when you learn that any uncountable basis (like the Fourier basis) always is composed elements outside of [math]L^2[/math].

Isn't this just the usual inner product of L^2?

It's just useful to have a notation just for quantum states. Parentheses are used in many situations but kets are not. Consider |+> and |->. Here's the info you get just based on notation. These two objects are states of some system, and they are dual to each other in some way. Now imagine if instead we didn't use kets and called them f and g, then also explicitly state all the info about them. It get's tedious very quickly.

How is that bad notation it's literally exactly what it looks like. It's an inner product. Have you never seen function spaces before?

>dimension is always well defined
Are you retarded?

>Are you retarded?
The context is vector spaces, so yes it's well-defined.

I assume by redpill you mean just show me some non bullshit souped up in gay notation way of representing this in terms of stuff you already know that's more or less conceptually right but in some way flawed.

Represent your dot product as a sum of components
[eqn]\langle u | v \rangle = \sum_{i=1}^3 u_i v_i[/eqn]

Except we also allow the components to be indexed by a continuous variable, so this is infinite dimensional

[eqn]\langle u | v \rangle = \int_0^1 u(x) v(x) dx[/eqn]

You can refine this, but conceptually this is where you need to be. If you ever "programmed", just consider it to be like a list comprehension or something, it can just go over anything it's not too serious. Just a big kid's dot product mostly with a billion autistic caveats.

For instance the Heisenberg uncertainty principle is really just a consequence of
[eqn] u\cdot v = |u| |v| \cos \theta[/eqn]

anyone who doesn't tell you this is jacking themselves off to their superior iq desu senpai

that's like saying the inner product really is a consequence of multiplication

so a quick question, M.E. here
bra-ket notation just showing how orthogonal two functions are, sort of like a cross product but for functions?
like if i have an function that is normalized so that if i integrate it with itself and it yields 1 is that the same as putting it in the bra-ket notation?

...

my point is that the notation is ugly and cumbersome, and it's become clear to me that physicists prefer it because they are gay, like yourself

>cumbersome
wat, its the same as the math notation for inner product, except with a line instead of a comma.

I haven't studied quantum mechanics and my knowledge about functional analysis is essentially nill. But, I know Linear Algebra pretty well which made me understood Riesz representation theorem from wikipedia since it is essentially the same with the corresponding theorem for finite dimensional spaces.
This bracket notation seems pretty neat, what exactly is the problem with it?

>Riesz representation theorem

OP here, I'm only reading about this just now. I never knew this about dual vector spaces, this is a really neat theorem.

misinterpreted quote.

I think what you are really trying to ask is "is the goofy looking integral thing an inner product?" and the answer is yes. Functions can be treated as vectors (verify against vector axioms) and the integral equation is an inner product (verify against inner product axioms). You can extend a lot of geometric intuition to functions in this way.

t. brainlets

>Putting dx on the outside
Pleb

[math]6^{2} \div 2(3) + 4
= 36 \div 2(3) + 4
= 36 \div 6 + 4
= 6 + 4
= 10 [\math]

I'm sorry, but that's retarded.

So if [math]\langle \psi_1 |[/math] is the bra-, and [math]| \psi_2 \rangle[/math] is the -ket, what's the -c-?