Quantum mechanics

I need quantum mechanics info (for beginners)

le cat is le dead and le alive at le same time
>bazoingo

quantum mechanics is a fucking meme, and a stale one at that

you do know there are other interpretations than Copenhagen, right?

Could you be a bit more specific?
What are you learning it for? What exactly in quantum mechanics do you want to learn about (it's a HUGE field)? What is your background? To what depth do you want to study?

Sure thing senpai.

The basic object of study in QM is the "state", normally denoted by a ket [math] | \psi \rangle [/math], this is a vector (often called an "abstract vector" in some introductory text, but I find this name retarded and insulting so it'll be avoided). the ket can define a matrix in a discrete case and an integral in the continuous case.

You might be wondering how we define an inner product with kets, well to do that we need the ket's dual, the bra [math] \langle \phi | [/math] we stick the two together as [eqn] \langle \phi | \psi \rangle = \left ( \langle \psi | \phi \rangle \right )^* \in \mathbb { C } [/eqn] Furthermore [math] | | \psi \rangle |^2 = \langle \psi | \psi \rangle [/math] but normalisations are physically meaningless, so we may as well choose one that makes our life easier, in that case we pick [math] \langle \psi | \psi \rangle = 1 [/math].

Now if we have state vector [math] | \psi \rangle [/math] we can define a basis for this in the following way [eqn] | \psi \rangle = \sum _{n} \psi _{n} | n \rangle [/eqn] and we want the basis to be orthonormal, so we enforce [math] \langle m | n \rangle = \delta _{m,n} [/math] so that we have [eqn] \langle m | \psi \rangle = \sum _n \psi _n \delta _{m,n} = \psi _m [/eqn] we call [math] \langle m | \psi \rangle [/math] the "probability amplitude", moreover we have the following relation, called the "completeness relation" [eqn] \hat { 1 } = \sum _{n} | n \rangle \langle n | [/eqn]

When it comes tp extracting values from QM we use operators called "observables" )or perhaps that would be more accurate to say that observable are operators with certain properties), so if we have an observable [math] \xi _n [/math] in the basis [math] | n \rangle [/math] the we can construct the "spectral decomposition as:

[eqn] \hat { \xi } = \sum _m \xi _m \langle m | n \rangle [/eqn] Which satisfies [eqn] 1.) ~ | n \rangle ~ \text { are the eigenvectors (often called eigenstates) and } ~ \xi _n ~ \text { are eigenvalues } \\ 2.) ~ \hat { \xi } ~ \text {are linear } \\ 3.) ~ \hat { \xi } ~ \text { is Hermitian } [/eqn]Hence we can think of the measurement process as projecting the state [math] | \psi \rangle [/math] onto a specific eigenstate [math] | n \rangle [/math], to this end we define the projection operator [eqn] \hat { P } _n = | n \rangle \langle n | [/eqn] And so after measurement the state is no longer [math] | \psi \rangle [/math] but [math] \hat { P } _n | \psi \rangle [/math].

Now lets define the average value of many measurements of the same state [eqn] \bar { \xi } = \sum _{n} | \langle n | \psi \rangle |^2 = \sum _n \langle \psi | n \rangle \langle n | \psi \rangle = \sum _n \langle \psi | ( \hat { \alpha } _n | n \rangle \langle n |) \psi \rangle = \langle \psi | \hat { \xi } | \psi \rangle [/eqn]

le alternate uniberses
le infinite superpositions

i don`t think there`s a more fedora tipping branch of science than quantum physics

In the continuous case we pretty much just swap the summation with an integral, The definition of ket then becomes [eqn] | \psi \rangle = \int _{a} ^{b} \phi (x) | x \rangle ~ dx [/eqn] and we can enforce by insisting that [math] \langle x | x' \rangle = \delta ( x - x') [/math] we can see that this produces the expected result [eqn] \langle x' | \phi \rangle = \int _{a} ^{b} \phi (x) \langle x' | x \rangle ~ dx = \int _{a} ^{b} \phi (x) \delta ( x -x') dx = \phi (x') [/eqn]Likewise we define a completeness relation by [eqn] \int _{a} ^{b} | x \rangle \langle x | dx = \hat { 1 } [/eqn]Lets use our machinery to define the position operator [eqn] \hat { x } = \int _{a} ^{b} dx x | x \rangle \langle x | \\ \implies \hat { x } | x \rangle = \int _{a} ^{b} dx' x' | x \rangle \langle x' | x \rangle = \int _{a} ^{b} dx' x' | x' \rangle \delta ( x' - x )=x | x \rangle [/eqn]Finally we can define the continuous analogue of the average value of many measurements as [eqn] \langle \phi | x | \phi \rangle = \int _{a} ^{b} dx \langle \phi | x \rangle x \langle x| \phi \rangle = \int _{a} ^{b} dx \phi ^{*} x \phi[/eqn]

And that (along with the commutator) is pretty much the heart of QM, would you like to know more?

Sure, i want to learn quantum computing and I want to learn the basics principles

Physicists are disgusting. No discussion on how/why these series actually converge.

whats the matter user? cant stand a lack of mathematical rigor?

It's simple user, it's physics so they all must converge. If it helps, just take convergence as an axiom :^)

Is for a work

Omitting a discussion of convergence leads to students having inherent misconceptions on basic properties of these series.

People end up thinking they always behave the same way as finite sums and fuck things up.

ty

>finite sum
>discussion of convergence

I'm pretty sure a finite sum always converge

can you give an example of a mistake one could make by doing that?

quantum physics doesn't care about any of that. it cares about the math.
what it actually means is a question for philosophy

HAHAHAHAHAHAHAHAHAHAHA

If you are assuming finite sums then you are assuming a finite dimensional hilbert space and therefore omitting a lot of possible QM systems.

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>what are integrals and L2 functions

This looks so cool.

>Every infinite dimensional Hilbert space is an L2 space

>implying I said that

What are you even objecting to? By definition a Hilbert space has a finite inner product...

Read any freshman physics textbook.

Einberg and Resnick is p good. Currently finishing up Ch 5, the examples are very easy to follow which is always good