Ciro Santilli @cirosantilli 34

Notation used in quantum mechanics.

Ket is just a vector. Though generally in the context of quantum mechanics, this is an infinite dimensional vector in a Hilbert space like $L^2$.

Bra is just the dual vector corresponding to a ket, or in other words projection linear operator, i.e. a linear function which can act on a given vector and returns a single complex number. Also known as... dot product.

For example:is basically a fancy way of saying:that is: we are taking the projection of $y$ along the $x$ direction. Note that in the ordinary dot product notation however, we don't differentiate as clearly what is a vector and what is an operator, while the bra-ket notation makes it clear.

The projection operator is completely specified by the vector that we are projecting it on. This is why the bracket notation makes sense.

It also has the merit of clearly differentiating vectors from operators. E.g. it is not very clear in $x\cdot y$ that $x$ is an operator and $y$ is a vector, except due to the relative position to the dot. This is especially bad when we start manipulating operators by themselves without vectors.

This notation is widely used in quantum mechanics because calculating the probability of getting a certain outcome for an experiment is calculated by taking the projection of a state on one an eigenvalue basis vector as explained at: Section "Mathematical formulation of quantum mechanics".

Making the projection operator "look like a thing" (the bra) is nice because we can add and multiply them much like we can for vectors (they also form a vector space), e.g.:just means taking the projection along the $x+y$ direction.

Ciro Santilli thinks that this notation is a bit over-engineered. Notably the bra's are just vectors, which we should just write as usual with $\overrightarrow{v}$... the bra thing makes it look scarier than it needs to be. And then we should just find a different notation for the projection part.

Maybe Dirac chose it because of the appeal of the women's piece of clothing: bra, in an irresistible call from British humour.

But in any case, alas, we are now stuck with it.

The Fourier series of an $L^2$ function (i.e. the function generated from the infinite sum of weighted sines) converges to the function pointwise almost everywhere.

The theorem also seems to hold (maybe trivially given the transform result) for the Fourier series (TODO if trivially, why trivially).

Only proved in 1966, and known to be a hard result without any known simple proof.

This theorem of course implies that Fourier basis is complete for $L^2$, as it explicitly constructs a decomposition into the Fourier basis for every single function.

TODO vs Riesz-Fischer theorem. Is this just a stronger pointwise result, while Riesz-Fischer is about norms only?

One of the many fourier inversion theorems.

A set of theorems that prove under different conditions that the Fourier transform has an inverse for a given space, examples:

Approximates an original function by sines. If the function is "well behaved enough", the approximation is to arbitrary precision.

Fourier's original motivation, and a key application, is solving partial differential equations with the Fourier series.

Can only be used to approximate for periodic functions (obviously from its definition!). The Fourier transform however overcomes that restriction:

The Fourier series behaves really nicely in $L^2$, where it always exists and converges pointwise to the function: Carleson's theorem.

Most notable example: $L^2$.

Key for quantum mechanics, see: mathematical formulation of quantum mechanics, the most important example by far being $L^2$.

$L^p$ for $p==2$.

$L^2$ is by far the most important of $L^p$ because it is quantum mechanics states live, because the total probability of being in any state has to be 1!

$L^2$ has some crucially important properties that other $L^p$ don't (TODO confirm and make those more precise):

$L^p$ is:

And then this is why quantum mechanics basically lives in $L^2$: not being complete makes no sense physically, it would mean that you can get closer and closer to states that don't exist!

TODO intuition

Some sources say that this is just the part that says that the norm of a $L^2$ function is the same as the norm of its Fourier transform.

Others say that this theorem actually says that the Fourier transform is bijective.

The comment at math.stackexchange.com/questions/446870/bijectiveness-injectiveness-and-surjectiveness-of-fourier-transformation-define/1235725#1235725 may be of interest, it says that the bijection statement is an easy consequence from the norm one, thus the confusion.

TODO does it require it to be in $L^1$ as well? Wikipedia en.wikipedia.org/w/index.php?title=Plancherel_theorem&oldid=987110841 says yes, but courses.maths.ox.ac.uk/node/view_material/53981 does not mention it.

A measurable function defined on a closed interval is square integrable (and therefore in $L^2$) if and only if Fourier series converges in $L^2$ norm the function:

The wave equation contains the entire state of a particle.

From mathematical formulation of quantum mechanics remember that the wave equation is a vector in Hilbert space.

And a single vector can be represented in many different ways in different basis, and two of those ways happen to be the position and the momentum representations.

More importantly, position and momentum are first and foremost operators associated with observables: the position operator and the momentum operator. And both of their eigenvalue sets form a basis of the Hilbert space according to the spectral theorem.

When you represent a wave equation as a function, you have to say what the variable of the function means. And depending on weather you say "it means position" or "it means momentum", the position and momentum operators will be written differently.

This is well shown at: Video "Visualization of Quantum Physics (Quantum Mechanics) by udiprod (2017)".

Furthermore, the position and momentum representations are equivalent: one is the Fourier transform of the other: position and momentum space. Remember that notably we can always take the Fourier transform of a function in $L^2$ due to Carleson's theorem.

Then the uncertainty principle follows immediately from a general property of the Fourier transform: en.wikipedia.org/w/index.php?title=Fourier_transform&oldid=961707157#Uncertainty_principle

In precise terms, the uncertainty principle talks about the standard deviation of two measures.

We can visualize the uncertainty principle more intuitively by thinking of a wave function that is a real flat top bump function with a flat top in 1D. We can then change the width of the support, but when we do that, the top goes higher to keep probability equal to 1. The momentum is 0 everywhere, except in the edges of the support. Then:

Bibliography: