Ciro Santilli @cirosantilli 37

Basically the operators are just analogous to the classical ones e.g. the classical:becomes:

Besides the angular momentum in each direction, we also have the total angular momentum:

Then you have to understand what each one of those does to the each atomic orbital:

There is an uncertainty principle between the x, y and z angular momentums, we can only measure one of them with certainty at a time. Video 1. "Quantum Mechanics 7a - Angular Momentum I by ViaScience (2013)" justifies this intuitively by mentioning that this is analogous to precession: if you try to measure electrons e.g. with the Zeeman effect the precess on the other directions which you end up modifing.

TODO experiment. Likely Zeeman effect.

If Ciro Santilli were to write a book about quantum mechanics as of 2020 (before OurBigBook.com went live), he would upload an OurBigBook Markup website to GitHub Pages.

But there is one major problem with that: the entry barrier for new contributors is very large.

If they submit a pull request, Ciro has to review it, otherwise, no one will ever see it.

Our amazing website would allow the reader to add his own example of, say, The uncertainty principle, whenever they wants, under the appropriate section.

Then, people who want to learn more about it, would click on the "defined tag" by the article, and our amazing analytics would point them to the best such articles.

In this solution of the Schrödinger equation, by the uncertainty principle, position is completely unknown (the particle could be anywhere in space), and momentum (and therefore, energy) is perfectly known.

The plane wave function appears for example in the solution of the Schrödinger equation for a free one dimensional particle. This makes sense, because when solving with the time-independent Schrödinger equation, we do separation of variable on fixed energy levels explicitly, and the plane wave solutions are exactly fixed energy level ones.

Quantum superposition is really weird because it is fundamentally different than "either definite state but I don't know which", because the superposition state leads to different measurements than the non-superposition state.

Examples:

There is also a time-energy uncertainty principle, because those two operators are also complementary.

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: