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.
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 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:
  • to localize the wave in space at position 0 to reduce the space uncertainty, we have to reduce the support. However, doing so makes the momentum variation on the edges more and more important, as the slope will go up and down faster (higher top, and less x space for descent), leading to a larger variance (note that average momentum is still 0, due to to symmetry of the bump function)
  • to localize the momentum as much as possible at 0, we can make the support wider and wider. This makes the bumps at the edges smaller and smaller. However, this also obviously delocalises the wave function more and more, increasing the variance of x
One of the main reasons why physicists are obsessed by this topic is that position and momentum are mapped to the phase space coordinates of Hamiltonian mechanics, which appear in the matrix mechanics formulation of quantum mechanics, which offers insight into the theory, particularly when generalizing to relativistic quantum mechanics.
One way to think is: what is the definition of space space? It is a way to write the wave function such that:
  • the position operator is the multiplication by
  • the momentum operator is the derivative by
And then, what is the definition of momentum space? It is of course a way to write the wave function such that:
  • the momentum operator is the multiplication by
physics.stackexchange.com/questions/39442/intuitive-explanation-of-why-momentum-is-the-fourier-transform-variable-of-posit/39508#39508 gives the best idea intuitive idea: the Fourier transform writes a function as a (continuous) sum of plane waves, and each plane wave has a fixed momentum.
A way to write the wavefunction such that the position operator is:
i.e., a function that takes the wavefunction as input, and outputs another function:
If you believe that mathematicians took care of continuous spectrum for us and that everything just works, the most concrete and direct thing that this representation tells us is that:
the probability of finding a particle between and at time
equals:
This operator case is surprisingly not necessarily mathematically trivial to describe formally because you often end up getting into the Dirac delta functions/continuous spectrum: as mentioned at: mathematical formulation of quantum mechanics
One dimension in position representation:
In three dimensions In position representation, we define it by using the gradient, and so we see that
Video 1. Position and Momentum from Wavefunctions by Faculty of Khan (2018) Source. Proves in detail why the momentum operator is . The starting point is determining the time derivative of the expectation value of the position operator.
Appears directly on Schrödinger equation! And in particular in the time-independent Schrödinger equation.
There is also a time-energy uncertainty principle, because those two operators are also complementary.
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.
Video 1. Quantum Mechanics 7a - Angular Momentum I by ViaScience (2013) Source.

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