Change of basis between symmetric matrices by
Ciro Santilli 35 Updated 2025-04-24 +Created 1970-01-01
When we have a symmetric matrix, a change of basis keeps symmetry iff it is done by an orthogonal matrix, in which case:
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 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
Bibliography:
- www.youtube.com/watch?v=bIIjIZBKgtI&list=PL54DF0652B30D99A4&index=59 "K2. Heisenberg Uncertainty Relation" by doctorphys (2011)
- physics.stackexchange.com/questions/132111/uncertainty-principle-intuition Uncertainty Principle Intuition on Physics Stack Exchange
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