What's your opinions on the Axiom of Choice?

[Hurkyl]

Now, to try an answer to your question I would say that the eigenvectors of the position operator in the position representation are:

$$\hat{x} = x \rightarrow eigenVectors(\hat{x}) = \{\phi_{x'}:\mathbb{R} \rightarrow \mathbb{C} | \phi_{x'}(x) \mapsto \delta(x - x') , x'\in \mathbb{R}\}$$

$$\hat{p} = i \frac{\partial}{\partial x} \rightarrow eigenVectors(\hat{p}) = \{\phi_{k}:\mathbb{C} \rightarrow \mathbb{R} | \phi_{k}(x) \mapsto e^{i k x} , k\in \mathbb{R}\}$$

In physics we say that either of these sets of eigenvectors are an uncountable basis!

The linear combinations become definite integrals covering the range of the eigenvalues:

[tex]f(x) = \int_{-\infty} ^{\infty} c_{x'} \phi_{x'}(x) dx' [/itex]<br /> <br /> where the coefficients are given by Fourier's trick:<br /> <br /> [tex]c_{x'} = \int_{-\infty} ^{\infty} f(x) \phi_{x'}(x) dx [/itex]<br /> <br /> I'm interested to learn more about what's really going on here.[/tex][/tex]

[tex]$$Well, the thing here is that neither of these "bases" contain a single element of your Hilbert space! <img src="https://www.physicsforums.com/styles/physicsforums/xenforo/smilies/oldschool/bugeye.gif" class="smilie" loading="lazy" alt=":bugeye:" title="Bug Eye :bugeye:" data-shortname=":bugeye:" /> The position and momentum operators, acting on our Hilbert space, actually do not have any nonzero eigenvectors.<br /> <br /> There are at least two ways to make sense of this. One is via a "rigged Hilbert space", where in addition to our Hilbert space, we consider a smaller subspace of "test functions", its dual space of "generalized functions". The position and momentum operators have eigenvectors in this larger space of generalized functions. (distribution is another buzzword to look for)<br /> <br /> Another is through a "direct integral" of Hilbert spaces, which generalizes the (finite) direct sum of Hilbert spaces in a way similar to how the ordinary integral can be viewed as generalizing finite sums of numbers. We can write our Hilbert space as a direct integral of a copy of <b>C</b> at each point in the real line. Now, your "basis vectors" are not elements of our Hilbert space of interest, but instead a choice of basis vectors for these individual Hilbert spaces.$$[/tex]

 

[Hurkyl]

now i haven't read the whole thread but why again are we discussing the truth value of an axiom?

Because people aren't used to separating the notion of a mathematical theory and their favorite interpretation of that theory.