Hurkyl
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<br /> Well, the thing here is that neither of these &quot;bases&quot; 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 &quot;rigged Hilbert space&quot;, where in addition to our Hilbert space, we consider a smaller subspace of &quot;test functions&quot;, its dual space of &quot;generalized functions&quot;. 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 &quot;direct integral&quot; 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 &quot;basis vectors&quot; are not elements of our Hilbert space of interest, but instead a choice of basis vectors for these individual Hilbert spaces.Crosson said: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:
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 /> c_{x&#039;} = \int_{-\infty} ^{\infty} f(x) \phi_{x&#039;}(x) dx [/itex]<br /> <br /> I&#039;m interested to learn more about what&#039;s really going on here.