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A quantum system is often given as a wavefunction of position. This is a vector in a continuous, infinite-dimensional vector space, with uncountable dimension.

However we also know that the energy eigenkets are complete, which means that the dimension of the space should be able to be given by the number of energy eigenkets. How can this be, when the energy eigenvalues are often discrete, and hence of a countable number?

Qmechanic
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Wayne
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1 Answers1

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The completeness of the states comes from the fact that they form a solution to a Sturm-Liouville problem, that is independent of the coordinates you use to represent them. Imagine you have a set of states $| n \rangle $ that form a basis

$$ \sum_n |n\rangle \langle n| = \mathbf{1} $$

Now, one can represent them in space $\phi_n(x) = \langle x|n\rangle$ or in momentum $\varphi_n(p) = \langle p|n\rangle$, $\ldots$. But the set $| n \rangle$ actually never change when you do this.

danielsmw
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caverac
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