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This question already has an answer here:

The dimension of the Hilbert space is determined by the number of independent basis vectors. There is a infinite discrete energy eigenbasis $\{|n\rangle\}$ in the problem of particle in a box which can be used to expand a general state $|\psi\rangle$ as: $$|\psi\rangle=\sum\limits_{n=0}^{\infty} C_n |n\rangle$$ Here, the basis implies that the Hilbert space has countably infinite number of basis vectors. But one could well expand $|\psi\rangle$ in any other basis as well, say the continuous position basis. Then that superposition, $$|\psi\rangle=\int dx \psi(x)|x\rangle$$ is also possible. But this basis $\{|x\rangle\}$ is continuous and uncountably infinite. Such sets cannot be equal because they have different cardinality. Then my questions is how many independent basis vectors are really there in this Hilbert space?

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marked as duplicate by Luboš Motl, Kyle Kanos, John Rennie, Abhimanyu Pallavi Sudhir, jinawee Feb 11 '14 at 15:20

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The answer is that $\{|x\rangle\}$ is not a basis of $L^2(\mathbb R)$ which admits only countable basis. The point is that objects like $|x\rangle$ are not vectors in $L^2(\mathbb R)$. To provide them with a rigorous mathematical meaning one should enlarge $L^2(\mathbb R)$ into an extended (non-Hilbert) vector space structure including Schwartz distributions or adopt a viewpoint based on the so called direct integral of Hilbert spaces. These are structures quite complicated to use with respect to a standard Hilbert space. Nevertheless the practical use of formal objects like $|x\rangle$ is quite efficient in physics provided one is able to distinguish between problems arising from physics and false problems just due to a naive misuse of the formalism.

(When I was student I wasted time in discussing if identities like $A|\psi\rangle = |A\psi \rangle$ had any sense.)

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My naive guess is that the position "basis" vectors are not in fact in this Hilbert space.

My reasoning is that while you can write any function with a finite number of jump discontinuities in the basis $| n\rangle$, the position basis vectors $\delta(x - x_0) = |x\rangle$ cannot be written as such. Hence they are not in that Hilbert space.

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