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Consider the position and momentum vector sets

$$X= \{|x\rangle |\ x \in \Bbb R^3\}$$

$$P= \{|p\rangle |\ p \in \Bbb R^3\}$$

By the assumption of quantum mechanics, both $X$ and $P$ are total orthonormal sets. In addition to that, $\langle x|p\rangle = e^{ip \cdot x} \neq 0$, but this conflicts with Kreyszig's functional analysis Lemma 3.5-3 which states: a vector in an inner product space can only has countably many non-zero Fourier coefficients with respect to an orthonormal family.

How's this possible?

Is the position/momentum vector representation in quantum mechanics incorrect?

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An arbitrary basis $\{v_\alpha\}_{\alpha\in A}\subset \mathscr H$ may be uncountable, but $S=\{\alpha\in A:h\in \mathscr H;\ \langle h,v_\alpha\rangle \neq 0 \}$ is countable, and in this case, $h=\sum_{\alpha\in S}\langle h,v_\alpha\rangle v_\alpha. $

I am not a physicist, but isn't this fact exactly why $p$ and $x$ are quantized?

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  • $\begingroup$ But S in my case is uncountable since both $X$ and $P$ are uncountable and their mutual product are all non zero $\endgroup$ – JokingBear Oct 29 '18 at 2:44
  • $\begingroup$ If you are working in a Hilbert Space, $S$ can not be uncountable. There is a good introduction to these ideas in Rudin's Real and Complex Analysis. $\endgroup$ – Matematleta Oct 29 '18 at 2:48
  • $\begingroup$ Then how can you explain the inner product of $|p>$ and $|x>$ $\endgroup$ – JokingBear Oct 29 '18 at 4:05
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    $\begingroup$ Please define 1). the Hilbert Space you are working in. 2). The inner product you are using $\endgroup$ – Matematleta Oct 29 '18 at 4:16
  • $\begingroup$ Consider the $L^2(R)$ space and vector $|x>$ is eigen vector of $x$ operator, and |p> is eigen vector of $-i\partial_x$ operators with their inner product $<x|p> = e^{ip \cdot x}$ $\endgroup$ – JokingBear Oct 29 '18 at 4:35

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