# Quantum mechanics: total orthonormal sets & position/momentum space

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?

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?
• But S in my case is uncountable since both $X$ and $P$ are uncountable and their mutual product are all non zero – JokingBear Oct 29 '18 at 2:44
• 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. – Matematleta Oct 29 '18 at 2:48
• Then how can you explain the inner product of $|p>$ and $|x>$ – JokingBear Oct 29 '18 at 4:05
• 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 $<x|p> = e^{ip \cdot x}$ – JokingBear Oct 29 '18 at 4:35