Using integrals to expand a vector in continuous basis

Question

I am new to quantum mechanics. I have been trying to understand why when we want to represent a function $$\psi(x)$$ as a ket in continuous basis |x> we us the integral:

$$\vert \psi(x)\rangle =\int\psi(x)\vert x\rangle dx$$

where in non-continuous basis it is :

$$\sum\psi(x)\vert x\rangle $$ clearly the $dx$ gives different units here so I am not sure if integrals make sense to use to expand the vector in these basis. Also, I have heard that continuous means uncountable which I am not sure how that is uncountable, can't we just index all the basis with natural numbers since last time I checked we have infinity of them?

Answer 1

You want to think of $$ \psi(x)=\langle x\vert\psi\rangle $$ as a (complex) number interpreted as the “component” of $\vert\psi\rangle$ on the basis vector $\vert x\rangle$, with $\langle x\vert\bar x\rangle=\delta(x-\bar x)$. This way \begin{align} \vert \psi\rangle &= \int\,dx\, \psi(x) \vert x\rangle \, ,\\ \psi(\bar x)=\langle \bar x\vert\psi\rangle &= \int \,dx\, \psi(x)\langle \bar x\vert x\rangle \end{align} is basically the (continuous) generalization of $$ \vec r = \sum_{i} {\hat \iota} \,\left({\hat \iota}\cdot \vec r \right). $$ where the resolution of the identity \begin{align} \hat I&=\sum_i {\hat \iota}\ {\hat \iota}\cdot\\ \hat I\vec r=\vec r&= \sum_i {\hat \iota}\ {\hat \iota}\cdot \vec r \end{align} is replaced by the continuous $$ \hat I= \int dx \vert x\rangle\langle x\vert\, . $$