# Using integrals to expand a vector in continuous basis

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?

• A countable set is one that can be indexed by the natural numbers $\{0, 1, 2, \ldots \}$. The real numbers cannot be indexed by the naturals. – DanielSank Nov 19 '17 at 5:22

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 $$\hat I= \int dx \vert x\rangle\langle x\vert\, .$$
• so it is not actually component? I mean what kind of space is this is it still Hilbert space, if so is there a name for it?. how is the inner product defined here? also I am still confused of how the integral is considered an expansion why go through this path where you use the Dirac delta function to project. i.e why can't we just expand by the sum: $$\lim_{n\rightarrow \infty}(\sum_{i=1}^{ \infty} \psi(x+i(1/n))|x+i(1/n))>)$$ or is the integral just a matter of notation convenience? – NegativeTension Nov 19 '17 at 2:27
• @NegativeTension I’m not sure I understand your comment. You can think of $\psi(x)$ as a component. The kets live in a Hilbert space, and the inner product $\langle \phi\vert\psi\rangle=\int dx \phi^*(x)\psi(x)$. – ZeroTheHero Nov 19 '17 at 2:36
• @NegativeTension because $x$ is continuous rather than discrete? – ZeroTheHero Nov 19 '17 at 2:51