On Dirac notation: inner product vs base representation

Question

The Dirac notation $\langle a | b \rangle$ seems somewhat ambiguous.

1. On one hand, it can be seen as inner product of elements $a(x)$ and $b(x)$ of the Hilbert space $\scr H$, namely: $$\langle a | b \rangle ={\displaystyle}\int_\mathbb R a^*(x) \ b(x) \ dx.\tag{1}$$

2. On the other hand, it's the evaluation of $b$ at its $a$th component, with respect to a particular orthonormal base for $\scr H$.

   • discrete case. $$\langle n | b\rangle = b_n, \tag{2d}$$ where $\displaystyle \sum_n |n\rangle\langle n | = \mathbb I\ $ and $\langle n | m \rangle = \delta_{nm}.$

   • continuous case. $$\langle x | b\rangle = b(x), \tag{2c}$$ where $\displaystyle \int_\mathbb R |x\rangle\langle x | = \mathbb I\ $ and $\langle x | x' \rangle = \delta(x- x').$

The obvious conclusion is that you are free to see $\langle a | b \rangle$ in both ways, that is, 1. and 2. are equivalent.

But it can't be! For instance:

$$b(x) \stackrel{(2c)}{=} \langle x | b\rangle \stackrel{(1)}{\neq} \displaystyle \int _\mathbb R x b(x) dx.\tag{3}$$

So? How can I choose the right way to see it a priori?

Answer 1

Given $$ \mathbb{I} = \int |x\rangle\langle x| \,\mathrm{d}x $$ You can rewrite the scalarproduct $\langle x | b\rangle = b(x)$ by inserting a 'one' in the middle \begin{align*} \langle x | b \rangle &= \langle x | \mathbb{I} | b\rangle = \int \langle x | x'\rangle \langle x'| b\rangle \,\mathrm{d}x' \\ &= \int \delta(x-x')b(x') \,\mathrm{d}x' \\ &= b(x) \end{align*}

This explains how to get the right result, but I'm honestly not sure where you made a mistake.