Decomposition of Linear Operators in Hilbert Spaces Using Dirac’s formalism of Quantum Mechanics, if a complex Hilbert space $\mathscr{H}$ is given, the elements of $\mathscr{H}$ are denoted by $|x\rangle$ and elements of its dual $\mathscr{H}’$ are denoted by $\langle x|$. Sometimes, if a countable basis $\{ |n\rangle : n \in \Bbb N\}$ is given, a formula like the following holds
$$ \mathbf{1} = \sum_{n \in \Bbb N} |n\rangle\langle n|.$$
This is quite an obvious formula if $\mathscr{H}$ is a separable space (i.e. with a countable basis). Often, however, if there is an uncountable basis $\{ |x\rangle : x \in \Bbb R\}$, a similar formula is used:
$$ \mathbf{1} = \int |x\rangle\mathrm{d}x\langle x | $$
which is understandable in the Dirac’s notation, but I don’t understand it formally.
In the theory of Hilbert spaces,

*

*What are the projection operators $|x\rangle\langle x|$ and what is their meaning?

*In which sense the integral holds?

Thank you for any help with this.
 A: The continuous case is based around the Fourier transform, as opposed to the Fourier series. For example, consider the case of the Fourier transform on $\mathbb{R}$:
$$
         f = \frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{f}(s)e^{isx}ds
       = \int_{-\infty}^{\infty}\langle f,e_s\rangle e_sds,\;\;\; e_s(x)=\frac{1}{\sqrt{2\pi}}e^{isx}.
$$
Notice the similarity to the formalism of the Fourier series
$$
                     f = \sum_{s=-\infty}^{\infty}\langle f,e_s\rangle e_s,\;\;\; e_s(x)=\frac{1}{\sqrt{2\pi}}e^{isx}.
$$
Dirac's treatment is over-simplified in the continuous case, and the notation is not so nice when there is a mix of continuous and discrete spectrum. But it's the thought that counts when it comes to Dirac's formalism, and the thought is an elegant extension of the Fourier series to a Fourier integral and/or a mixed Fourier integral and Fourier series expansion.
General expansions are not fully handled by a "discrete" and an "absolutely continuous" expansion, but Physicists rarely need to consider a measure that is "singular" with respect to Lebesgue measure, but which is not discrete. The problems Dirac considers do not require it.
