# 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,

1. What are the projection operators $$|x\rangle\langle x|$$ and what is their meaning?
2. In which sense the integral holds?

Thank you for any help with this.

• 1. It's the map $|y\rangle\mapsto |x\rangle\langle x|y\rangle$ which is the orthogonal projection to the span (=line) of $x$, provided that $\|x\|=1$. Commented Aug 26, 2020 at 13:34

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.
• @Logos : The discrete Fourier transform is straightforward, and the Dirac notation agrees with Fourier transform notation. However, the "continuous" transforms are different. For example $\langle f,e_s\rangle$ has no pointwise meaning in $s$ for the $L^2$ Fourier transform. There $L^2$ would is the correct setting, and pointwise is replaced by $L^2$ functions with respect to a discrete and/or continuous measure. That sort of defect exists in the general continuous case for Dirac. There are subtleties that Dirac notation does not properly address. But the intuitive nature makes up for a lot. Commented Aug 28, 2020 at 1:38