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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.

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    $\begingroup$ 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 $\|x\|=1$. $\endgroup$
    – Berci
    Commented Aug 26, 2020 at 13:34

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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.

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  • $\begingroup$ Thank you, this looks really useful. However, some books say that the basis of Dirac’s notation is made of eigenvectors of the operator, and I suppose that the basis for the Fourier transform are not actually eigenvectors of the operator. Do you know what is the analogy in this case? $\endgroup$
    – Logos
    Commented Aug 27, 2020 at 8:38
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    $\begingroup$ @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. $\endgroup$
    – Disintegrating By Parts
    Commented Aug 28, 2020 at 1:38
  • $\begingroup$ Thank you, I understand it now. That is a very nice and thorough explanation. $\endgroup$
    – Logos
    Commented Aug 28, 2020 at 6:43
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    $\begingroup$ @Logos : You're welcome. I am a fan of Dirac. His book on Quantum is very readable and enjoyable. $\endgroup$
    – Disintegrating By Parts
    Commented Aug 29, 2020 at 2:28

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