# Dirac-delta-functions as eigenbasis of the position operator - pure nonsense? Or can more be said?

I remember overthinking equations like $$$$\mathbf{1}=\int dx\ |x\rangle\langle x|\tag{1}$$$$ and $$$$X=\int dx\ |x\rangle\langle x|x\tag{2}$$$$ when I had my introductory QM lecture ($$\mathbf{1}$$ is the identity on $$L^2(\mathbf{R})$$ and $$|x\rangle$$ is "defined" as the Dirac-delta function that vanishes everywhere except in $$x\in\mathbb{R}$$).

Surprisingly, this formalism is self-consistent if one uses the "axiom" $$\langle a|x\rangle=\delta(x-a)$$ and "linearity", e.g. $$$$\langle a|X|\psi\rangle=\langle a|X|\int dx\ |x\rangle\langle x|\psi\rangle=\int dx\ \langle a|X|x\rangle\langle x|\psi\rangle \\ =\int dx\ x\langle a|x\rangle\langle x|\psi\rangle=\int dx\ x~\delta(x-a)\psi(x)=a\psi(a).$$$$

I am now wondering why the nonsense above is commonly presented in introductory QM courses. Is there more to the story? Here are some ideas:

• Equations $$(1)$$ and $$(2)$$ remind me a bit of the spectral theorem - is there a connection?
• I've had a very brief introduction to rigorous distribution theory (I've learned how to solve inhomogeneous ODEs using distributions) and I'd say $$\langle x|$$ can be regarded as a distribution/linear map $$\mathcal{L}^2(\mathbf{R})\ni f\mapsto f(x)\in\mathbf{R}$$.$$^1$$ But apart from that, I don't see how distribution theory helps to make $$(1)$$ and $$(2)$$ meaningful...maybe someone who knows more about distributions does. :)

$$^1$$ $$\mathcal{L}^2$$ is the set of square-integrable functions, $$L^2$$ is the set of equivalence classes.

• Maybe helpful physics.stackexchange.com/q/604762/255696 Jan 11, 2021 at 11:29
• Actually, the $\delta$ is a distribution, the Dirac distribution. Remember that linear forms can be written as a bra vector since they are elements of the dual space. Then, you can write $<\delta_a|\phi>=\int dx \delta(x-a)\phi(x):=\phi(a)$ for a test function $\phi$. But you are also right to say that any bra vector $<x|$ is a distribution, since it is an element of the dual space, i.e. the set of linear maps acting on your vector space. The definition for $\delta$ usually given in physics classes however is wrong. Jan 11, 2021 at 11:30
• I think your lectures did not include the RHS formulation of QM, which is, when properly presented by the lecturer, illuminating about distributions and QM. Search this wonderful site about the key words "rigged Hilbert space". Jan 11, 2021 at 12:47
• Can you post a more specific question? It seems the question now is "how distribution theory justifies Dirac's formalism in quantum theory", which is too general and too broad. Jan 11, 2021 at 15:20
• Also this Answer. In ordinary Hilbert space, (2) is an integral over a projection valued measure. Jan 11, 2021 at 18:38

I have asked you to search phys.SE for the "rigged Hilbert space" words. There should be a lot of returns, the most recent which addresses your second point is the one linked here on this page (right side): Conceptual question on eigenvectors in quantum mechanics. It is pretty good, apart from a technical detail that is not so important to you.

As to point 1. They are "Dirac formal expressions" (completeness of a system of generalized eigenvectors and expansion of an operator in a basis formed by its generalized eigenvectors) which can be derived, under certain simplifying assumptions, from the spectral theorem in a rigged Hilbert space (Gelfand-Kostyuchenko-Maurin or Gelfand-Maurin, depending on the source). This is a very involved result, which needs more than "rigorous distribution theory" (which can be seen as an application of general rigged Hilbert spaces to only particular examples, such as $$\mathcal D\subset L^2 \subset \mathcal D'$$ and $$\mathcal S\subset L^2 \subset \mathcal S'$$, see the first four volumes of Gelfand and coworkers).

A good textbook of QM which goes at length in rigor including a part of RHS is Galindo and Pascual (2 vols, 2nd Ed. in Spanish or its translation into English by Springer Verlag). If this is out of reach, then Capri "Nonrelativistic QM" and Manoukian "QM" also have sections on it.

• Thank you very much for taking the time to write an answer (and also for improving the question). The references are really appreciated. Jan 11, 2021 at 18:09
• @Filippo All good, RHS in QM and QFT is my favorite topic in all mathematical physics. Jan 11, 2021 at 18:23

I just want to add my 2¢ and provide an alternative formalization of the equations in question. Since understanding QM through the lens of rigged Hilbert spaces is in my opinion a minority view and most mathematical physicists formulate the theory simply on (unequipped) Hilbert spaces, this might be a more accessible view.

Let $$M$$ be a self-adjoint operator and $$E_M$$ its spectral measure. Then the notation $$A = \int_{\sigma(M)} f(m) \;\; \left| m \right>\!\left< m \right| \, \mathrm{d}m$$ means precisely $$A = \int_{\sigma(M)} f(m) \;\; \mathrm{d}E_M(m) \: .$$ In other words, the whole symbol “$$\left|m\right>\!\left” can be understood as the spectral measure of $$M$$.

Then $$(2)$$ follows trivially from the Spectral theorem and $$(1)$$ is a consequence of $$E_M(\sigma(M)) = \mathcal{H}$$.

• Thank you for your 2 cents ;) What do you mean by an unequipped Hilbert space? Feb 1, 2021 at 14:12
• The common names for $\Phi \subset \mathcal{H} \subset \Phi'$ are “rigged Hilbert space”, “equipped Hilbert space” or “Gelfand triple”. I wanted to stress that we only work with the Hilbert space $\mathcal{H}$, without the additional structure that comes with $\Phi, \Phi'$ :)
– m93a
Feb 1, 2021 at 15:27
• @m93a, so even though for the basis $|x\rangle$, we have $\int dx|x\rangle\langle x| = \hat{1}$, for a general basis vector $|u\rangle$, which is not orthonormalized, the resolution of Identity would be $\int du\space h(u)|u\rangle\langle u| = \hat{1}$ instead. Here $h(u)$ can be thought of the measure or some weight function. Oct 26, 2021 at 20:23
• @m93a is that correct ? If it is, we could then write $\langle u|u'\rangle = \frac{\delta(u-u')}{h(u)}$, where $h(u)$ is the same measure. Oct 26, 2021 at 20:25
• @NakshatraGangopadhay I'm sorry, I don't understand your question or how it is related to my answer. Yes, you can scale elements of the rigged Hilbert space – it's still a vector space. What you generally can't do, however, is the inner product of two elements of $\Phi'$ – ie. for fixed $u, u' \in \Phi'$, $\langle u \vert u' \rangle$ is not well-defined.
– m93a
Oct 28, 2021 at 14:58