# Proof an infinite dimensional/continous completeness relation $\int|x\rangle \langle x| dx=1$

My question is how to prove $$\int|x\rangle \langle x| dx=1$$ where $|x\rangle$ is a eigenstate of a self-adjoint operator $X$ whose spectrum is continuous?

I want to have a rigorous mathematical proof. Any book recommendation is also appreciated, but please indicate the page of the proof relate to the above fact.

• Such an operator does, rigorously speaking, not have eigenstates. This makes a rigorous proof a bit difficult ;) You might want to learn about the spectral theorem for unbounded operators, but I can't really point to just one page in a book. Mar 22, 2018 at 22:28
• @Noiralef Thanks for your comments. However, could you tell me what does the above theorem look like in math. I have some trouble to clarify its mathematical background. Could you be more detailed? Rather than saying spectral theory, could you say some theorem which may contribute to the result. Thanks! Mar 22, 2018 at 23:11
• One possible starting point can be chapter VIII of Reed and Simon's book, Methods of Modern Mathematical Physics, Vol 1. I think the statement you are looking for is somehow discussed after theorem VIII.5. You can also try a shorter exposition (without proof) in appendix C.3 of Galindo and Pascual's book, Quantum Mechanics, Vol 1. Mar 22, 2018 at 23:51

Let $$A$$ be a generic observable, that is, a self-adjoint operator on a Hilbert space $$\mathcal H$$. In general, there might be many values $$\lambda$$ in the spectrum of $$A$$ which do not have a corresponding eigenvector in $$\mathcal H$$. However, physicists often still choose to write the corresponding “eigenket” $$|\lambda\rangle$$, even if it's just a formal expression. For example, one often sees the integral: $$\begin{equation} \int f(\lambda) \; |\lambda\rangle\langle\lambda| \; \mathrm{d}\lambda \tag{1}\label{int1} \end{equation}$$ Although the individual symbols “$$|\lambda\rangle$$” and “$$\mathrm{d}\lambda$$” aren't well-defined, the integral as a whole surprisingly is!

It is a deep result in functional analysis, the so called Spectral theorem, that for every normal operator $$A$$ there is a (operator-valued) measure $$E_A$$ with these properties:

1. If $$\Omega$$ is a subset of the spectrum of $$A$$, then $$E_A(\Omega)$$ is a projector to the corresponding “eigen-subspace”
2. If $$\Omega$$ is the entire spectrum of $$A$$, then $$E_A(\Omega) = I$$
3. If $$\Omega$$ and the spectrum of $$A$$ are disjoint, then $$E_A(\Omega) = 0$$
4. The Lebesgue integral $$\int_{\mathbb C} \lambda \, \mathrm{d}E_A(\lambda)$$ is equal exactly to $$A$$

For every operator $$A$$ there is precisely one such $$E_A$$ and we call it the spectral measure of $$A$$.

If you take an “infinitesimal slice” of the spectrum $$\Omega = [\lambda, \; \lambda + \mathrm{d}\lambda]$$, then the measure will return a projector to that “infinitesimal eigenspace” $$E_A(\Omega) = |\lambda\rangle\langle\lambda|$$. Therefore it makes sense to identify the integral \eqref{int1} with the rigorously defined Lebesgue integral with measure $$E_A$$: $$\begin{equation} \int f(\lambda) \; |\lambda\rangle\langle\lambda| \; \mathrm{d}\lambda \quad := \quad \int_{\mathbb C} f(\lambda) \; \mathrm{d}E_A(\lambda) \end{equation}$$ Now, it should be obvious from the property “2.” of the spectral measure, that the integral $$\int |\lambda\rangle\langle\lambda|\mathrm{d}\lambda$$ is equal to the identity for every normal operator. Your question is just a specific case of this integral, with $$A=\hat x$$. Since the math can be a little too abstract, you'd maybe want to see the spectral measure for $$\hat x$$ – well, in position representation it's just the characteristic function: $$\begin{equation} \big( E_{\hat x}(\Omega) \; \psi \big)(x) = \chi_\Omega(x) \, \psi(x) = \begin{cases} \psi(x) \text{ for } x \in \Omega \\ \hspace{7pt} 0 \hspace{8pt} \text{ for } x \notin \Omega \end{cases} \end{equation}$$

• So there is no way to understand it properly without functional analysis and measure theory ? That seems a bit daunting. I know from experience that these topics aren't touched in chemistry, but how is it in physics. Are these topics part of an undergraduate course in physics ? Apr 7, 2021 at 7:39
• Sadly, functional analysis is required to understand the completeness relation – it's a claim about the spectral properties of an operator, after all. Physicists often try to “sidestep” the problem by using rigged Hilbert space where $|x\rangle$ is an actual object, but truly proving the completeness relations for rigged Hilbert spaces is even more nuanced and requires more knowledge of FA, ironically. At my uni I had to take a specialised course, Mathematical methods of QM, to learn the required parts of FA. The standard QM course for physics undergrads is very non-rigorous there, too.
– m93a
Apr 7, 2021 at 12:44

As explained in the other answer, your "completeness relation" can be interpreted as the spectral theorem. Here's another approach:

## $$1$$ Momentum operator

The equation $$\begin{equation} \frac{1}{(2\pi)^n}\int|k\rangle\langle k|\,\mathrm{d}k=1 \end{equation}$$ is usually used as a shorthand for the Fourier inversion theorem: $$\begin{equation} \psi(x)=\langle x|\psi\rangle=\frac{1}{(2\pi)^n}\int\langle x|k\rangle\langle k|\psi\rangle\,\mathrm{d}k=\frac{1}{(2\pi)^n}\int\mathrm{e}^{\mathrm{i}kx}\langle k|\psi\rangle\,\mathrm{d}k \end{equation}$$

## $$2$$ Position operator

We can give a precise meaning to the equation $$\begin{equation} \int|x\rangle\langle x|\,\mathrm{d}x=1 \end{equation}$$ using Gelfland triples (see the answer to this question). However, this is not so popular. But $$\begin{equation} \langle\phi|\psi\rangle:=\int\overline{\phi(x)}\psi(x)\,\mathrm{d}x=\int\langle\phi|x\rangle\langle x|\psi\rangle\,\mathrm{d}x \end{equation}$$ for all square integrable functions $$\phi,\psi$$.

$$^1$$In addition, if we regard the vector space of differentiable functions as the domain of the momentum operator $$P$$, $$|k\rangle$$ is a genuine eigenvector of $$P$$: $$P|k\rangle=\hbar k|k\rangle$$.

• 1. I would argue that “nothing to do with the spectral theorem” is a too harsh and radical statement :) It is a formal expression without any rigorous a priori meaning, therefore several rigorous meanings can be given to it. For example you could have $|k\rangle$ to be a rescaled momentum eigenket, then you'd also have a rescaled momentum spectral measure and you could do integrals with it. That said, the Fourier inversion theorem is a good alternative view on it and it also formalizes the beast which is $\langle x|p\rangle$.
– m93a
Apr 11, 2021 at 19:30
• 2. If you formalize the integral using a Gelfand triple $\Phi \subset \mathcal H \subset \Phi'$, the integrand is a map between $\Phi \to \Phi'$, therefore the result only holds for the test functions $\phi \in \Phi$, not for all square integrable functions in $\mathcal H$. To see this is the case, try to apply the delta eg. to the characteristic function of $\mathbb R \setminus \mathbb Q$, which is a square-integrable function on a bounded interval. Only the result of the integral is a bounded operator on $\mathcal H$ (the identity), so it has a unique extension to the whole $\mathcal H$.
– m93a
Apr 11, 2021 at 19:30
• @m93a Thank you very much for the comments :). What exactly do you mean by a rescaled momentum eigenket (since $\langle k|k\rangle=\infty$)? Apr 11, 2021 at 21:39
• On a vector space that doesn't have a norm, you can still scale vectors (ie. multiply them by a scalar). The spaces $\Phi, \Phi'$ from the Gelfand triple are examples of such non-metrizable spaces. If you let $|k\rangle = (2\pi)^{n/2} \, |p\rangle$, then the integral in your answer follows trivially from $\int |p\rangle\langle p| \mathrm dp \equiv \int_{\mathbb R} \mathrm dE_p = 1$, ie. the “spectral” version of the completeness relation. The rescaled metric would then be $E_k(\Omega) := (2\pi)^n \, E_p(\Omega)$
– m93a
Apr 11, 2021 at 22:45
• @m93a With $E_p$ you mean the PVM associated to the momentum operator? Apr 12, 2021 at 7:04