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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.

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    $\begingroup$ 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. $\endgroup$
    – Noiralef
    Mar 22, 2018 at 22:28
  • $\begingroup$ @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! $\endgroup$ Mar 22, 2018 at 23:11
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    $\begingroup$ 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. $\endgroup$
    – secavara
    Mar 22, 2018 at 23:51

2 Answers 2

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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}

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    $\begingroup$ 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 ? $\endgroup$
    – Hans Wurst
    Apr 7, 2021 at 7:39
  • $\begingroup$ 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. $\endgroup$
    – m93a
    Apr 7, 2021 at 12:44
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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 $P$: $P|k\rangle=\hbar k|k\rangle$.

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  • $\begingroup$ 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 $\langle x|p\rangle$. $\endgroup$
    – m93a
    Apr 11, 2021 at 19:30
  • $\begingroup$ 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 $\mathcal H$. $\endgroup$
    – m93a
    Apr 11, 2021 at 19:30
  • $\begingroup$ @m93a Thank you very much for the comments :). What exactly do you mean by a rescaled momentum eigenket (since $\langle k|k\rangle=\infty$)? $\endgroup$
    – Filippo
    Apr 11, 2021 at 21:39
  • $\begingroup$ 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 $E_k(\Omega) := (2\pi)^n \, E_p(\Omega)$ $\endgroup$
    – m93a
    Apr 11, 2021 at 22:45
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    $\begingroup$ @m93a With $E_p$ you mean the PVM associated to the momentum operator? $\endgroup$
    – Filippo
    Apr 12, 2021 at 7:04

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