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

Question

I remember overthinking equations like \begin{equation} \mathbf{1}=\int dx\ |x\rangle\langle x|\tag{1} \end{equation} and \begin{equation} X=\int dx\ |x\rangle\langle x|x\tag{2} \end{equation} 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. \begin{equation} \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). \end{equation}

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. :)

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

Answer 1

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.

Answer 2

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 “$\left|m\right>\!\left<m\right| \mathrm{d}m$” can be understood as the spectral measure of $M$.

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

Answer 3

I begin with a definition of the $f$ representation and $q$ representation of a wave function [1]. Next, in Proposition 1, I use the definition to show that the use of Dirac-delta distributions as eigenfunctions is self consistent with the definition of wave function in both the $f$ representation and the $q$ representation. To be clear, in Proposition 1, the Dirac-delta distributions are not shown to be solutions of any particular eigen-value equation. Finally, I show that Dirac-delta distributions are the eigenfunctions of the position operator since they yield the correct value of the average position.

Definition [Wave function in the $f$ representation and in the $q$ representation] $\,$

(1) The function $a(f)$ is called a wave function in the $f$ representation.

• $a(f) = \int \Psi(q)\,\Psi_f^*(q)\,dq$ with normalization $\int\Psi_{f}^*(q^\prime)\, \Psi_f(q)\,df = \delta(q^\prime - q)$ is an expansion of $a$ in terms of the functions $\Psi_f^*$ along with the expansion coefficient $\Psi(q)= \int a(f) \, \Psi_f (q)\,df$.
• $\left|a(f)\right|^2$ determines the probability for the physical quantity $f$ to lie in a given interval $df$.
• The functions $\Psi^*_f$ are the eigenfunctions of the coordinate $q$ in the $f$ representation.

(2) The function $\Psi(q)$ is called a wave function in the $q$ representation.

• $\Psi(q) = \int a_f\,\Psi_f(q)\,df$. with normaliation $\int\Psi_{f^\prime}(q)\, \Psi_f^*(q)\,dq = \delta(f^\prime - f)$ is an expansion of $\Psi$ in terms of the functions $\Psi_f$ along with the expansion coefficient $a_f= \int\Psi(q)\,\Psi_f^*(q)\,dq$.
• $\left|\Psi(q)\right|^2$ determines the probability for the system to have coordinates lying in a given interval $dq$.
• The functions $\Psi_f(q)$ are the eigenfunctions of the quantify $f$ in the $q$ representation.

Now, I show that wave functions constructed with Dirac-delta distributions are consistent with the definition above.

Proposition 1 Let $q_f\in\mathbb{R}$ and Let $\theta\in [-\pi,\pi]$. Suppose the functions $\Psi^*_{q_f} = e^{-i\theta}\,\delta(q-q_f)$ are the eigenfunctions of the coordinate $q$ in the $f$ representation. Then:

(1) The wave function in the $f$ representation is an expansion of $a$ in terms of the functions $\Psi_{q_f}^*$ where $$a(q_f) = \int_{-\infty}^\infty \Psi(q)\,e^{-i\theta}\,\delta(q-q_f)\,dq = \Psi(q_f)\,e^{-i\theta}, $$ and with normalization $$\int e^{-i\theta}\,\delta(q^\prime-q_f)\, e^{+i\theta}\,\delta(q-q_f)\,dq_f = \delta(q^\prime - q) = \delta(q - q^\prime)\tag{2}$$ and with expansion coefficient $$\Psi(q)= \int_{-\infty}^\infty \, \Psi(q_f)\,e^{-i\theta}\, e^{+i\theta}\,\delta(q-q_f)\,dq_f = \Psi(q).$$ Note that $\left|a(q_f)\right|^2= \left|\Psi(q_f)\right|^2$ determines the probability for the physical quantity $q$ to lie in the interval $dq_f$.

(2) The wave function in the $q$ representation is an expansion of $\Psi$ in terms of the functions $\Psi_f$ along with the expansion coefficient $a_f$ where $$ a_{q_f}= \int\Psi(q)\,e^{-i\theta}\,\delta(q-q_f)\,dq = \Psi(q_f)\,e^{-i\theta} , $$ and with normalization $$ \int_{-\infty}^\infty e^{ i\theta}\,\delta(q-q_f^\prime)\, e^{-i\theta}\,\delta(q-q_f)\,dq = \delta(q^\prime-q_f) = \delta(q_f^\prime - q_f)= \delta(q_f - q_f^\prime). \tag{4}$$ Note that $$ \left|\Psi(q)\right|^2 = \left|a_{q_f}\right|^2 $$ determines the probability for the system to have coordinates lying in a given interval $dq$. Also note that $$\Psi(q) = \int \Psi(q_f)\,e^{-i\theta}\,e^{+i\theta}\,\delta(q-q_f) \,dq_f = \Psi(q ). $$

Now, Landau explains [1] that we define the integral operator $\hat{q}$ in such a way that the integral product of $\hat{q}\Psi$ and the complex conjugate function $\Psi$ is equal to the mean value $\overline{q}$: $$ \overline{q} = \int_{-\infty}^\infty \Psi^*(q) \left(\hat{q}\Psi \right)(q)\,dq. \tag{6} $$ He further states that the form of the operators for various physical quantities can be determined from direct physical considerations. Well, the physical consideration offered is that $$ \overline{q} = \int_{-\infty}^\infty \Psi^*\!(q)\,q\,\Psi \!\left(q\right)\,dq. \tag{8} $$ Comparing (6) and (8), we deduce that $$ \hat{q} = q. $$

Now, we speak here of a physical quantity $q$, which is the coordinate. To this physical quantity there corresponds its operator $\hat{q}$. The eigenfunctions $\Psi_{q_f}$ corresponds to states in which the physical quantity $q$ has a definite value. In other words, the eigenfunctions $\Psi_{q_f}$ corresponds to a definite eigenvalue $q_f$. Suppose that $\Psi$ is in the definite state $\Psi_{q_f}$. Then I expect that \begin{align*} \overline{q} &= q_f \,\delta(0)\tag{10} \end{align*} To understand how come a distribution is given on the right-hand side of (10), the interested reader should see note [2] below.

All that is left to show is that when the operator is applied to a wave function in a definite state that the associated eigenfunction produces the anticipated mean value of the position. In other words that $\Psi_{q_f} = e^{i\theta}$ produces $q_f\,\delta(0)$. In fact, \begin{align*} \overline{q} & = \int \Psi_{q_f}^*(q) \left(\hat{q}\Psi_{q_f}\right)(q)\,dq. \\ & = \int \Psi_{q_f}^*(q) \,q \,\Psi_{q_f}\!\left(q\right)\,dq. \\ & = \int e^{ -i\theta}\,\delta(q-q_f^\prime) \,q \,e^{ i\theta}\,\delta(q-q_f^\prime) \,dq. \\ & = \int \delta(q-q_f^\prime) \,q \, \delta(q-q_f^\prime) \,dq. \\ & = q_f^\prime \,\delta(q_f^\prime-q_f^\prime). \\ & = q_f \,\delta(0) . \end{align*}

We have shown what we wish to show: the usage of Dirac-distributions are consistent with representations of wave functions, and when applied to a wave function in a definite state, the associated eigenfunction produces the anticipated mean value of the position.

Bibliography

[1] Landau and Lifshitz, Coure on Theoretical Physics, Volume 3, pp. 10-11, 18.

[2] What I propose in the body of my answer may seem counterintuitive. For, in the case of a discrete spectrum of eigenvalues one might justifiable propose that \begin{align*} \overline{q} &= q_f .\tag{20} \end{align*} Many students of quantum physics seem to use (20) as well for the continuous spectrum $q$ as well. Yet, for a continuous spectrum of eigenvalues, as given in (2) and (4) we have that $$\int_{-\infty}^\infty\Psi_{q_f^\prime}(q)\,\Psi_{q_f }(q)\,dq =\delta(q_f^\prime-q_f) $$ and that
$$ \int_{-\infty}^\infty\Psi_{q_f}(q^\prime) \Psi_{q_f }(q) =\delta(q_f^\prime-q_f)\,dq_f = \delta (q^\prime - q). $$ Therefore to propose (20) appears unjustifiable to this author. Rather, what appears justifiable is more like \begin{align*} \overline{q} &= \int_{-\infty}^\infty\Psi_{q_f }^*(q)\,\left(\hat{q} \Psi_{q_f }\right)(q) \,dq \\ &= \int_{-\infty}^\infty\Psi_{q_f }^*(q)\, q_f\, \Psi_{q_f }(q) \,dq \\ &= q_f\,\delta (q_f - q_f) \\ &= q_f\,\delta (0) . \end{align*}