# On the physical interpretation of Dirac delta distribution

The original purpose of Dirac was to make up the eigenstate of the position operator $$\hat X$$. Now, quantum states are complex-valued functions in the Schwartz space $$\mathscr{S}(\mathbb{R}^n)$$ The main point is... $$\delta$$ is not a function. It can't be associated with a (unique) norm, since Schwartz space is just a Fréchet space, not Banach.

Therefore the probabilistic interpretation of it as a wave function, $$|\delta|^2$$, is non-sense, and still at some point I will for sure see a particle somewhere.

So... what is the correct way to look at it?

• I would suggest looking up rigged Hilbert space and the references given there. Commented Sep 15, 2021 at 13:02

Operators with purely continuous spectra - such as $$\hat X$$ - do not have any eigenstates. The $$\delta$$-distribution serves as a generalized eigenstate, providing a formalization of $$|x\rangle$$ and $$\langle x|$$ which are extremely useful computational tools, despite the fact that they do not themselves correspond to physically realizable states.

[...] at some point I will for sure see a particle somewhere.

This is reflected in the fact that the probability of measuring the particle somewhere is $$1$$; this doesn't require the existence of position eigenstates.

Using Born's rule the probability of obtaining the result $$a$$ from observable $$A$$ over a state $$\psi$$ is $$\langle \psi|P_a|\psi\rangle$$, where $$P_a$$ projects in the $$a$$-eigenspace of $$A$$. But in this case there is no eigenspace, for there are no proper eigenstates, so... ?

The map $$\mu_A$$, which associates a measurement outcome to the corresponding projection operator, is called a projection-valued measure. By measurement outcome, I mean a (Borel-measurable) subset $$E\subseteq \mathbb R$$ which could consist of a single value - e.g. $$\mu_A(\{a\})= P_a$$, in your notation - or multiple distinct values, e.g. $$\mu_A(\{a,b,c\}) = P_a+P_b+P_c$$.

If the spectrum of the operator $$A$$ is purely continuous, then the projection $$\mu_A(\{a\})=0$$ for any individual point $$a\in \mathbb R$$. Case in point, the PVM associated to the position operator is simply $$\mu_X(E) = \chi_E \mathbb I \qquad \chi_E(x)= \begin{cases} 1 & x\in E \\ 0 & \text{else}\end{cases}$$

where $$\chi_E$$ is the indicator function on $$E$$ and $$\mathbb I$$ is the identity operator. As a result, if the state of the system corresponds to some wavefunction $$\psi\in L^2(\mathbb R)$$, the probability of measuring $$\hat X$$ to lie in $$E$$ is given by

$$\mathrm{Prob}_\psi(\hat X,E) := \frac{\langle \psi|\mu_X(E)|\psi\rangle}{\langle\psi|\psi\rangle} = \frac{\int_\mathbb R \psi^*(x) \chi_E(x) \psi(x) \mathrm dx}{\int_\mathbb R \psi^*(x)\psi(x)\mathrm dx}$$ $$= \frac{\int_E |\psi(x)|^2 \mathrm dx}{\int_\mathbb R |\psi(x)|^2\mathrm dx}$$

• Ok but then... what gives me the probability of finding the particle at some point if I can't compute expectation values? Commented Sep 15, 2021 at 13:14
• @ric.san I'm not sure what you mean. Why can't you compute expectation values? Commented Sep 15, 2021 at 13:20
• Using Born's rule the probability of obtaining the result $a$ from observable $A$ over a state $\psi$ is $\langle \psi | P_a | \psi \rangle$, where $P_a$ projects in the $a$-eigenspace of $A$. But in this case, for $A = \hat X$ there is no eigenspace, for there are no proper eigenstates, so... ? Commented Sep 15, 2021 at 13:30
• @ric.san I have expanded my answer to cover your follow-up question. Commented Sep 15, 2021 at 13:52