Delta functions as eigenstates

Question

In Quantum mechanics, it's standard to say that the eigenstate of the position operator in 1D is the Dirac delta function.

More formally, define a linear map $\hat x$ on some enhanced version of $L^2(\mathbb R)$ by $f \mapsto xf$. Then $f_{x_0}$ is an eigenfunction of $\hat x$ with eigenvalue $x_0$ if

$$\hat x f_{x_0} = x_0 f_{x_0}$$

How we get from here to $f$ necessarily being a Dirac delta function? Two starting observations:

1. $f$ is normalized in the sense that $\int_{\mathbb R} f^2 \ dx = 1$. Hence from the eigenvalue equation $x_0 = \int_{\mathbb R} x_0f^2(x) \ dx = \int_{\mathbb R} xf^2(x) \ dx$
2. $\hat x$ is symmetric/Hermitian and hence if $f_{x_0}$ is an eigenfunction for eigenvalue $x_0$ and $f_{x_1}$ is an eigenfunction for $x_1$, then those two functions are orthogonal and $\int_{\mathbb R} f_{x_0}(x)f_{x_1}(x) \ dx = 0$.

It is clear that if we have the right version of $L^2(\mathbb R)$ then the Dirac delta function $\delta(x-x_0)$ is such an eigenfunction. But how do we show it is the only eigenfunction? Orthogonality seems helpful, but I don't know how to formalize it.

Or is it the case that we can't prove this and that the eigenspace of $\hat x$ is not unidimensional?

Any suggestions?

Answer 1

This may not be a complete, 100% rigorous answer, but we can get somewhere just by using some regular analysis.

Suppose $x_0 \neq x_1$ are two real numbers and assume without loss of generality that $f$ is real valued and $x_1>x_0$. Let $0<\epsilon<(x_1-x_0)$. If we assume on the contrary that $f_{x_0}$ has a nonzero integral in the neighborhood $(x_1-\epsilon, x_1+\epsilon)$, we can perform the following computations: $$ \int_{x_1-\epsilon}^{x_1+\epsilon} f_{x_0}(t) \,dt = \frac{1}{x_1-x_0} \int_{x_1-\epsilon}^{x_1+\epsilon} (x_1-x_0)f_{x_0}(t)\,dt = \frac{1}{x_1-x_0} \int_{x_1-\epsilon}^{x_1+\epsilon} (x_1-t)f_{x_0}(t)\, dt, $$ where I used the fact that $x_0 f_{x_0}(t) = tf_{x_0}(t)$. But then by the choice of integration region, we see that $(x_1-t)\leq |x_1-t| \leq \epsilon$, and hence $$ \int_{x_1-\epsilon}^{x_1+\epsilon} f_{x_0}(t) \,dt \leq \frac{1}{x_1-x_0}\epsilon \int_{x_1-\epsilon}^{x_1+\epsilon} f_{x_0}(t)\,dt < \int_{x_1-\epsilon}^{x_1+\epsilon} f_{x_0}(t)\,dt$$ since $\epsilon/(x_1-x_0) < 1$. We assumed that this integral was nonzero, and hence we arrive at the statement $1<1$. This is nonsense, so we must have that $$ \int_{x_1-\epsilon}^{x_1+\epsilon} f_{x_0}(t) \,dt = 0 $$ for every small enough $\epsilon$. This shows that $f_{x_0}$ integrates to $0$ on every open neighborhood not containing $x_0$, which in turn implies that $f_{x_0}(t)=0$ for $t\neq x_0$.

I will note that I am not sure on how to make an argument involving eigenfunctions of the position operator rigorous since no function $f \in L^2$ will be an eigenfunction, but for a plausibility argument this might be good enough to convince someone that Dirac delta functions are the only kind of functions you can get.

Answer 2

For any non-atomic measure the δ “function” can’t be an element of L². It isn’t an element of any $L^p$, even L¹ (although the latter is a finer fact than δ ∉ L²). By no means isn’t δ “L²-normalized” because “δ²” is a nonsense. Any attempt to reach δ with (a sequence of) continuous functions will lead to increasing L² norm up to infinity. One can’t square a distribution, or (generally) multiply two distributions. A generalized functions teacher should inform students about it.

Speaking of such “eigenfunctions” is an abuse of language motivated by projection of finite-dimensional intuition (and notation) to Hilbert spaces and operator algebras.

Gelfand proposed a theory of “rigged” Hilbert spaces to formalize this stuff, but it didn’t gain much acceptance. Spectral theory provides an alternative approach; it shifts emphasis from “eigenstates” to “eigenspaces”, that makes more sense with continuous spectrum.