This question comes from the Complexification section of Thomas Thiemann's Modern Canonical Quantum General Relativity, but I believe it just deals with the foundations of quantum mechanics.

The basic idea is you want to include distributional measures like the Dirac delta into your Hilbert space. So you start with a Hilbert space $\mathcal{H}=L_2(\mathcal{C},d\mu)$ as the square integrable functions on a configuration space $\mathcal{C}$ with respect to a measure $d\mu$. Now, if you want to try and include the Delta function defined with respect to a probability measure $d\mu'$ via

$$\int \delta(q-q') f(q')d\mu'=f(q),$$

You need a distributional extension (his exact phrase for this) of the configuration space, $\bar{\mathcal{C}}$. He goes on to say then you might be able to make a distributional state like

$$\psi_{q'}=\exp(-\phi) \delta_{q'}$$ for an appropriate smoothing function $\phi$ be square-integrable if you picked the smoothing function carefully (this all has to do with coherent states, which again, I think is not overly important).

The problem I have is I don't understand how this can be a Hilbert space. Specifically:

  1. It looks like we have two measures, $d\mu$ for the original function on $\mathcal{C}$ and $d\mu'$ for the measure on $\bar{\mathcal{C}}$. Is this possible, or have I misunderstood and we are just extending $d\mu$ to also apply to distributions, so like $\bar{d\mu}$? Is this always possible?
  2. In addition, this should still be a vector space so how am I supposed to make sense of the additive property? The distrubutions are dual to the functions on $\mathcal{C}$, so can I just add such duals together like $$f(q)+\delta(q-q')$$ and expect them to make sense? Sure they make some sense in the inner product, but shouldn't this be well-defined in the underlying vector space?
  • 3
    $\begingroup$ The extension is not a Hilbert space because it is not self-dual/has no inner product (what would the inner product of $\delta$ with itself be?). This construction is known as rigged Hilbert space/Gel'fand triple. In particular, you define the "distributation extension" as the dual on some subspace of your Hilbert space $\endgroup$
    – ACuriousMind
    Commented Jun 4, 2015 at 15:55
  • 1
    $\begingroup$ Side note on question 2: Addition of distributions is not a problem, they certainly form a vector space (by definition a the dual of a vector space is a vector space). $\endgroup$ Commented Jun 4, 2015 at 15:59
  • $\begingroup$ @ACuriousMind: Ok, I think I see. The deltas are part of the dual space to some subspace of the Hilbert space, equiped with their own measure so that $\delta^*(f)=<\delta,f>$ (or some correct notation!). So the distributions are not expected to be square-integrable, but if you're tricky you might be able to construct square-integrable functions which include deltas. $\endgroup$
    – levitopher
    Commented Jun 4, 2015 at 16:29

1 Answer 1


I am not at all an expert of quantum gravity, but I think you have misunderstood the point.

As I understand it, the point is not necessarily having distributions as quantum Hilbert space vectors, but having a distributional "configuration space", i.e. distributions as the domain of the function(al)s that are the quantum vectors.

While in QM (i.e. for particles) the configuration space (and thus the related phase space) is finite dimensional, in QFT the configuration space is usually a real infinite dimensional Hilbert space. There is a result due to Segal, where it is proved that the Fock space constructed over the one-particle space $H$ is isomorphic to an $L^2(H_r,d\mu_g)$ space of square integrable functionals $\Psi(f)$, where the argument (the object of the configuration space) is an element $f\in H_r$ (the real Hilbert space with the same elements as $H$ and scalar product the real part of the scalar product of $H$); and the measure $\mu_g$ is a Gaussian measure over the space $H_r$. This means that when you integrate the functional $\bar{\Psi}\Psi$ over the measure $d\mu_g(f)$, you are doing an integration over the whole possible $f\in H_r$: $$\int_{H_r}\bar{\Psi}(f)\Psi(f)d\mu_g(f)\; .$$ This is the quantum field analogue of the usual $L^2(\mathbb{R}^d,dx)$ space of square integrable functions with Lebesgue measure commonly used in particles' QM (and it is isomorphic to the more common Fock space formulation for the scalar field).

Now the configurations space $H_r$ is already a space of distributions (if as usual it is separable, then it is isomorphic to some $L^2(\mathbb{R}^d,dx)$ and that is a subspace of $\mathscr{S}'(\mathbb{R}^d)$), and not only of smooth functions. A similar procedure is attempted in considering the configuration space for the gravitational field. The point is that it is not sufficient to consider smooth configurations: it is necessary, to suitably define a measure $\mu$, to take the closure (in a suitable topology I imagine) of the smooth configurations, therefore obtaining also distributional objects. I don't know the details, but it seems that this can be done in a sufficiently rigorous fashion. Anyways, the space of quantum states is still the space of square integrable functionals $\Psi(f)$ of these configurations $f$, simply the latter may be more general/singular than only smooth configurations.

  • $\begingroup$ I think you have it exactly right - in LQG the Hilbert space is $L_2(\bar{\mathcal{A}},\mu)$ over the space of distributional connections $\bar{\mathcal{A}}$. They must be doing something similar here for the coherent states, but the general language was confusing me. Thanks! $\endgroup$
    – levitopher
    Commented Jun 9, 2015 at 16:32

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