# Is the Hilbert space of quantum physics the space of tempered distributions?

I am currently studying quantum mechanics and I was introduced to Dirac notation (bra-ket notation). I have read multiple times that quantum states are represented by vectors called kets ($$|\psi\rangle$$) which belong to an "abstract Hilbert space".

However, I wasn't satisfied with this approach, since it doesn't give us information about which Hilbert space we are talking about.

Reading the Wikipedia page for the "position operator", I learned that this operator is usually defined on the space of tempered distributions $$\mathcal{S}'(\mathbb{R}^N)$$, which is the continuous dual space of $$\mathcal{S}(\mathbb{R}^N)$$, the space of Schwartz functions.

I also found an article written by David Carfi, which says that we can indeed use tempered distributions as our quantum state vectors, but I do have some questions about this space :

• is there an inner product $$(.,.)$$ we can define on $$\mathcal{S}'(\mathbb{R}^N)$$ ?
• is there an inner product $$(.,.)$$ we can define on $$\mathcal{S}'(\mathbb{R}^N)$$ such as $$\big(\mathcal{S}'(\mathbb{R}^N), \, (.,.)\big)$$ is a Hilbert space ?
• if not, how can we define the bra corresponding to a $$|\psi\rangle$$ ket ?

Thank you.

• There is no sensible inner product between a tempered distribution and another tempered distribution. Tempered distributions act on Schwartz class functions; this is an asymmetric dual pairing, in contrast to the situation of Hilbert spaces.
– Ian
Aug 2 at 16:03
• If I recall correctly, the way to deal with this is through rigged hilbert spaces, en.m.wikipedia.org/wiki/Rigged_Hilbert_space Aug 2 at 16:16

For a particle which moves in a space of $$N$$ dimensions, one has $$N$$ position operators $$\hat{q}_1,\ldots,\hat{q}_N$$ and also $$N$$ momentum operators $$\hat{p}_1,\ldots,\hat{p}_N$$. These are unbounded operators (i.e., not everywhere defined) on the Hilbert space $$\mathcal{H}=L^2(\mathbb{R}^N)$$. For a function $$\psi(q)=\psi(q_1,\ldots,q_N)$$ in this space, the action of these operators is given by $$\hat{q}_j(\psi)(q)=q_j\psi(q)$$ and $$\hat{p}_j(\psi)(q)=-i\frac{\partial\psi}{\partial q_j}(q)\ .$$ Here $$j$$ is an index while $$i$$ is the square root of $$-1$$, and I set Plank's constant equal to $$1$$ for simplicity. A good domain on which to define these operators is the space of test functions $$\mathscr{S}(\mathbb{R}^N)$$ which is inside $$L^2$$. Now if one wants eigenvectors for these operators, unfortunately they are not to be found in $$L^2$$, but in the bigger space of distributions $$\mathscr{S}'(\mathbb{R}^N)$$. As mentioned in a comment, there is a theory for that called the theory of rigged Hilbert spaces, but I would not recommend learning it.
So the answer to the question is: no, the physical Hilbert space is not the space of temperate distributions. Note however that for a scalar Bosonic quantum field theory in $$d$$-dimensional spacetime, the physical Hilbert space is $$\mathcal{H}=L^2(\mathscr{S}'(\mathbb{R}^{d-1}),d\nu)$$ for some Borel probability measure $$\nu$$ on a $$\mathscr{S}'(\mathbb{R}^{d-1})$$. So there, distributions play a role in defining the Hilbert space, albeit with one extra layer of abstraction since $$\mathscr{S}'(\mathbb{R}^{d-1})$$ replaces what before was $$\mathbb{R}^N$$.
• The way I see it, Gelfand's rigged Hilbert space approach is just a general machine for proving isomorphisms of spaces of test functions/temperate distributions with much simpler sequence spaces of rapid decay/at most polynomial growth, see for example galaxy.cs.lamar.edu/~rafaelm/webdis.pdf Now I find these isomorphisms extremely useful, but I only need that for $\mathscr{S}'(\mathbb{R}^N)$ where I can do it by hand. If working with distributions on more general Nash manifolds, then I guess rigged Hilbert spaces might be really needed. Aug 2 at 21:48
• Thank you for your answer. The issue I have when using $L^2$ as our physical Hilbert space is that operators such as the position operator have no eigenvectors in this space. Therefore, it seems to me that there is a contradiction with one of the postulates of quantum physics, which says that the set of eigenvectors of all observables form a basis of $\mathcal{H}$. Aug 3 at 8:09