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.
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 $\mathbb{R}^N$.