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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – Ian
    Aug 2 at 16:03
  • 4
    $\begingroup$ If I recall correctly, the way to deal with this is through rigged hilbert spaces, en.m.wikipedia.org/wiki/Rigged_Hilbert_space $\endgroup$
    – Nick Alger
    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 $\mathbb{R}^N$.

  • $\begingroup$ Yes, and/but an expression of the Schwartz space of tempered distributions as the limit of Hilbert spaces related to the quantum harmonic oscillator, to show that it's nuclear Frechet... and then to see that its dual is a colimit of Hilbert spaces, is useful. I have the impression that Dirac operated in a fashion that somehow took these things for granted... $\endgroup$ Aug 2 at 21:36
  • $\begingroup$ 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. $\endgroup$ Aug 2 at 21:48
  • $\begingroup$ It is certainly true, in this situation, as in many, that the thing that "must be done", is in fact correct, although the certification of its correctness may only be possible at a more sophisticated level... I myself am a bit amused by this sort of disconnect between "factual correctness" and "absolute truth". :) $\endgroup$ Aug 2 at 22:02
  • $\begingroup$ 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 $\mathcal{H}$. $\endgroup$
    – Loris
    Aug 3 at 8:09
  • $\begingroup$ The postulates of quantum physics should not be taken too literally. In particular, the notion of "basis" it calls for is not that of elementary linear algebra courses. $\endgroup$ Aug 3 at 17:33

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