# Constructing quantum mechanics in a topological vector space endowed with a sesquilinear form

Generally it is said that quantum mechanics can be constructed in a topological vector space $V$ and its dual $V^*$, endowed with a sesquilinear form.

The topology is required for convergence of a set of vectors $\{\psi_n\}_{n\in \mathbb{R}}$, and the sesquilinear form is required for defining transition amplitude

$$(\psi,\phi) , \phi \in V , \psi \in V^*.$$

Now I want to define adjoint (dual or conjugate) operator for a linear operator $A$ with domain $D(A)\subseteq V$. However, as I've seen, this operator is defined over Hilbert space, Banach space, and the weaker one, locally convex topological vector space with strong dual.

So my question is basically that if the space $V$ above is indeed sufficient for defining adjoint operator?

If not, what are the weakest conditions on the vector space that allow defining adjoint operator?

Any comment or suggesting reference would be appreciated.

• Is this not better suited for MathSE? – ZeroTheHero May 3 '17 at 8:30
• I also asked a similar question in MathSE. But, as I saw, mathematical physicists claim the first sentence of the text. So I hope I can find them here. – HR-Physics May 3 '17 at 8:36

There are some strong motivations, in my opinion, to set a quantum theory on Hilbert spaces (natural framework for the representation of involutive algebras of observables).

Nonetheless, the adjoint, or transposed, of a continuous linear map from a top vec space $V$ to a top vec space $W$ can be defined given any $\mathcal{V}$ and $\mathcal{W}$ in compatible duality with $V$ and $W$ respectively (with sesquilinear forms $\sigma_V$ and $\sigma_W$, antilinear in the first argument, that separate points in all spaces). The transpose of a continuous map $u:W\to V$ is a map $^{\mathrm{t}}u:\mathcal{V}\to \mathcal{W}$ (iirc continuous with respect to the weak topologies) defined by $$\sigma_{W}(\,^{\mathrm{t}}u(\psi), \xi)=\sigma_V(\psi,u(\xi))\; ,\; \psi\in \mathcal{V}\;,\; \xi\in W\; .$$ You can easily verify that this definition coincides with the usual one for (bounded) operators in Hilbert spaces. Dealing with unbounded operators in this setting is not so usual, but not impossible with suitable generalizations.

• Thanks for your answer. As you mentioned many operators in QM are unbounded. For dealing with such operators, is it enough to restrict our attention only to domain of those operators, and again the definition above makes sense? If this is not correct, I would be thankful if you tell me what you meant by "suitable generalizations". – HR-Physics May 3 '17 at 9:04
• It is not clear what "unbounded" means if there is no measure of "length" (i.e. a norm). In other words, what is infinite in a top vec space without additional structure? Anyways, if you have a not everywhere defined linear map, you can probably still give meaning to a closed operator $A$ defined on $D(A)\subset V$ imposing that $D(A)\times V$ is a closed subset of $V\times V$ (with the product topology). If $V$ is in compatible duality with itself, you may also give meaning to self-adjoint operators, etc. However, let me stress again that there are many reasons for which this may not be seen – yuggib May 3 '17 at 10:37
• as convenient to model quantum mechanics, opposed to the standard theory of operators in Hilbert spaces. Also, there surely will be a lot less theorems and results holding in this more general framework. – yuggib May 3 '17 at 10:39
• You're right. You pointed out the key notion of norm. To define boundedness, the sesquilinear form should also define norm of vectors ( that does not so in general). With this additional structure on top vec space, we define norm topology on the space. On the other hand, every normed vector space is locally convex ( I'm not sure but I remember sth), and therefore the adjoint of linear map (bounded or unbounded) can be defined. Thank you very much. – HR-Physics May 3 '17 at 11:25