Let us consider a Hamiltonian $\hat{H}$ for a certain system and suppose that I would like to know if it is possible to define it (i.e. its domain) in such a way that it results self-adjoint. In principle, computation of the deficiency indices and the use of the relative Von Neumann theorem should be enough to establish whether or not $\hat{H}$ is self-adjoint or if self-adjoint extensions exist. Nevertheless, in order to compute the indices, it is necessary to solve two differential equations.

Now, suppose my operator $\hat{H}$ - and so the adjoint $\hat{H}^{\dagger}$- is so complicated that it is impossible to find solutions of the above-mentioned differential equations.

In order to proceed in the task, a possibile way could consist in the study of the time evolution operator generated by $\hat{H}$ itself.

In particular it should be possible to write, at the first order in t, what follows, e.g. in position representation: \begin{equation} \langle{x}| \hat{U}_t |{\psi} \rangle=\int _{\mathbb{R}} dx' \langle{x}| \hat{U}_t | {x'}\rangle \langle{x'}| {\psi} \rangle \approx \langle{x}|{\psi}\rangle-\frac{\mathcal{i}}{\hbar} t \int _{\mathbb{R}} dx' \langle{x}| \hat{H} |{x'} \rangle \langle{x'}|{\psi}\rangle \end{equation}

Clearly this expression is useful only if the "matrix element" $\langle{x}| \hat{H} |{x'} \rangle$ is known.

Let us suppose that this is the case. Since the unitarity condition on $\hat{U}_t$ implies for the kernel of the integral the condition: \begin{equation} \biggl(\langle{x}| \hat{H} |{x'}\rangle \biggr)^{*}=\langle{x'}| \hat{H} |{x} \rangle, \end{equation} to test this property should be enough to establish if $\hat{H}$ is self-adjoint or not.


I pretty much believe that what I derived above is roughly correct. Nevertheless what puzzles me is that in the above procedure it is not clear to me where the domain in which I define $\hat{H}$ comes into play and, ultimately, beside the explicit form of the operator, is the domain that plays a crucial role with respect to the self-adjointness property.

So, in the end, I guess my questions are: what am I missing to complete/correct my demonstration in order to take into account the domain of $\hat{H}$? And how this domain is related to the domain of the relative operator $\hat{U}_t$? Is it possible the that kernel I am dealing with is not well-defined since it involves position "eigenstates"?


1 Answer 1


I cannot understand well your issue, since the formulation (the proposed solution in particular) is really too vague and based on formal notions close to manipulations by physicists, where similar issues are really too subtle to be addressed. Instead your question concerns mathematics. Indeed, you seem to appreciate the difference between Hermitian, symmetric, and selfadjoint operator. Unfortunately, the formalism you are handling is not able to account for these fine distinctions.

The only comments of mine, in some way in the spirit of your attempt, are the following ones.

  1. The evolutor $U_t:= e^{itH}$ is unitary if and only if $H$ is selfajoint. Actually it is even difficult to define $e^{itH}$ if $H$ is not selfadjoint. That is because the definition is based on the spectral calculus which is at disposal if $H$ is selfadjoint (At least it must be normal, but then $e^{itH}$, with $t$ real, is unitary if and only if $H$ is selfadjoint).

  2. If conversely you have at your disposal a one-parameter group of unitary operators $U_t$, $t\in \mathbb{R}$, which is strongly continuous (i.e. $\mathbb{R} \ni t \mapsto U_t\psi$ is continuous for every given $\psi \in {\cal H}$), then according to the celebrated Stone theorem, there is a unique way to write it as $U_t = e^{itH}$, where $H$ is a selfadjoint operator. The domain $D(H)$ of $H$ is given by $$D(H) := \{\psi \in {\cal H}\:|\: t^{-1}(U_t\psi -\psi)\quad \mbox{converges in ${\cal H}$ for $t\to 0$}\}\:.$$ Obviously, $$H\psi = -i \lim_{t\to 0} \frac{1}{t}(U_t\psi -\psi)\:.\tag{1}$$

  3. A useful fact (a consequence of the Stone theorem) is that of a dense subspace $S \subset {\cal H}$ is invariant under the action of the strongly-continuous unitary group $U_t$, namely $U_t(S) \subset S$ for every $t\in \mathbb{R}$, then the unique selfadjoint generator $H$ can be obtained as the closure of $H|_S$. In other words $H$ is essentially selfadjoint on $S$.

EXAMPLE. If you consider the family of operators $(U_t\psi)(x) = \psi(x-t)$ for $\psi \in L^2(\mathbb{R}, dx)$, you see that they define a unitary strongly continuous one-parameter group of unitaries. Therefore there is a unique selfadjoint generator $H$. If $S$ is the Schwartz space, you see that $U_tS \subset S$ and that, if $\psi \in S$, the derivative in (1) yields $$(H|_S\psi)(x) = i\frac{d}{dx} \psi(x)$$ This implies that the selfadjoint generator $H$ of $U_t$ is just the closure of the symmetric operator $id/dx$ acting on the Schwartz space.

  • $\begingroup$ Part I. Thanks prof. Moretti for your answer, precious as always. I was aware that in some way I was "mixing" different languages in the formulation of my problem: it was just a reflection of my limits in treating the issue in a full rigorous manner. Mine was just a rough attempt to find a way to demonstrate the (non-)self-adjointness of H through the (non-)unitarity of U. From your answer it is now clear to me that I cannot assume the existence of U as an exponentiated version of H, if H is not at least normal. That's probably the breaking point. $\endgroup$
    – RH_ss
    Commented May 17, 2023 at 16:14
  • $\begingroup$ Part 2. Nevertheless I am still struggling. For example, let's suppose that I have a self-adjoint operator H, so I can construct my unitary operator U. At this point what is wrong, at a rigorous level, with my expansion in t? Is the misuse of completeness relation that physicists do? I am asking this because I guess that in some way sloppy physical notation has to be reconciled with the rigorous formal one. And also, to come back to the core of my question: what is left to do if I am not able to compute deficiency indices and I do not have at my disposal a family of unitary operators? $\endgroup$
    – RH_ss
    Commented May 17, 2023 at 16:16
  • 2
    $\begingroup$ The point is that things like $\langle x|$ or $| x\rangle$ are just formal objects. A rigorous interpretation needs an extension of the theory to a non-Hilbert space context. There everything becomes even more obscure. $\endgroup$ Commented May 17, 2023 at 17:38

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