This question mainly arises from thoughts concerning the quantum system of a single particle in the real half-line.

As it is known from the literature, a careful analysis of the position and momentum operator for this system in the position representation, where the operator $\hat{x}$ acts as a multiplicative operator and the operator $\hat{p}$ acts as a differential operator, each in a proper domain, leads to conclude that the momentum operator $\hat{p}$ is not essentially self-adjoint and it does not admit any self-adjoint extensions. In other words, the momentum, defined as above, cannot be an observable for the system.

(I summarized this part very briefly, assuming all the details are known, but to be clear the analysis I am referring to is analogue to the one discussed here: What's the deal with momentum in the infinite square well?)

This also should suggest that the p-representation is not a physical representation and hence it is not viable.

Nevertheless, if I start again from scratch, assuming that the only content of my knowledge is the algebra $[\hat{x},\hat{p}]=i\hbar$ and the configuration space represented from the positive real half-line, in principle I could make use of the p-representation. But in this representation, where the operator $\hat{p}$ acts just as a multiplicative operator in a proper domain, I cannot see how I can obtain the same conclusion about its self-adjointness. And if this is true, that is that in this representation p can be self-adjoint, how it is possible to conciliate this result with the former one about the observability of momentum?

I can see that I am handling two different operators in different spaces and from this perspective, it could be reasonable to obtain that one does not admit self-adjoint extensions while the other does. But from a physical point of view concerning the status of observable of a theory I am lost.

So, there should be a loophole somewhere for sure, but I am not able to find it.

Thanks all in advance.

  • $\begingroup$ Yes, if you unitary transform the Hilbert space (by means of a Fourier transformation, for example), coordinate and momentum "switch places". Selfadjointness of $x$ in its "own-representation" converts to selfadjointness of $p$ in its own. And the same for the non-selfadjointness. $\endgroup$
    – DanielC
    Commented Nov 15, 2022 at 20:36
  • $\begingroup$ I can understand this, but what about the physical observability of an operator? It seems in this way it is not possible to state "the momentum is not an observable for this system" or something similar. $\endgroup$
    – RH_ss
    Commented Nov 15, 2022 at 23:01
  • $\begingroup$ Just because the momentum operator isn't self-adjoint on a bounded domain doesn't make it suspect on the entire real line. $\endgroup$ Commented Nov 15, 2022 at 23:55
  • $\begingroup$ There's an old AJP article on this doi.org/10.1119/1.1328351 that clearly distinguishes the importance of the domain. I read a recent paper on this also but I need to find it... $\endgroup$ Commented Nov 15, 2022 at 23:57
  • 1
    $\begingroup$ @RH_ss the point of the domain is that it breaks the equivalence between position and momentum. If you have an infinite square well, position is well defined but there are multiple ways of defining the momentum operator. So you could think about having an infinite square well in momentum space, where the potential goes to infinity if the momentum crosses some boundaries, and then momentum would be well defined and position would have multiple possible definitions. For the full real line, both position and momentum are equivalent, so each representation is healthy $\endgroup$ Commented Nov 16, 2022 at 14:38

1 Answer 1


By the standard axioms of the textbook Quantum Mechanics, momentum in a direction, coordinate, spin, and Hamiltonian are fundamental observables, which means that in whatever physical system we should try to realize them as self-adjoint operators, ideally all. In systems where coordinate and momentum along the coordinate are "conjugated" observables, one is free to choose a Hilbert space (since all of them are isomorphic to one another) as $L^2$.

We thus have: $$ \displaystyle{\left\{x, p_x, L^2 \left(\Omega\subseteq\mathbb{R}, dx\right)\right\} \longleftrightarrow\left\{\tilde{x}, \tilde{p_x}, L^2 \left(\tilde{\Omega}\subseteq\mathbb{R}, dp\right)\right\}} $$

So in the coordinate representation, $x$ could easily be self-adjoint, but $p_x$ could only be symmetric (with no self-adjoint extensions). Change the Hilbert space by a Fourier transformation and "in momentum space" $p_x$ could easily be rendered self-adjoint, but the $x$ would be again symmetric (with no self-adjoint extensions).

So the correct steps for any problem are:

  • describe the system (for example a spinless particle in a certain potential).
  • according to the potential, determine the concrete form of the Hamiltonian as a function of the other observables.
  • after the Hamiltonian is determined, you need to choose the Hilbert space appropriate to solve the spectral problem for the Hamiltonian. This usually means playing with a differential equation and boundary conditions.
  • with the chosen Hilbert space, ensure that the Hamiltonian is self-adjoint, i.e. the "physical boundary conditions" in the ODE make the Hamiltonian self-adjoint as a differential operator. See if the self-adjointness of the Hamiltonian automatically entails the self-adjointness of the other fundamental observables (coordinate, momentum, angular momentum, spin). If not all, tough luck.
  • $\begingroup$ Thanks for your detailed answer. In my reasoning, I was thinking about the matter only at a kinematic level, let's say, but I guess that in the end - as you pointed out - only the Hamiltonian could provide a physical guide. Just to be sure about a technical detail: in the specific case of the particle in the half-line (in position) the space $\bar{\Omega}$ mapped through the Fourier transform should be the whole real line now, am I right? $\endgroup$
    – RH_ss
    Commented Nov 16, 2022 at 19:24
  • $\begingroup$ @RH_ss No, it should still be the open half-line. $\endgroup$
    – DanielC
    Commented Nov 16, 2022 at 20:39
  • $\begingroup$ Ok, I believe this is exactly the problematic point. My naive reasoning was that, since the momentum of a particle in the half-line (say a free particle) can take any value, in switching in p-representation the configuration space should have been the whole real line, but according to what you are saying this is not the case. So this also should mean that the Fourier transform cannot map the whole real line in the half-line and vice versa, I guess. Maybe this is the issue in my argument. $\endgroup$
    – RH_ss
    Commented Nov 16, 2022 at 21:53

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