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Not all Hilbert space vectors are generally considered physical, due to various reasons. A particular example (as found in Hall's "Quantum mechanics for mathematicians") is the classic particle in a box example : only states obeying the boundary conditions $\psi(0) = \psi(L) = 0$ are considered.

He later mentions that states not obeying this condition are not part of $\text{Dom}(\hat H)$, which I'm guessing might be the theoretical ground for dismissing them.

Is there a generic criterion for qualifying which states are physical? To which operators should they be part of the domain to qualify as such?

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  • $\begingroup$ I do not know the precise statement you mention from Hall's book, but as it stands it does not make sense. The only view to understand the statement is exactly that you propose: we are speaking of vectors in the domain of some relevant operator like $H$, and not of the vectors representing pure states, which are all (barring superselection rules). $\endgroup$ – Valter Moretti Dec 13 '17 at 15:28
  • $\begingroup$ @Slereah: Which page? $\endgroup$ – Qmechanic Dec 13 '17 at 15:31
  • $\begingroup$ The statement would be "The domain of $\hat H$ should be chosen in such a way that $\hat H$ is essentially self-adjoint and, thus, symmetric, meaning that $\langle \phi, \hat H \psi \rangle = \langle \hat H \phi, \psi \rangle$ for all $\phi, \psi$ in $\text{Dom}(\hat H)$. For this to hold, $\phi$ and $\psi$ must satisfy appropriate boundary conditions, which will allow the boundary term in the integration by part to be zero.", p. 82 $\endgroup$ – Slereah Dec 13 '17 at 15:32
  • $\begingroup$ Thanks for the complete quotation. It does not seem to me that here the text tries to select "physical" states but just, say, "physical domain" of certain operators to make them essentially self-adjoint. Notice that the true observables are the self-adjoint extensions of these operators (not the initial operators) and the vectors in their domains generally do not satisfy the said boundary conditions. However this is not the right way to look at QM. The fact that a vector does not belong to a domain does not imply that this vector is unphysical. $\endgroup$ – Valter Moretti Dec 13 '17 at 15:42
  • $\begingroup$ Wouldn't a quantum state that cannot be measured be considered unphysical, though? A quantum state for which the energy or momentum isn't self-adjoint doesn't sound like a state that could be produced in the real world. $\endgroup$ – Slereah Dec 13 '17 at 15:44
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Consider the free particle on the real line with standard Hamiltonian $$H = \frac{1}{2m}P^2\:.$$ Next consider a state $\psi=\psi(x)$ consisting of a compactly-supported smooth function attaining constantly the value $c\neq 0$ in the interval $[a,b]$ and smoothly vanishing outside that interval.

$\psi$ seems a good candidate for a physically meaningful wavefunction. I stress that it simultaneously belongs to the domains of $H$, $X$, and $P$. (The Fourier transform $\hat{\psi}$ of $\psi$ belong in the Schwartz space so that $\psi$ is an element of $D(P^k)$ for every $k=1,2,\ldots$)

Now perform a measurement of position and suppose that the particle is found in $[a',b'] \subset [a,b]$.

In view of that standard projection postulate the wavefunction immediately after the measurement is $$\psi'(x) = Nc \quad\mbox {if $x\in [a',b']$, $\quad \psi(x) =0\quad$ if $x\not\in [a',b']$}$$ where $N$ is the normalization constant.

This new function does not belong to the domain of $H$. Indeed, $$D(H) = \left\{\psi \in L^2(\mathbb R)\:\left|\: \int_{\mathbb R} p^4|\hat{\psi}(p)|^2 dp <+\infty\right.\right\}$$ In the considered case, for some positive constant $C$, $$p^4|\hat{\psi'}(p)|^2 = C p^2 (1-\cos(p(b'-a'))\:.$$ The integral of this function diverges (I used an online integrator to be sure completely). Thus $\psi' \not \in D(H)$.

This elementary and very idealized example shows how the existence of incompatible observables ($X$ and $H$ here) and the collapse postulate make very difficult to define a class of physically sensible states as belonging to a suitable class of domains of physically meaningful observables.

There are actually several ways out from my no-go example. One may argue that the collapse postulate is oversimplified and physical measurement procedures of continuous spectrum observables admit a more accurate description in terms of POVM and quantum operations. This is true, my intention was just to show how the problem is subtle.

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A physical (pure) state is the full unit ray of any solution to the Schroedinger equation:

$$ \frac{d\Psi (t)}{dt} = \frac{1}{i\hbar} H(t) \Psi (t) $$

subject to an initial condition $\Psi (t=0) = \psi_0 $, where all psi and Psi are unit norm vectors in a complex separable Hilbert space. This equation is readily solved for a time-independent Hamiltonian and the particular set of states called https://en.wikipedia.org/wiki/Stationary_state. Schroedinger's equation is reduced to the spectral equation for the Hamiltonian operator which can very well have no solution in the Hilbert space, for example for a free Galilean particle of spin=0 in $\mathbb{R}^n$. By definition, the set of eigenstates of the time-independent Hamiltonian is a subset of the domain of self-adjointness of H.

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  • $\begingroup$ The joy of downvoting without telling why :) $\endgroup$ – DanielC Dec 14 '17 at 0:05

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