Implications of unbounded operators in quantum mechanics Quantum mechanical  observables  of  a  system  are  represented  by  self - adjoint  operators  in  a  separable  complex  Hilbert  space  $\mathcal{H}$. Now  I understand a lot  of operators  employed  in  quantum  mechanics  are  unbounded  operators, in  nutshell these  operators cannot  be  defined for  all  vectors  in $\mathcal{H}$. For  example  according  to "Stone - von Neumann", the canonical commutation  relation $[P, Q] =-i\hbar I$ has no  solution for $P$ and  $Q$  bounded ! My basic question  is : 


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*If  the  state  of  our  system  $\psi$  is  for  example not in the domain of $P$ (because  $P$  is  unbounded), i.e., if $P\cdot \psi$ does not, mathematically, make sense, what  does  this  mean ? Does it  mean  we  cannot  extract  any  information  about  $P$  when  the  system  is  in  state  $\psi$ ?

 A: It doesn't really mean anything bad at all, in spite of the confused answers one sometimes receives about this matter. The big point is: there is no very direct physical meaning to applying an operator $H$ or $Q$, even if it is an observable, to a wave function $\psi$, even if $\psi$ is in the domain of $H$ or $Q$. (See also Operator versus Linear Transformation,  https://physics.stackexchange.com/a/18933/6432 .
In my opinion, the only operator with very direct physical meaning is the time evolution operator $e^{iHt}$ for some $t$, or for all $t$, and as you seem to realise, by the Stone-von Neumann theorem, this exists even for discontinuous $\psi$ even for unbounded $H$.  
But, also this is my opinion, the exact definition of Hilbert space is not that important and many mathematical physicists who worry (too much) about whether discontinuous wave functions, which are obviously outside the domain of $H$, are physical or not, manage to formulate Quantum Mechanics just fine on a dense subspace of a Hilbert space.  Other, with the opposite worry, again, in my opinion, worrying too much about something which isn't really that bad, talk about rigged Hilbert spaces or nuclear spaces in order to somehow include infinite norm states and exclude non-differentiable wave functions....see Sudbery, Quantum Mechanics and the Particles of Nature. for this.  
There is a mathematical reason for thinking that the exact choice of what domain or what space, whether the Hilbert space, or a subspace of smooth wavefunctions inside the Hilbert Space, or the Schwartz space of rapidly decreasing and smooth functions, or an extension of the Hilbert space to include some dual objects such as distributions, to think of as the domain of these operators, is....unimportant because no matter what space you choose, you get the same physics, and that reason is a theorem of Wilfrid Schmid, Henryk Hecht, and Dragan Milicic, or at least somebody or other, which says that if you have a semi-simple Lie group operating on the space, (if the QM is going to made relativistic you eventually have to assume the Lorentz group acts) and if the representation has a finite composition series (this excludes quantum fields), then the algebraic structure of that rep. is independent of which space you consider (within broad limits).  Earlier versions of such results were proved by Nelson, Garding, and Harish-Chandra, and gave a very pleasant surprise to Hermann Weyl and everyone else involved at the time...
Now very concretely, even if $\psi$ is not in the domain of $H$, or any other observable $Q$, it is still true that the Hilbert space has a Hilbert basis of eigenstates of $H$ and hence even if $\psi$ is horrible and discontinuous and everything bad, it still holds that
$$\psi = \lim_{n\rightarrow \infty}\sum_{i=0}^{i=n} c_i v_i$$
where $v_i$ is the normalised eigenstate of $H$ (or $Q$) with eigenvalue, say, $\lambda_i$, and $c_i$ are complex numbers, the so-called Fourier--Bessel coefficients of $\psi$, and the convergence is not pointwise but in the L$^2$ norm.  Now notice: each finite sum 
$$\sum_{i=0}^{i=n} c_i v_i$$ is an analytic function, if $H$ is hypo-elliptic, as is often true, e.g., the harmonic oscillator, and is at any rate smooth and in the domain of $H$.  
And it is still true, by the axioms of QM, that the probability that $H$ (or $Q$) will, if measured, take the value of $\lambda_i$, is $\vert c_i \vert ^2$ whether or not $\psi$ is in the domain of $H$ (or $Q$).
Pedagogically, there is this widespread confusion that an observable, since it is an operator, ought to be applied to a function since it is an operator, but this is just a naive confusion.  If anything should be applied to the wave function as an operator, it is the exponential of $iH$, which is always bounded.  To repeat: just because $H$ is an operator, and $\psi$ is a function, doesn't mean you should apply $H$ to $\psi$.  Although when you can, that may be a useful shortcut, it is not necessary to ever do it, and the axioms of QM, when stated carefully, never ask you to apply $H$ to $\psi$.  What they ask you to do is, for the unitary time evolution to apply the exponential of $iH$, and for the Born rule probabilities, expand $\psi$ to get its Fourier--Bessel coefficients.  The sloppy way of thinking, which one often sees, works fine for many simple QM problems, but leads to people asking precisely this OP, precisely since it is sloppy.  The careful axiomatisation states things the way I formulated them.
A: It usually just means that the expectation of the energy of the state is infinity (and so could never arise from a state with finite average energy by time evolution or measurements), so that the physical state space does not need these states, except as limits.
An example for P is the state
$$\psi(p) = {1\over\sqrt{p^2+1}} $$
This is normalizable (i.e., square integrable), but multiplying by $p$ takes you out of the normalizable states---so it takes you outside the Hilbert space---so $\psi$ is not, strictly speaking, in the domain of P. The Hamiltonian has a P$^2$ term, so that the expectation of the energy of this state is infinite, since
$$ \langle \psi |P P|\psi\rangle =\infty$$
means that $\langle P^2\rangle=\infty$, so that $\langle H\rangle=\infty$.
An analogous $\psi$ for x would be defined by replacing $p$ above by $x$:
$$\phi(x) = {1\over\sqrt{x^2+1}} $$ Such a state is not as localized, even though zero is its average position. But it has an infinite variance in position--- if you measure the position again and again, you will not have an average deviation. This wavefunction is infinitely spread out, so it is not really useful for describing a particle which is somewhere. It is an idealization, like a plane-wave state.
These mathematical subtleties are never all that interesting outside of pure mathematics--- they are eliminated completely by working in a finite lattice of a bounded size, and this procedure must preserve all physical behavior for a large enough lattice with small enough spacing. The regulator cannot wreck the physics.
See also here: Regularisation of infinite-dimensional determinants
A: Faced with some of the confusing aspects of the « bad » wave functions, such as those which are discontinuous or whose expectations are undefined, some people might wish to adopt the attitude that there is something unphysical about an unbounded operator.  This is the opposite point of view to the more common one that the unbounded operators which represent observables are physical but the « bad » wave functions, i.e., those which are normalisable but are not in the domain of the observable, are unphysical.
There are physical reasons against considering unbounded observables, except for the Hamiltonian itself (which, as often remarked, is a case apart from other observables).  The analysis of the measuring process by Wigner, and by Araki and Yanase, famously shows that unless an observable $Q$ commutes with the Hamiltonian, no measurement apparatus of finite size can implement exact measurements of the observable.  But since observables are supposed to be connected with measurements, this could suggest that such a $Q$ is not exactly physical.  AT least this could justify the attitude that the exact properties of $Q$ are not physical.
One could replace $Q$ by a truncated and smoothed or cut-off version, which might avoid this, but then it would be a bounded operator, and it would wreck all the usual calculations of the textbooks making them unmanageable.  But it would mean that these « bad » wave functions were no longer so bad: they would have finite expectations and so on.
This was not exactly your question, but it is somewhat relevant to some of the more confusing aspects of unbounded observables, discontinuous wave functions, etc.  This line of thought leads to formulating Quantum Mechanics in terms of algebras of bounded operators only, as done by Irving Segal, Streater, Wightman, and many others, and has led to interesting treatments of Quantum Field Theory and Statistical Mechanics by Araki, van Hove, Ruelle, and many others. 
These confusing aspects can also be avoided in other ways, which have already been mentioned.
