$\psi(x)=e^{-x^2} \sin (e^{x^2})$ is a square-integrable function, but why isn't the expectation value of the energy well-defined?

Question

As far as I know, a Hilbert space consists of all square-integrable function.

So, $\psi(x)$ defined as $\psi(x)=e^{-x^2} \sin (e^{x^2})$, is in a Hilbert space, because it is square-integrable: $ \int_{-\infty}^{\infty}dx\psi^2(x)= \int_{-\infty}^{\infty}dx e^{-2x^2}\sin^2(e^{x^2})< \int_{-\infty}^{\infty}dx e^{-2x^2}<\infty$

And, when $x \rightarrow \infty$, then the leading term of second derivative is:

$\frac{d^2}{dx^2}\psi(x)\approx x^2 e^{x^2} \sin (e^{x^2})$

Then, $\psi(x)\frac{d^2}{dx^2}\psi(x)\approx x^2 \sin (e^{x^2})$

So, the expectation value of kinetic energy T does not converge: $T(\psi) \approx \int_{-\infty}^{\infty}dx \psi(x)\frac{d^2}{dx^2}\psi(x) \approx \int_{-\infty}^{\infty}dxx^2 = \infty$

(This function seems to be unphysical because it oscillates so fast when x increase.)

Is there a function which does not have expectation value for an operator?, or The definition of Hilbert space is not ‘set of all square-integrable function’?

Answer 1

You are seeing a direct consequence of the Hellinger-Toeplitz theorem.

The theorem states, essentially, that an unbounded, symmetric operator (like the kinetic energy operator) cannot be well-defined on the entire Hilbert space. It must have a domain that is smaller than the whole Hilbert space. In this case, you have selected a state which falls outside the kinetic energy operator's domain - indeed, if you apply the operator to the state in question, you will find that the resulting state is not square-integrable! You have taken a state inside the Hilbert space and returned a state outside the Hilbert space, which means that the action of the kinetic energy operator is simply not well-defined for this state. Considering this, it would make sense to conclude that, since any physical state must have a finite kinetic energy, this is not a physical state.

Answer 2

The problem you are facing here is that the wave function you are considering is not inside the domain of the Hamiltonian. Operators that are not defined on the whole Hilbert space are known as unbounded operators, both the Position and the Momentum operator are unbounded for example. For the momentum operator you can easily see this if you think of a function that is square integrable but not differentiable. The reason these operators appear naturally in the formulation of quantum mechanics is due to the Stone-von Neumann theorem regarding the canonical commutation relations.