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

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’?

• Doesn't the spatial frequency of a wave function roughly indicate its energy? A wave function with diverging frequency presumably corresponds to a system with divergent energy? Commented Dec 13, 2019 at 16:44