I'm having trouble with the explanation of the kinetic energy on the classically forbbiden region on a step potential ($V=0$ for $x<0$, $V=V_0$ for $x>0$ and $E<V_0$).
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On the classically forbidden region, the wavefunction looks like $$\psi(x)=Ae^{-\alpha x},$$ where $\alpha=\frac{\sqrt{2m(V_0-E)}}{\hbar^2}$.
As you have already mentioned, but I want to emphasize, the wave function you have written above is not normalizable in the entire region of interest, since the region of interest also includes $x<0$.
If you apply the kinetic energy operator to $\psi$. $$\hat{K}[\psi(x)]=\frac{-\hbar^2}{2m}\alpha^2\psi(x)$$ Which implies that the observable for the kinetic energy is $K=\frac{-\hbar^2}{2m}\alpha^2<0$.
You applied the kinetic energy operator to the segment of the wavefunction that looks like $e^{-\alpha x}$, and you found that the result is the same function multiplied by a negative number. This is indeed true. Is this really so surprising? Since, after all, you were the one who asserted that $E<V_0$ in this region in the first place.
However, this does not mean that the kinetic energy can be observed to be negative. (Explained further below.)
Negative kinetic energy is absurd, right?
What you have found is that you applied the kinetic energy operator to a segment of a wavefunction and found a negative number times that same function. But, what you have not done, is observe a negative kinetic energy. (Explained further below.)
What's wrong with this calculation? Any clarification on the kinetic energy on this region?
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Which implies that the observable for the kinetic energy is... $<0$.
The observable values for an operator are its eigenvalues. You have found something that looks like an eigenfunction with a negative eigenvalue. But this can't be the function in the entire space, since it would blow up and not be normalizable for $x<0$, so is not a proper eigenfunction.
You may have learned that the plane waves form a complete set. In the planewave basis the matrix elements of the kinetic energy operator are:
$$
K_{p,q} = \delta_{p,q}\frac{\hbar^2 p^2}{2m}\;,
$$
which shows that the observable values of the kinetic energy are non-negative.