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I'm doing an introductory course on quantum mechanics. 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 $E<V_0$). I've seen explanations on this website, my text book and something my professor said in relation with the uncertainty principle, but I really don't get it. Specifically, I want to know what's wrong with the following reasoning:

On the classically forbidden region, the wavefunction looks like $$\psi(x)=Ae^{-\alpha x},$$ where $\alpha=\frac{\sqrt{2m(V_0-E)}}{\hbar^2}$. 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 $K=\frac{-\hbar^2}{2m}\alpha^2<0$.

Negative kinetic energy is absurd, right? What's wrong with this calculation? Any clarification on the kinetic energy on this region? Thanks in advance.

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  • $\begingroup$ As far as I understand you, you consider something like $\psi(x):=\theta(x) Ae^{-\alpha x}$ (?). Roughly speaking: This function is in $L^2(\mathbb R)$, but its derivatives are not, and hence it is not in the domain of the momentum/kinetic energy operator; there hence is no meaning to $\hat K\psi$ (and neither its expectation value $K_\psi=\langle \psi,\hat K\psi\rangle$, which I suppose is what you indicate with $K$ in your last equation (?)). $\endgroup$ Commented Jul 10 at 8:24

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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 $E<V_0$).

... 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?

...

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.

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Negative kinetic energy is absurd, right? What's wrong with this calculation?

Kinetic energy value we assign to the particle, based on measurement or the psi function, cannot be negative. When you use the whole psi function, not just its part, then the expected average

$$ \langle \psi |\hat{K}|\psi\rangle = \int \psi^* \hat{K} \psi~ dq $$ for any normalized function $\psi$ is always positive or zero.

However, since this expectation value is a sum of contributions over the whole configuration space, in parts of this space, the contribution can be negative. E.g. in hydrogen atom, far enough from the proton, the region is classically inaccessible, and the contribution to the above expected average is negative there. This is purely a mathematical feature of Schroedinger's theory of $\psi$, a function of position in the configuration space; it does not mean that when one finds particle in such region (e.g. far enough from the proton), that the electron has negative kinetic energy there. Such measurement of position changes the system, the previous $\psi$ function is no longer appropriate, and we do not know electron's energy or momentum or kinetic energy at that point of time; we only know its position.

If you worry about conservation of energy when the electron is found in the forbidden region, there is no direct problem: such finding can only happen if other matter is present, and this interacts with the system, so there is no basis for expecting conservation of energy of the electron in proton's field, only for conservation of energy of the combined system, involving the measuring body. If net energy of the electron in proton's field found after the position measurement is higher than energy it had in its stationary state, the difference in energy can be always attributed to the measuring body/environment supplying it, sometime before or during the observation.

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Expectation values obey classical rules (Ehrenfest's theorem) and to be in such a region where the potential energy is positive and higher than your total energy you would need to have a negative kinetic energy in the classical sense. This isn't much of a problem- plenty of aspects of quantum mechanics are at odds with classical mechanics.

In terms of your issue with the observable kinetic energy being negative: You'd be hard pressed to observe a particle during it's transition in a classically forbidden zone without significantly altering the potential it's in.

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