# Minimum possible Kinetic Energy of a confined electron

The problem is this:

Consider an electron confined in a region of nuclear dimensions (about 5 fm). Find its minimum possible kinetic energy in MeV. Treat this problem as one-dimensional, and use the relativistic relation between E and p.

From Heisenberg's uncertainty relation for position and momentum $\Delta x \Delta p\geq\frac{ℏ}{2}$, $\Delta pc= \frac{ℏc}{2\Delta x}=39.4\text{ MeV}$ where $2\Delta x=5\text{ fm}$ is the width of the region. Plugging this into the relativistic energy-momentum relation $E^2=(mc^2)^2+(pc)^2$, $E=39.4\text{ MeV}$ which is the correct answer.

However, in my book there is also an equation for the zero-point energy which is defined as the lowest possible kinetic energy for a quantum particle confined in a region (one-dimensional) of width $a$ and is given by $\langle K \rangle= \frac{ℏ^2}{2ma^2}$ and the answer I get here is about $1.52\text{ GeV}$.

Why are/should these answers be so different?

Also the questions says "The large value you will find is a strong argument against the presence of electrons inside nuclei, since no known mechanism could contain an electron with this much energy."

How is this argument made exactly?

-

The form of your book's value $\frac{\hbar^2}{2ma^2}$ makes it clear that that is a fully classical (non-relativistic) limit ($\frac{p^2}{2m}$, right?), while yours is a fully relativistic one (after all $E,p \gg m_e$).

The argument about the nuclear confinement is simply that both the gravitational and elctromagnetic potentials on the electron due to the nucleus are many orders of magnitude smaller than 40 MeV (the weak nuclear force too, but you may not know how to compute this), and the electron is not affected by the strong nuclear force.

-
+1, but in my opinion one should say that 40 MeV is the right answer, that the other answer is wrong because the energy grows quadratically in nonrelativistic approximation, and only linearly in the ultra-linear regime (I know you said it, but am not sure if it is clear enough for the pedagogical stage of this question). – Ron Maimon Oct 30 '12 at 4:55

Actually, the derivation you did to get the "correct" answer isn't valid, although it's quite possible that you've never been taught this; even in the world of physics, many people don't know it. What the uncertainty principle tells you is the "spread" in possible values of momentum. If you measure the electron's momentum many times, this uncertainty is the minimum possible standard deviation of the results.

But knowing the uncertainty tells you nothing about the actual minimum value. For that you have to figure out the energy eigenvalues of the system, and then you can pick the lowest one. That is the minimum energy that you can possibly measure the electron to have, and in a case like this where the potential is zero within the region of interest, the same value is the minimum kinetic energy. It will generally be much larger than the spread you would calculate from Heisenberg's uncertainty principle.

So in order to properly do this problem, you will need the Hamiltonian operator so that you can find its eigenvalues. In this case, the Hamiltonian is $H = \frac{p^2}{2m}$ if you restrict the problem to the well, and if you find the eigenvalues of that operator on that region, the lowest one is $\frac{\hbar^2\pi^2}{2ma^2}$ (according to Wikipedia), where $a$ is the width of the region.

The above assumes that you're working in nonrelativistic quantum mechanics, of course; as you've noticed, the energy is much higher than the mass of the electron, and so if you wanted to make a realistic calculation, you would have to use the proper relativistic Hamiltonian. But I'm guessing that is beyond the scope of your class.

-
Erp...yes. I neglected to pay attention to the "minimum" in the text of the question. Using the uncertainty relationship to find the expectation for the RMS momentum is acceptable because the system is bound (and therefore the mean is zero), but for the minimum, you obviously need to proceed along these lines. None-the-less the OP says that 40 is "correct", so the question may expect the HUP version and just be poorly worded. – dmckee Oct 31 '12 at 23:56
@dmckee ok, one can indeed use HUP to find the expectation for RMS momentum, but I doubt that this is what people usually mean when they say "minimum," and in any case doing so would require a proper calculation of the expectation of RMS position. In my experience questions which expect this sort of solution are usually based on an incorrect understanding by the question's author of what HUP can and can't be used for - it is a commonly taught misconception. But of course it could be bad wording. – David Z Nov 1 '12 at 17:26
Yes, plugging the uncertainty of the momentum into the equation for K did seem off. I hope to learn more about the Hamiltonian that I see so much of soon. – photon Nov 6 '12 at 23:53
@simplysimple you will learn about that when you take a class on Lagrangian and Hamiltonian mechanics (also known as "analytical mechanics" or sometimes "classical mechanics".) It's usually a sophomore-level class in a typical university physics curriculum. – David Z Nov 7 '12 at 17:22