Sunday, April 14, 2013

Why Heisenberg can't stop atomic collapse


A heuristic argument to show that hydrogen atoms are stable and have a minimum energy level is wrong. I will assume undergraduate level quantum mechanics in the discussion.


There's a popular argument used to explain why atoms are stable. It shows there is a lower bound on the energy level of an electron in the atom that makes it impossible for electrons to keep "falling" forever all the way down to the nucleus. You'll find it not only in popular science books but in courses and textbooks on quantum mechanics.

A rough version of the argument goes like this: the closer an electron falls towards the nucleus the lower its potential energy gets. But the more closely bound to the nucleus it is, the more accurately we know its position and hence, by Heisenberg's uncertainty principle (HUP), the less accurately we know its momentum. Increased variance in the momentum corresponds to an increase in kinetic energy. Eventually the decrease in potential energy as the electron falls is balanced by an increase in kinetic energy and the electron has reached a stable state.

The problem is, this argument is wrong. It's wrong related to the kind of heuristic reasoning about wavefunctions that I've talked about before.

Before showing it's wrong, let's make the argument a bit more rigorous.

Bounding wavefunctions

The idea is to show that for any possible normalised wavefunction ψ of an electron in a Coulomb potential, the expected energy is bounded below by some constant. So we need to show that
is bounded below where
and p is momentum.
Consider a wavefunction that is confined mainly around the nucleus so

The first fact we need is that Heisenberg uncertainty principle tells us that 
(assuming we're in a frame of reference where the expected values of p and x are zero).

If the wavefunction is spread out with a standard deviation of a then the electron is mostly around a distance a from the nucleus. So the second fact is that we can roughly approximate the expected value of 1/r as 1/a.

Combine these two facts and we get, roughly, that
I hope you can see that the right hand side, as a function of a, is bounded below. The graph of the right hand side as a function of a looks like:
It's now an exercise in calculus to find a lower bound on the expected energy. You can find the details in countless places on the web. Here a link to an example from MIT, which may have come directly from Feynman's Lectures on Physics.

The problem

The above discussion assumes that the wavefunction is basically a single packet confined around a distance a from the nucleus, something like that graphed above. But if a lower energy state can be found with a different wavefunction the electron will eventually find it, or an even lower energy state. In fact, by using a wavefunction with multiple peaks we will find that the Heisenberg uncertainty principle doesn't give a lower bound at all.

We'll use a wavefunction like this:
It has a packet around the origin just like before but it also has a sharp peak around r=l. As I'm showing ψ as a function of r this means we have a shell of radius l.

Let's say

where ψ1 is normalized and peaked near the original and ψis our shell of radius l. Assume no overlap between ψ1 and ψ2.

In this case you can see that we can make
as large as we like by making l as large as we like while still leaving us free to make the central peak whatever shape we want. This means that the estimate of 
coming from HUP can be made as small as we like while making the central peak as close to a Dirac delta as we want. Informally, HUP controls of the overall spread of the wave function but not the spread of individual peaks within it.

For a large enough shell, ψcontributes little to the total expected potential energy, but ψ1 can contribute an arbitrarily low amount because we can concentrate it in areas where 1/r is as large as we want. So we can make the total expected potential energy as low as we like. And yet we can also keep the estimate of the kinetic energy given by HUP as close to zero as we like. So contrary to the original argument, the Heisenberg uncertainty principle doesn't give us a lower bound on the energy at all. The argument is wrong.

But wait, we know there is a lowest energy state...

Yes, the energy of a wavefunction in a Coulomb potential is in fact bounded below. After all, atoms are stable. But the Heisenberg uncertainty principle doesn't show it. The inequality in HUP becomes an equality when the wavefunction is a Gaussian function. It provides a good bound for functions that are roughly Gaussian, ie. that form a single "lump". But it provides only weak bounds for wavefunctions with multiple peaks and in this case it's not the appropriate tool to use.

The Heisenberg uncertainty principle is an inequality about ordinary functions interpreted in the context of quantum mechanics (QM). The field of functional analysis provides many such inequalities. A great paper by Lieb, The Stability of Matter, gives an inequality due to Sobolev that can also be interpreted in the context of QM. Sobolev's inequality is more appropriate when considering the hydrogen atom and it gives a good lower bound, demonstrating that the hydrogen atom is stable after all.

But wait, the Heisenberg uncertainty principle argument gives the right energy...

Getting a correct answer doesn't always justify the methods. I can give at least two reasons why the original method appears to work.

1. The HUP gives a good bound for wavefunctions that are roughly Gaussian. The lowest energy level for the hydrogen atom is given (very roughly) by such a function. So an estimate based on HUP should be roughly correct. However, HUP alone can't tell us that the lowest energy state is Gaussian. The argument is only useful if we can get this information from somewhere else.

2. You can get an estimate for the lowest energy level of the hydrogen atom (assuming it exists) by dimensional analysis. Invalid physical arguments that are dimensionally correct will often give the correct result because there is only one dimensionally correct expression possible.

But wait, it's just a heuristic argument...

Heuristic arguments are crucial to physics. But when similar heuristic arguments give opposite results they become problematic. In particular, it's no good saying an argument is inexact or qualitative when it gives a bound on the energy that isn't just off by an order of magnitude, but completely fails to give a bound at all. Part of the issue here is that the Coulomb potential goes to infinity as r goes to zero and so more care is required. The HUP argument above can be adapted to give good results when the potential is bounded below, for example it gives a reasonable estimate for square wells.

But there may be a clever way of using HUP to bound the energy that I haven't seen. If you can see it, please tell me.

The source

Most of what I said above I learnt from the excellent paper on the Stability of Matter by Lieb that I mentioned above.