Tuesday, September 26, 2006

Admitting my ignorance

I don't know what is you attitude towards rigorous functional analysis but mine could be summarised as "I know it exists. There are subtleties but they don't bite as long as you are not asking for it. So, for everyday quantum mechanics, it's enough to remember that operators are not just matrices and there might be convergence issues (otherwise taking the trace would show that [x,p]=i cannot work)".

I knew that most of the time we are dealing with unbounded operators which are thus not continuous and mathematicians might be worried about their domains of definition (which can only be a proper subset of the Hilbert space) but if you do the natural things (and implicitly work on the proper dense subset of the Hilbert space), you will be ok. Furthermore, the spectrum of an operator can be a bit tricky as everybody knows that the 'eigenfunctions' for example of the momentum operator are plane waves which are not square integrable and similarly, eigenfunctions of x are 0 as elements in L^2. But every child knows that the proper definition of the spectrum of A are those z for which (A-z) is not invertible and any physics argument involving eigenfunctions can be made precise using wave packets which are not exactly eigenfunctions but if one wanted to one could control the error and after a long and messy argument you could prove what the physicist had known right from the beginning.

But, as I have learned, sometimes the subtleties are also physically relevant: The first time I realised this was in my oral diploma exam: I was asked to discuss the particle in a piecewise constant potential (and compute reflection and transmission coefficients etc). I was asked why I picked particular boundary conditions of my wave function at the jumps of the potential. Luckily, instead of parroting what I had read in some textbook ('the probability current has to be continuous so no probability gets lost') I had one of my very few bright moments and realised (I promise I came up with this myself, I had not heard or read it before) that this comes from requiring the Hamiltonian (esp. the kinetic term) to be self-adjoined: If you check this property, you have to integrate by parts and the boundary terms vanish exactly if you assume the appropriate continuity conditions of the wave function.

More recently I learned when the distinction between continuous and point spectrum is physically important: Long ago, in some advanced quantum mechanics class, we were shown some strange, seemingly unmotivated calculation with a random potential which after some time showed that that the eigenfunctions of the Hamiltonian have exponential decay. And "thus, even with arbitrarily small randomness the conductor turns into an isolator." I had never quite understood how this calculation was supposed to imply this conclusion. Only a few months ago, I understood in a seminar by Hajo Leschke that what was really meant was "with probability 1 the spectrum is a pure point spectrum and thus there are no scattering states". For more information check out a PhysRept by Fröhlich and Spence or, if you are particularly brave, the discussion of the RAGE-theorem in Reed Simon vol. III.

But this is not what I came to tell you about. I came to tell you that yesterday over lunch I was reading quant-ph/0609163 by H. Nicolic about myths and facts about quantum mechanics. I could comment on many sections but one particular argument stroke me. It goes back to Pauli and shows that if your Hamiltonian is bounded from below there is no time operator.

Here I will give you a slightly modified version: Consider quantum mechanics on the half line . In the zeroth approximation you would take as your Hilbert space. Obviously, in this space x is a positive operator. From the above reasoning it follows that you probably want to ask your wave functions to vanish at 0 as otherwise p is not symmetric:.

Now take some wave function
such that the expectation value of x is finite, say . Now, you can convice yourself that by applying a translation operator
you produce a new state for which x has the negative expectation !

How did that happen, wasn't x supposed to be a positive operator? The solution can be found in chapter 2.5 of Thirring's text book vol. 3 (no link from Amazon) as pointed out to me by Wolfgang Spitzer. The solution comes really from the functional analysis fine print: p is only symmetric but not self-adjoined. The domain of definition of is strictly larger as it does not require the vanishing condition at 0! There is no self-adjoined extension of p which is still hermitean. And therefore you cannot form the translation operator: It is not defined.

Another way to see this is to realise that for
you need all powers of p. And those are only symmetric (vanishing boundary terms) if actually all derivatives of the wave function vanish at 0. And as the translation operator works nicely only on analytic functions (after all it's just the Taylor series) that requirement does not leave us with too many functions.

Therefore you really have to worry about the finer points of functional analysis not to translate wave packets to where they should not be!

Friday, September 15, 2006

While you wait: web 2.0

There is some physics in the pipe but unfortunately not yet in a state to be discussed at large. So while you wait (hopefully), let me point out an article in this week's Zeit (in German of course) which I find a nice summary of implications of web 2.0. And this wouldn't be web 2.0 if I would not offer you some music videos from youtube to listen to while you read (mainly Michael Brecker, Mike Stern and Keith Jarret).