Wednesday, February 10, 2010

How to obtain a polymer Hilbert space

On Monday, I will be at HU Berlin to give a seminar on my loop cosmology paper (at 2pm in case you are interested and around). Preparing for that I came up with an even more elementary derivation of the polymer Hilbert space (without need to mention C*-algebras, the GNS-construction etc). Here it goes:

Let us do quantum mechanics on the line. That is, the operators we care about are x and p. But as you probably know, those (more precisely, operators with the commutation relation [x,p]=i) cannot be both bounded. Thus there problems of domains of definition and limits. One of the (well accepted) ways to get around this is to instead work with Weyl operators U(a)=\exp(iax) and V(b)=\exp(ibp). As those will be unitary, they have norm 1 and the canonical commutation relations read (with the help of B, C and H) U(a)V(b)=V(b)U(a)e^{iab}. If you later want, you can go back to x=dU(a)/da|_{a=0} and similar for p.

Our goal is to come up with a Hilbert space where these operators act. In addition, we want to define a scalar product on that space such that U and V act as unitary operators preserving this scalar product. We will deal with the position representation, that is wave functions \psi(x). U and V then act in the usual way, V(b) by translation (V(b)\psi)(x)=\psi(x-b) and U(a) by multiplication (U(a)\psi)(x)=e^{iax}\psi(x). Obviously, these fulfil the commutation relation. You can think of U and V as the group elements of the Heisenberg group while x and p are in the Lie algebra.

Here now comes the only deviation from the usual path (all the rest then follows): We argue (motivated by similar arguments in the loopy context) that since motion on the real line is invariant under translation (at least until we specify a Hamiltonian) is invariant under translations, we should have a state in the Hilbert space which has this symmetry. Thus we declare the constant wave function |1\rangle=\psi(x)=1 to be an element of the Hilbert space and we can assume that it is normalised, i.e. \langle 1|1\rangle=1.

Acting now with U(a), we find that linear combinations of plane waves e^{ikx} are then as well in the Hilbert space. By unitarity of U(a), it follows that \langle e^{ikx}| e^{ikx}\rangle =1, too. It remains to determine the scalar product of two different plane waves \langle e^{ikx}|e^{ilx}\rangle. This is found using the unitarity of V and sesquilinearity of the scalar product: \langle e^{ikx}|e^{ilx}\rangle = \langle V(b) e^{ikx}|V(b)e^{ilx}\rangle = e^{ib(l-k)}\langle e^{ikx}|e^{ilx}\rangle. This has to hold for all b and thus if k\ne l it follows that the scalar product vanishes.

Thus we have found our (polymer) Hilbert space: It is the space of (square summable) linear combinatios of plane waves with a scalar product such that the e^{ikx} are an orthonormal basis.

Now, what about x and p? It is easy to see that p when defined by a derivative as above acts in the usual way, that is on a basis element pe^{ikx}=ke^{ikx} which is unbounded as k can be arbitrarily large. The price for having plane waves as normalisable wave functions is, however, that x is not defined: It would be xe^{ikx} = \lim_{\epsilon\to 0}\frac{e^{i(k+\epsilon}x}-e^{ikx}}{\epsilon}. But for \epsilon\ne 0 the two exponentials in the denominator are always orthogonal and thus not "close" as measured by the norm. The denominator always has norm 2 and thus the limit is divergent. Another way to see this is to notice that x would of course act as multiplication by the coordinate x, but x times a plane wave is no longer a linear combination of plane waves.

To make contact with loop cosmology one just has to rename the variables: What I called p for a simplicity of presentaion is the volume element v in loop cosmology while the role of x is played be the conjugate momentum \beta.

If you want you can find my notes for the blackboard talk at HU here (pdf or djvu