Why are polymer representations called "polymer representations"?

Polymer representations deal with non-continuous unitary representations of groups acting on nonseparable Hilbert spaces (see e.g. http://arxiv.org/abs/0704.0007).

But what has this to do with polymers?


I can see this thread is almost four years old., and the OP probably has got the answer. But here's help to the community.

Polymer quantization has nothing to do with "polymers" from Material Science. Rather, this name stems from the fact that with respect to a given lengthscale, the space this describes looks exactly like a 1-D lattice, or a 1-D polymer chain, if you will.

Seeing as how this is a whole new field and all, I am going to give only a brief but technical description to explain the nomenclature. Interested people may follow the literature.

Consider a Hilbert space defined over a discrete position space, in the sense that one chooses a countably infinite selection of $\{x_k\}$ points from the real line. A particle wavefunction can be written in the form:

$\psi(x) = \sum_k C_k e^{ipx_k}$

Define the following inner product as:

$<x|x'> = \lim_{T\rightarrow\infty}\int_{-T}^{T}e^{-ipx}e^{ipx'} = \delta_{xx'}$

Now consider the translation operator $U_\lambda = e^{ip\lambda/\hbar}$, and take its action on a state $|x>$. According to our definition of the inner product:

$<x|U_\lambda|x'> = \lim_{T\rightarrow\infty}\int_{-T}^{T}e^{-ipx}e^{ip(x'+\lambda)}$

which is nothing but the quantity $<x|x'+\lambda>$, which again follows from the definition. Therefore:

$U_\lambda |x>= |x + \lambda>$

The translation operator only generates finite translations in the position space. This paints a picture of a lattice with seperation constant $\lambda$, when a lengthscale $\lambda$ has been fixed. Hence the name "Polymer" quantization.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.