Re: metric tensor 'operator' does not commute...

From: Hontas F. Farmer III (hfarmer_at_niu.edu)
Date: 01/27/05 

Date: Thu, 27 Jan 2005 15:45:14 +0000 (UTC)

I have not done the math to confirm what you have written. Takeing you at
your word....

The exercise you have performed is a confirmation of Heisenberg's
uncertainty principle. I ask you what is the differece between the metric
and position. In a sense they are the same thing. So of course they
commute.

qmagick@yahoo.com wrote:

> Now comes
> the fun part: commutation, well, fun for QM type people anyways.
> If p_n is the momentum operator (i.e. p_n = -i h d/dx_n), then
>
> [G_mn, p_j] = G_mn p_j - p_j G_mn = p_j g_mn (6)
>
> Since [G,p] != 0 this means that (if you believe my formulas)
> you can not simultaneously measure the metric and the momentum...
> Now, I am not 100% sure on how to interpret that but I will leave
> it to you to mull over. I also computed the results for
> [G_mn, G_jk] = [G_mn, x_k] = 0. This means you can measure the
> position and metric simultaneously no problem. It is curious that
> both the metricand position commute but that each both do not
> commute with the momentum.

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