Previously, we discussed the indeterminate result of Australia’s federal election and how that left us in a very rare state of macroscopic quantum superposition. Well, I discussed it, and if you did too then you’re probably wondering how far the analogy can be pushed in light of this week’s Oakeshott-Windsor led resolution.
The answer is: just a bit further. Instead of a normal election result, where the quantum system collapses into a classical state either one way or the other, we have a more fragile equilibrium that still has a chance of fluctuating, even if only on individual pieces of legislation. This close-as-you-can-get-to-classical-while-still-being-a-bit-fuzzy-around-the-edges situation is, I declare, a coherent state.
To save me having to go into the details, please read the linked Wikipedia article if you want to find out more. I usually find Wikipedia to be quite good on matters of quantum physics, and this article is no exception. One of its gems is to point out that a coherent state is not the same thing as a Fock state, which is a state with a definite quantum number of particles.
But then again, some would say we really are in a totally focked state.
(Thank you, thank you. It took two blog posts and two weeks of hung parliament to build up the gag, but I think you’ll agree the punchline was worth it.)
One thought on “Schrödinger’s parliament – update”
I’m still voting for Katter for Speaker. Anyone but Oakeshott, actually, that would totally blow the budget for the extra electricity and salary.